Logic of Inorganic Reactions 1
SECTION - I VALENCY/OXIDATION NUMBER/BALANCING EQUATIONS Modern Definition of Valency Valency is the power of an atom of an element to combine with other atoms measured by the number of electrons which an atom or radical will lose, gain or share to form a chemical compound. Types of Valences:
(i)Electrovalency( shown in ionic compounds eg. NaCl, K2SO4 etc) (ii)Covalency (shown in covalent compounds eg SO2, NH3, H2SO4 etc) Electrovalency Radicals: Radical is an atom or a group of atoms which form an ion (positive or negative) by loss or gain of electrons. It is part of an ionic compound. Two radicals form a compound. Types of Radicals: (i)Basic Radical: This forms the +ve ion or cation part of the compound. Usually + + 2+ 3+ + 2+ 3+ + + metals e.g Na , K , Ca , Al , Ag , Hg , Fe etc. (exception: NH4 , H ). This part comes from a base. (ii)Acid Radical: This forms the –ve ion or anion part of the compound. Usually it is made up of nonmetals. Some acid radicals contain both metal and nonmetal atoms. (i)monoatomic acid radical: Cl–, Br–, S2-, O2-, N3-, P3-, H– etc 2- 3- 2- 2- – – – (ii)Compound acid radical : SO4 , PO4 , CO3 , SO3 , OH , NO2 , NO3 – 2- – etc. (exceptions: MnO4 , CrO4 , AlO2 etc which contain both metal and non-metal) Charge of Basic Radical = + valency of the radical, Charge of Acid Radical = - velency of the radical Task: Write the ionic representations of the following radicals: Al, Fe(ous), Cu(ic), Sn(ous),
NH4, Br, O, SO4, NO2, MnO4, Cr2O7, C, N, PO4 Definition of Electrovalency: Electrovalency is the number of electrons lost or gained (not shared)by a radical. The valency tables(table 1 and 2) given later are electrovalencies. IMPORTANT: In fact all the radicals listed in table 1 under basic radicals are positive ions which are formed by the loss of electron(s). Likewise all the radicals listed in table 2 under acid radicals are negative ions which are formed by the gain of electrons. The basic and acid radicals are never to be considered as neutral species. Na is a metal which is neutral. But Na as a radical is present in a compound like NaCl. Here Na is not neutral although we conventionally write as Na. Actually it is Na+. Similarly iron(Fe) is a metal which is neutral. But Fe as a radical is present in compounds in two forms i.e ferrous(eg. ferrous sulphate) in which it remains as Fe++ or Fe2+ ions and the other ferric( e.g ferric sulphate) in which +++ 3+ it remains as Fe or Fe ions. Similarly chlorine as a gas is neutral and has formula Cl2. But Cl as a radical(chloride) present in any compound like NaCl is actually not Cl, instead it is Cl– (a -ve ion). Conclusion: All basic radicals form +ve ions and all acid radical form -ve ions. The magnitude of charge is equal to the valency of the radical. Dr. S. S.Tripathy 2 Concepts in Chemistry
Covalency : The number of electrons which an atom shares in forming a covalent bond with other atoms in a molecule or ion is called the covalency of the atom. There is nothing called covalency of a group of atoms(radical) like that we learnt in electrovalency(sulfate, carbonate etc.). Covalency is determined for an atom in a covalent molecule or ion. This type of valencies are found in binary covalent compounds like PCl5, NH3, CO2 etc. and covalent ions 2- - like SO4 , NO3 etc. The details on covalency will be discussed later. In compound like aluminium sulphate there are two types of valencies- electrovalency between 3+ 2- 2- Al and SO4 and covalency within S and O atoms in forming SO4 . Similarly in compound like Na2CO3 the valency of Na and CO3 are 1+ and 2- respectively while in CO32- the valency of C is 4 as it is sharing four of its elelctrons with the three oxygen atoms. Note that each covalent bond is made up of two electrons contributed one each by the the two linking atoms.
2- O Na CO + 2 3 Na + CO3 ; C (covalency of C=4) electrovalency O O Note that both electrovalency and covalency are expressed by a single term called Oxidation Number(ON) or Oxidation State(OS). The only difference between valency and ON is that there is a sign in ON(+ve and -ve) while there no sign in valency. These things will be made more clear in later sections.
CHEMICAL FORMULAE
Molecular Formula: A molecular formula is a concise way of expressing information about the atoms that constitute a particular chemical compound. It identifies each type of element by its chemical symbol and identifies the number of atoms of such element to be found in each individual molecule of the compound. The number of atoms(if greater than one) is indicated as a subscript. Molecular formula is valid for covalent molecular substances like CO2, NH3, H2SO4,
N2O5 etc. which exist in the form of discrete molecules. Empirical Formula: Empirical formula of a compound is a simple expression of the relative number of each type of atom or ratio of the atoms of different elements present in it. Empirical formula is valid for non-molecular substances which remain as network solids. All ionic compounds like NaCl, CaSO4, KNO3 etc. and a few covalent network solids like SiO2 fall into this category of non-molecular substances. In any ionic compound, say for example NaCl, there is no individual NaCl molecule. Na+ and Cl- ions are arranged alternately in a repeated manner in three dimension to form a gigantic network. In such cases molecular formula cannot be known rather their empirical formula which gives a simple ratio of the elements present in the compound can be known.
Dr. S. S.Tripathy Logic of Inorganic Reactions 3
Crossover Rule: While writing chemical formula of an ionic compound, valency of basic and acid radicals are crisscrossed. In other words the valency of basic radical becomes the subscript of acid radical and vice versa. The following examples will suffice.
32 Al (SO ) Al (SO4) 2 4 3
2 2 Zn (CO3) ZnCO3 Note that parenthesis is used for a compound radical which contains more than one element such as CO3, SO4 etc. only when there is any subscript associated with it. In the first example, parenthesis is used for SO4 but not in the second example for CO3. For monoatomic radicals containing one element such as Al, Cl, O etc parenthesis is not used even if there is a coefficient [e.g. AlCl3 and not Al(Cl)3] Question arises why does this crisscrossing of valencies are done? This is done to equalise the total positive and negative charge as the molecule is neutral i.e net charge of the molecule should be zero.
2- SO4 3+ Al
2- SO4
3+ Al 3 2 2- SO4 Al (SO4) Al2(SO4)3
The hollows(depressions) in the above picture of the basic radical is the site from which electrons have been lost and the mount(bulging out) portions in the acid radical are the sites at which electron have been accepted. There must be a complete matching between the hollows with the mounts. In aluminium sulphate, there are two Al3+ ions, so the total +ve charge is 2 X (+3) = +6; and 2- that is why there are three SO4 ions, so that the total -ve charge is 3 X (-2) = -6. In zinc 2+ 2- carbonate, one Zn ion has +2 charge and one CO3 ion has -2 charge and thus the charge balance occurs. While writing formula for such compounds in which the valencies of acid and basic radicals are same or are simple multiples of each other, then the valencies are simplified. In zinc carbonate the valency 2 for each radical get cancelled while writing the formula. In cupric ferrocyanide, the valencies get simplified by dividing with a factor 2. Hence to conclude, empirical formula gives the simplest whole number ratio of basic and acid radicals.
4 2 2 2 4 Cu [Fe(CN)6] Cu [Fe(CN) ] Cu [Fe(CN)6] 2 6 IMPORTANT: In ionic compounds the formula we talk of are empirical formula not molecular formula. Dr. S. S.Tripathy 4 Concepts in Chemistry
VALENCY TABLES AND FORMULA WRITING TIPS
The formula writing is the most vital part in the study of chemical sciences. Unless and until you write formula correctly it is not worthwhile to proceed further. So practise adequately to master over it. Students often neglect this part of chemistry and become permanently weak in the subject. You should not try to cram the valencies of different basic and acid parts(radicals) given in the tables below rather just read them and practise writing formulae of compounds for a large number of times. Check them yourself and know your mistakes. Within a few days of regular practice and self check, these will automatically be stored in your memory. No special effort will be necessary for the purpose. BASIC RADICALS (TABLE-I)
Valency=1 Valency=2 Valency=3 Valency=4 Valency=5 sodium(Na) magnesium aluminium(Al) stannic arsenic(As) potassium(K) (Mg) chromic (Sn) antimonic mercurous(Hg)* barium(Ba) (Cr) plumbic (Sb) cuprous(Cu) zinc(Zn) antimonous (Pb) vanadium(V)
ammonium(NH4) cupric(Cu) (Sb) platinum aurous(Au) mercuric(Hg) auric(Au) (Pt) silver(Ag) cobaltous arsenous(As) titanium hydrogen(H) (Co) cobaltic(Co) (Ti) lithium(Li) nickel(Ni) bismuth(Bi) rubidium(Rb) lead(Plumbous) ferric(Fe) cesium(Cs) (Pb) thallium(Tl)(ous) ferrous(Fe) manganic antimonyl(SbO) srontium(Sr) (Mn) bismuthyl(BiO) stannous(Sn) scandium(Sc) chromous(Cr) thallium(Tl)(ic) manganous (Mn) berryllium(Be) cadmium(Cd) calcium(Ca) palladium(Pd) platinum(Pt)
2+ Dr. S. S.Tripathy * Mercurous- Hg2 Logic of Inorganic Reactions 5 ACID RADICALS(TABLE-II) Valency = 1 Valency = 2 Valency = 3 Valency = 4
fluoride(F) carbonate(CO3) nitride(N) carbide(C) chloride(Cl) sulphate(SO4) phosphate or ferrocyanide bromide(Br) oxide(O) orthophosphate [Fe(CN)6] iodide(I) sulphite(SO3) (PO4) nitrate(NO3) sulphide(S) ferricyanide dioxide(O2) nitrite(NO2) thiosulphate(S2O3) [Fe(CN)6] pyrophosphate bicarbonate or manganate(MnO4) borate or (P2O7) or hydrogen chromate(CrO4) orthoborate carbonate(HCO3) dichromate(Cr2O7) (BO3) bisulphite or oxalate(C2O4) oxychloride(OCl) hydorgen sulphite(HSO3) zincate(ZnO2) phospide(P) hypochlorite(OCl or ClO) stannite(SnO2) arsenite(AsO3) permanganate(MnO4) stannate(SnO3) arsenate(AsO4) chlorite(ClO2) plumbite(PbO2) cobaltinitrite bisulphate or plumbate(PbO3) [Co(NO2)6] hydorgen sulphate(HSO4) berrylate(BeO2) chlorate(ClO3) peroxide(O2) bromate(BrO3) tetrathionate(S4O6) iodate(IO3) hydorgen phosphite meta aluminate (HPO3) (AlO2) peroxydisulphate superoxide(O2) or persulphate(S2O8) hypophosphite(H2PO2) carbide(C2) perchlorate(ClO4) hydrogen phosphate perbromate(BrO4) (HPO4) periodate(IO4) molybdate(MoO4) hydride(H) dihydrogen pyroantimonate
cyanide(CN) (H2Sb2O7) isocyanide(NC) silicate or metasilicate
cyanate(OCN) (SiO3) thiocyanate(SCN) tungstate(WO4) isocyanate(NCO) titanate(TiO3) isothiocyanate(NCS) tetraborate(B4O7) acetate(CH3COO) nitroprusside[Fe(CN)5NO] hydroxide(OH) dihydrogen phosphate
(H2PO4) metaphosphate(PO3) chromite(CrO2) metaborate(BO2) vanadate(VO3) azide(N3) bisulphide(HS)
argentocynide[Ag(CN)2] antimonite(SbO2) antimonate(SbO3) bismuthate(BiO3) perrhenate(ReO4) ferrite(FeO2)
Dr. S. S.Tripathy 6 Concepts in Chemistry
SELF ASSESSMENT QUESTIONS:(SAQ) While reading the text you will find Self Assessment Question(SAQ) very frequenty. You are advised not to skip over the SAQs. SAQs are meant for giving better understanding and build up a good foundation on the particular aspect discussed. So answer SAQs at each step before proceeding further, check with the correct answer(response) given at the end. Rectify your mistakes if any and then proceed further. SAQ 1: Write down the formula of the following: (i)magnesium chloride, (ii)sodium sulphate, (iii)calcium nitrate, (iv)ferric sulphide, (v)cupric oxide, (vi)potassium sulphite, (vii)sulphuric acid, (viii)aluminium phosphate, (ix)barium peroxide, (x)ammonium cyanide SAQ 2: Write the formula of zinc carbonate, cuprous sulphide, ammonium sulphate, potassium oxide, mercuric sulphite, ferrous sulphide, sodium carbonate, calcium nitrite, stannous chloride, nitric acid
SAQ 3: Write the name of the following (NH4)3PO4, FeCl3, K2SO3, Mg3N2, Na2O2,
Hg2CO3, AlP, Ca(HCO3)2, HgS, H3PO4 SAQ 4: Write the formula of the following: Practise each set at different time and check. your mistakes youself. Set-I: calcium acetate, potassium manganate, magnesium hypochlorite, ferric sulphate, mercuric nitrate, ammonium bicarbonate, cuprous sulphide, aluminium bromide, zinc sulphite, nitrous acid, Set-II: sodium chromate, magnesium chloride, potassium phosphate, calcium phosphide, stannous sulphide, plumbous(lead)carbonate, ferrous nitrite, mercurous hydrogen carbonate, ammonium chlorate, hydrobromic acid Set-III: sodium phosphate, calcium sulphate, potassium dichromate, magnesium chlorite, stannic chloride, aluminium carbide, silver nitrite, cobalt(ous) sulphide, ferric sulphite, calcium manganate Set-IV: magnesium bicarbonate, sodium cyanide, ferrous sulphate, chromic hydroxide, zinc thiosulphate, potassium ferrocyanide, ammonium dichromate, arsenous nitrate, cupric(copper)bicarbonate, hypochlorous acid Set-V: strontium nitrate, potassium sulphite, cupric ferrocyanide, aluminium phosphide, lead(plumbous) acetate, cuprous carbonate, sodium bicarbonate, chromic oxide, magnesium perchlorate, phosphoric acid(orthophosphoric acid) Set-VI: potassium hydrogen phosphite, sodium thiosulphate, ferrous phosphate, calcium ferricyanide, stannous oxide, plumbic oxide(lead dioxide), mercuric dichromate, calcium carbide, ammonium sulphate, chloric acid Set-VII: sodium meta-aluminate, calcium hydride, barium peroxide, manganous phosphate, potassium chromate, magnesium permanganate, chromous oxide, ferrous hydroxide, potassium zincate, perchloric acid Set-VIII: potassium dihydrogen pyroantimonate, ammonium oxalate, sodium tetraborate,
Dr. S. S.Tripathy Logic of Inorganic Reactions 7
potassium cobaltinitrite, calcium silicate, cupric pyrophosphate, sodium tetrathionate, calcium acetate, cuprous peroxydisulphate, potassium hypophosphite SET-IX: sodium tungstate, potassium nitroprusside, permanganic acid, ammonium isothiocyanate, manganous bromite, potassium chromite, antimonyl chloride, mercuric periodate, ferrous vanadate, sodium plumbate, SET-X: aluminium hydride, bismuth hypoiodite, ferrous ferricyanide, chromic isocyanide, barium dihidrogen phosphate, rubium superoxide, ammonium molybdate, arsenic chloride, cobaltic arsenite, auric perchlorate SET-XI: potassium argentocyanide, sodium bismuthate, ammonium metaborate, potassium stannate, bismuthyl chloride, sodium ferrite, ferric oxide, vanadium pentoxide, plumbic sulphate, potassium antimonite
Dr. S. S.Tripathy 8 Concepts in Chemistry
IONIC EQUATIONS
You already know that a compound has two parts or radicals namely basic radical and acid radical . For example in K2SO4, K is the basic part and SO4 is the acid part.
Potassium sulphate
K2SO4 K SO4 (Basic part) (Acid part) The basic part has come from the base(in this case KOH) and the acid part has come from the acid(in this case H2SO4). It has already been discussed before that these are ionic compounds and these radicals are really ions, not neutral species. In the above example, they are K+ and 2- SO4 , not K and SO4 as written above. Aqueous Solution: Many compounds which are soluble in water form free ions on dissolution. Such compounds are mostly are ionic in nature. For example when solid K2SO4 is dissolved in water one molecule of K2SO4 forms two potassium ions (cations), each carrying a charge of +1 and one sulphate ion (anion) carrying a charge of 2-. The basic part forms the cation and the acid part forms the anion.
+ 2- K2SO4 2 K + SO4 (Basic part) (Acid part) The magnitude of charge which each ion carries is equal to the valency of the part (radical) with appropriate sign. Acid part carries -ve and basic part +ve charges. The valency of potassium is 1 and since it is basic radical, it will carry +1 charge. The valency of sulphate(SO4) is 2 and since it is acid radical, it will carry 2- (or --)charge. Solid inorganic compounds which are soluble in water dissociate into free ions. All solid inorganic compounds whether are highly soluble in water or not have one basic and one acid part. Silver chloride(AgCl) is feebly soluble in water(practically insoluble), but it contains the basic radical Ag and acid radical Cl. All inorganic acids and bases also contain positive and negative ions. The following SAQ will make the idea clear.
SAQ 5: Write down the acid and basic radicals with appropriate charge shown with the symbol/formula. Also indicate how many ions of each type will be formed fromt hem on dissociation in water. Set-I: zinc carbonate, calcium phosphate, ammonium nitrate, hydrogen sulphide, calcium acetate, sodium chlorate, sulphuric acid, aluminium carbide, ferrous nitrite Set-II: potassium permanganate, ferric sulphate, nitric acid, magnesium carbonate, aluminium hydroxide, sodium dichromate, ammonium perchlorate, mercurous nitrate, cupric thiosulphate, hydrochloric acid
Dr. S. S.Tripathy Logic of Inorganic Reactions 9
ACIDS AND BASES Acids: substances which produce H+ ions in aqueous solution are called acids. They are of the following types:
(i) Monoprotic: These possess one replaceable H atom (e.g. HCl, HBr, HI, HNO3 etc.),
(ii) Diprotic: These contain two replaceable H atoms(e.g H2SO4, H2CO3 etc),
(iii) Triprotic: These contain three replaceable H atoms(e.g H3PO4) On the basis of their strength, acids are classified into two types. (a) Strong Acid: which ionises almost completely in aqueous solution. For example, hydrochloric acid(HCl) when put in water ionises (dissociates) almost completely forming H+ and Cl- ions and very little unionised HCl remains after dissolution.
H2O + - HCl(g) H (aq) + Cl (aq) (Very little) (Very large) (Very large) (b) Weak Acid: A weak acid though completely soluble in water like strong acid , ionises only slightly or feebly in aqueous solution. HCN (hydrogen cyanide) for example when put in water completely dissolves but ionises to a very small extent. So that very little H+ and CN- exist in solution and unionised HCN molecules exist in large number.
H2O + - HCN(g) H (aq) + CN (aq) (very large) (Very little) (Very little) Examples:
Strong Acids: HCl, HBr, HI, HNO3, HClO4, HClO3, H2SO4.These are the few strong acids.
Weak Acids: HF, HCN, H2CO3, HNO2, H2SO3, H3PO4, H3PO3, H3PO2, HClO, etc. and almost all organic acids such as CH3COOH(acetic acid), HCOOH(formic acid), citric acid etc. Bases: The oxides and hydroxides of metals are commonly called bases. (a) Soluble Strong Bases: A strong base is highly soluble in water and ionises almost completely in water producing large number of OH- ions. The solution of the soluble base is called an ALKALI.
2+ + - - NaOH(s) Na (aq) + O H (aq) ; CaO + H2O Ca(OH)2 Ca + 2 OH Oxides and hydroxides of Group 1(alkali metals e.g Li, Na, K, Rb, Cs) and three heavier elements of Group 2(alkaline earth metals -Ca, Sr and Ba) belong to this category. Strictly speaking only the alkali metal oxides and hydroxides are strong bases. Oxides and hydroxides of Ca, Sr and Ba are less basic and less soluble than alkali metal oxides and hydroxides. (b) Insoluble or Weak bases: They are slightly soluble and weakly ionisable in water. Hydroxides and oxides of all metals other than those cited under soluble strong base category fall into this class. e.g Fe(OH)3, MgO etc.
Exception: NH3 is highly soluble in water but forms a weak base. when NH3 gas dissolves + in water a weak base NH4OH is formed which ionises very little in water and forms NH4 and OH- ions to a very small extent.
Dr. S. S.Tripathy 10 Concepts in Chemistry
SAQ 6 : Indicate which one are strong acids (S) and which are weak acids(W) among the following:
H3PO4, H2SO4, HCN, CH3COOH, HNO3, HF, HCl, HCOOH, HBr, HNO2, H2CO3, HI,
H2SO3, HBrO, HClO4,
Ionic Equations: Let us first know how ionic equations are written. For that purpose let us take an acid-base reaction. You know that an acid reacts with a base to produce a salt and water. Acid-Base Reactions:
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O (l) First, all the species which are marked with the symbol (aq) are converted to their ions(cation and anion). The symbol 'aq' stands for aqueous and such species are soluble in water and undergo ionisation to produce their respective cations(basic part) and anions(acid part). Note that the species marked with the symbol (s), (l) or (g) representing the species existing as solid, liquid and gas respectively are written as such without converting them into their ions. These species are insoluble and do not dissolve in water to produce their respective ions. Total Ionic Equation(TIE:
+ - + - + - H + Cl ++ Na OH Na ++ Cl H2O(l)
When all the species associated with (aq) mark are converted to their ions we get the equation called Total Ionic Equation as shown in the above example. After that we find out which ions appear common on both LHS and RHS. These are the ions which do not actually take part in the reaction and are called spectator ions. Just like when we view a cricket match in a stadium or television as spectators, we do not take part in the game. It is the the players who play and are important persons and not we, who merely watch the game. In the same manner, the ions which do not take part in the reaction are not important. In the above example, Na+ and Cl- are the spectator ions. These are to be cancelled from both the sides to get a clean equation containing only the involved ions(the players in the chemical game). This simple ionic equation is called Net Ionic Equation. Net Ionic Equation(NIE):
+ - H + OH H2O(l)
In this example, actually the H+ ion from an acid reacts with OH- ions from a base to form undissociated water molecule. Other ions are simply not important and do not contribute to the reaction. So they have been removed from the equation. Note that we shall use the abbreviations TIE for Total Ionic Equation and NIE for Net Ionic Equation henceforth.
Dr. S. S.Tripathy Logic of Inorganic Reactions 11
SAQ7:Find the net ionic equations(NIE) in case of the following
(i) Ba(OH)2(aq) + H2SO4(aq) ------> BaSO4(s) + H2O(l) (ii) CH3COOH(aq) + NaOH(aq) ------> CH3COONa(aq) + H2O(l) (iii) HNO3(aq) + Cu(OH)2 (s)------> Cu(NO3)2(aq) + H2O(l)
How shall we know which solid species is insoluble and hence to be associated with a mark (s) and which are soluble so as to be associated with a mark (aq). Of course, there is no difficulty in detecting liquid or gaseous species for example Cl2(g), N2(g), NO2(g), H2O2(l), Br2(l) etc. The real trouble lies with the solid species. The following tables give the solubility rules for some common solids at room temperature. If a solid has a minimum solubility of 0.1 mole per litre of solution at room temperature, it is regarded as a soluble solid. If the solubility is less than 0.001 mole per litre at room temperature it is regarded as insoluble solid. Solubility lying between 0.001 to 0.1 mole per litre comes under slightly soluble category. In the following tables slighly soluble solids(eg. CaSO4, PbCl2, PbBr2 etc.) have been kept under insoluble category. Do not try to cram the table, rather try to refer it at the time of your need.
Solubility Rules(Table-I) (A) Soluble Category Exceptions:(Insoluble) 1. Inorganic acids (ALL)_
(HCl, H2SO4, HNO3 etc.) 2. alkali metal (Li+, Na+, K+ , Rb+, Cs+, Fr+) + and NH4 salts (ALL)_ - 3. NO3 (nitrates) (ALL)_ - - 4. ClO3 (chlorates), ClO4 (perchlorates) (ALL)_ - 5. CH3COO (acetates) (ALL)_ 2- 2+ 2+ 2+ 2+ 2+ 2+ + 6. SO4 (sulphates) Pb , Ca ,Sr , Ba , Hg2 , Hg , Ag 7. Cl-(chloride),Br-(bromide)* - 2+ + 2+ + + I (iodide)* Hg2 (ous), Ag , Pb , Cu , Tl
Solubility Rules(Table-II) (B) Insoluble Category Exceptions(Soluble) - + + + 1. F (fluorides) alkali metal(Na , K , NH4 , Rb+,Cs+, Fr+), Ag+, Tl+ flourides 2. O2-(oxides), OH-(hydroxides) alkali metal (Li+, Na+, K+, Rb+,Cs+, Fr+), alkaline earth metal (only 2+ 2+ 2+ 2+ + + Ca ,Sr ,Ba , Ra ), Tl & NH4 oxides and hydroxides 2- 2- + + + + 3. CO3 (carbonates), SO3 (sulphite) alkali metal (Li , Na , K , Rb , 3- + + PO4 (phosphates), Cs , Fr ), alkaline earth metal 3- 2+ + AsO4 (arsenates) and (only Be ) & NH4 salts 2- CrO4 (chromates) 4. S2-(sulphides) alkali metal (Li+, Na+, K+, Rb+, Cs+, Fr+)alkaline earth metal (Be2+, Mg2+,Ca2+,Sr2+,Ba2+,Ra2+ )& + NH4 sulphides + 2+ 2+ - - 5. Ag , Hg2 and Pb salts acetate(CH3COO ) & nitrate(NO3 ) * HgBr and HgI are insoluble. Dr. S. S.Tripathy2 2 12 Concepts in Chemistry
SAQ 8: Indicate whether the following are soluble(S) or insoluble(I) in water.
(i)HCl (ii)NH4Cl (iii)PbSO4 (iv)Ca(NO3)2 (v)Hg2Cl2 (vi)Na2SO4 (vii)BaSO4,
(viii)K3PO4 (ix)H3PO4 (x)CH3COONa (xi)AgBr (xii)Mg(ClO3)2 (xiii)AgCl,
(xiv)NaClO4 (xv)Bi(NO3)3 SAQ 9: Indicate which are insoluble(I) and which are soluble(S) in water.
(i)CaF2 (ii)Fe(OH)3 (iii)BeCO3 (iv)CaCO3 (v)CaS (vi)NaOH
(vii)NH4F (viii)Ca3 (PO4)2 (ix)KF (x)CuS
Note that the solubility rules given in the above tables are never to be memorised. For the beginners, there will be always an indication in the question about species which are solid(s), liquid(l), gas(g) or aqueous(aq.). In case such indications are absent, then you are advised to refer the solubility rules and indicate which are soluble(aq) and which are not (s). For gases and liquids there will be no difficulty in identifying.
SAQ 10:Write down the total and net ionic equations(TIE and NIE) for the following. The equations are not to be balanced. Do not write the number of +ve and -ve ions produced from each molecule while writing TIE. → (i) H2SO4(aq) + Ca(OH)2 (aq) CaSO4(aq) + H2O(l) → (ii) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) → (iii) NaOH(aq) + H3PO4(aq) Na3PO4(aq) + H2O(l) → (iv) BaCl2(aq) + K2CrO4(aq) BaCrO4(s) + KCl(aq) → (v) Pb(NO3)2(aq.)+ HCl(aq) PbCl2(s) + HNO3(aq) SAQ 11: Write down the Net ionic equations(NIE) for the following. Don't balance. → (i) NH4OH(aq) + AlCl3(aq) Al(OH)3(s) + NH4Cl(aq) → (ii) Fe(OH)2(s)+ HNO3(aq) Fe(NO3)2 (aq) + H2O(l) → (iii) Na2CO3(aq) + CaCl2(aq) CaCO3(s) + NaCl(aq) → (iv) Fe2(SO4)3(aq) + NaOH(aq) Fe(OH)3(s) + Na2SO4(aq) → (v) HBr(aq) + Ca(OH)2(aq) CaBr2(aq) + H2O(l) SAQ 12: Write down the Net Ionic Equations for the following. Don't balance the equations. → (i) Zn(s) + HCl(aq) ZnCl2(aq) + H2(g) → (ii) Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g) → (iii) CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq) → (iv) Al(s) + NaOH(aq) NaAlO2(aq) + H2(g) → (v) NaOH(aq) + H3PO4(aq) NaH2PO4(aq) + H2O(l) SAQ 13:Write the total and net ionic equations → (i)K2Cr2O7(aq) +H2SO4(aq) +FeSO4(aq) Cr2(SO4)3(aq)+K2SO4(aq)+Fe2(SO4)3 (aq)+H2O(l) → (ii)Cl2(g) + NaOH(aq) NaClO3(aq) + NaCl(aq) + H2O(l) → (iii)KMnO4(aq) + H2SO4(aq) + H2C2O4(s) K2SO4(aq) + MnSO4(aq) + CO2(g) + H2O(l)
Dr. S. S.Tripathy Logic of Inorganic Reactions 13
OXIDATION NUMBER(O.N) or OXIDATION STATE (O.S) At this stage it is strongly felt that the students should know about an important concept, called Oxidation Number(ON) also called Oxidation State(OS). Although this concept is mostly theoretical in nature, it is of great significance and has tremendous application. You will know about this very soon.
Let us imagine a molecule of H2O. In water molecule, oxygen atom is connected with two hydrogen atoms by two covalent bonds. Each covalent bond consists of two electrons. Imagine that all the atoms in the water molecule quarrel with each other and would like to be separated from each other such that the electron pair constituting each covalent bond will go to that atom which is more electronegative. O HH Thus oxygen atom gets two additional electrons from the two hydrogen atoms and hence carries a charge of -2 over its head after the separation. Each hydrogen atom loses one electron and carries a charge of +1 over its head. Thus the oxidation number(ON) of oxygen is
-2 and each hydrogen is +1. We can similarly analyse for a molecule of NH3. N H H In NH , when three pairs of electrons of the three covalent bonds go to nitrogen H 3 atom, it gets three additional electrons and hence acquire a charge of -3 and hence its oxidation number(ON) is -3. Each hydrogen loses one electron and acquire a charge of +1 and hence its ON is +1. To sum up, Oxidation Number is the theoretical charge which an atom will acquire if it is hypothetically separated from all other atoms in a molecule in such a manner that the bond pairs of electrons are transferred to the more electronegative atom. If the two atoms are identical ( e.g -O-O- or -S-S- etc.), then the bond pair of electrons is equally distributed(one electron each) by the two atoms. Thus each atom gets back its own electron and therefore does not acquire any charge due to that. To find ON of an atom in a molecule in a manner described above, it is necessary to have a knowledge of the bond structure of the molecule and electronegativity of elements. Often it is not easy for a beginner like you to have a thorough idea on all these now. So, we have to depend on some rules for finding the ON of an atom in a molecule or an ion. It is rather better to use these rules than to write the bond structure of the molecule and find the ON by way of analysis as described before. Rules for determining O.N (1) The O.N of any element in the elemental or uncombined state is zero. Polyatomic
elements such as H2, O2, P4, S8 etc. are also included in it.
SAQ 14: What is ON of the following species: Zn, Na, Cl2, P4, Fe (2) The O.N of element existing as simple monoatomic ion in the compound is equal to the charge of the ion(e.g. Na+1,Cl-1). The charge of the ion is nothing but equal to valency with appropriate sign. For basic part(radical) +ve and acid part(radical) -ve signs are used with the the respective valences.
Dr. S. S.Tripathy 14 Concepts in Chemistry
SAQ 15: What is ON of the underlined element:
(i) CaCl2, (ii)NH4Cl, (iii)FeSO4, (iv)Cu2Br2, (v)ZnS , (vi)Sn(NO3)2, (vii)Mg3N2, (viii)HgI2 SAQ 16: Find the ON of the underlined elements
FeSO4, Mn3(PO4)2, Zn3P2, As2O3, SnCl4, Be2C , KMnO4, H2SO4, MgO
(3) The ON of oxygen(O) is always -2 excepting in peroxides in which it is -1,
superoxides in which it is -½ and OF2 in which it is +2.
SAQ 17: Find the ON of oxygen in the following: ZnO, O2, H2O, H2O2, KO2, Al2O3
(4) The ON of hydrogen(H) is always +1 excepting metal hydrides in which it is -1.
SAQ 18: Find the ON of hydrogen in the following: H2O, NH3, H2SO4, NaH, CaH2, AlH3
(5) In a neutral compound the sum of O.Ns of all the atoms of all the elements is zero.
Example: Let us take the molecule SO3. What is the ON of S in this molecule? Let us assume the ON of S to be x. The sum of the ONs of all the atoms in the molecule is zero as per this rule. x + 3(-2) =0 , So x = +6. For one oxygen atom the ON is -2 and therefore for 3 oxygen
atom it is 3(-2)=-6; The ON of S in SO3 is +6.
SAQ 19: Find the ON of underlined elements.(i)SO2, (ii)HNO3, (iii)H2SO4, (iv)KMnO4,
(v)P2O5, (vi)K2Cr2O7, (vii)HClO4
(6)The sum of ONs of all the atoms in a polyatomic ion is equal to the charge of the ion, while the charge of the ion is equal to the valency of the ion with the appropriate sign(acid radical -ve and basic radical +ve) 2- Example: SO4 . Since the valency of sulphate radical is 2, its charge will be -2 as it is an acid radical. Let us find the ON of S in the ion. According to the above rule the sum of the ONs of all the atoms should be equal to the charge of the ion. x + 4(-2) = -2 ⇒ x = +6. So the ON of S in sulphate ion is +6. NB: Conventionally while writing ON, the sign comes first followed by the numerical value. But when we talk of real ion the sign comes after the numerical value. For example 2- when we say ON of S in SO4 , it is +6 but when we say the charge of the sulphate ion, it is written as 2- or - - . This is merely a matter of convention used to distinguish between the two concepts; while ON is the theoretical charge which one atom will carry, an ion has the actual charge which the whole ion carries. Note that often students don't give +ve sign in ONs and write merely 7 instead of +7. This is an error. Always a sign + or - has to be placed before the numerical value of ON. Charge O.N +6 2- SO4 Also note that the ON is to be placed just above the atom(not as superscript written for charge). Dr. S. S.Tripathy Logic of Inorganic Reactions 15
COVALENCY: The covalency of an atom in a covalent molecule or ion is the number of electrons which the atom shares with other atoms while forming the covalent bonds. The covalency of an atom is equal to oxidation number in magnitude. There is no sign(+ve or -ve) in covalency. It is a pure 2- number. For example, the ON of S in SO4 is +6 and so its covalency is 6. SAQ 20: (A) Find the O.N of the underlined atom - 2- (i) NH3, (ii) KNO3, (iii) NO2 , (iv) NO2, (v) N2, (vi) N2O5, (vii) KMnO4, (viii) Cr2O7 2- - + - (ix) S2O3 , (x) ClO3 , (xi) NH4 , (xii)NO3 , (xiii) NH4NO3,, (xiv) MnO2, (xv) Cr2(SO4)3, 2- (xvi) HClO, (xvii) SO3 , (xviii) HCl, (xix) SbCl3, (xx) PH3 (B) Find the covalency of the underlined atoms in case of the following. - - (i) NO3 (ii) SO3 (iii) ClO3
(iv) NH3 (v)N2O5
Oxidation and Reduction (Redox Reaction) This type of reaction is called oxidation-reduction(or redox) reaction, which involves both oxidation as well as reduction. You already know about some primitive definitions of oxidation and reduction; oxidation as the addition of oxygen or removal of hydrogen and reduction as addition of hydrogen or removal of oxygen. These old definitions have been outdated. Now you will be knowing the most modern definitions of oxidation and reduction. Oxidation: A process in which ON of certain element is increased is called oxidation. This takes place by the loss of electron. Thus oxidation can also be defined as a process in which electron(s) is(are) lost. Reduction: A process in which ON of certain element is decreased is called reduction. This takes place by the gain of electron. Thus reduction can also be defined as a process in whcih electron(s) is(are) gained. Look to the following example.
[O]
0 +1 +2 0 Zn + HCl ZnCl2 + H2 (RA) (OA) [R]
In this example, the ON of Zn was zero(uncombined state) in LHS and it becomes +2 in ZnCl2 in RHS. Thus the ON of Zn has increased from 0 to +2 due to loss of 2 electrons by neutral Zn 2+ atom to form Zn ion. This process is oxidation. We say that Zn is oxidised to ZnCl2. On the other hand, the ON of H in HCl is +1 in LHS and in H2 it is 0 in the RHS. Thus the ON of H has decreased from +1 to 0 due to gain of one electron. This is the reduction process. We say that
HCl is reduced to H2. We found also that wherever there is oxidation, there is a reduction. Whenever there is a loss there is definitely a gain. Just like, if you lose a 100 rupee note on the road, it is your loss, but it is the gain of a person who finds it. Like loss and gain, oxidation and reduction also go simultaneously. There cannot be an oxidation without having a reduction or vice versa.
Dr. S. S.Tripathy 16 Concepts in Chemistry
The substance which is oxidised is called Reducing Agent(RA) and the substance which is reduced is called OxidisingAgent(OA). In the above reaction, Zn is oxidised, hence it + is the reducing agent(RA) as it reduces H to H2(ON=0). HCl is reduced, hence it is the oxidising agent(OA) as it oxidises Zn(ON=0) to Zn2+. SAQ 21: Show by indicating the ON which is oxidised and which reduced. Also indicate which is Oxidising Agent(OA) and which Reducing Agent(RA). The elements which have undergone changes in ON have been underlined.The equations are not balanced. Do not also try to balance them. → (i) Na + H2O NaOH + H2 → (ii) N2 + H2 NH3 → (iii) P4 + Cl2 PCl3 → (iv) NH3 + CuO Cu + N2 + H2O → (v) H2 + I2 HI → (vi) Mg + H2SO4 MgSO4 + H2 → (vii) N2 + O2 NO → (viii) Cu + HNO3 Cu(NO3)2 + NO2 + H2O → (ix) MnO2 + HCl MnCl2 + H2O + Cl2 → (x) KMnO4 + H2SO4 + FeSO4 K2SO4 + MnSO4 + Fe2(SO4)3 +H2O Broad Classification of Inorganic Reactions: All inorganic reactions belong to the following two types. (i)Redox type: Already discussed above (ii)Metathesis type(Non-redox): No change of ON occurs in respect of any element. (i) Redox Reactions: Change in ON takes place in this case. One is oxidised(ON increased) and the other is reduced(ON is decreased). We have already discussed this before. [O] 0 +1 +2 0 ex. Zn + HCl ZnCl2 + H2 [R] (ii) Metathesis Reactions: No change in ON takes place for any element. The hundreds of double replacement reactions(also called as double decomposition) that you know belong to this category. Look to the following reactions. +1-1 +1 -1 eg. AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)
+1 +1+1 +1 NaOH ++ H2SO4 Na2SO4 H2O(l) In such reactions no element undergoes change in ON. You may calculate the ON of each element and find for yourself that no change in ON has taken place. You are advised to avoid calculation of ON once you see a reaction belonging to double displacement reaction type. In such reactions, change in ON does not take place as the same acid and basic radical(ions) are present on both the sides although their partners are changed.
Dr. S. S.Tripathy Logic of Inorganic Reactions 17
Caution: Remember that there cannot be any reaction involving only oxidation and no reduction or vice versa. In other words, the reaction cannot be an intermediate between redox and metathesis types. The following example will explain it more clearly. +1 0 +1 +1 (wrong) Na2O + H2O NaOH + H2 In this case only H has undergone reduction from +1 to 0 and there is no oxidation. This is not possible. So the products have been wrongly predicted. The correct reaction is
+1 +1 +1 +1 Na2O + H2O NaOH (It is a metathesis reaction, not redox) SAQ 22: Indicate by assigning ON which reactions are redox and which metathesis. If you find a reaction to be double resplacement type(metathesis), don't find the ONs of atoms. Leave them as such. There is no change in ON in those reactions.
(i) H2 + Cl2 ------> HCl
(ii) Zn(NO3)2 + NH4OH ------> Zn(OH)2 + NH4NO3
(iii) Cu + HNO3 ------> Cu(NO3)2 + NO2 + H2O
(iv) Na + H2O ------> NaOH + H2
(v) NaOH + HNO3 -----> NaNO3 + H2O
(vi) AgNO3 + HCl -----> AgCl + HNO3 PRACTICE QUESTIONS 1. Using solubility rules, predict which of the following are insoluble(I) and whcih are soluble(S) in water.
NaNO3, KBr, MgF2, FeCl2, MgCO3, BaSO4, MgS, Na3PO4, Na2CrO4, Ba(ClO4)2,
Ba(OH)2, Hg2Cl2, NH4F, BeCO3, PbI2, Ag2SO4, LiBr, K2Cr2O7, AgCl, Mg3(PO4)2 2. Write down the anions produced from the following acids in water.
HClO3, H3PO4, CH3COOH, HCN, HOCl, H2SO4, HNO2, HMnO4, H3PO3, H3PO2,
HIO4, HClO2, 3. Find the total and net ionic equations in case of the following.Balancing is not necessary.
(i) HI(aq) + KOH(aq) ------> KI(aq) + H2O(l)
(ii) NaOH(aq) + HClO4(aq) ------> NaClO4(aq) + H2O(l)
(iii) HNO3(aq) + NH3(g) ------> NH4NO3 (aq)
(iv) CuSO4(aq) + H2S(g) ------> CuS(s) + H2SO4(aq)
(v) H2SO4(aq) + MgO(s) ------> MgSO4(aq) + H2O(l)
(vi) Zn(s) + NaOH(aq) ------> Na2ZnO2(aq) + H2(g)
(vii) Al(OH)3(s) + HBr(aq) ------> AlBr3(aq) + H2O(l)
(viii) Mn(NO3)2(aq) + Na2S(aq) ------> MnS(s) + NaNO3(aq)
(ix) Ba(OH)2(aq) + H2SO4(aq) ------> BaSO4(s) + H2O(l)
(x) H3PO4(aq) + Ca(OH)2 (aq)------> Ca3(PO4)2 (s)+ H2O(l)
(xi) MnO2(s) + HCl(aq) ------> MnCl2(aq) + Cl2(g) + H2O(l)
(xii) KBr(aq) + H2SO4(aq) ------> K2SO4(aq) + Br2(l) + SO2(g) + H2O(l)
Dr. S. S.Tripathy 18 Concepts in Chemistry
(xiii) Cu(s) + HNO3(aq) ------>Cu(NO3)2 + H2O(l) + NO(g)
(xiv) K2Cr2O7(aq) +H2SO4(aq) + KNO2(aq) ------> K2SO4 + Cr2(SO4)3(aq) + KNO3(aq)+ H2O(l)
(xv) I2(s) + Na2S2O3(aq) -----> Na2S4O6(aq) + NaI(aq) → (xvi) HI +HNO2 NO + H2O + I2 → (xvii) KOH + KMnO4 K2MnO4 + O2 + H2O 4. Predict the products of the following neutralisation reactions and also write the total and net ionic equations.These are the simple double replacement(ion exchange) reactions that you know. Refer the tables on solubility rules to know whether the salt produced is soluble(aq) or insoluble(s). (i) NaOH(aq) + HBr(aq) → ______+ ______→ (ii) Ca(OH)2(aq) + HNO3(aq) ______+ ______→ (iii) H2SO4(aq) + KOH(aq) ______+ ______→ (iv) Ba(OH)2(aq) + HCl(aq) ______+ ______→ (v) Mg(OH)2 (s) + H2SO4(aq) ______+ ______→ (vi) MnO(s) + H3PO4 (aq) ______+ ______→ (vii) Al2O3 (s)+ H2SO4 (aq) ______+ ______→ (viii) Fe(OH)3 (s)+ HBr(aq) ______+ ______→ (ix) NaOH(aq) + HClO4 (aq) ______+ ______→ (x) Cr2O3(s) + H2SO4(aq.) ______+ ______
5. Assign the Oxidation Number(ON) to element underlined in the following.
NaCl, BaF2, NH3, H2O, NaH, SO3, CO2, P2O3, HClO, MnCl2, As(NO3)3, SnSO4, 2- 2- + - - - Hg2Cl2, H2SO4, BCl3, SF6, HClO3 , SO3 , S4O6 , NH4 , NO3 , NO2 , H2CO3, ClO2 ,
K2Cr2O7, Na2O2, MnO2, SnCl4, FeBr3, Cu2Cl2, NH4NO3, NO, N2O, N2, ZnO, K2O, 2- - 2- Ca(OH)2, Al, Al2(SO4)3, OF2, N2O5, S2O3 , OCN , HCN, S2O8 , H2SO5, CrO5, KO2, 2- 4- Ni(CO)4, CaS5, [SnS3] ,[Fe(CN)6] , - - 2- + XeOF4, NaN3, SCN , O2F2, CN , N3H, [Ni(CN)4] , [Cr(H2O)4(CN)2] , C3H8, C6H12O6, 2- BrCl3, HIO3, H2[SiF6], CaNCN, NaCN, Ca(OCl)Cl, S2O3 6. Indicate in the following reactions which is oxidized and which reduced. Also identify which is Oxidising Agent(OA) and which Reducing Agent(RA). Assign the ON of the involved atoms and show the Oxidation and Reduction processes by arrow mark. → (i) K + H2O KOH + H2 → (ii) NH3 + O2 N2 + H2O → (iii) Na + NH3 NaNH2 + H2 → (iv) Ca + H2O Ca(OH)2 + H2 → (v) Al + NaOH + H2O NaAlO2 + H2 → (vi) Zn + H2SO4 ZnSO4 + H2 → (vii) CuO + NH3 Cu + H2O + N2 → (viii) P4 + O2 P2O5 → (ix) H2 + Cl2 HCl → (x) H2O2 + H2S S + H2O
Dr. S. S.Tripathy Logic of Inorganic Reactions 19 → (xi) FeBr3 + Cl2 FeCl3 + Br2 → (xii) I2 + Na2S2O3 NaI + Na2S4O6 → (xiii) SnCl2 + HgCl2 SnCl4 + Hg2Cl2 → (xiv) Cu + HNO3 Cu(NO3)2 + NO2 + H2O
(xv) KMnO4 + H2SO4 + FeSO4 ------> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O
(xvi) K2Cr2O7 + H2SO4 + H2S ------> K2SO4 + Cr2(SO4)3 + S + H2O 7. Indicate in the following reactions which is oxidized and which reduced. Also identify which is Oxidising Agent(OA) and which Reducing Agent(RA). Assign the ON of the involved atoms and show the Oxidation and Reduction processes by connecting lines. → (i) Fe2O3 + CO CO2 + Fe 2- → 2- - (ii) I2 + S2O3 S4O6 + I → (iii) Fe2(SO4)3 + H2O2 FeSO4 + O2 + H2SO4 → (iv) Ca(OCl)2 + KI + HCl I2 + CaCl2 + H2O + KCl → (v) PbO2 + HBr PbBr2 + Br2 + H2O → (vi) HNO3 + Zn Zn(NO3)2 + NH4NO3 + H2O → (vii) NH4NO3 N2O + H2O → (viii) Pb(NO3)2 PbO + NO2 + O2 → (ix) P4 + H2SO4 SO2 + H3PO4 + H2O → (x) Cl2 + NaOH NaCl + NaClO3 + H2O 8. Show by indicating the ON which is oxidised and which reduced.
(i) CuS + HNO3 ------> Cu(NO3)2 + S + H2O + NO
(ii) Zn + HNO3 ------> Zn(NO3)2 + H2O + NH4NO3
(iii) MnO + PbO2 + HNO3 -----> HMnO4 + Pb(NO3)2 + H2O
(iv) KMnO4 + KCl + H2SO4 ------> MnSO4 + K2SO4 + Cl2 + H2O
(v) K2C2O4 + K2Cr2O7 + H2SO4 ------> K2SO4 + Cr2(SO4)3 + CO2 + H2O
(vi) HNO3 + HI -----> NO + I2 + H2O
(vii) NH3 + O2 ------> NO + H2O
(viii) CuO + NH3 -----> N2 + Cu + H2O
(ix) SO2 + H2S ----> S + H2O
(x) NaNO3 -----> NaNO2 + O2
(xi) KClO3 ----> KCl + O2
(xii) Pb(NO3)2 ------> PbO + NO2 + O2
(xiii) Zn + NaOH ------> Na2ZnO2 + H2
(xiv) KClO3 + H2SO4 -----> KHSO4 + O2 + ClO2 + H2O
(xv) Sn + HNO3 -----> SnO2 + NO2 + H2O
(xvi) I2 + HNO3 -----> HIO3 + NO2 + H2O
(xvii) KBr + H2SO4 -----> K2SO4 + Br2 + SO2 + H2O
(xviii) Ca3(PO4)2 + SiO2 + C ------> CaSiO3 + P4 + CO
(xix) Na + NH3 -----> NaNH2 + H2
(xx) H2O2 + H2S -----> S + H2O
(xxi) Cr(OH)3 + Na2O2 ------> Na2CrO4 + H2O
(xxii) O3 + Hg ------> Hg2O + O2
Dr. S. S.Tripathy 20 Concepts in Chemistry
(xxiii) K4[Fe(CN)6] + O3 + H2O -----> K3[Fe(CN)6] + O2 + KOH
(xxiv) NaBrO3 + NaBr + HCl -----> NaCl + Br2 + H2O
(xxv) KMnO4 ------> K2MnO4 + MnO2 + O2 → (xxvi) MnSO4 + (NH4)2S2O8 + H2O MnO2 + H2SO4 + (NH4)2SO4 → (xxvii) Zn + H2SO4 + As2O3 AsH3 + ZnSO4 + H2O → (xxviii) ClO2 + H2O2 + NaOH NaClO2 + O2 + H2O - → - (xxix) N2H4 + BrO3 N2 + Br + H2O → (xxx) CuSO4 + KI Cu2I2 + K2SO4 + I2 → (xxxi) Fe + H2O Fe3O4 + H2 → (xxxii) H2O2 + NaOCl NaCl + O2 + H2O
RESPONSE TO SAQS SAQ 1: 2 1 1 2 (i) Mg Cl = MgCl2 (ii) Na (SO4) = Na2SO4 2 1 3 2 (iii)Ca (NO3) = Ca(NO3)2 (iv) Fe S = Fe2S3 2 2 1 2 (v) Cu O = CuO (vi) K (SO3) = K2SO3 3 3 (vii)H2SO4 (viii) Al (PO4) = AlPO4 2 2 1 1 (ix) Ba (O2) = BaO2 (x) (NH4) (CN) = NH4CN
Note that the parenthesis used for SO4 in (ii) has been removed as there is no coefficient for it. But in (iii) the parenthesis is retained for NO3 as there is a coefficient 2 for it. Similarly the parentheses have been removed in (vi), (viii), (ix) and (x). If you are not thorough in writing the formula you first practise in the way shown above i.e first place the valencies at the top (superscript) and simplify if required and then crisscross them as subscripts to write the formula.
SAQ 2: ZnCO3, Cu2S, (NH4)2SO4, K2O, HgSO3, FeS, Na2CO3, Ca(NO2)2, SnCl2,
HNO3
SAQ 3: ammonium phosphate, ferric chloride, potassium sulphite, magnesium nitride, sodium peroxide, mercurous carbonate, aluminium phosphide, calcium bicarbonate (or calcium hydrogen carbonate), mercuric sulphide, phosphoric acid
SAQ 4: Set-I: Ca(CH3COO)2, K2MnO4, Mg(OCl)2, Fe2(SO4)3, Hg(NO3)2, NH4HCO3,
Cu2S, AlBr3, ZnSO3, HNO2,
Set-II: Na2CrO4, MgCl2, K3PO4, Ca3P2, SnS, PbCO3, Fe(NO2)2, HgHCO3, NH4ClO3, HBr
Set-III: Na3PO4, CaSO4, K2Cr2O7, Mg(ClO2)2, SnCl4, Al4C3, AgNO2, CoS, Fe2(SO3)3,
CaMnO4
Dr. S. S.Tripathy Logic of Inorganic Reactions 21
Set-IV: Mg(HCO3)2, NaCN, FeSO4, Cr(OH)3, ZnS2O3, K4[Fe(CN)6], (NH4)2Cr2O7,
As(NO3)3, Cu(HCO3)2, HOCl(HClO)
Set-V: Sr(NO3)2, K2SO3, Cu2[Fe(CN)6], AlP, Pb(CH3COO)2, Cu2CO3, NaHCO3, Cr2O3,
Mg(ClO4)2, H3PO4
Set-VI: K2HPO3, Na2S2O3, Fe3(PO4)2, Ca3[Fe(CN)6]2, SnO, PbO2, HgCr2O7, CaC2,
(NH4)2SO4, HClO3
Set-VII: NaAlO2, CaH2, BaO2, Mn3(PO4)2, K2CrO4, Mg(MnO4)2, CrO, Fe(OH)2,
K2ZnO2, HClO4
Set-VIII: K2H2Sb2O7, (NH4)2C2O4, Na2B4O7, K3[Co(NO2)6], CaSiO3, Cu2P2O7,
Na2S4O6, Ca(CH3COO)2, Cu2S2O8, KH2PO2
SET-IX: Na2WO4, K2[Fe(CN)5(NO)], HMnO4, NH4NCS, Mn(BrO2)2, KCrO2, SbOCl,
Hg(IO4)2, Fe(VO3)2, Na2PbO3
SET-X: AlH3, Bi(OI)3, Fe3[Fe(CN)6]2, Cr(NC)3, Ba(H2PO4)2, RbO2, (NH4)2MoO4,
AsCl5, CoAsO3, Au(ClO4)3
SET-XI: K[Ag(CN)2], NaBiO3, NH4BO2, K2SnO3, BiOCl, NaFeO2, Fe2O3, V2O5,
Pb(SO4)2, KSbO2 SAQ 5: 2+ 2- 2+ 3- + - + 2- 2+ Set-I:Zn and CO3 , 3Ca & 2PO4 , NH4 & NO3 , 2H & S , Ca & - + - + 2- 3+ 4- 2+ - 2CH3COO , Na & ClO3 , 2H & SO4 , 4Al & 3C , Fe & NO2 + - 3+ 2- + - 2+ 2- 3+ - Set-II:K & MnO4 , 2Fe & 3SO4 , H & NO3 ,Mg & CO3 , Al & 3OH , + 2 + - + 2+ - - 2+ 2Na & Cr2O7 , NH4 & ClO4 , Hg (or Hg2 ) & NO3 (2 NO3 ) Cu & 2- + - S2O3 , H and Cl (Note that the charges of the acid and basic parts have been given to all without considering whether they are soluble in water or not)
SAQ 6: H3PO4(W), H2SO4(S), HCN(W), CH3COOH(W),HNO3(S), HF(W),
HCl(S),HCOOH(W), HBr(S), HNO2(W), H2CO3(W),HI(S), H2SO3(W), HBrO(W),
HClO4(S) 2+ - + 2- SAQ7: (i) TIE:Ba + OH + H + SO4 ------> BaSO4(s) + H2O(l)
Since BaSO4 is associated with a (s) mark, it is insoluble and hence it has not been ionised. H2O also is associated with (l) mark, it has also not been ionised. Note that in the above equation we should have written 2 OH- and 2 H+ in the LHS which we did not. This is because we are not intending to balance the equation now and as a matter of fact the equation given in the question is not a balanced equation. Our sole purpose is to write the ions with their appropriate charges. That is all. We don't have to bother how many of each ion is produced in the LHS and RHS at this moment. In other words, we are not interested in balancing the equation now. 2+ - + 2- NIE:Ba + OH + H + SO4 ------> BaSO4(s) + H2O(l) Since no ion is found to be common in both sides, no ion is a spectator. All ions are involved and hence the net ionic equation is same as total ionic equation.
Dr. S. S.Tripathy 22 Concepts in Chemistry
- + - + + - + + (ii) TIE: CH3COO + H ++ Na OH CH3COO Na H2O(l) + - NIE:H +OH ------> H2O(l) In this case, like the first example shown in the text, H+ ion from acid reacts with OH- ion from base to form undissociated H2O molecule. In fact this is the net ionic equation for almost all acid- base(neutralisation) reactions. + - 2+ - (iii) TIE:H + NO3 + Cu(OH)2(s) ------> Cu + NO3 + H2O(l) - In this case, NO3 is the spectator ion which is cancelled from both the sides to get NIE. + 2+ NIE: H + Cu(OH)2(s) ------> Cu + H2O(l) SAQ 8: Refer the solubility rules to verify answer.
(i) HCl(S): All inorganic acids are soluble; (ii)NH4Cl(S): All ammonium salts are soluble
(iii) PbSO4-(I): All sulphates are soluble excepting a few such as PbSO4, SrSO4, BaSO4,
HgSO4, Ag2SO4 which are insoluble.
(iv) Ca(NO3)2- (S): All nitrates are soluble: (v)Hg2Cl2-(I): All chorides are soluble excepting
a few such as Hg2Cl2, AgCl and PbCl2 which are insoluble.
(vi) Na2SO4- (S): All Na salts are soluble; (vii)BaSO4-(I): (viii)K3PO4- (S): All
K salts are soluble; (ix)H3PO4-(S): All inorganic acids are soluble;
(x) CH3COONa-(S): All Na salts and all acetates are soluble. + 2+ 2+ (xi) AgBr-(I): All bromides are soluble excepting Ag , Hg2 and Pb .
(xii) Mg(ClO3)2- (S): All chlorates are soluble; (xiii)AgCl-(I)
(xiv) NaClO4-(S): All perchlorates are soluble.
(xv) Bi(NO3)2 - (S): All nitrates are soluble SAQ 9:
+ + + + + (i) CaF2-(I): All fluorides are insoluble excepting those of Na , K , NH4 , Ag , Tl which are soluble. + + + 2+ (ii) Fe(OH)3 -(I): All hydroxides are insoluble excepting those of Na , K , NH4 , Ca , Sr2+, Ba2+ which are soluble. + + + 2+ (iii) BeCO3-(S): All carbonates are insoluble excepting those of Na , K , NH4 , Be which are soluble.
(iv) CaCO3- (I) + + + 2+ 2+ 2+ 2+ (v) CaS-(S): All sulphides are insoluble excepting Na , K , NH4 , Mg ,Ca ,Sr ,Ba which are soluble. (vi) NaOH-(S)
(vii) NH4F-(S) + + + 2+ (viii) Ca3(PO4)2-(I): All phosphates are insoluble excepting those of Na , K , NH4 , Be which are soluble. (ix) KF-(S) (x) CuS-(I)
Dr. S. S.Tripathy Logic of Inorganic Reactions 23
SAQ 10:
+ 2- 2+ - 2+ 2- (i) TIE: H + SO4 + Ca + OH ------> Ca + SO4 + H2O(l) + - NIE: H + OH ------> H2O + - + - + - (ii) TIE: Ag + NO3 + Na + Cl ------> AgCl(s) + Na + NO3 NIE: Ag+ + Cl------> AgCl(s) + - + 3- + 3- (iii) TIE: Na + OH + H + PO4 ------> Na + PO4 + H2O(l) + - NIE: H + OH ------> H2O 2+ - + 2- + - (iv) TIE: Ba + Cl + K + CrO4 ------> BaCrO4(s) + K + Cl 2+ 2- NIE: Ba + CrO4 ------> BaCrO4(s) 2+ - + - + - (v) TIE: Pb + NO3 + H + Cl ------> PbCl2(s) + H + NO3 2+ - NIE: Pb + Cl ------> PbCl2(s) Note that in all these equations, balancing has not been done. SAQ 11: 3+ - (i) Al + OH ------> Al(OH)3(s) + 2+ (ii) Fe(OH)2(s) + H ------> Fe + H2O(l) 2+ 2- (iii) Ca + CO3 ------> CaCO3(s) 3+ - (iv) Fe + OH ------> Fe(OH)3(s) + - (v) H + OH ------> H2O(l) In all the above equations the students are advised to write first the total ionic equation(TIE) and then write net ionic equations(NIE) after cancelling the spectator ions. SAQ 12: + 2+ (i) Zn(s) + H ------> Zn + H2(g) + 2+ (ii) Mg(s) + H ------> Mg + H2(g) (iii) Cu2+ + Zn(s) ------> Cu(s) + Zn2+ - - (iv) Al(s) + OH ------> AlO2 + H2(g) - + (v) OH + H ------> H2O(l) SAQ 13:
+ 2- + 2- 2+ 2- 3+ 2- + 2- 3+ 2- (i) TIE: K + Cr2O7 + H + SO4 + Fe + SO4 ------> Cr + SO4 + K + SO4 + Fe + SO4 + H2O(l) 2- + 2+ 3+ 3+ NIE: Cr2O7 +H + Fe ------> Cr + Fe + H2O(l) + 2- Note that we have cancelled all K ions and all SO4 ions from LHS and RHS without considering 2- whether such ions appear same number times in both sides or not. SO4 ions appeared two times in the LHS while the same appeared three times in the RHS. Even then we cancelled all of them. This is because we have not balanced the equation and had we balanced the equation, + 2- the number of K ions and SO4 ions would have definitely appeared same number of times in both the sides. Since we are interested only to write the net ionic equations, we tactfully detected the unimportant(spectator) ions which are not involved in the reaction and cancelled them. + - + - + - (ii) TIE: Cl2(g) + Na + OH ------> Na + Cl + Na + ClO3 + H2O(l) - - - NIE: Cl2(g) + OH ------> Cl + ClO3 + H2O(l) Here also we cancelled one Na+ in LHS and two Na+ in the RHS. To remind you again that we have not done anything wrong as we merely want the net ionic equation which is not balanced. It may be remembered here that while cancelling the spectator ions from the two sides the
Dr. S. S.Tripathy 24 Concepts in Chemistry students should find first the ions which do not take part in the reaction. Then remove them all without considering how many times they appear in which side.
+ - + 2- + 2- 2+ 2- (iii) TIE: K + MnO4 + H + SO4 + H2C2O4(s) ------> K + SO4 + Mn + SO4 + CO2(g) + H2O(l) - + 2+ NIE: MnO4 + H + H2C2O4(s) ------> Mn + CO2(g) + H2O(l) 2- Here also we have cancelled SO4 one time from LHS and two times from RHS. Never mind, we are not going to get a balanced ionic equation now. SAQ 14: Zero. All of them are in the uncombined elementary state. SAQ 15: (i) Ca = +2(Since the valency of Ca is 2), (ii) Cl= -1(since the valency of Cl is 1) (iii) Fe = +2(since valency of ferrous iron is 2), (iv) Cu = +1(since valency of cuprous is 1) (v) S = -2(since valency of sulphide is 2), (vi) Sn = +2(since valency of stannous is 2) (vii)Mg = +2(valency of Mg is 2) and N = -3(valency of nitride is 3) (viii) Hg =+2(valency of mercuric is 2) and I = -1(valency of iodide is 1) Note that the basic parts have been given +ve charge and acid parts -ve charge. SAQ 16: +2, +2, -3, +3, +4, -4, +1, +1, -2
SAQ 17: -2, 0 (uncombined state), -2, -1(since it is peroxide: O2 has a charge of -2, so for one oxygen atom it is -1, refer the valency table), -½ (since it is superoxide: O2 has a charge of -1, so for one oxygen atom it is -½, refer the valency table),-2 SAQ 18: +1, +1, +1, -1, -1, -1 (since the last three compounds are metallic hydrides)
It is to to be noted that ON always is assigned for one atom, for H2O, the ON of hydrogen is +1, not +2, similarly for NH3, the ON of H is +1 not +3. But for calculation purpose when we find the sum of ONs we take +2 and +3 respectively. We shall see this in the SAQ 19. SAQ 19: (i) x +2(-2) =0, ⇒ x = +4 (ii) +1+x +3(-2) =0, ⇒ x = +5, (iii) +2 +x +4(-2)=0, ⇒ x =+6, Note that for one H atom the ON is +1 and so for 2 H atoms the total ON is +2. (iv) +1 + x + 4(-2) = 0 ⇒ x = +7 (v) 2x + 5(-2) =0 ⇒ x=+5(note that since there are two P atoms we wrote 2x but ON of P=x). (vi) +2 +2x + 7(-2) =0 ⇒ x= +6,(there are two K atoms and each has an ON of +1, again there are two Cr atoms each has ON of x) (vii) +1+x+4(-2)=0 ⇒ x= +7 SAQ 20:(A) (i) x+3=0, ⇒ x= -3 (ii) +1+x+3(-2) = 0, ⇒ x=+5, (iii)x+2(-2)=-1 ⇒ x=+3, (iv) x+2(-2)=0, ⇒ x=+4, (v) 0(uncombined state) (vi) 2x+5(-2)=0, ⇒ x=+5, (vii) +7(see SAQ 19), (viii) 2x-14=-2, ⇒ x=+6, (ix) 2x-6=-2, ⇒ x=+2, (x) x-6=-1, ⇒ x=+5 ⇒ ⇒ (xi) x+4=+1, x=-3, (xii) x-6=-1, x=+5 (xiii) For NH4NO3, we have to find separately + - for the two N atoms, once taking NH4 (already found in bit xi as -3) and and then taking NO3 (already found in bit xii as +5). Note that in this compound you should not find a single ON for N
Dr. S. S.Tripathy Logic of Inorganic Reactions 25 by taking 2x for N, (xiv) x-4=0, ⇒ x=+4, (xv) +3, since the valency of chromic(Cr), a monoatomic ion is 3, hence its ON has to be +3, (xvi) +1+x-2 =0, ⇒ x=+1, (xvii) x-6=-2, ⇒ x=+4, (xviii) -1(Cl is a monoatomic ion having valency 1, so ON is -1) (xix) +3, Sb(ous) is a monoatomic ion having valency 3, hence ON +3. (xx) x+3 =0, ⇒ x= -3 - - (B) (i)(NO3 ) 5 (ii) (SO3) 6 (iii)(ClO3 ) 5 (iv) (NH3) 3 (v) (N2O5) 5 SAQ 21: Student is advised to find the ON of the elements marked with asterisk by x method explained before and then proceed to analyse which is oxidation and reduction.
(i) LHS: Na=0(uncombinded state), H in H2O=+1, RHS: Na in NaOH=+1 and H in
H2=0(uncombinded state) [O]
0 +1 +1 0 Na ++ H2O NaOH H2 (RA) (OA) [R]
In this case Na is oxidised to NaOH and hence acts as reducing agent(RA) while H2O is reduced to H2 and hence acts as oxidising agent(OA). (ii) LHS: N in N2 =0(uncombinded state), H in H2=0,RHS: N in NH3= -3, H in NH3=+1 [O] 0 0 -3 +1 N2 + H2 NH3 (OA) (RA) [R]
N2 is reduced to NH3 as its ON decreased from 0 to -3 and hence acts as OA while H2 is oxidised to NH3 as its ON increased from 0 to +1. So it acts as RA. (iii) LHS: P in P4=0(uncombined state), Cl in Cl2= 0, RHS: P in PCl3=+3, Cl in PCl3= -1 [O] 00 +3 -1 P + Cl PCl3 4 2 [R] (RA) (OA)
(iv) LHS: N in NH3=-3, Cu in CuO = +2, RHS: Cu =0(uncombined state), N in N2=0 [O] +2 -3 00 CuO + NH3 Cu ++ N2 H2O (OA) (RA) [R]
CuO is reduced to Cu as the ON of Cu decreased from +2 to 0 and hence acts as OA while NH3 is oxidised to N2 as the ON of N increased from -3 to 0, and hence acts as RA.
(v) LHS: H in H2=0, I in I2=0, RHS: H in HI= +1 and I in HI= -1, [O] 00 +1 -1 HI H2 + I2 [R] , (RA) (OA)
Dr. S. S.Tripathy 26 Concepts in Chemistry
In this case H2 is oxidised to HI hence it is the RA while I2 is reduced to HI so that it is OA.
(vi) LHS: Mg =0, H in H2SO4=+1, RHS: Mg in MgSO4=+2, H in H2=0 [O] 0 +1 +2 0 Mg ++ H2SO4 MgSO4 H2 (RA) (OA) [R]
In this case Mg is oxidised to MgSO4 and hence is RA while H2SO4 is reduced to H2 and hence is the OA.
(vii) LHS: N in N2=0, O in O2=0, RHS: N in NO=+2, O in NO= -2 [O] 0 0 +2 -2 N2 + O2 NO (RA) (OA) [R]
(viii) LHS: Cu =0, N in HNO3 =+5, RHS: Cu in Cu(NO3)2=+2, N in NO2=+4 [O] 0 +5 +2 +4 Cu + HNO3 Cu(NO3)2 ++ NO2 H2O (RA) (OA) [R]
(ix) LHS: Mn in MnO2= +4, Cl in HCl =-1, RHS: Mn in MnCl2=+2, Cl in Cl2=0 [O] +4 -1 +2 0 MnO2 + HCl MnCl2 + Cl2 + H2O (OA) (RA)[R]
In this case MnO2 is reduced to MnCl2 as the ON of Mn decreased from +4 to +2, hence it is
OA. HCl is reduced to Cl2 as the ON of Cl increased from -1 to 0, hence HCl is RA.
(x) LHS: Mn in KMnO4=+7, Fe in FeSO4 =+2(valency of ferrous is 2),
RHS: Mn in MnSO4=+2(valency of Manganous is 2), Fe in Fe2(SO4)3=+3(valency of ferric is 3) [O] +3 +7 +2 +2 K SO MnSO Fe (SO ) H O KMnO4 + H2SO4 + FeSO4 2 4 ++4 2 4 3 +2 (OA) (RA) [R]
In this case, FeSO4 is oxidised to Fe2(SO4)3 as the ON of Fe increased from +2 to +3, hence
FeSO4 is the RA. KMnO4 is reduced to MnSO4 as the ON of Mn decreased from +7 to +2, hence KMnO4 acts as OA. SAQ 22: (i) Redox: The ON of H changes from 0 to +1 while Cl changes from 0 to -1 (ii) Metathesis: It is double replacement reaction i.e partners(acid and basic radicals) have been exchanged. In such reactions, no element undergoes any change in ON. (iii) Redox: The ON of Cu changes from 0 to +2 while of N changes from +5 to +4. (iv) Redox: The ON of Na changes from 0 to +1 while of H changes from +1 to 0. (v) Metathesis: It is also a double replacement reaction: Neutralisation reaction. (vi) Metathesis: Double replacement reaction. Dr. S. S.Tripathy Logic of Inorganic Reactions 27
ANSWERS TO PRACTICE QUESTIONS
1. NaNO3-S(All nitrates are soluble), KBr-S(All K salts are soluble), MgF2-I(All fluorides + + + + + are insoluble excepting Na , K , NH4 , Ag , Tl ), FeCl2-S(All chlorides are soluble + 2+ 2+ + excepting Ag , Hg2 and Pb ), MgCO3-S(All carbonates are insoluble excepting Na , + + 2+ 2+ 2+ 2+ K , NH4 , Be ,Mg ), BaSO4-I(All sulphates are soluble excepting Pb , Sr , Ba2+,Hg2+,Ag+), MgS-S(All sulphides are insoluble excepting the sulphides of Na+, K+, + 2+ 2+ 2+ 2+ NH4 , Mg ,Ca ,Sr ,Ba ), Na3PO4-S(All Na salts are soluble), Na2CrO4-S(all Na
salts are soluble), Ba(ClO4)2-S(All perchlorates are soluble), Ba(OH)2-S(All hydroxides + + + 2+ 2+ 2+ are insoluble excepting those of Na , K , NH4 , Ca , Sr , Ba ), Hg2Cl2-I(All chlorides + 2+ 2+ are soluble excepting Ag , Hg2 and Pb ), NH4F-S(All flourides are insoluble excepting + + + + + + + Na , K , NH4 , Ag , Tl ), BeCO3-S(All carbonates are insoluble excepting Na , K , + 2+ 2+ + 2+ NH4 , Be ,Mg ), PbI2-I(All iodides are soluble excepting those of Ag , Hg2 and 2+ 2+ 2+ 2+ 2+ + Pb ), Ag2SO4-I(all sulphates are soluble excepting those of Pb , Sr , Ba ,Hg ,Ag ), + 2+ 2+ LiBr-S(all bromides are soluble excepting those of Ag , Hg2 and Pb ), K2Cr2O7-S(all + 2+ 2+ K salts are soluble), AgCl-I(All Chlorides are solulbe excepting Ag , Hg2 and Pb ), + + + 2+ 2+ Mg3(PO4)2-S(all phosphates are insoluble excepting those of Na , K , NH4 , Be ,Mg ). - 3- - - - 2. ClO3 (chlorate), PO4 (phosphate), CH3COO (acetate), CN (cyanide), OCl 2- - (hypochlorite), SO4 (sulphate), NO2 (nitrite). 3. Net Ionic equations(NIE) are given below. The readers are to first write the TIE for each. + - (i) H + OH ------> H2O + - (ii) H + OH ------> H2O + + (iii) H + NH3(g) ------> NH4 2+ + (iv) Cu + H2S(g) ------> CuS(s) + H + 2+ (v) H + MgO(s) ------> Mg + H2O(l) - 2- (vi) Zn(s) + OH ------> ZnO2 + H2(g) + 3+ (vii) Al(OH)3(s) + H ------> Al + H2O(l) (viii) Mn2+ + S2------> MnS(s) 2+ - + 2- (ix) Ba + OH + H + SO4 ------> BaSO4(s) + H2O(l) + 3- 2+ - (x) H + PO4 + Ca + OH ------> Ca3(PO4)2(s) + H2O(l) + - 2+ - (xi) TIE: MnO2(s) + H + Cl ------> Mn + Cl + Cl2(g) + H2O(l) + - 2+ NIE: MnO2(s) + H + Cl ------> Mn + Cl2(g) + H2O(l) Note that in this case Cl- has to be removed from RHS but not from LHS. If you remove Cl- - from LHS, then where from Cl2(g) is produced in RHS? So in such case, you consider Cl to be an unimportant species in the RHS and is to be removed from it. But Cl- is the involved species - in LHS which produces Cl2 in the RHS. So tactfully keep Cl in LHS and remove from RHS. Same type of reaction is given in the next question. + - + 2- + 2- (xii) TIE: K + Br + H + SO4 ------> K + SO4 + Br2(l) + SO2(g) + H2O(l) - + 2- NIE: Br + H + SO4 ------> Br2(l) + SO2(g) + H2O(l) 2- 2- Here also you found that SO4 has been removed from RHS but not from LHS. It is SO4 of 2- LHS which has resulted SO2 in the RHS. SO4 is the spectator ion in the RHS and so deleted from that side.
Dr. S. S.Tripathy 28 Concepts in Chemistry
+ - 2+ - (xiii) TIE: Cu(s) + H + NO3 ------> Cu + NO3 + NO(g) + H2O(l) + - 2+ NIE: Cu(s) + H + NO3 ------> Cu + NO(g) + H2O(l) - Here also NO3 is deleted from RHS but kept in LHS for the same reason explained before. 2- + - 3+ - (xiv) NIE: Cr2O7 + H + NO2 ------> Cr + NO3 + H2O(l) 2- 2- - (xv) I2(s) + S2O3 ------> S4O6 + I → (xv) HI(aq) +HNO2(aq) NO(g) + H2O(l) + I2(s) + - + - TIE: H + I + H + NO2 ------> NO(g) + H2O(l) + I2(s) + - - NIE: H + I + NO2 ------> NO(g) + H2O(l) + I2(s) → (xvi) KOH(aq) + KMnO4(aq) K2MnO4(aq)+ O2 (g) + H2O (l) - - 2- NIE: OH + MnO4 ------> MnO4 + O2 + H2O + - 4. (i) NaBr(aq) + H2O(l); NIE: H + OH ------> H2O + - (ii) Ca(NO3)2(aq) + H2O; NIE: H + OH ------> H2O + - (iii) K2SO4(aq) + H2O(l); NIE: H + OH ------> H2O + - (iv) BaCl2(aq) + H2O(l); NIE: H + OH ------> H2O + 2+ (v) MgSO4(aq) + H2O(l); NIE: Mg(OH)2(s) + H + ------> Mg + H2O(l) + 3- (vi) Mn3(PO4)2(s) + H2O(l); NIE: MnO(s) + H + PO4 -----> Mn3(PO4)2(s)+ H2O(l) + 3+ (vii) Al2(SO4)3(aq) + H2O(l); NIE: Al2O3(s) + H -----> Al + H2O(l) + 3+ (viii) FeBr3(aq) + H2O(l); NIE: Fe(OH)3(s) + H -----> Fe + H2O(l) + - (ix) NaClO4(aq) + H2O(l); NIE: H + OH ------> H2O + 3+ (x) Cr2(SO4)3(aq) + H2O(l); NIE: Cr2O3(s) + H ------> Cr + H2O(l) 5. The beginner is advised to calculate by x method.
NaCl: +1(valency of Na is 1), BaF2: -1(valency of F is 1), NH3: -3, H2O: -2, NaH:-1(ON of
H in metallic hydrides is -1), SO3: +6, CO2: +4, P2O3: +3, HClO: +1, MnCl2: +2(valency of
Manganous is 2), As(NO3)3: +3(valency of arsenous is 3), SnSO4:+2(valency of stannous is 2),
Hg2Cl2: +1(valency of mercurous is 1), H2SO4: +6, BCl3: +3, SF6: +6, HClO3: +5, 2- 2- + - - SO3 : +4, S4O6 : 2.5(note that ON can be a fraction), NH4 : -3, NO3 : +5, NO2 : +3, - H2CO3: +4, ClO2 : +3, K2Cr2O7: +6, Na2O2: -1(the ON of oxygen in peroxide is -1),
MnO2: +4, SnCl4: +4(valency of stannic is 4), FeBr3: +3(valency of ferric iron is 3),
Cu2Cl2: +1(valency of cuprous is 1), NH4NO3: -3 and +5(to be found out seprately, refer SAQ
20), NO: +2, N2O: +1, N2: 0(uncombined state), ZnO: +2(valency of Zn is 2), K2O: +1(valency of K is 1), Ca(OH)2: +2(valency of Ca is 2), Al: 0(uncombined state), Al2(SO4)3: +3(valency 2- of Al is 3), OF2: +2(F is -1, so O has to be +2), N2O5 : +5, S2O3 : +2 OCN- : +4(C will lose 3 electrons to N and 1 electron to O); HCN: +2(C will lose 3 electrons to 2- 2- N and gain 1 electron from H); S2O8 : +6( there are 6 normal oxide and 1 peroxide [S2O6(O2)] ⇒ 2- , 2x-12-2=-2 x =+6); H2SO5: +6 (there are 3 normal oxide and 1 peroxide[SO3(O2)] , x-6-2 ⇒ =-2 x = +6) CrO5: +6 (there are 1 normal oxide and two peroxide groups [CrO(O2)2], x-2- ⇒ 4=0 x =+6) KO2 : -1/2 (superoxide) Ni(CO)4: 0(since CO is a neutral molecule) CaS5: -2/5 2- 2- 4 (since it is a polysulphide ion S5 ); [SnS3] : -2 ; [Fe(CN)4] : +2 and +2 respectively(since CN has -1 charge, ON of C is +2 as C loses 3 electrons to N and gains one electron due to its negative charge)
- - 2- XeOF4: +6; NaN3 : -1/3; SCN : -2; O2F2 : +1; CN : -3; N3H: -1/3; [Ni(CN)4] : +2 (x-4=-2, + CN is taken as -1); [Cr(H2O)4(CN)2] : +3 (x-2=+1); C3H8 : -8/3; C6H12O6 : 0 (note that ON Dr. S. S.Tripathy Logic of Inorganic Reactions 29 concept is irrelevant for organic compounds although it can be calculated like inorganic compounds);
BrCl3 : +3( Br is less electronegative than Cl); HIO3: +5; H2[SiF6]: +4; CaNCN: +4(C will lose two electrons to each N) NaCN: +2 (the ON of C in any cyanide ion is +2) Ca(OCl)Cl : +1 and -1 respectively from OCl- and Cl-( Note that the average ON of Cl in 2- CaOCl2 is 0 although in reality there are two types of Cl atoms in the molecule); S2O3 : +2 (but in reality there are two types of S atoms having ON +4 and 0 respectively so that the average ON is 0. Note that in the structure of thiosulphate ion there is one S-S linkage which breaks homolytically to make one S acquire 0 state. More you can know when you study the chapter chemical bonding)
6.(i) K goes from 0(K) to +1(KOH) and is oxidised while H goes from +1(H2O) to 0(H2) and
thus H2O is reduced. OA: H2O and RA: K
(ii) N goes from -3(NH3) to 0(N2), thus NH3 is oxidised, while O goes from 0(O2) to
-2(H2O), thus O2 is reduced. OA: O2 and RA: NH3
(iii) Na goes from 0(Na) to +1(NaNH2), thus Na is oxidised, while H goes from +1(NH3) to
0(H2), thus NH3 is reduced. OA: NH3, RA: Na
(iv) Ca goes from 0 to +2 while H goes from +1 to 0. Thus OA:H2O, RA:Ca
(v) Al goes from 0(Al) to +3(NaAlO2), thus Al is oxidised, while H goes from +1(NaOH) to
0(H2), thus NaOH is reduced. OA: NaOH, RA: Al
(vi) Zn goes from 0(Zn) to +2(ZnSO4), thus Zn is oxidised, while H goes from +1(H2SO4) to
0(H2), thus H2SO4 is reduced. OA: H2SO4, RA: Zn
(vii) Cu goes from +2(CuO) to 0(Cu), thus CuO is reduced while N goes from -3(NH3) to
0(N2), thus NH3 is oxidised. OA: CuO, RA: NH3
(viii) P goes from 0 to +5 while O goes from 0 to -2.OA: O2 and RA: P4
(ix) H goes from 0 to +1 while Cl goes from 0 to -1. OA: Cl2 and RA: H2
(x) O goes from -1(H2O2) to -2(H2O), thus H2O2 is reduced while S goes from -2(H2S) to
0(S), thus H2S is oxidised. OA: H2O2, RA: H2S
(xi) Br goes from -1 to 0 while Cl goes from 0 to -1. OA: Cl2, RA: FeBr3
(xii) I goes from 0(I2) to -1(NaI), thus I2 is reduced while S goes from +2(Na2S2O3) to
2.5(Na2S4O6), thus Na2S2O3 is oxidised. OA: I2, RA: Na2S2O3
(xiii) Sn goes from +2 to +4 while Hg goes from +2 to +1. OA: HgCl2, RA: SnCl2
(xiv) Cu goes from 0(Cu) to +2[Cu(NO3)2], thus Cu is oxidised while N goes from +5(HNO3)
to +4(NO2), thus HNO3 is reduced. OA: HNO3, RA: Cu
(xv) Mn goes from +7(KMnO4) to +2(MnSO4), thus KMnO4 is reduced while Fe goes from
+2(FeSO4) to +3[Fe2(SO4)3], thus FeSO4 is oxidised. OA: KMnO4, RA: FeSO4
(xvi) Cr goes from +6(K2Cr2O7) to +3[Cr2(SO4)3], thus K2Cr2O7 is reduced, while S goes
from -2(H2S) to 0(S), thus H2S is oxidised. OA: K2Cr2O7, RA: H2S 7. Only the changes in ON have been given. The reader is to assign the OA and RA. (i) Fe from +3 to 0 and C from +2 to +4 (ii) I from 0 to -1 and S from +2 to +2.5 (iii) Fe from +3 to +2, O from -1 to 0 (iv) Cl from +1 to -1, I from -1 to 0 (v) Pb from +4 to +2, Br from -1 to 0
Dr. S. S.Tripathy 30 Concepts in Chemistry
+ (vi) N from +5 to -3(NH4 in NH4NO3), Zn from 0 to +2 + - (vii) N from -3(NH4 ) to +1, N from +5(NO3 ) to +1(N2O).
In this case same substance(NH4NO3) is oxidised as well as reduced. OA and RA are not assigned to such redox reactions. These are called disproportionation reactions. We shall study about this later. (viii) N from +5 to +4, O from -2 to 0. Same as (vii):disproportionation (ix) P from 0 to +5, S from +6 to +4.
(x) Cl from 0 to -1(NaCl), Cl from 0 to +5(NaClO3). same as above: disproportionation 8. (i) S from -2 to 0 and N from +5 to +2. + (ii) Zn from 0 to +2 and N from +5 to -3 (NH4 )
(iii) Mn from +2 to +7(HMnO4) and Pb from +4 to +2. (iv) Mn from +7 to +2 and Cl from -1 to 0.
(v) C from +3(K2C2O4) to +4 and Cr from +6 to +3. (vi) N from +5 to +2 and I from -1 to 0. (vii) N from -3 to +2 and O from 0 to -2. (viii) Cu from +2 to 0 and N from -3 to 0. (ix) S from +4 to 0 and S from -2 to 0. (x) N from +5 to +3 and O from -2 to 0. (xi) Cl from +5 to -1 and O from -2 to 0. (xii) N from +5 to +4 and O from -2 to 0. (xiii) Zn from 0 to +2 and H from +1 to 0.
(xiv) Cl from +5 to +4(ClO2) and O from -2 to 0. (xv) Sn from 0 to +4 and N from +5 to +4. (xvi) I from 0 to +5 and N from +5 to +4. (xvii) S from +6 to +4 and Br from -1 to 0. (xviii) P from +5 to 0 and C from 0 to +2. (xix) Na from 0 to +1 and H from +1 to 0. (xx) O from -1 to -2 and S from -2 to 0.
(xxi) Cr from +3 to +6 and O from -1(Na2O2) to -2. (xxii) O from 0 to -2 and Hg from 0 to +1. (xxiii) Fe from +2 to +3 and O from 0 to -2. (xxiv) Br from +5 to 0 and Br from -1 to 0.
(xxv) Mn from +7 to +6(K2MnO4) and +4(MnO2), two reductions, O from -2 to 0. Note that a redox reaction can have more than one oxidation and reduction reaction, but at least one of each type of reaction has to be there. 2- (xxvi) Mn from +2 to +4 and O from -1(peroxydisulphate= [S2O6(O2)] ) to -2 (xxvii) Zn from 0 to +2 and H from +1 to -1 (xxviii) Cl from +4 to +3 and O from -1 to 0 (xxix) N from -2 to 0 and Br from +5 to -1 (xxx) Cu from +2 to +1 and I from -1 to 0 (xxxi) Fe from 0 to +-8/3 and H from +1 to 0 (xxxii) O from -1 to 0 and Cl from +1 to -1
Dr. S. S.Tripathy Logic of Inorganic Reactions 31
BALANCING CHEMICAL EQUATIONS
Balancing of an equation means placing suitable coefficients before the reactants and products so that the number of atoms of every element in the LHS is same as that in the RHS.
Example: Al(OH)3 + H2SO4 ------> Al2(SO4)3 + H2O (not balanced) 2Al(OH)3 + 3H2SO4 ------> Al2(SO4)3 + 3H2O (balanced) The second equation is balanced, because there are same number of Al atoms, same number of O, H and S atoms on the two sides. Metathesis reactions: You already know that a metathesis reaction is that reaction in which no change in ON takes place for any element. The balancing of such reaction is done by hit and trial or the inspection method. (1) Hit and Trial Method (Inspection Method) To balance a metathesis reaction (reaction in which ONs are not changed) such as double replacement reaction of the type shown before in the previous section is very easy. This is done by hit and trial (inspection) method. Mere looking at both the sides gives us clue as to what coefficients are to be placed before the reactants and products to get a balanced equation. Answer the following SAQs. SAQ 1: Balance the following reactions:
(i) Ca(OH)2 + H3PO4 ------> Ca3(PO4)2 + H2O (ii) ZnO + H2SO4 ------> ZnSO4 + H2O (iii) FeCl2 + H2S ------> FeS + HCl (iv) P2O5 + H2O ------> H3PO4 (v) NH4Cl + Ca(OH)2------> CaCl2 + NH3 + H2O REDOX REACTIONS: As you must have realised that balancing metathesis reactions( double replacement type) requires no special technique. Only hit and trial method (inspection) solves the purpose. However it is often difficult to balance a redox equation by hit and trial method. It is very much time consuming particularly for complicated reactions and moreover there is a possibility of arriving at a wrong balanced equation in hit and trial method. Therefore hit and trial method is not recommended for such reactions. For simple redox reactions, hit and trial method is a good choice. The following examples will illustrate the point.
Na + H2SO4 ------> Na2SO4 +H2 (not balanced) 2Na + H2SO4 ------> Na2SO4 +H2 (balanced) N2 + H2 ------> NH3 (not balanced) N2 + 3H2 ------> 2NH3 (balanced) P4 + Cl2 ------> PCl5 (not balanced) P4 + 10Cl2 ------> 4PCl5 (balanced) However it is difficult to balance other redox equations which are not simple type as shown above. The following examples will explain this.
4NH3 + 5O2 ------> 4NO + 6H2O No doubt one can balance the above equation by hit and trial method, but it is time taking. We therefore discuss a new method called Oxidation Number(ON) method to balance redox equations.
Dr. S. S.Tripathy 32 Concepts in Chemistry
(2) OXIDATION NUMBER(ON) METHOD: Let us take the following examples: +2 -3 00 (i) CuO + NH3 N2 ++ Cu H2O In this equation we find that the ON of Cu goes from +2 to 0. So what is the change in ON that the oxidising agent(CuO) has undergone? . It is |+2-0|=2. This 2 is placed as the coefficient before the reducing agent NH3. Note that change is merely the magnitude of difference between the two ONs.
CuO + 2 NH3 N2 ++ Cu H2O Look to the first equation. The ON of N goes from -3 to 0. What is the change of ON that the reducing agent has undergone? It is |-3-0| =3. This 3 is placed as coefficient before the oxidising agent CuO.
3 CuO + 2 NH3 N2 ++ Cu H2O After that, it becomes damn easy to balance by hit and trial method. We have to place 3 before Cu in RHS to equalise Cu. To equalise H we have to place 3 before H2O in RHS,and by doing so we find that the O is automatically balanced.
3 CuO + 2 NH3 N2 ++ 33 Cu H2O The beauty of this method is that while balancing for the last but one element, the last element is automatically balanced. In the above example, we found that while balancing for H, oxygen atoms were automatically balanced. Let us look to another example.
(ii) NH3 + O2 ------> NO + H2O (R) -3 0 +2 -2 NH3 + O2 NO + H2O (O)
In this case we found that the change in ON in the RA(NH3) is |-3-(+2)|=5, so 5 is placed as coefficient before OA(O2)
NH3 + 5 O2 ------> NO + H2O Look again to the original equation. O goes from 0 to -2. The change is |0-2|=2. This change is for one oxygen atom. But the OA contains two O atoms(O2), so the total change is 2X2=4.
Thus 4 is placed as coefficient before the RA(NH3). Note that in NH3 there is one N atom, so the net change is 5. But in O2 there are two O atoms, so the total change is not 2, it is 4.
4NH3 + 5O2 ------> NO + H2O After this, it becomes extremely easy task to balance by inspection. 4 is to be placed before NO in RHS to equalise N on both sides. To equalise H, we have to place 6 before H2O. In doing so we find that O is automatically balanced(10 on either side).
4NH3 + 5O2 ------> 4NO + 6H2O Now let us know a few rules for balancing an equation by ON method. Dr. S. S.Tripathy Logic of Inorganic Reactions 33
RULES FOR BALANCING BY ON METHOD: (i) First the ON of the atoms undergoing oxidation-reduction are found out and placed above the respective atoms. (ii) The total change of ON in the OA is placed as the coefficient before RA and the total change of ON in RA is placed as coefficient before OA. (iii) Then hit and trial procedure is adopted to balance the equation.
Another example is given below for a better understanding.
Cl2 + NH3 ------> HCl + N2 0 -3 -1 0 + + Cl2 NH3 HCl N2 ( Cl2 is the OA and NH3 is RA) The change in ON of one Cl =|0-(-1)| =1 For two Cl atoms present in the OA, the total change in ON =2, so 2 is placed as coefficient before NH3(RA).
Cl2 + 2NH3 ------> HCl + N2
The change in ON of in RA(NH3) = |-3-0| = 3. Since there is one N atom in the RA, the total change of ON in the OA is also 3, so 3 is placed as coefficient before Cl2(OA)
3Cl2 + 2NH3 ------> HCl + N2 (Note that for the purpose of finding out the total change in ON, the number of atoms present in products i.e RHS is not considered. As for example we have N2 in RHS, but we do not make the total change 3X2=6 for N atoms to place before Cl2. Since NH3 has one N atom, the change is 3. For that purpose, only the reactant side, i.e LHS is considered) After that, mere hit and trial (inspection) completes the process of balancing in a second. 6 to be placed before HCl to equalise Cl and by that we see that H is automatically balanced (6H on either side).
3Cl2 + 2NH3 ------> 6HCl + N2 (Balanced equation)
SAQ 2: Balance the following by ON method.
(i) KI + H2SO4 ------> I2 + K2SO4 + H2S + H2O
(ii) I2 + HNO3 ------> HIO3 + NO2 + H2O
(iii) Cu + HNO3 ------>Cu(NO3)2 + NO + H2O
(3) PARTIAL EQUATION METHOD: Let us now learn another simple method to balance a redox reaction. This is applicable particularly when the oxidising agents (OA) are KMnO4, K2Cr2O7, HNO3, H2SO4 etc. Let us take the reaction
KMnO4 + H2SO4 + FeSO4 ------> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O It is a tough task to balance this big equation by hit and trial method. We can balance this equation by ON method as we have just now learnt. But there is another simpler method to
Dr. S. S.Tripathy 34 Concepts in Chemistry balance this type of equation. This is called partial equation method. In this method, the
OA(KMnO4) is split into a few smaller fragments.
KMnO4 ------> K2O + MnO + [O]
KMnO4 breaks down to oxide of potassium, oxide of manganese (ous) and nascent oxygen[O]. The nascent means freshly born. That means this nascent oxygen is now born freshly to die in the second step. Note that all the three smaller fragments that are produced from the OA will die i.e will be used up in the subsequent steps. So that we write a few steps or partial equations. Note that in each step, the equation has to be balanced by hit and trial method. Then we shall perform algebraic operation to add all the partial equations. We shall thus cancel those species which are used i.e which appear in both LHS and RHS by doing algebraic manipulations and finally get the balanced equation of the original reaction. Let us look to the following equations.
(a) Reactions involving KMnO4
Reactions involving KMnO4 and K2Cr2O7 often take place in presence of an acid, preferably
H2SO4. In the example below, FeSO4 (ferrous sulphate) reacts with acidified KMnO4 to produce a mixture of potassium sulphate, manganous sulphate, ferric sulphate and water. We know that here FeSO4 is the reducing agent (RA) and KMnO4 is the oxidising agent(OA). Let us balance the following reaction by partial equation method.
KMnO4 + H2SO4 + FeSO4 K2SO4 + MnSO4 + Fe2(SO4)3 + H2O
2 KMnO4 K2O + 2 MnO + 5 [O] (Step 1) K2O + H2SO4 K2SO4 + H2O (Step 2) 2 X MnO + H2SO4 MnSO4 + H2O (Step 3) (Step 4) 5 X 2 FeSO4+ H2SO4 + [O} Fe2(SO4)3 + H2O
2 KMnO4 + 8 H2SO4+ 10 FeSO4 K2SO4 + 2 MnSO4 + 5 Fe2(SO4)3 + 8 H2O (balanced equation)
In this the fragments K2O and MnO produced in the first step are consumed in the second and third step by reacting with the acid(H2SO4) present in the medium to produce salt and water. In the fourth step FeSO4(RA) is oxidised to Fe2(SO4)3 by nascent oxygen. Remember that whenever
FeSO4 is oxidised to Fe2(SO4)3, it takes the help of H2SO4. Otherwise the equation cannot be balanced. Note that before adding all the step equations, we have to see that each individual step equation is balanced. While adding the four steps, K2O gets cancelled. To cancel 2MnO from RHS we have to multiply the whole equation of step(3) by 2. Likewise to cancel 5[O] from
RHS, we have to multiply the whole equation of step(4) by 5. After cancellation of K2O, 2MnO and 5[O] from both the sides, all the species of LHS and RHS are added separately and written below. Further simplification of this equation, if required, is done like algebraic equation. If H2O appears on both the sides, then it is removed from one side and excess H2O is only written on the other side. Similarly any other species, if appears both in LHS and RHS is brought to one side.
Let us take another example on KMnO4.
Dr. S. S.Tripathy Logic of Inorganic Reactions 35
(ii)
KMnO4 + H2SO4 + H2S K2SO4 + MnSO4 + S + H2O
2 KMnO4 K2O + 2 MnO + 5 [O] (Step 1) K2O + H2SO4 K2SO4 + H2O (Step 2) 2 X MnO + H2SO4 MnSO4 + H2O (Step 3) + (Step 4) 5 X H2S + [O] SH2O
2 KMnO4 + 3 H2SO4 + 5 H2S K2SO4 + 2 MnSO4 + 5 S + 8 H2O (balanced equation)
In this case, the first three steps are same as the first example. Only the forth step is different.
The RA is H2S in this case which is oxidised to S. You can always look at the RHS of the original equation to know what would be the products of each step. While adding the four steps, K2O, 2MnO and 5[O] are cancelled by suitable algebraic manipulations and we get the balanced equation. Do you think, that you could have balanced these equations by hit and trial method easily? The answer is NO. So to conclude, we say that a redox reaction is balanced either by ON method described before or partial equation method described just now. Note that the partial equation method is merely a simple technique devised to balance a complex equation in a simple and economical manner. The actual reaction does not take place in the successive steps as discussed before. There is nothing called nascent oxygen in this reaction. The nascent atom concept in the solution is very primitive and has been outdated. The redox reactions truly take place by the loss and gain of electrons and not by way of the generation of nascent oxygen atoms. But students are advised to follow this technique to get the balanced equations easily.
(b) Reactions involving K2Cr2O7
K2Cr2O7 + H2SO4 + SO2 K2SO4 + Cr2(SO4)3 ++ H2SO4 H2O
K2Cr2O7 K2O ++ Cr2O3 3 [O] (Step 1) K2O + H2SO4 K2SO4 + H2O (Step 2) Cr2O3 + 3 H2SO4 Cr2(SO4)3 + 3 H2O (Step 3) 3 X SO2 + [O] SO3 (Step 4) (Step 5) 3 X SO3 + H2O H2SO4
K2Cr2O7 + 4 H2SO4 + 3 SO2 K2SO4 + Cr2(SO4)3 + 3 H2SO4 + H2O
K2Cr2O7 +++ 3 SO2 + H2SO4 K2SO4 Cr2(SO4)3 H2O
Dr. S. S.Tripathy 36 Concepts in Chemistry
In this case K2Cr2O7 is broken into three fragments namely K2O, Cr2O3 and nascent oxygen. This equation is immediately balanced. Then in step 2 and step 3, the oxides of K and Cr react with H2SO4 present in the medium to produce the respective salt and water. In step 4, SO2(RA) reacts with nascent oxgyen[O] to produce SO3. But SO3 is unstable. It immediately reacts with
H2O in the step 5 to produce the corresponding acid, H2SO4. So we find that H2SO4 was used as a reactant and it is formed also in the product. So we add the five steps by multiplying suitable coefficients (3) to the step equations 4 and 5, to cancel 3[O] and 3SO3 from the two sides. K2O and Cr2O3 are also cancelled as such. On addition it gives an equation, which carries H2O and
H2SO4 on either side. So we have to simplify further to get the balanced equation. We find that in the net balanced equation, H2SO4 has been wiped out from RHS. Don't worry for that. At times, it happens like that at the last step of simplification. SAQ 3: Show which is oxidised and which reduced by assigning ONs. Also balance the equations by partial equation methods.
(i)KMnO4 + H2SO4 + H2C2O4 ------> K2SO4 + MnSO4 + CO2 + H2O
(ii)K2Cr2O7 + H2SO4 + H2O2 ------> K2SO4 + Cr2(SO4)3 + O2 + H2O
(c) Reactions involving HNO3:
Equations involving HNO3 can also be balanced by partial equation method. Let us take an example.
HNO3 + Cu ------> Cu(NO3)2 + N2O + H2O We can balance this equation by Oxidation Number Method as explained before. In partial equation method, HNO3(OA) is broken down to small fragments i.e an oxide of nitrogen (in this case N2O), H2O and nascent oxygen[O]. While breaking HNO3, the student is advised to look to the RHS to know which oxide of nitrogen is to be written.
HNO3 ------> N2O + H2O + [O] Let us immediately balance it and use [O] to react with Cu(RA) in the second step to give CuO.
Again since the medium is acidic(HNO3), CuO being a basic substance reacts with the acid to produce salt and water. The partial equations are added up and nascent oxygen atoms are cancelled to get the balanced equation.
2HNO3 ------> N2O + H2O + 4[O] (i) 4X [Cu + [O] ------> CuO] (ii) (4 Multiplied to cancel 4[O])
4X [CuO + 2HNO3 ------> Cu(NO3)2 + H2O](iii) (4 multiplied to cancel 4CuO) ______
4 Cu + 10HNO3 ------> N2O + 4Cu(NO3)2 + 5H2O` (balanced equation)
Note that the oxide of nitrogen produced in reactions involving HNO3 depends on the type of
HNO3 used. The following gives a list of products formed from HNO3.
Conc. HNO3:NO2
Moderately concentrated HNO3 :NO
Dilute HNO3:N2O or N2
Very dilute HNO3:NH3( NH3 + HNO3 ------> NH4NO3)
Dr. S. S.Tripathy Logic of Inorganic Reactions 37
This means that concentrated HNO3 forms nitrogen dioxide gas(NO2), moderately concentrated
HNO3 forms nitric oxide(NO), dilute HNO3 forms nitrous oxide(N2O) or nitrogen(N2) and very dilute HNO3 first forms NH3 which reacts with HNO3 to form ammonium nitrate(NH4NO3). In the example given above, the nitric acid used must be dilute as the oxide of nitrogen produced is nitrous oxide(N2O) For the purpose of balancing, you do not have to bother all about this. You can know what is the product just by looking to RHS of a given equation. SAQ 4: Balance the following by partial equation method.
(i) HNO3 + Zn ------> NH4NO3 + Zn(NO3)2 + H2O
(ii) HNO3 + Mg ------> Mg(NO3)2 + NO + H2O
(d) Reactions involving conc. H2SO4
Reactions involving concentrated H2SO4 can also be balanced by partial equation method.
H2SO4 is broken into SO2 gas, H2O and nascent oxgyen[O]. The process is similar to the balancing discussed for HNO3 above. Let us take an example.
Cu + H2SO4(conc) ------> CuSO4 + SO2 + H2O
H2SO4 ------> SO2 + H2O + [O] (i) Cu + [O] -----> CuO (ii) ([O] is cancelled)
CuO + H2SO4 ------> CuSO4 + H2O (iii) (CuO is cancelled) ______
Cu + 2 H2SO4 ------> CuSO4 + SO2 + 2 H2O (balanced equation) SAQ 5: Balance the following by partial equation method.
Al + H2SO4(conc.) ------> Al2(SO4)3 + SO2 + H2O
IMPORTANT: After balancing an equation by any method, check whether the balancing is correct or not by tallying two sides(LHS and RHS) element by element and atom by atom. If you find that somewhere it is not tallying, then you have done the steps somewhere wrong. In such case you revise again to detect your mistake. (4) ION-ELECTRON METHOD: This is another interesting technique of balancing redox reactions. This is applied to the ionic equations(not molecular equations). So if a molecular equation is given to you, first you have to find the net ionic equation from it and then balance the ionic equation by this technique. Then you can balance the molecular equation by looking to the coefficients in the balanced ionic equation. Do you remember how to find the net ionic equation from a molecular equation that you have studied before? Example-1
KMnO4(aq) + H2SO4(aq) + FeSO4(aq) ------> K2SO4(aq) + MnSO4(aq) + Fe2(SO4)3(aq) + H2O(l) - + 2+ 2+ 3+ Net ionic equation: MnO4 + H + Fe ------> Mn + Fe + H2O Now the job is to balance this ionic equation. There are rules which we have to follow for doing that. Carefully read the following.
Dr. S. S.Tripathy 38 Concepts in Chemistry
(A) Acidic Medium The following rules are applicable for reactions carried out in acidic medium. If you find H+ ion present either in LHS or RHS, then the reaction has definitely been carried out in acidic medium and in such case, you follow this method. - + 2+ 2+ 3+ MnO4 + H + Fe ------> Mn + Fe + H2O From this you separate the reduction and oxidation processes by taking one reactant and the corresponding product from the net ionic equation at a time. Build up two separate step equations, one for reduction and the other for oxidation. Then add the two to get the balanced ionic equation. - 2+ In the above reaction, MnO4 (LHS) has been reduced to Mn (RHS). Can you not say that the ON of Mn changes from +7 to +2 in this case.? Copy the reduction step from the original equation as follows. Reduction step: - 2+ MnO4 ------> Mn Rules: (i) Equalise the number of atoms other than O and H on both sides by placing suitable coefficient. In the above example the number of Mn atoms on either side is same (one). So we will not place any coefficient on any side. (ii) Count how many O atoms are in excess in any particular side and add same
number of H2O molecules on the opposite side.
In the above example, four O atoms are in excess in LHS, so we have to add 4 H2O molecules in RHS. - 2+ MnO4 ------> Mn + 4H2O (iii) Now count how many H atoms are in excess in any particular side and add same number of H+ ions on the opposite side. In this case there are 8 H atoms excess in RHS. So we have to add 8 H+ ions in LHS. - + 2+ MnO4 + 8 H ------> Mn + 4H2O (iv) Then we have to equalise the charge on both sides by adding electrons. Note that while you add one electron to a particular side, charge-wise it is -1, as electron carries one unit of negative charge. - + In this case, the total charge in LHS is +7( -1 for MnO4 and +8 for eight H ) and the total charge in RHS is +2(from one Mn2+). So we have to add 5 electrons in LHS so as to equalise the charge on both sides to +2. - + - 2+ MnO4 + 8 H + 5e ------> Mn + 4H2O Thus we have completed balancing of the reduction step. Note that in the reduction step, electrons appear in LHS as reduction is the process in which electron is gained. Let us take the oxidation step. Oxidation Step: In the above example Fe2+ has been oxidised to Fe3+. Let us write that. Fe2+ ------> Fe3+
Dr. S. S.Tripathy Logic of Inorganic Reactions 39
Since the number of Fe atoms is same on either side, we do not have to put any coefficient on + any side. Since there is no O and H atoms on any side, the addition of H2O and H ions are not to be done. Thus rules (i), (ii) and (iii) are not applicable in this case. Only we shall apply the last rule(iv), i.e equalisation of charge by adding electrons. In this case one electron has to be added in the RHS to equalise the charge on both sides(+2). Fe2+ ------> Fe3+ + e- Thus we found that in oxidation process, electron appears in RHS as oxidation is the process in which electron is lost. Now we have to add the two steps (reduction and oxidation) and cancel the electrons from both the sides by doing suitable algebraic manipulations and get the balanced ionic equation. - + - 2+ MnO4 + 8 H + 5e ------> Mn + 4H2O 5X [Fe2+ ------> Fe3+ + e-] (5e are cancelled on both sides) ______
- + 2+ 2+ 3+ MnO4 + 8H + 5Fe ------>Mn + 5Fe + 4 H2O (balanced equation) Now you can check whether the balancing is correct or not by tallying two sides(LHS and RHS) element-wise, atom-wise and charge-wise. If you find that somewhere it is not tallying,then you have done the steps somewhere wrong. In such case you revise again to detect your mistake. Balancing of the molecular equation from balanced ionic equation: Now we can balance the original molecular equation by looking to the coefficients of the balanced ionic equation. → KMnO4 + 4 H2SO4 + 5 FeSO4 MnSO4 + 1/2 K2SO4 + 5/2 Fe2(SO4)3 + 4 H2O + 3+ We needed 8 H ions, so we placed the coefficient 4 with H2SO4. We needed 5 Fe , so we placed the coefficient 5/2 with Fe2(SO4)3. In this way the whole equation is balanced. In order to eliminate fraction in the balanced equation, the factor 2 is multiplied throughout the equation(both LHS and RHS). → 2KMnO4 + 8H2SO4 + 10 FeSO4 2MnSO4 + K2SO4 + 5 Fe2(SO4)3 + 8 H2O Example 2 : 2- + - 3+ - Let us balance the ionic equation: Cr2O7 + H + NO2 ------> Cr + NO3 + H2O first and after that balance the molecular equation → K2Cr2O7 + H2SO4 + KNO2 K2SO4 + Cr2(SO4)3 + KNO3 + H2O In this reaction, the ON of Cr changes from +6 to +3 and N from +3 to +5. Check by x method 2- 3+ - - of finding the ON. So Cr2O7 is reduced to Cr and NO2 has been oxidised to NO3 . Reduction step: 2- 3+ Cr2O7 ------> Cr First we have to equalise Cr atoms by placing 2 before Cr3+ in RHS 2- 3+ Cr2O7 ------> 2Cr
Since there are 7 O atoms excess in LHS, we have to add 7 H2O molecules in RHS. 2- 3+ Cr2O7 ------> 2Cr + 7 H2O
Dr. S. S.Tripathy 40 Concepts in Chemistry
Now there are 14 H atoms excess in RHS, we have to add 14 H+ ions in LHS. 2- + 3+ Cr2O7 + 14 H ------> 2Cr + 7 H2O LHS contains +12 charge(-2+14) and RHS contains +6(2X3) charge. So we have to add 6 electrons on LHS to equalise the charge on both sides(+6). 2- + - 3+ Cr2O7 + 14 H + 6e ------> 2Cr + 7 H2O Oxidation step:
- - NO2 has been oxidised to NO3 . Let us write them and apply the same set of rules for balancing them. - - NO2 ------> NO3 N atoms are same in number, so we do not have to do anything. There is one excess O atom in
RHS, so we have to add one H2O molecule in LHS. - - NO2 + H2O ------> NO3 By doing so, now 2 H atoms are in excess in LHS, and so we have to add 2 H+ ions in RHS. - - + NO2 + H2O ------> NO3 + 2 H Now equalisation of charge is made by adding 2 electrons in the RHS, so that the net charge is -1 on both the sides. - - + - NO2 + H2O ------> NO3 + 2 H + 2e
Now we have to add the two steps and cancel the electrons.
2- + - 3+ Cr2O7 + 14 H + 6e ------> 2Cr + 7 H2O - - + - 3X [NO2 + H2O ------> NO3 + 2 H + 2e ] ( electrons get cancelled) ______2- + - 3+ - + Cr2O7 + 14 H + 3NO2 + 3H2O ------>2Cr + 3NO3 + 6H +7H2O + We can further simplify by taking H2O and H to one side. 2- + - 3+ - Cr2O7 + 8 H + 3NO2 ------>2Cr + 3NO3 +4 H2O (balanced equation)
Balancing of the molecular equation from balanced ionic equation; In the same manner as we did for the previous example, we can balance the molecular equaiton by looking to the coefficents of balanced ionic equation. → K2Cr2O7 + 4 H2SO4 + 3 KNO2 K2SO4 + Cr2(SO4)3 + 3 KNO3 + 4 H2O
Example 3 : + - 2+ Cu2O + H + NO3 ------> Cu + NO + H2O 2+ First identity which is reduced and which is oxidised. Cu2O is oxidised to Cu as the ON of Cu 2+ - increases from +1(Cu2O) to +2 in Cu . NO3 is reduced to NO as the ON of N is decreased - from +5(NO3 ) to +2(NO).So let us start with oxidation step. Note that you can start from any step you like, not necessarily always with reduction or oxidation. Oxidation step: 2+ Cu2O ------> Cu Equalise the Cu atoms by placing the coefficient 2 in RHS. Dr. S. S.Tripathy Logic of Inorganic Reactions 41
2+ Cu2O ------> 2Cu
Then addition of water is to be done. Since one O atom is in excess in LHS, one H2O molecule is to be added in RHS. 2+ Cu2O ------> 2Cu + H2O Now since 2 H atoms are in excess in RHS, we have to add 2H+ in LHS. + 2+ Cu2O + 2 H ------> 2Cu + H2O Finally equalisation of charge is done by adding 2 electrons in the RHS, so that charge on either side becomes +2. + 2+ - Cu2O + 2 H ------> 2Cu + H2O +2e Thus we found that in the oxidation step, electron has appeared in RHS. Reduction step: - NO3 ------> NO
Since N atoms are same on either side(one), we start by adding 2H2O molecules on RHS as there are two excess O atoms in LHS. - NO3 ------> NO + 2 H2O Now we have to add 4H+ ions in LHS as there are 4 excess H atoms in RHS. - + NO3 + 4 H ------> NO + 2 H2O Charge is made equal by adding 3 electron on LHS, so that the net charge is zero on either side.We find that in reduction step, electron appears in the LHS. - + - NO3 + 4 H + 3e ------> NO + 2 H2O Then we have to add the two steps. Let us copy two steps again one above the other.
+ 2+ 3X [Cu2O + 2 H ------> 2Cu + H2O +2e- - + - 2X [NO3 + 4 H + 3e ------> NO + 2 H2O] (6 electrons are cancelled) ______+ - 2+ 3Cu2O + 14H + 2 NO3 ------> 6Cu + 2NO + 7 H2O (balanced equation)
Balanced molecular equation: 3Cu2O+2KNO3+7H2SO4---->6CuSO4+K2SO4+ 2NO+7H2O. Let us see how we can balance a redox reaction taking place in basic(alkaline) medium. Note that a redox reaction can take place either in acidic medium as discussed before or in in basic(alkaline) medium. (B) Basic or Alkaline Medium: Let us take one example to understand the rules.
Br2(l) + KIO3(aq) + KOH(aq) ------> KBr(aq) + KIO4(aq) + H2O(l) Find the net ionic equation as you were doing before by cancelling the spectator ions. - - - - Net ionic equation: Br2 + IO3 + OH ------> Br + IO4 + H2O - - - Br2( ON=0) is reduced to Br (ON=-1), while IO3 (ON of I=+5)is oxidiesd to IO4 (ON of I=+7)
0 +5 - - - +7 - Br2 + IO3 +++ OH Br IO4 H2O This reaction has been carried out in alkaline medium as we find OH- in LHS. Note that in alkaline medium, OH - has to appear either in LHS or RHS. Rules: Let us take the oxidation step first and see the rules which are to be applied.
Dr. S. S.Tripathy 42 Concepts in Chemistry
Oxidation step:
- - IO3 ------> IO4 (i) Equalise the element other than H and O by placing suitable coefficient as you did in acidic medium. In this case number of I atoms is same on both the sides. (ii) Count how many O atoms are in excess in a particular side and add same number of
H2O molecules on the opposite side. This is same as the rule (ii) for balancing in acidic medium.
In this case one O atom is in excess in RHS, so we have to add one H2O molecule in LHS. - - IO3 + H2O ------> IO4 (iii) Count how many H atoms are in excess in a particular side. Add same number of - OH ions on the same side and same number of H2O molecules on the opposite side. - In this case, LHS contains 2 excess H atoms, so we have to add 2 OH ions on LHS and 2 H2O molecules on RHS. - - - IO3 + H2O + 2OH ------> IO4 + 2 H2O Note that in the alkaline medium H2O is added two times, in rule (ii) and (iii). (iv)Finally equalisation of charge is made by adding electrons to a particular side exactly in the same way as in acidic medium. - - - - IO3 + H2O + 2OH ------> IO4 + 2 H2O +2e So the charge on either side becomes -3. Let us now take the reduction step and apply the same rules to it. Reduction step: - Br2 ------>Br Applying the first rule, we place coefficient 2 in RHS to equalise Br atom. - Br2 ------> 2 Br Rules (ii) and (iii) are not applicable in this case as there is no H and O. So we straightway come to rule(iv) i.e equalisation of charge. - - Br2 + 2e ------> 2 Br Now the charge on either side becomes -2. Finally two steps are added and electrons are cancelled. - - - - IO3 + H2O + 2OH ------> IO4 + 2 H2O +2e - - Br2 + 2e ------> 2 Br (2 electrons are cancelled) ______- - - - IO3 + Br2 + 2OH + H2O ------>IO4 + 2Br + 2H2O On further simplification, we get, - - - - IO3 + Br2 + 2OH ------>IO4 + 2Br + H2O (balanced ionic equation) Balancing of the molecular equation from balanced ionic equation: → KIO3 + Br2 + 2KOH KIO4 + 2 KBr + H2O Try the following SAQ in alkaline medium. SAQ 6: Deduce the net ionic equation and balance the ionic equation by ion-electron method.
K2CrO4(aq) + Cu2O(s) + H2O(l) ------> Cu(OH)2(s) + KCrO2(aq) + KOH(aq) ELECTRON BALANCE DIAGRAMME METHOD This is the alternative method of balancing by oxidation number(ON) method. Whenever the usual ON method as explained before fails while balancing an equation, this method is adopted. In this case, the procedure is similar to balancing by ion-electron method discussed just before.
Dr. S. S.Tripathy Logic of Inorganic Reactions 43
By this method, we can know the coefficients of both LHS and RHS species which we cannot know by the usual ON method discussed earlier. Example: → Cl2 + NaOH NaCl + NaClO3 + H2O (i) First assign the ON of atoms undergoing changes. (ii) Select only the atoms(not the molecule or ion) of each side with their ONs which suffer change and develop oxidation and reduction steps as you did in ion-electron method. + - Here addition of H , OH or H2O is not done. 0 +5 Oxidation : Cl Cl
0 -1 Reduction: Cl Cl (iii) Balance the charge on the two sides of each equation by adding required number of electrons(e-) on one of the sides as we did in the last step for balancing by ion-electron method. Then add the two equations by cancelling the electrons from both the equation. Thus we get the electron- balance diagramme.
0 +5 -1 Cl Cl + 5 e 0 -1 -1 Cl + e Cl X 5 0 +5 -1 6 Cl Cl + 5 Cl (electron balance diagramme) (iv) Now we have to balance the molecular equation by looking to the coefficients in the electron balance diagramme. Let us develop a part balanced molecular equation from this.
3 Cl2 ------> NaClO3 + 5 NaCl Then the total equation can be balanced by mere inspection.
3 Cl2 + 6 NaOH ------> NaClO3 + 5 NaCl + 3 H2O SAQ 7: Balance by electron-balance diagramme technique of ON method.
(i) K2Cr2O7 + HCl ------> KCl + CrCl3 + Cl2 + H2O
(ii) FeS2 + O2 -----> Fe2O3 + SO2
PRACTICE QUESTIONS 1. Balance the following ionic equations by ion-electron method. Wherever
necessary add H2O molecule in the equation. - 2+ + 3+ (i) NO3 + Fe + H ------> Fe + NO2 + H2O - - (ii) NH3 + MnO4 + OH ------> MnO2 + NO2 3+ 2+ (iii) Fe + H2O2 ------> Fe + O2 (acidic medium) - - iv) Cr(OH)2 + I2 + OH ------> Cr(OH)3 + I 2- + 2- 3+ (v) C2O4 + H + Cr2O7 ------> CO2 + Cr + H2O - + - 2+ (vi) MnO4 + H + I ------> Mn + I2 + H2O 2- 2- - (vii) S2O3 + I2 ------> S4O6 + I - + - - (viii) OCl + H + I ------> I2 + Cl - - - (ix) Cl2 + IO3 ----> IO4 + Cl (alkaline medium) Dr. S. S.Tripathy 44 Concepts in Chemistry
2. Balance the following by Oxidation Number method.
(i)NH3 + O2 ------> NO + H2O
(ii)H2S + SO2 ------> S + H2O
(iii)Fe2O3 + CO ------> Fe + CO2
(iv)SnCl2 + O2 + HCl ------> SnCl4 + H2O
(v)NaClO3 + KI + HCl ------> NaCl + I2 + KCl + H2O
(vi)SbCl5 + KI ------> SbCl3 + I2 + KCl
(vii)NaPO3 + BrF3 ------> NaPF6 + Br2 + O2
(viii)Sn + HNO3 ------> SnO2 + NO + H2O
(ix)Cr2O3 + HNO3 + Na2CO3 ------> Na2CrO4 + CO2 + HNO2
(x) Cu + HNO3 ------> Cu(NO3)2 + N2O + H2O
(xi)Cu + HNO3 ------> Cu(NO3)2 + NO + H2O
(xii)BaCrO4 + KI + HCl ------> BaCl2 + I2 + KCl + CrCl3 + H2O
(xiii)Ca3(PO4)2 + SiO2 + C ------> CaSiO3 + P4 + CO
(xiv)Zn + HNO3 ------> Zn(NO3)2 + NH4NO3 + H2O
(xv)K2Cr2O7 + KI + H2SO4 ------> K2SO4 + Cr2(SO4)3 +I2 +H2O
(xvi)KMnO4 + H2SO4 + H2S ------> K2SO4 + MnSO4 + S + H2O
(xvii)Cl2 + KIO3 + KOH ------> KCl + KIO4 + H2O → (xviii)N2H4 + KBrO3 N2 + KBr + H2O
(xix)Sn + NaOH ------> Na2SnO3 + H2
(xx)Al + NaOH ------> NaAlO2 + H2 3. Balance the following by partial equation method.
(i) KMnO4 + H2SO4 + HI ------> K2SO4 + MnSO4 + I2 + H2O
(ii) KMnO4 + H2SO4 + H2O2 ------> K2SO4 + MnSO4 + O2 + H2O
(iii) KMnO4 + H2SO4 + KNO2 ------> K2SO4 + MnSO4 + KNO3 + H2O
(iv) KMnO4 + H2SO4 + H2S ------> K2SO4 + MnSO4 + S + H2O
(v) KMnO4 + H2SO4 + H2C2O4 ------> K2SO4 + MnSO4 + CO2+ H2O
(vi) K2Cr2O7 + H2SO4 + HI ------> K2SO4 + Cr2(SO4)3 + I2 + H2O
(vii) K2Cr2O7 + H2SO4 +H2O2 ------> K2SO4 + Cr2(SO4)3 + O2 + H2O
viii) K2Cr2O7 + H2SO4 +KNO2 ------> K2SO4 + Cr2(SO4)3 + KNO3 + H2O
(ix) K2Cr2O7 + H2SO4 +H2S ------> K2SO4 + Cr2(SO4)3 + S + H2O
(x) K2Cr2O7 + H2SO4 +H2C2O4 ------> K2SO4 + Cr2(SO4)3 + CO2 + H2O
(xi) Cu + HNO3 ------> Cu(NO3)2 + NO + H2O
(xii) Fe + HNO3 ------> Fe(NO3)2 + N2O + H2O
(xiii)Mg + H2SO4 ------> MgSO4 + SO2 + H2O
Dr. S. S.Tripathy Logic of Inorganic Reactions 45
4. Balance the following by ion electron method. 2- 3+ 4+ 2- (i)S2O8 + Ce ------> Ce + SO4 (acidic medium) + - 2+ (ii)Cu2O + H + NO3 ------> Cu + NO + H2O - 2- 2- (iii)MnO4 + SO3 + OH------> MnO2 + SO4 + H2O 2+ 2- - 2- (iv)Mn + S2O8 ------> MnO4 + SO4 (Acidic medium) 2- + 3+ (v)Cr2O7 + C2H4O + H ------> Cr + C2H4O2 + H2O - 2- 2- (vi)S + OH ------> S + S2O3 - - + - (vii)ClO3 + I + H ------> Cl + I2 + H2O + (viii)Ag + AsH3 ------> H3AsO4 + Ag (alkaline medium) - 2+ (ix)CuS + NO3 ------> Cu + S8 + NO (acidic medium) - - - - (x)ClO3 + SbO2 ------> ClO2 + Sb(OH)6 (basic medium) - 2+ (xi)Zn + NO3 ------> Zn + NH3 (alkaline medium) - 2- - 2- (xii)BrO3 + S2O3 ------> Br + S4O6 + H2O(acidic medium) - 2- 2- (xiii)MnO4 + SnO2 + H2O ------> MnO2 + SnO3 (basic medium)
5. Get the net ionic equation and balance by ion-electron method. From the balanced ionic equation, balance the molecular equation.
(i)K4[Fe(CN)6](aq) + HCl(aq) + H2O2(l) ------> K3[Fe(CN)6](aq) +
KCl(aq) +H2O(l)
(ii)N2H4 (l)+ K3[Fe(CN)6](aq)------> N2 (g)+ K4[Fe(CN)6](aq) + H2O(l) (alkaline medium)
(iii)C2H2 (g)+ KMnO4(aq)------> MnO2(s)+ K2C2O4(aq)+ KOH(aq) +H2O(l)
(iv)HgCl2(aq)+ SnCl2(aq)------> Hg2Cl2(s)+ SnCl4(aq)
(v)NaNO3(aq) + Zn(s) + NaOH(aq) -----> NH3(g) + Na2ZnO2(aq) + H2O(l)
(vi)HgS(s) + HNO3(aq) + HCl(aq)------> S + H2[HgCl4](aq) + NO(g) + H2O(l)
(vii) K2Cr2O7(aq) + H2SO4(aq)+KNO2(aq) ------> K2SO4(aq)+ Cr2(SO4)3(aq)+
KNO3(aq)+ H2O(l)
(viii) K2Cr2O7(aq)+ H2SO4(aq)+H2S(g) ------>K2SO4(aq)+ Cr2(SO4)3(aq)+ S + H2O(l)
(ix) K2Cr2O7(aq)+ H2SO4(aq)+H2C2O4(s)-----> K2SO4(aq)+ Cr2(SO4)3(aq)+ CO2(g)+ H2O(l)
(x) Cu(s) + HNO3(aq) ------> Cu(NO3)2(aq)+ NO(g) + H2O(l)
(xi) Fe(s) + HNO3(aq)------> Fe(NO3)2(aq)+ N2O(g)+ H2O(l)
Supplimentary Questions: 1. Balance by any one method: (either by ion-electron or ON method)
(i) CuS(s) + HNO3(aq) ------> Cu(NO3)2(aq) + S + H2O(l) + NO(g)
(ii) As2S5(s)+ HNO3(aq) ------> H3AsO4(aq)+ H2SO4(aq)+ H2O(l) + NO2(g)
(iii) Zn(s) + HNO3(aq) ------> Zn(NO3)2(aq) + H2O(l) + NH4NO3(aq)
(iv) Na2C2O4(aq) + KMnO4(aq) + H2SO4(aq) -----> K2SO4(aq) +MnSO4(aq) +
CO2(g)+ Na2SO4(aq) + H2O(l) Dr. S. S.Tripathy 46 Concepts in Chemistry
(v) MnO + PbO2 + HNO3 ------> HMnO4 + Pb(NO3)2 + H2O
(vi) Na2HAsO3(aq) + KBrO3(aq) + HCl(aq) -----> NaCl(aq) + KBr(aq) +
H3AsO4(l)
(vii) FeS2 + O2 ------> Fe2O3 + SO2
(viii) Ca(OCl)2(aq) + KI(aq) + HCl(aq) ------> I2(s) + CaCl2(aq) + H2O(l) + KCl(aq)
(ix) NaOCl(aq) + NaOH(aq) + Bi2O3(s) ----> NaBiO3(aq) + H2O + NaCl(aq)
(x) Sn + HNO3 ------> SnO2 + NO2 + H2O → (xi) CrI3 + KOH + Cl2 K2Cr2O7 + KIO4 + KCl + H2O → (xii) Na2TeO3 + + NaI + HCl NaCl + Te + I2 + H2O → (xiii) K2Cr2O7 + SnCl2 + HCl CrCl3 + SnCl4 + KCl + H2O → (xiv) K3[Fe(CN)6] + Cr2O3 + KOH K4[Fe(CN)6] + K2CrO4 + H2O → (xv) NaHSO4 + Al + NaOH Na2S + Al2O3 + H2O → (xvi) KI + H2SO4 K2SO4 + I2 + H2S + H2O → (xvii) Cr2O3 + Na2CO3 + KNO3 Na2CrO4 + CO2 + KNO2 → (xviii) KClO3 + H2SO4 KHSO4 + ClO2 + O2 + H2O → (xix) P2H4 PH3 + P4H2 → (xx) CoCl2 + KNO2 + HCl K3[Co(NO2)6] + NO + KCl + H2O - → 2- (xxi) MnO4 MnO4 + O2 (alkaline medium) Also balance the molecular equation - → - (xxii) Au + CN + O2 [Au(CN)4] + H2O (alkaline medium). Also balance the molecular equation → 2+ (xxiii) Zn + As2O3 AsH3 + Zn (acid medium) Also balance the molecular equation - → - - (xxiv) Cl2 + IO3 IO4 + Cl (alkaline medium) Also balance the molecular equation - → 2+ + (xxv) Zn + NO3 Zn + NH4 (acidic medium) Also balance the molecular equation → (xxvi) V + NaOH + H2O Na3HV6O17 + H2 → (xxvii) MnBr2 + PbO2 + HNO3 HMnO4 + Pb(NO3)2 + Pb(BrO3)2 + H2O
2. Complete and balance the following: 2+ + - 4+ (i) Sn + H + NO3 -----> Sn + NO 2- - 3+ (ii) Cr2O7 + I ----> I2 + Cr (acidic medium) + - 2+ + (iii) Zn + H + NO3 -----> Zn + NH4 2- + 2- (iv) S2O3 + Ag ------> Ag + S4O6 - - (v) P4 + OH ----> PH3 + H2PO2 - - - (vi) Br2 + IO3 ----> Br + IO4 (basic medium) - 2+ 2+ 4+ (vii) MnO4 + Sn ----> Mn + Sn (acidic medium) 2- - - (viii) S2O8 + I ----> I3 + SO2 (acidic medium) 2- - (ix) CrO4 + Cu2O ----> Cu(OH)2 + [Cr(OH)4] (alkaline medium)
Dr. S. S.Tripathy Logic of Inorganic Reactions 47
RESPONSE TO SAQs
SAQ 1: Balance the following reactions:
(i) 3Ca(OH)2 + 2H3PO4 ------> Ca3(PO4)2 + 6 H2O (ii) ZnO + H2SO4 ------> ZnSO4 + H2O (iii) FeCl2 + H2S ------> FeS + 2HCl (iv) P2O5 + 3H2O ------> 2H3PO4 (v) 2NH4Cl + Ca(OH)2------> CaCl2 + 2NH3 + 2H2O SAQ 2: -1 +6 0 -2 (i) KI + H2SO4 I2 + K2SO4 ++ H2S H2O
(KI is RA and H2SO4 is OA)
The change in ON in RA = |-1-0| = 1, so 1 is the coefficient before OA(H2SO4). 1 is not written because if no other coefficient is placed, then the coefficient 1 is understood to be present. The change in ON in OA = |+6-(-2)| = 8, so 8 is to be placed before KI(RA)
8KI + H2SO4 ------> I2 + K2SO4 + H2S + H2O
Then hit and trial. To equalise I, we have to place 4 before I2 in RHS and to equalise K, we have to place 4 before K2SO4.
8KI + H2SO4 ------> 4I2 + 4K2SO4 + H2S + H2O Let us equalise S now. The number of S atoms in RHS is 5, so 5 is placed as coefficient before
H2SO4 in LHS. Note that the coefficient 1 that was found in the first case is now revised to 5.
8KI + 5H2SO4 ------> 4I2 + 4K2SO4 + H2S + H2O (Revised coefficient of H2SO4)
To equalise H now, we have to place 4 before H2O. Thus we see that by doing so, O is automatically balanced(20 O in either side).
8KI + 5H2SO4 ------> 4I2 + 4K2SO4 + H2S + 4H2O (balanced equation) Just imagine that had you not adopted this technique, could you have balanced this equation by hit and trial method? The answer may be YES. But in that case you would have taken a very long time to do so. Therefore you are advised not to take an attempt to balance a redox reaction by hit and trial method if it appears to be little tough. Always make use of this ON method.
0 +5 +5 +4 ++ (ii) I2 + HNO3 HIO3 NO2 H2O (I2 is RA and HNO3 is OA)
The change in ON per one I = |0-5| =5, so for two I atoms present in the RA(I2), the total change is 10. So 10 is placed as coefficient before HNO3(OA).
The change in ON of N = |+5-(+4)| =1, and since there is one N atom in HNO3, the total change is also 1. So 1 is to be placed as the coefficient of I2 and therefore it is left as such.
I2 + 10HNO3 ------> HIO3 + NO2 + H2O
To equalise N, we place 10 before NO2 in RHS and to equalise I, we place 2 before HIO3 in RHS.
Dr. S. S.Tripathy 48 Concepts in Chemistry
I2 + 10HNO3 ------> 2HIO3 + 10NO2 + H2O
To equalise H atoms, we place 4 before H2O, and by doing so, it is seen that O is automatically balanced(30 O on either side).
I2 + 10HNO3 ------> 2HIO3 + 10NO2 + 4H2O (balanced equation) What a fantastic method!!!! Could you have balanced it by hit and trial method? The answer is again, NO. 0 +5 +2 +2 + Cu(NO ) ++ NO H O (iii) Cu HNO3 3 2 2 (RA is Cu and OA is HNO3)
The change of ON in Cu(RA) is 2 and in HNO3(OA) is 3. So first 2 is placed before HNO3 and
3 placed before Cu. Then hit and trial starts. To equalise Cu, we have to place 3 before Cu(NO3)2.
32Cu + HNO3 3Cu(NO3)2 ++ NO H2O By doing so the total number of N atoms in RHS is 7. So we have to revise the coefficient of 7 HNO3 in the LHS and place 7 in place of 2. Then to equalise H, we have to place 2 before
H2O.
++7 33Cu + 7 HNO3 Cu(NO3)2 N O 2 H2O In so doing we found that O could not be balanced. 21 O are on LHS and 22.5 on RHS. So in such case we have to re-revise the coefficient of NO. Note that we cannot do any change in Cu and Cu(NO3)2. So we place 2 before NO and accordingly change HNO3 and H2O. For that we revise the coefficient of HNO3 from 7 to 8 to equalise 8 N atoms on either side. Then to equalise
H we place 4 before H2O. Now we find that O atoms are automatically balanced on either side. So the balanced equation is
3 Cu + 8 HNO3 3 Cu(NO3)2 + 2 NO + 4 H2O (balanced equation) SAQ 3: (i)
KMnO4 + H2SO4 + H2C2O4 K2SO4 + MnSO4 ++ CO2 H2O
22KMnO4 K2O + MnO + 5 [O] (Step 1) K2O + H2SO4 K2SO4 + H2O (Step 2) 2 X MnO + H2SO4 MnSO4 + H2O (Step 3) (Step 4) 5 X H2C2O4 + [O] 2CO2+ H2O 10 2 KMnO4 + 3 H2SO4 + 5 H2C2O4 K2SO4 + 2 MnSO4 + CO2 + 8 H2O (balanced equation)
In this case also the first three steps are same as done in the example in the text. In the forth step
H2C2O4(oxalic acid) reacts with nascent oxygen to form CO2. The question is how shall you know this? This is very simple. You merely look to the RHS, you can know the products
Dr. S. S.Tripathy Logic of Inorganic Reactions 49 immediately. Can you indicate now which is oxidised and which reduced. Mn has ON +7 in
KMnO4 and +2 in MnSO4, so KMnO4 has been reduced. While the ON of C in H2C2O4 is
+3(find it by x method)and it is +4 in CO2. So H2C2O4 has been oxidised. So which is OA and RA? You can better say that now.
(ii)
K2Cr2O7 + H2SO4 + H2O2 K2SO4 + Cr2(SO4)3 ++ O2 H2O
K2Cr2O7 K2O + Cr2O3 + 3 [O] (Step 1) K2O + H2SO4 K2SO4 + H2O (Step 2) Cr2O3 ++ 33 H2SO4 Cr2(SO4)3 H2O (Step 3) 3 X H2O2+ [O] O2 + H2O (Step 4) (Step 5) K2Cr2O7 + 4 H2SO4 + 3 H2O2 K2SO4 + Cr2(SO4)3 + 3 O2 + 7 H2O
In this case too, the product in the 4th step(H2O2 reacting with [O]) is known just by looking at the RHS. It is O2. The other product has to be H2O to balance H. This is known from mere common sense.
In this case K2Cr2O7 is the OA as the ON of Cr is +6 and it is reduced to +3 in Cr2(SO4)3. H2O2 is the RA as the ON of O is -1(peroxide) and it is oxidised to 0 in O2 of RHS.
SAQ 4: (i) In this case the HNO3 used must be very dilute type as NH4NO3 is produced.
HNO3 + H2O ------> NH3 + 4[O] (i)
[NH3 + HNO3 ------> NH4NO3 4X [Zn + [O] ------> ZnO] (ii) (4 multiplied to cancel [O])
4X [ZnO + 2HNO3 ------> Zn(NO3)2 + H2O] (iii) (4 multiplied to cancel 4 ZnO) ______
4Zn + 10HNO3 ------> 4Zn(NO3)2 + NH4NO3 + 3H2O (balanced equation)
Note that in step (i) H2O appears in LHS, otherwise you cannot balance the equation,remember this.
(ii) In this case the HNO3 used must be moderately concentrated as NO is produced.
2HNO3 ------> 2NO + H2O + 3[O] 3X[Mg + [O]------> MgO ]
3X[MgO + 2HNO3 ---> Mg(NO3)2 + H2O ] ______
3Mg + 8HNO3 ------> 3Mg(NO3)2 + 2NO + 4H2O (balanced equation) SAQ 5:
3X [H2SO4 ------> SO2 + H2O + [O]] (3 multiplied to cancel 3[O])
2Al + 3[O] -----> Al2O3
Al2O3 + 3H2SO4 ------> Al2(SO4)3 + 3H2O (Al2O3 get cancelled) ______
2Al + 6H2SO4 ------> Al2(SO4)3 + 3SO2 + 6H2O (balanced equation) Dr. S. S.Tripathy 50 Concepts in Chemistry
SAQ 6: 2- - - Net ionic equation: CrO4 + Cu2O(s) + H2O(l) ------> Cu(OH)2(s) + CrO2 + OH Note that this reaction has been carried out in alkaline medium as OH- is present in the RHS. - Could you find which is oxidised and which reduced? K2CrO4 has been reduced to CrO2 because the ON of Cr had decreased from +6 to +3. Cu2O has been oxidised to Cu(OH)2 as the ON of Cu has increased from +1 to +2. So let us take the steps separtely and apply the rules. Reduction step: 2- - CrO4 ------> CrO2 Cr atoms are same, so no coefficient is to be placed in any side.
Then 2 H2O molecules are to be added on RHS as LHS has 2 excess O atoms. 2- - CrO4 ------> CrO2 + 2H2O - Then 4OH are to be added on RHS and 4 H2O to be added on LHS as RHS has 4 excess H atoms. 2- - - CrO4 + 4 H2O ------> CrO2 + 2H2O + 4OH Then equalisation of charge is done by adding 3 electrons in LHS so that the charge on either side becomes -5. 2- - - - CrO4 + 4 H2O + 3e ------> CrO2 + 2H2O + 4OH Oxidation step:
Cu2O ------> Cu(OH)2 First we have to place a coefficient of 2 in RHS to equalise Cu atoms.
Cu2O ------> 2Cu(OH)2 Since in LHS there is one O atom and in RHS there are 4 O atoms, RHS has 3 excess O atoms.
So we have to add 3 H2O molecules on LHS
Cu2O + 3 H2O ------> 2Cu(OH)2 Now let us find how many H atoms are excess in which side. LHS contains 2 excess H atoms. - So we have to add 2 OH on LHS and 2 H2O molecules on RHS. - Cu2O + 3 H2O + 2OH ------> 2Cu(OH)2 + 2H2O Then charge equalisation is made by adding 2 electrons on RHS, so that both sides have -2 charge. - - Cu2O + 3 H2O + 2OH ------> 2Cu(OH)2 + 2H2O + 2e Let us add now the two steps. 2- - - - 2 X [CrO4 + 4 H2O + 3e ------> CrO2 + 2H2O + 4OH ] - - 3X [Cu2O + 3 H2O + 2OH ------> 2Cu(OH)2 + 2H2O + 2e ] (6 electrons are cancelled) ______2- - - - 2CrO4 + 17H2O + 6OH + 3Cu2O ------> 2CrO2 + 10 H2O + 6Cu(OH)2 +8OH On further simplification we get, 2- - - 2CrO4 + 7H2O + 3Cu2O ------> 2CrO2 + 6Cu(OH)2 + 2 OH
SAQ 7:
+6 -1 +3 0 (i) K2Cr2O7 + HCl KCl + CrCl3 + Cl2 + H2O
Dr. S. S.Tripathy Logic of Inorganic Reactions 51
+6 +3 Reduction Cr + 3 e-1 Cr -1 0 Oxidation Cl Cl + e-1 X 3 +6 -1 +3 0 Cr + 3 Cl Cr + 3 Cl (electron balance diagramme)
Looking to 2 Cr atoms in K2Cr2O7, let us multiply the above equation by 2 throughout +6 -1 +3 0 2Cr + 6 Cl 2Cr + 6 Cl So we can develop a part balanced molecular equation,
K2Cr2O7 + 6 HCl ------> 2 KCl + 2 CrCl3 + 3 Cl2
But equation is not yet balanced and one more product is missing(H2O). Since HCl is used both as a reducing agent(RA) and an acid to form salt, more number of Cl- ions are required and so the coefficient of HCl has to be changed. First for equalising O atoms on both the sides,
7 is placed as the coefficient of H2O and accordingly to balance H atoms, the coefficient of HCl is changed to 14.
K2Cr2O7 + 14 HCl -----> 2 KCl + 2CrCl3 + 3 Cl2 + 7 H2O (ii)
+2 -1 0 +3 -2 +4 FeS2 + O2 Fe2O3 + SO2 Here there are two oxidations i.e Fe2+ is oxidised to Fe3+ and S-1 is oxidised to S+4. Note that in iron disulphide(called iron pyrite or fool's gold) Fe is in the +2 state and sulphide is a disulphide(belongs to polysulphide category) in which S is in -1 state. We get the net oxidation step by adding the following two oxidation steps. Oxidation:
+2 +3 Fe Fe + e-1 -1 +4 2S 2S + 10e-1 -1 +2 +3 +4 -1 Fe + 2S Fe + S + 11 e Note that we multiplied the 2nd oxidation step by 2 to get the ratio of Fe and S equal to 1:2 to agree with the formula of iron dilphide. 0 -2 Reduction: O + 2 e-1 O Adding the oxidaton and reduction steps by cancelling 22 electrons on either side, we get, 0 +2 -1 -2 +3 +4 11 O + 2 Fe + 4 S 11 O + 2 Fe + 4 S (electron balance diagramme) Looking to the electron balance diagramme, we can balance the molecular equation,
2 FeS2 + 11/2 O2 ------> Fe2O3 + 4 SO2 To eliminate fraction, we multiply the equation by 2 throughout
4 FeS2 + 11 O2 ------> 2 Fe2O3 + 8 SO2
Dr. S. S.Tripathy 52 Concepts in Chemistry
ANSWERS TO PRACTICE QUESTIONS - 2+ + 3+ 1. (i) NO3 + Fe + H ------> Fe + NO2 + H2O (Acidic medium) - + - Reduction step: NO3 + 2H + e ------> NO2 + H2O Oxidation step: Fe2+ ------> Fe3+ + e- ______2+ - + 3+ Fe + NO3 + 2H ------> NO2 + Fe + H2O - - (ii) NH3 + MnO4 + OH ------> MnO2 + NO2 (Alkaline medium) - - - Reduction step: 7X[MnO4 + 4H2O + 3e ------> MnO2 + 2H2O + 4OH ] Oxidation step: 3X[NH + 2H O + 7OH------> NO + 7H O + 7e-] 3______2 2 2 - - 7MnO4 + 28H2O + 3NH3 + 6H2O + 21OH ------> - 7MnO2 + 14H2O + 28OH + 3NO2 + 21H2O - - On simplifying we get: 7MnO4 + 3NH3 ------> 7MnO2 + 3NO2 + 7 OH + H2O 3+ 2+ (iii) Fe + H2O2 ------> Fe + O2 (acidic medium) + Oxidation step: H2O2 ------> O2 + 2H + 2e Here H2O addition is not necessary Reduction step: 2X[Fe3+ +e ------> Fe2+ ] ______3+ 2+ + H2O2 + 2Fe ------> O2 + 2Fe + 2H - - (iv) Cr(OH)2 + I2 + OH ------> Cr(OH)3 + I (alkaline medium as OH- is present) - - Reduction step: [I2 + 2e ------> 2I ] - - Oxidation step: 2X [Cr(OH)2 + H2O + OH ------> Cr(OH)3 + H2O + e ] ______- - 2Cr(OH)2 + I2 + 2OH ------> 2Cr(OH)3 + 2I 2- + 2- 3+ (v) C2O4 + H + Cr2O7 ------> CO2 + Cr + H2O (acidic medium) 2- + - 3+ Reduction step: Cr2O7 + 14H + 6e ------> 2Cr + 7H2O 2- - Oxidation step: 3X [C2O4 ------> 2CO2 + 2e ] ______2- 2- + 3+ Cr2O7 + 3C2O4 + 14H ------> 2Cr + 6CO2 + 7H2O - + - 2+ (vi) MnO4 + H + I ------> Mn + I2 + H2O - + - 2+ Reduction step: 2X [MnO4 + 8H + 5e ------> Mn + 4H2O] - - Oxidation step: 5X [2I ------> I2 + 2e ] ______- + - 2+ 2MnO4 + 16H + 10I ------> 2Mn + 5I2 + 8H2O 2- 2- - (vii) S2O3 + I2 ------> S4O6 + I - - Reduction step: I2 + 2e ------> 2I 2- 2- - Oxidation step: 2S2O3 ------> S4O6 + 2e ______2- - 2- I2 + 2S2O3 ------> 2I + S4O6 In this case neither acidic nor basic medium is required. As you see, in the steps, there is no need of H+ or OH- ions for balancing. - + - - (viii) OCl + H + I ------> I2 + Cl (acidic medium) - + - - Reduction step: OCl +2H + 2e ------> Cl + H2O (Cl changed from +1 to -1) Oxidation step: 2I------> I + 2e- (I from -1 to 0) ______2 - - + - OCl + 2I + 2H ------> Cl + I2 + H2O
Dr. S. S.Tripathy Logic of Inorganic Reactions 53
- - - (ix) Cl2 + IO3 ----> IO4 + Cl (alkaline medium) - - Reduction step: Cl2 + 2e ------> 2Cl - - - - Oxidation step: IO3 + H2O + 2OH ------> IO4 + 2H2O + 2e ______- - - - Cl2 + IO3 + 2OH ------> IO4 + 2Cl + H2O 2.
(i) NH3 + O2 ------> NO + H2O ON of N goes from -3 to +2. So the change in ON =|-3-(+2)| =5, and change in ON in two atoms of O= 2X|-2|=4
4NH3 + 5O2 ------> 4NO + 6H2O
(ii) H2S + SO2 ------> S + H2O
ON of S goes from -2 in H2S to 0 in S, so change in ON is 2. The ON of S change from +4 in
SO2 to 0 in S, so the change is 4.
4H2S + 2SO2 ------> 6 S +4 H2O
This is further simplified as : 2H2S + SO2 ------> 3 S +2 H2O
(iii) Fe2O3 + CO ------> Fe + CO2 The ON of Fe goes from +3 to 0, so the change per atom is 3 and for 2 Fe atoms present in
Fe2O3, the total change is 2X3=6, The ON of C changes from +2 to +4. So the change is 2.
2Fe2O3 + 6CO ------> 4 Fe + 6CO2
Further simplification results: Fe2O3 + 3CO ------> 2 Fe + 3CO2
(iv)SnCl2 + O2 + HCl ------> SnCl4 + H2O
ON of Sn goes from +2 to +4, so the change in ON is 2 which is placed before O2(OA). The change in ON per one O atom is |0-2|=2 and for 2 atoms present in O2 the total change is 2X2=4, which is to be placed before SnCl2(RA). Then hit and trial.
4SnCl2 + 2O2 +8 HCl ------> 4SnCl4 +4 H2O On simplification, we get
2SnCl2 + O2 + 4HCl ------> 2SnCl4 +2 H2O
(v) NaClO3 + KI + HCl ------> NaCl + I2 + KCl + H2O
The ON of Cl goes from +5(NaClO3) to -1(NaCl) and the change is +5-(-1)=6, which is to be placed before KI(RA). The change in ON of I is |-1-0|=1, so no other coefficient is to be placed before NaClO3(OA). Then hit and trial gives the balanced equation.
NaClO3 + 6KI + 6HCl ------> NaCl +3 I2 + 6KCl + 3H2O
(vi) SbCl5 + KI ------> SbCl3 + I2 + KCl The change in ON of Sb is |+5-3|=2, which is to be placed before KI. The change in ON in I is
|-1-0|=1, so no other coefficient is to be placed before SbCl5. Then hit and trial.
SbCl5 + 2KI ------> SbCl3 + I2 + 2KCl
(vii) NaPO3 + BrF3 ------> NaPF6 + Br2 + O2
The ON of Br goes from +3(BrF3) to 0(Br2) and hence the change is 3 which is to be placed before NaPO3(RA). The ON of O changes from -2(NaPO3) to 0(O2) and hence the change is 2 per atom. For 3 O atoms present, the total change is 2X3=6, which is to be placed before
BrF3(OA). Then hit and trial. 9 3NaPO3 + 6BrF3 ------> 3NaPF6 + 3Br2 + 2 O2 To eliminate the fractional coefficient, the whole equation is multiplied by 2.
6NaPO3 + 12BrF3 ------>6 NaPF6 + 6Br2 + 9O2 Dr. S. S.Tripathy 54 Concepts in Chemistry
Now we can simplify by dividing the equation by 3 througout.
2NaPO3 + 4BrF3 ------>2 NaPF6 + 2Br2 + 3O2
(viii) Sn + HNO3 ------> SnO2 + NO + H2O The Change in ON of Sn is |0-4|=4 and change in ON of N is |5-2|=3, so
3Sn + 4HNO3 ------> SnO2 + NO + H2O Then hit and trial gives
3Sn + 4HNO3 ------> 3SnO2 + 4NO + 2H2O (balanced equation)
(ix) Cr2O3 + HNO3 + Na2CO3 ------> Na2CrO4 + CO2 + HNO2
The ON of one Cr changes by |+3-6| =3, so for two Cr atoms present in Cr2O3(OA), the total change is 6. The change in ON of N is |+5-3|=2, so we get first,
2Cr2O3 + 6HNO3 + Na2CO3 ------> Na2CrO4 + CO2 + HNO2 Hit and trial gives the following
2Cr2O3 + 6HNO3 + 4Na2CO3 ------> 4Na2CrO4 + 4CO2 + 6HNO2 Further simplification gives
Cr2O3 +3 HNO3 + 2Na2CO3 ------> 2Na2CrO4 + 2CO2 + 3HNO2
(x) Cu + HNO3 ------> Cu(NO3)2 + N2O + H2O The change in ON of Cu =2 and N=|5-1)=4, so first we write as follows.
4Cu + 2HNO3 ------> Cu(NO3)2 + N2O + H2O Hit and trial gives
4Cu + 2HNO3 ------> 4Cu(NO3)2 + N2O + H2O
But now RHS contains 10 N atoms, so we have to revise the coefficient of HNO3 in LHS and accordingly balance the rest.
4Cu + 10HNO3 ------> 4Cu(NO3)2 + N2O + 5H2O
(xi) Cu + HNO3 ------> Cu(NO3)2 + NO + H2O The change in ON of Cu=2 and N=3, so we first write
3Cu + 2HNO3 ------> Cu(NO3)2 + NO + H2O Hit and trial gives,
3Cu + 2HNO3 ------> 3Cu(NO3)2 + NO + H2O
Since RHS now contains 7 N atoms, we have to revise the coefficient of HNO3 from 2 to 7 to equalise the N atoms.
3Cu + 7HNO3 ------> 3Cu(NO3)2 + NO + H2O 7 To balance H, we have to place 2 before H2O. But we find that O is not balanced by that. 7 3Cu + 7HNO3 ------> 3Cu(NO3)2 + NO + 2 H2O So we have to again revise the number of N atoms by balancing 2 before NO(RHS) and accordingly change the coefficient of HNO3. After that H is adjusted again .
3Cu + 8HNO3 ------> 3Cu(NO3)2 + 2NO + 4H2O (balanced equation)
(xii) BaCrO4 + KI + HCl ------> BaCl2 + I2 + KCl + CrCl3 + H2O The ON of Cr changes from +6 to +3, so the change is 3. The change in ON of I is 1.So we first write
BaCrO4 + 3KI + HCl ------> BaCl2 + I2 + KCl + CrCl3 + H2O Further hit and trial gives 3 BaCrO4 + 3KI + 8HCl ------> BaCl2 + 2 I2 + 3KCl + CrCl3 + 4H2O
Dr. S. S.Tripathy Logic of Inorganic Reactions 55
To eliminate fraction, we multiply by 2 throughout
2BaCrO4 + 6KI + 16HCl ------> 2BaCl2 + 3I2 + 6KCl + 2CrCl3 + 8H2O
(xiii) Ca3(PO4)2 + SiO2 + C ------> CaSiO3 + P4 + CO
The ON of P change from +5 to 0 for one P atoms. For two P atoms present in Ca3(PO4)2, the total change is 10. The change in ON of C is 2. So first we write
2Ca3(PO4)2 + SiO2 +10 C ------> CaSiO3 + P4 + CO Hit and trial gives
2Ca3(PO4)2 + 6SiO2 +10 C ------> 6CaSiO3 + P4 + 10CO
(xiv) Zn + HNO3 ------> Zn(NO3)2 + NH4NO3 + H2O + The change of ON of Zn is 2 and the ON of N changes from +5(HNO3) to -3(NH4 of NH4NO3). So the change in ON=|+5-(-3)|=8, so first we write,
8Zn + 2HNO3 ------> Zn(NO3)2 + NH4NO3 + H2O Hit and trial results
8Zn + 2HNO3 ------> 8Zn(NO3)2 + NH4NO3 + H2O
RHS now contains 18 N atoms, so we have to modify the coefficient of HNO3 and accordingly H is adjusted.
8Zn + 18HNO3 ------> 8Zn(NO3)2 + NH4NO3 + 7H2O We find that O is not balanced. So again we have to modify the N atoms by first placing coefficient
2 before NH4NO3 and then adjusting HNO3 and then other atoms.
8Zn + 20HNO3 ------> 8Zn(NO3)2 + 2NH4NO3 + 6H2O Now interestingly we find that O is balanced.
(xv) K2Cr2O7 + KI + H2SO4 ------> K2SO4 + Cr2(SO4)3 +I2 +H2O The ON of Cr changes from +6 to +3, so the change is 3 per atom. But for 2 Cr atoms prsent in
K2Cr2O7(OA), the total change in ON=6. The change in ON of I is 1. So first we write,
K2Cr2O7 + 6KI + H2SO4 ------> K2SO4 + Cr2(SO4)3 +I2 +H2O Further hit and trial gives
K2Cr2O7 + 6KI + 7H2SO4 ------> 4K2SO4 + Cr2(SO4)3 +3I2 + 7H2O(balanced)
(xvi) KMnO4 + H2SO4 + H2S ------> K2SO4 + MnSO4 + S + H2O The change in ON of Mn =|7-2|=5, and that of S=|-2-0|=2, so first we write,
2KMnO4 + H2SO4 + 5H2S ------> K2SO4 + MnSO4 + S + H2O, Further hit and trial gives,
2KMnO4 + 3H2SO4 + 5H2S ------> K2SO4 + 2MnSO4 + 5S + 8H2O
(xvii) Cl2 + KIO3 + KOH ------> KCl + KIO4 + H2O
The change in ON of Cl is 1 for one Cl atom. So for 2 Cl atoms present in the OA(Cl2), the total change is 2. The change in ON of I=|5-7|=2, so first we write,
2Cl2 + 2KIO3 + KOH ------> KCl + KIO4 + H2O, Hit and trial gives,
2Cl2 + 2KIO3 + 4KOH ------> 4KCl + 2KIO4 + 2H2O On simplification, it gives,
Cl2 + KIO3 + 2KOH ------> 2KCl + KIO4 + H2O → (xviii) N2H4 + KBrO3 N2 + KBr + H2O N goes from -2 to 0, so for 2 N atoms the total change of ON is 4. Br goes from +5 to -1 and so the change is 6.
6N2H4 + 4KBrO3 -----> 6N2 + 4KBr + 12H2O
Dr. S. S.Tripathy 56 Concepts in Chemistry
(xix)Sn + NaOH ------> Na2SnO3 + H2 Sn goes from 0 to +4, so the change is 4. H goes from +1 to 0, so the change is 1.
H2O has to be added in LHS for balancing. By adding one H2O molecule there will be 4H atoms in LHS to account for the total change of ON in H to balance with the total change in ON in Sn(4).
Sn + 2NaOH + H2O ------> Na2SnO3 + 2H2
(xx)Al + NaOH ------> NaAlO2 + H2
Al goes from 0 to +3 while H goes from +1 to 0. H2O molecule is to be added in LHS for balancing. By adding one H2O molecule there will be 3 H atoms in LHS to account for the total change of ON in H to balance with the total change in ON in Al(3).
Al + NaOH + H2O ------> NaAlO2 + 3/2 H2
2Al + 2NaOH + 2H2O ------> 2NaAlO2 + 3H2 3.(i)
+ + 2KMnO4 K2O 25 MnO [O] K2O + H2SO4 K2SO4 + H2O 2 X MnO ++ H2SO4 MnSO4 H2O 5 X 2HI + [O] I2 + H2O + 2 KMnO4 + 3 H2SO4 + 10 HI K2SO4 ++ 258 MnSO4 I2 H2O
(ii)
+ + 2KMnO4 K2O 25 MnO [O] + K2O + H2SO4 K2SO4 H2O 2 X MnO ++ H2SO4 MnSO4 H2O + 5 X H2O2 + [O] H2O O2 + + 2KMnO4 ++ 3 H2SO4 58 H2O2 K2SO4 + 2 MnSO4 5 O2 H2O
(iii)
+ + 2KMnO4 K2O 2 MnO 5 [O] K2O + H2SO4 K2SO4 + H2O 2 X MnO + H2SO4 MnSO4 + H2O + 5 X KNO2 [O] KNO3 + 2KMnO4 + 3 H2SO4 + 52 KNO2 K2SO4 + MnSO4 5 KNO3 + 3H2O
(iv) Try this by partial equation method in the similar manner as shown above.
2KMnO4 + 3H2SO4 + 5H2S ------> K2SO4 + 2MnSO4 + 5S + 8H2O
(v) 2KMnO4 + 3H2SO4 + 5H2C2O4 ----> K2SO4 + 2MnSO4 + 10CO2 + 8H2O (vi)
+ + K2Cr2O7 K2O Cr2O3 3 [O] + + K2O H2SO4 K2SO4 H2O + + Cr2O3 3 H2SO4 Cr2(SO4)3 3 H2O + + 3 X 2HI [O] I2 H2O + + + + + K2Cr2O7 4 H2SO4 6 HI K2SO4 Cr2(SO4)3 3 I2 7 H2O Dr. S. S.Tripathy Logic of Inorganic Reactions 57
(vii) Try this by partial equation method of your own and check with the answer.
K2Cr2O7 + 4H2SO4 + 3H2O2 ------> K2SO4 + Cr2(SO4)3 + 3O2 + 7H2O
(viii) K2Cr2O7 + 4H2SO4 + 3KNO2 ------> K2SO4 + Cr2(SO4)3 + 3KNO3 + 4H2O
(ix) K2Cr2O7 + 4H2SO4 + 3H2S ------> K2SO4 + Cr2(SO4)3 + 3S + 7H2O
(x) K2Cr2O7 + 4H2SO4 + 3H2C2O4------> K2SO4 + Cr2(SO4)3 + 6CO2 + 7H2O (xi)
+ + 2HNO3 23NO H2O [O] 3 X Cu + [O] CuO + 3 X CuO + 2 HNO3 Cu(NO3)2 H2O + 3 Cu 8 HNO3 3 Cu(NO3)2 ++ 24 NO H2O
(xii) Try yourself and check. 4Fe + 10 HNO3 ------> 4Fe(NO3)2 + N2O + 5H2O (xiii)
H2SO4 SO2 + H2O + [O] Mg + [O] MgO + MgO + H2SO4 MgSO4 H2O Mg + 2 H2SO4 MgSO4 + SO2 + 2 H2O
Note that whenever you find a reaction involving conc. H2SO4, the first step is the breaking down of H2SO4 giving SO2, H2O and nascent oxygen atom.
2- 2- 4.(i) Reduction step: S2O8 + 2e ------> 2SO4 (O changes from -1 to -2) Oxidation step: 2X[Ce3+ ------> Ce4+ +e] ______2- 3+ 2- 4+ S2O8 +2Ce ------> 2SO4 + 2Ce - + (ii) Reduction step: 2X[NO3 + 4H + 3e------> NO + 2H2O] + 2+ Oxidation step: 3X[ Cu2O + 2H ------> 2Cu + H2O +2e] ______+ - 2+ 3Cu2O + 14H +2NO3 ------> 6Cu + 2NO + 7H2O - - (iii) Reduction step: 2X[MnO4 +4H2O +3e------> MnO2 + 2H2O + 4OH ] 2- - 2- 3 X[SO3 +H2O+2OH ------> SO4 + 2H2O+2e] ______- 2- - 2- 2MnO4 + H2O + 3SO3 ------> 2MnO2 + 2OH + 3SO4 2- 2- (iv) Reduction step: 5X[S2O8 + 2e------> 2SO4 ] 2+ - + Oxidation step: 2X[ Mn +4H2O ------> MnO4 + 8H +5e] ______2- 2+ 2- - + 5S2O8 + 2Mn + 8H2O ------> 10SO4 + 2MnO4 + 16H 2- + 3+ (v) Reduction step: Cr2O7 +14H + 6e------> 2Cr + 7H2O + Oxidation step: 3X[ C2H4O +H2O------> C2H4O2 + 2H +2e] ______2- + 3+ Cr2O7 + 3C2H4O+8H ------> 2Cr + 3C2H4O2 + 4H2O
Dr. S. S.Tripathy 58 Concepts in Chemistry
(vi) Reduction step: 2X[S + 2e ------> S2-] - 2- Oxidation step: 2S+3H2O +6OH ------> S2O3 + 6H2O +4e ______- 2- 2- 4S + 6OH ------> 2S +S2O3 + 3H2O - + - (vii) Reduction step: ClO3 + 6H +6e------> Cl + 3H2O - Oxidation step: 3X[ 2I ------> I2 + 2e] ______- + - - ClO3 + 6H + 6I ------> Cl + 3I2 + 3H2O
(viii) Reduction step: 8X[Ag+ + e ------> Ag] - Oxidation step: AsH3 +4H2O+8OH ------> H3AsO4 + 8H2O +8e ______+ - 8Ag + AsH3 + 8OH ------> H3AsO4 + 8Ag + 4H2O - + (ix) Reduction step: 16X [NO3 +4H +3e------> NO + 2H2O] 2+ Oxidation step: 3X[ 8CuS ------> 8Cu + S8 + 16e] ______- + 2+ 24CuS + 16NO3 + 64H ------> 16NO + 24Cu + 3S8 + 32H2O
- - - (x) Reduction step: ClO3 + 2H2O +2e------> ClO2 + H2O + 2OH - - - Oxidation step: SbO2 + 4H2O +2OH ------> Sb(OH)6 + 2H2O + 2e ______- - - - ClO3 + SbO2 +3H2O ------> ClO2 + Sb(OH)6 - - (xi) Reduction step: NO3 + 9H2O +8e------> NH3 + 3H2O + 9OH Oxidation step: 4X [Zn ------Zn2+ + 2e] ______- 2+ - 4Zn + NO3 + 6H2O ------> 4Zn + NH3 + 9OH - + - (xii) Reduction step: BrO3 +6H +6e------> Br +3H2O 2- 2- Oxidation step: 3X [2S2O3 ------> S4O6 + 2e] ______- 2- + - 2- BrO3 + 6S2O3 + 6H ------> Br + 3S4O6 + 3H2O - - (xiii) Reduction step: 2X[MnO4 + 4H2O + 3e------> MnO2 + 2H2O + 4OH ] 2- - 2- Oxidation step: 3X[ SnO2 + H2O + 2OH ------> SnO3 +2H2O + 2e] ______- 2- 2- - 2MnO4 + 3SnO2 + H2O ------> 2MnO2 + 3SnO3 +2OH 4- + 3- 5. (i) [Fe(CN)6] + H + H2O2 ------> [Fe(CN)6] + H2O + - Reduction step: H2O2 + 2H +2e ------> H2O + H2O 4- 3- - Oxidation step: 2X{[Fe(CN)6] ------>[Fe(CN)6] +e } ______4- + 3- 2[Fe(CN)6] + H2O2 + 2H ------> 2[Fe(CN)6] + 2H2O Looking to the coefficients of the balanced ionic equation, the original molecular equation is blananced. See this.
2K4[Fe(CN)6](aq) + 2HCl(aq) + H2O2(l) ------> 2K3[Fe(CN)6](aq) +
2KCl(aq) +2H2O(l)
Dr. S. S.Tripathy Logic of Inorganic Reactions 59
3- 4- (ii) N2H4 + [Fe(CN)6] ------> N2 + [Fe(CN)6] + H2O (alkaline medium) 3- 4- Reduction step: 4X{[Fe(CN)6] + e ------>[Fe(CN)6] } - Oxidation step: N2H4 + 4OH ------> N2 + 4H2O + 4e ______3- - 4- 4Fe(CN)6] + N2H4 +4OH ------> 4Fe(CN)6] + 4H2O + N2
N2H4 (l)+ 4K3[Fe(CN)6](aq)+4KOH------> N2 (g)+ 4K4[Fe(CN)6](aq) + +4H2O - 2- - (iii) C2H2 + MnO4 ------> MnO2 + C2O4 + OH + H2O (alkaline medium) - - Reduction step: 8X[MnO4 + 4H2O + 3e------> MnO2 + 2H2O + 4OH ] - 2- Oxidation step: 3X [C2H2 + 4H2O +10OH ------> C2O4 + 10H2O + 8e] ______- 2- - 8MnO4 +3C2H2 ------> 8MnO2 + 3C2O4 + 2H2O + 2OH
3C2H2 (g)+ 8KMnO4(aq)------> 8MnO2(s)+ 3K2C2O4(aq)+ 2KOH(aq) +2H2O(l) 2+ - 2+ 4+ (iv) Hg + C l + Sn ------> Hg2Cl2 + Sn 2+ - Reduction step: 2Hg + 2Cl + 2e------> Hg2Cl2 Oxidation step: Sn2+ ------> Sn4+ + 2e ______2+ - 2+ 4+ 2Hg + 2Cl + Sn ------> Hg2Cl2 + Sn
2HgCl2(aq)+ SnCl2(aq)------> Hg2Cl2(s)+ SnCl4(aq) - - 2- (v) NO3 + Zn + OH ------> NH3 + ZnO2 +H2O - - Reduction step: NO3 +9H2O +8e------> NH3 + 3H2O + 9OH - 2- Oxidation step: 4X [Zn + 2H2O + 4OH ------> ZnO2 + 4H2O + 2e] ______- - 2- 4Zn + NO3 + 7OH ------> NH3 + 4ZnO2 + 2H2O
NaNO3(aq) + 4Zn(s) + 7NaOH(aq) -----> NH3(g) + 4Na2ZnO2(aq) + 2H2O(l) - + - 2- (vi) HgS + NO3 + H + Cl ------> S + [HgCl4] + NO + H2O - + Reduction step: 2X[NO3 + 4H +3e------> NO + 2H2O] - 2- Oxidation step: 3X{HgS + 4Cl ------> S + [HgCl4] + 2e} ______- + - 2- 3HgS + 2NO3 + 8H + 12Cl ----> 2NO + 3S + 3[HgCl4] +4H2O
3HgS(s) + 2HNO3(aq) + 12HCl(aq)------> 3S + 3H2[HgCl4](aq) + 2NO(g) + 4H2O(l) 2- + - 3+ - (vii) Cr2O7 + 8H + 3NO2 ------> 2Cr + 3NO3 + 4H2O
K2Cr2O7 + 4H2SO4+3KNO2 ------> K2SO4+ Cr2(SO4)3+ 3KNO3+ 4H2O 2- + 3+ (viii) Cr2O7 + 8H + 3H2S ------> 2Cr + 3S + 7H2O
K2Cr2O7 + 4H2SO4 +3H2S ------> K2SO4 + Cr2(SO4)3 + 3S + 7H2O 2- + 3+ (ix) Cr2O7 + 8H + 3H2C2O4 ------> 2Cr + 6CO2 + 7H2O
K2Cr2O7 + 4H2SO4 +3H2C2O4 ------> K2SO4 + Cr2(SO4)3 + 6CO2 + 7H2O - + 2+ (x) 3Cu + 2NO3 + 8 H ------> 3Cu + 2NO + 4H2O
3Cu + 8HNO3 ------> 3Cu(NO3)2 + 2NO + 4H2O - + 2+ (xi) 4Fe + 2NO3 + 10H ------> 4Fe + N2O + 5H2O
4Fe + 10HNO3 ------> 4Fe(NO3)2 + N2O + 5H2O
Dr. S. S.Tripathy 60 Concepts in Chemistry
Supplimentary Questions: 1. Do these preferably by ON method and check with the answers given below.
(i) 3CuS + 8HNO3 ------> 3Cu(NO3)2 + 3S + 4H2O + 2NO
(ii) As2S5+ 40HNO3------> 2H3AsO4+ 5H2SO4+ 12H2O + 40NO2
(iii) 8Zn + 20HNO3 -----> 8Zn(NO3)2 + 6H2O+ 2NH4NO3
(iv) 5Na2C2O4 + 2KMnO4 + 8H2SO4 -----> K2SO4 +2MnSO4 +
10CO2+ 5Na2SO4 + 8H2O
(v) 2MnO + 5PbO2 + 10HNO3 -----> 2HMnO4 + 5Pb(NO3)2 + 4H2O
(vi) 6Na2HAsO3 + 2KBrO3 + 12HCl -----> 12NaCl + 2KBr + 6H3AsO4
(vii) 4FeS2 + 11O2 ------> 2Fe2O3 + 8SO2
(viii) Ca(OCl)2 + 4KI + 4HCl ----> 2I2 + CaCl2 + 2H2O + 4KCl
(ix) 2NaOCl + 2NaOH + Bi2O3 ----> 2NaBiO3 + H2O + 2NaCl
(x) Sn + 4HNO3 ----> SnO2 + 4NO2 + 2H2O +3-1 0 +6 +7 -1 (xi) CrI3 ++ KOH Cl2 K2Cr2O7 + KIO4 + KCl + H2O There are two oxidations and one reduction in this reaction. Let us balance it by electron- balance diagramme method.
+3 +6 oxidation: 22Cr Cr + 6 e- -1 +7 6 I 6+I 48 e- +3 -1 +6 +7 2Cr + 6 I 2 Cr + 6 I + 54 e-
0 -1 reduction: Cl + e- Cl X 54
+3 -1 0 +6 +7 -1 2 Cr + 6 I + 54 Cl 2 Cr + 6 I + 54 Cl
2CrI3 + 27Cl2 + 62KOH ------> K2Cr2O7 + 6KIO4 + 54KCl + 31H2O → (xii) Na2TeO3 + + 4NaI + 6HCl 6NaCl + Te + 2 I2 + 3H2O → (xiii) K2Cr2O7 + 3SnCl2 + 14HCl 2CrCl3 + 3SnCl4 + 2KCl + 7H2O → (xiv) 6K3[Fe(CN)6] + Cr2O3 + 10KOH 6K4[Fe(CN)6] + 2K2CrO4 + 5H2O → (xv) 3NaHSO4 + 8Al + 3NaOH 3Na2S + 4Al2O3 + 3H2O → (xvi) 8KI + 5H2SO4 4K2SO4 + 4I2 + H2S + 4H2O → (xvii) Cr2O3 + 2Na2CO3 + 3KNO3 2Na2CrO4 + 2CO2 + 3KNO2 → (xviii) 4KClO3 + 4H2SO4 4KHSO4 + 4ClO2 + O2 + 2H2O → (xix) 5P2H4 6PH3 + P4H2 → (xx) CoCl2 + 7KNO2 + 2HCl K3[Co(NO2)6] + NO + 4KCl + H2O Dr. S. S.Tripathy Logic of Inorganic Reactions 61
- - → 2- (xxi) 4OH + 4MnO4 4MnO4 + O2 + 2H2O
4KOH + 4KMnO4 ----> 4K2MnO4 + O2 + 2H2O - → - - (xxii) 4Au + 16CN + + 6 H2O + 3O2 4[Au(CN)4] + 12OH
4Au + 16KCN + 6 H2O + 3 O2 -----> 4K[Au(CN)2] + 12KOH + → 2+ (xxiii) 6Zn + 12H + As2O3 2AsH3 + 3H2O + 6Zn
6Zn + 12HCl + As2O3 ------> 2AsH3 + 3H2O + 6ZnCl2 - - → - - (xxiv) Cl2 + 2OH + IO3 IO4 + 2Cl + H2O
Cl2 + 2KOH + KIO3 ------> KIO4 + 2KCl + H2O + - → 2+ + (xxv) 4Zn + 10 H + NO3 4Zn + NH4 + 3H2O
4Zn + 10HNO3 ------> 4Zn(NO3)2 + NH4NO3 + 3H2O - → 3- (xxvi) 6V + 3OH + 14H2O HV6O17 + 15H2
6V + 3NaOH + 14HO ------> Na3HV6O17 + 15H2 (xxvii) There are two oxidations and one reduction in this case.
+2 +7 Oxidation: Mn Mn + 5 e -1 +5 2 Br 2 Br + 12 e +2 -1 +7 +5 Mn + 2 Br Mn + 2 Br + 17 e X 2
+4 +2 Reduction: Pb + 2 e Pb X 17
+7 +2 -1 +4 +5 +2 2 Mn + 4 Br + 17 Pb 2 Mn + 4 Br + 17 Pb
Balanced equation: → 2 MnBr2 + 17 PbO2 + 30 HNO3 2 HMnO4 + 2 Pb(BrO3)2 + 15 Pb(NO3)2 + 14 H2O
2. Do yourself and check with the following answers. 2+ + - 4+ (i) 3Sn + 8H + 2NO3 -----> 3Sn + 2NO + 4H2O 2- - + 3+ (ii) Cr2O7 + 6I 14H ----> 3I2 + 2Cr + 7H2O + - 2+ + (iii) 4Zn + 10H + NO3 -----> 4Zn + NH4 + 3H2O 2- + 2- (iv) 2S2O3 + 2Ag ------> 2Ag + S4O6 - - (v) 4P4 + 12OH + 12H2O----> 4PH3 + 12H2PO2 - - - - (vi) Br2 + IO3 + 2OH ----> 2Br + IO4 + H2O - 2+ + 2+ 4+ (vii) 2MnO4 + 5Sn + 16H ----> 2Mn + 5Sn + 8H2O 2- - + - (viii) S2O8 + 9I + 8H ----> 3I3 + 2SO2 + 4H2O 2- - - (ix) 2CrO4 + 3Cu2O +11H2O----> 6Cu(OH)2 + 2Cr(OH)4 + 2OH Dr. S. S.Tripathy SECTION-II LOGIC OF INORGANIC REACTIONS TYPES OF INORGANIC REACTIONS
All inorganic reactions fall into any one of the following two categories. (i) Non-Redox(Metathesis) (ii)Redox (A) NON-REDOX(METATHESIS) REACTIONS: The reaction in which there is no change of Oxidation Number(ON) of any element is called a metathesis reaction.Some common metathesis types of reactions are discussed below. (i) Double Replacement (Ion Exchange) Reaction: This involves the reaction between two substances (which are mostly soluble in water) in which the basic and acid parts (radicals) of two substances get exchanged. That means the basic part of one substance joins with the acid part of the other substance and vice versa. Double Replacement reaction is again of two types. (a)neutralisation(acid-base reaction) reaction (b)precipitation reaction Neutralisation: An acid reacts with a base(alkali) to produce a salt and water. The sour taste of acid and bitter(soda like) taste of alkali kill each other and a salty taste is found in the product.
NaOH + HCl ------> NaCl + H2O KOH + H2SO4 ------> K2SO4 + H2O Ca(OH)2 + HNO3 ------> Ca(NO3)2 + H2O In all these cases you find that the ON of none of the elements has undergone any change. So these are metathesis reactions. IMPORTANT: Note that in this chapter, you do not have to balance any equation. Our purpose is to know what products are formed during a reaction and why they are formed. In other words, we have to develop the logic of the reaction as to why a particula set of products are formed and why not any other products are formed. So you are categorically advised not to balance the equations throughout this chapter. Precipitation: When two soluble substances undergo double replacement reaction and one of the products is insoluble and appears as a precipitate, it is called precipitation reaction. Look to this example.
AgNO3 + HCl AgCl + HNO3 Here in this double replacement reaction, AgCl appears as a curdy white precipitate. Therefore it is called a precipitation reaction. Let us take some more examples.
Al2(SO4)3 + NH4OH Al(OH)3 + (NH4)2SO4
CaCl2 ++ (NH4)2CO3 CaCO3 NH4Cl CuSO4 ++ H2S CuS H2SO4 There are however some reactions which belong both to neutralisation as well as precipitation. See this example. 62 Concepts in Chemistry
Ba(OH)2 + H2SO4 BaSO4 + H2O (neutralisation and precipitation reaction) In this case, it is an acid-base reaction (neutralisation) and at the same time it is also a precipitation reaction, since BaSO4 is an insoluble white precipitate . If you are a beginner in chemical sciences, use the table on solubility rules given in Chapter-I to know which compounds are soluble in water which are not. In all these reactions, there is no change in ON of any element. SAQ 1: Detect wherever there is any mistake in the following equations. Correct them.(Do not balance)
(i)NaOH + H2SO4 -----> NaSO4 + H2O
(ii)Ba(OH)2 + H3PO4 ------> BaPO4 + H2O
(iii)CrCl3 + NaOH ------> CrOH + NaCl3
(iv)Bi(NO3)2 + H2S ------> BiS + H2NO3 While answering the SAQ you must have found that in writing the products of a reaction, you have to write the formula correctly and not merely exchange the radicals. Again you are made cautious at this stage that balancing equation in this program is not required. If you concentrate on balancing, then you cannot predict the products correctly. SAQ 2: Write the products of the following reactions. Indicate to which kind of double replacement reaction it belongs by giving the symbol, (N) for the neutralisation reaction, (P) for precipitation reaction and (B) for the reaction which falls into both the categories. Refer the table on solubility rules given in Chapter-I to know which one is an insoluble precipitate.
(i)Pb(NO3)2 + NaCl ------> ? (ii)BaCl2 + K2CrO4 ------>
(iii)KF + CaCl2 ------>? (iv)Pb(CH3COO)2 + H2S ---->?
(v)Ba(NO3)2 + K2CO3 ------>? (vi)FeCl3 + NH4OH ----->?
(vii)NaOH + H3PO4 ----->? (viii)HNO3 + Al2O3 ----->
(ix) MgO + H2SO4 ------>? (x) Cr2O3 + H3PO4 ------>? SAQ 3: What are the products of the following reactions? What type of reactions are these?
(i)NH4OH + H2SO4----->? (ii) NH3 + H2SO4 ------>?
(iii)NH4OH + HCl ----->? (iv)NH3 + HCl ------>?
(v)NH3 + HNO3------>?
(ii) Reaction of Nonmetallic Oxides with water: This is another type of metathesis reaction in which the ON does not change for any element. Nonmetallic oxides when react with water give the corresponding acids. eg. SO3 + H2O ------> H2SO4
Here SO3 is an oxide of the nonmetal sulphur. When sulphur trioxide reacts with water forms sulphuric acid (H2SO4). You know that in the periodic table nonmetals are situated in the right- upper part. Elements like C, N, O, P, S, Cl, Br, I etc. are called nonmetals. Their oxides are called acidic oxides, because these oxides produce acids when react with water. Note also that for the same reason, the nonmetallic oxides are called ACID ANHYDRIDES( anhydride means water removed. They are formed by the removal of water from acids). This means that when an acid anhydride reacts with water it gives back the same acid from which it is produced. The following table gives a list of some common nonmetallic oxides(acid anhydrides) and the corresponding acids.
Dr. S.S Tripathy Logic of Inorganic Reactions 63
non metallic Oxide acid nonmetallic Oxide acid
CO2 H2CO3 N2O5 HNO3 (carbon dioxide) (carbonic acid) (dinitrogen pentoxide) (nitric acid)
N2O3 HNO2 Br2O HOBr (dinitrogen trioxide) (nitrous acid) (dibromine oxide)(hypobromous acid)
SO2 H2SO3 Cl2O HOCl (sulphur dioxide) (sulphurous acid) (dichlorine oxide)(hypochlorous acid)
SO3 H2SO4 Cl2O3 HClO2 (sulphur trioxide) (sulphuric acid) (dichlorine trioxide)(chlorous acid)
P2O5 (P4O10)H3PO4 Cl2O5 HClO3 (diphosphorus pentoxide) (phosphoric acid) (dichlorine pentoxide)(chloric acid)
P2O3(P4O6)H3PO3 Cl2O7 HClO4 (diphosphorous trioxide) (phosphorous acid) (dichlorine heptoxide)(perchloric acid)
NO2* HNO2 + HNO3 ClO2* HClO2 + HClO3 (nitrogen dioxide) (nitrous acid + (chlorine dioxide) (chlorous acid + nitric acid) (chloric acid)
As2O3** H3AsO3 Mn2O7** HMnO4 (arsenous oxide) (arsenous acid) (dimanganese heptoxide) (permanganic acid)
As2O5** H3AsO4 CrO3** H2CrO4 (arsenic oxide) (arsenic acid) (chromium trioxide) (chromic acid)
* NO2 and ClO2 although are nonmetallic oxides form two acids each and therefore these oxides are called mixed acid anhydrides. These reactions are not metathesis reactions. They belong to redox type of reactions which we shall discuss later. **These are oxides of a metalloid(As) and metals (Mn, Cr) which are also acidic. These are given in the above table alongwith nonmetallic oxides
EXCEPTIONS: Note that there are few nonmetallic oxides like CO, NO, N2O which are neutral. They do not produce any acid when react with water. In fact they do not react with water. SAQ 4: Write the missing species of the following reactions. Also see that if the ON of some element is changing or not? Pick up the underlined element and find its ON on both the sides. Indicate the ON of that element on both the sides to confirm that it is a metathesis reaction.
(i)P2O5 + H2O ---->? (ii) N2O5 + H2O -----> ? (iii) ? + H2O ----> H3PO3
(iv) ? + H2O -----> HClO (v) N2O3 + H2O ------? (vi)Cl2O5 + H2O ----->?
(vii)? + H2O ---->HMnO4 (viii)Cl2O7 + H2O----->? (ix)CrO3 + H2O ------>? SAQ 5: Write the products of the following reactions and indicate the changes in ON. Are these metathesis reactions?
(i)NO2 + H2O ----->? (ii)ClO2 + H2O ------> ? SAQ 6: Write the anhydrides of the following acids:
H2SO4, HNO2, HClO, H2CO3, HClO4
Dr. S. S. Tripathy 64 Concepts in Chemistry
(iii) Reaction of metallic oxides with water: We had seen before that non-metallic oxides are mostly acidic in nature. Metallic oxides are mostly basic compounds (bases). There are some metallic oxides which are soluble in water.
Such examples are very rare. The common among them are Li2O, Na2O, K2O, Rb2O, Cs2O, CaO, SrO and BaO. When these oxides are treated with water, they dissolve forming their hydroxides. Such a solution is called an alkali. e.g Na2O + H2O ------> NaOH
CaO + H2O ------> Ca(OH)2 These are metathesis reactions in which there is no change in case of ON of any element. While writing the products of the above reaction some students wrongly write like this.
Na2O + H2O ------> NaOH + H2 (wrong) This is because in this equation H is reduced from +1 to 0(reduction) and there is no element which is oxidised (oxidation). This is not possible. Oxidation and reduction go together and one cannot happen without the happening of the other. So remember that whenever you find an oxide reacts with water by way of its dissolution in water, it will form only the hydroxide of the metal. If it does not dissolve, then there is no reaction. N.B: Note that MgO is slightly soluble on strong heating while other oxides are almost insoluble in water. SAQ 7: Write the products of the following:
(i)K2O + H2O ------> ? (ii)BaO + H2O ----->?
(iii) ? +H2O---->Sr(OH)2 (iv)Al2O3 + H2O ------>
(iv) Amphoterism: Substance which reacts both with acid and base is called an amphoteric substance. Oxides and hydroxides of a few metals such as Zn, Al, Sn, Pb, Cr, As, Co, Sb, Be, Si are amphoteric in nature. For example, ZnO can react with acid and in this reaction ZnO plays the role of a base to form salt and water.
ZnO(base) + HCl(acid) ------> ZnCl2 + H2O ZnO also can react with a strong base like NaOH or KOH to form salt and water. Here ZnO plays the role of an acid. Note that in such a case the acid radical of the salt contains Zn. Do you 2- remember an acid radical which contains Zn? Yes, it is zincate(ZnO2 ). The salt produced is sodium zincate.
NaOH(base) + ZnO(acid) ------> Na2ZnO2 + H2O The following table gives a list of the amphoteric metal ions and the names (and formulae) of the acid radicals formed from them.
Dr. S.S Tripathy Logic of Inorganic Reactions 65
metal/metalloid acid part formed alternative formula Ion with strong alkali
3+ - - Al AlO2 (meta-aluminate) [Al(OH)4] 2+ 2- 2- Zn ZnO2 (zincate) [Zn(OH)4] 2+ 2- 2- Sn SnO2 (stannite) [Sn(OH)4] 4+ 2- 2- Sn SnO3 (stannate) [Sn(OH)6] 2+ 2- 2- Pb PbO2 (plumbite) [Pb(OH)4] 4+ 2- 2- Pb PbO3 (plumbate) [Pb(OH)6] 2+ 2- 2- Co CoO2 (cobaltate) [Co(OH)4] 3+ - - Sb SbO2 (antimonite or stibite) [Sb(OH)4] 2+ 2- 2- Be BeO2 (berrylate) [Be(OH)4] 3+ - - Cr CrO2 (chromite) [Cr(OH)4] 4+ 2- 2- Si SiO3 (silicate) [Si(OH)6]
SAQ 8: Write the products of the following reactions. Use the above table for this. Give the ON of the metal on either side to confirm that the reaction is metathesis type.
(i) Al(OH)3 + KOH ------>? (ii)SnO + NaOH ------>?
(iii) PbO + NaOH ------>? (iv)SnO2 + KOH ------>? SAQ 9: Find the mistake in the following equations.(Hint: The formula is wrong somewhere)
(i) Zn(OH)2 + NaOH ------> NaZnO2 + H2O (ii)CoO + KOH ----->K3CoO2 + H2O
(iii) Al2O3 + NaOH ------> Na2AlO2 + H2O (iv)SnO + NaOH -----> Na2SnO3 + H2O (v) Reactions of carbonates, bicarbonates, sulphites, bisulphites, thiosulphates, sulphides and nitrites with dilute mineral acids When any one of the above mentioned salts reacts with dilute acids, effervescence takes place producing a gas. Effervescence is the process of evolution of gas bubbles with hissing sound. Carbonate and bicarbonate salts give CO2 gas; sulphite, bisulphite and thiosulphate salts give SO2 gas while sulphide salt gives H2S gas. Thiosulphate salt produce element sulphur alongwith SO2 gas. See these examples.
Na2CO3 + HCl ------> NaCl + CO2 + H2O
CaSO3 + H2SO4 ------> CaSO4 + SO2 + H2O
K2S2O3 + HCl ------> KCl + SO2 + S + H2O (Note that although the above reaction appears to be redox reaction, in the true sense it is a metathesis reaction. There are two types of S atoms in thiosulphate ion namely ON 0 and +4 which are retained in the products)
ZnS + H2SO4 ------> ZnSO4 +H2S In all these cases no element suffers any change in ON. These are therefore metathesis reactions.
Nitrite salts produce N2O3 gas which decomposes at room temperature to give both NO and NO2. While NO2 is a reddish brown gas, NO is colourless. Looking to the ultimate products formed at room temperature in this reaction it belongs to the redox type, although the orignial reaction at low temperature(<00C)is metathesis.
Dr. S. S. Tripathy 66 Concepts in Chemistry
NaNO2 + HCl ----low temp.----> NaCl + N2O3 + H2O (Metathesis)
NaNO2 + HCl ---room temp---> NaCl + H2O + NO + NO2 (Redox) SAQ 10: Predict the products of the following metathesis reactions.
(i)KHCO3 + H2SO4 ------>? (ii)Na2SO3 + HCl ------>?
(iii)Ca(HSO3)2 + HCl ----->? (iv)ZnCO3 + HBr ------>?
(v)Bi2S3 + HCl ------>? (vi)FeS + H2SO|(dil)------>?
(vii)CaS2O3 + H2SO4(dil) ------>? (viii)KNO2 + H2SO4(dil.)------>? (vi) Hydrolysis of nitrides, phosphides, carbides and sulphides Metallic nitrides, phosphides and carbides undergo easy hydrolysis to produce hydroxide of the metal and a gas. Hydrolysis is nothing but the reaction of any substance with water. A nitride salt gives ammonia gas(NH3), a phosphide salt gives phosphine(PH3) and carbide salt gives a hydride of carbon(methane or acetylene). Some metallic sulphides also undergo hydrolysis to produce hydrogen sulphide(H2S) gas. In all these cases, hydroxides of the metals are formed. Look at these examples.
Mg3N2 + H2O ------> NH3 + Mg(OH)2
AlP + H2O ------> PH3 + Al(OH)3 In all the cases the ON of elements do not change.
Al4C3 + H2O------> CH4 + Al(OH)3; CaC2 + H2O ------> C2H2 + Ca(OH)2
In the first example, we get methane(CH4) gas as the carbide has a valency 4(ON= - 4). The ON of C in both LHS and RHS are same(- 4) since it is a metathesis reaction. In the second example, we get acetylene(C2H2) as the calcium carbide is different from aluminium carbide. In this case 2- the ON of C is -1 (C2 ). In acetylene ON of carbon is also -1.
FeS + H2O ------> Fe(OH)2 + H2S Do you remember that when we add black salt(the salt that we use at home for palate) with water, we get a pungent rotten egg smelling gas(H2S)? It is because the black salt contains
FeS(alongwith NaCl) which is hydrolysed by water to produce H2S gas. SAQ 11: Write the products of the following hydrolysis reactions. (i)Aluminium Nitride + water------>? (ii)Calcium phosphide + water---->?
(iii)CaC2 + H2O ------>? (iv)ZnS + H2O ----->?
(vii) Thermal decomposition of carbonates and bicarbonates: Some metallic carbonates are not stable on heating and decompose on heating to produce carbon dioxide gas and a residue. For example, CaCO3 on heating gives CO2 gas and CaO as residue.
But NaHCO3 decomposes in a different way. It gives CO2 and H2O as gases and Na2CO3 is formed as residue. Note that Na2CO3 is very stable to heat and does not undergo decomposition. All these are metathesis reactions as no change in ON occurs in any element.
CaCO3 ------> CaO + CO2
NaHCO3------> Na2CO3 + CO2 + H2O Note that carbonates of Na, K, Rb and Cs are stable to heat and do not undergo any change..
Dr. S.S Tripathy Logic of Inorganic Reactions 67
SAQ 12: What are the products obtained by heating the following substances?
(i)MgCO3 ------>? (ii)KHCO3 ------>? (iii)Li2CO3 (viii) Decomposition of ammonium salts by a base(alkali):
When any ammonium salt is heated with a base(alkali), NH3 gas is evolved. Remember that you have to heat the reactants.
NH4Cl + NaOH -----heat------> NH3 + NaCl + H2O
(NH4)2SO4 + Ca(OH)2 ----heat------> NH3 + CaSO4 + H2O In all these cases the ON of N remains same(-3) so also the ON of all other elements. SAQ 13: What products will be obtained in the following reactions.
(i)NH4Cl + CaO --heat--->? (ii)(NH4)2CO3 + KOH --heat---->?
(iii)NH4Br + Ba(OH)2 --heat---->? (iv)(NH4)2SO4 + NaOH --heat--->? (ix) Thermal decomposition of hydroxides: Excepting hydroxides of Na, K, Rb and Cs, other hydroxides on strong heating decompose to form the corresponding oxides by losing water.
Al(OH)3 ------heat----> Al2O3 + H2O; Ca(OH)2 -----heat----> CaO + H2O
LiOH ------> Li2O + H2O SAQ 14: Write the products obtained by heating the following.
(i)Be(OH)2 (ii)Mg(OH)2 (iii)Cr(OH)3 (iv)KOH (x) Displacement of more volatile acid by a less volatile acid: Look to the following reactions.
NaNO3 + H2SO4 -----heat-----> Na2SO4 + HNO3 When a mixture of nitrate salt and conc. sulphuric acid is strongly heated and distilled we get nitric acid in the distillate. This is because, nitric acid is a more volatile acid(lower boiling point) and is displaced by a less volatile acid, sulphuric acid(higher boiling point) from the salt of the more volatile acid(nitrate). Sulphuric acid cannot be prepared by this method by heating a sulphate salt with conc. HNO3, because a more volatile acid HNO3 cannot displace a less volatile acid H2SO4 from the salt of the latter.
K2SO4 + HNO3 ------heat------> KNO3 + H2SO4 (not possible) SAQ 15: Give the products of the following:
(i)NaCl + H2SO4(conc.) ------>?(ii)KNO3 + H2SO4(conc.) ----->? (xi) Complexation reactions: 4- - + Radicals like [Fe(CN)6] , [Ag(CN)2] , [Ag(NH3)2] etc. are called complex ions which are parts of compounds called complex or coordination compounds. Complex radicals, by convention, are written inside a square bracket. There are two types of species inside the complex ion- (i) one is the metal ion (ii) and the others are ligands. In ferrocyanide ion, Fe2+ is the metal ion and 6 CN- ions are ligands. Hence the net charge of the complex is 4-. Similarly in diammine
Dr. S. S. Tripathy 68 Concepts in Chemistry
+ + silver(I) complex[Ag(NH3)2] , Ag is the metal ion and two neutral NH3 molecules are ligands. The detailed study on complex compounds will not be done in this book. All complexation reactions belong to metathesis type in which there is no change in ON of any element. Look to these examples.
AgCl + 2 NH3 ( 2NH4OH) ------> [Ag(NH3)2]Cl (silver diammine chloride) (+ 2H2O)
Fe(CN)2 + 4 KCN ------> K4[Fe(CN)6](potassium ferrocyanide)
Fe(CN)3 + 3 KCN ------> K3[Fe(CN)6](potassium ferricyanide)
HgI2 + 2KI ------> K2[HgI4] (potassium mercuric iodide)
AgCN + KCN ------> K[Ag(CN)2] (potassium argentocyanide)
CuSO4 + 4NH3 (4NH4OH) ------> [Cu(NH3)4]SO4 (+4 H2O) (cuprammine sulphate)
Ni + 4 CO------> [Ni(CO)4] (nickel tetracarbonyl)
Cu2(CN)2 + 6KCN -----> 2K3[Cu(CN)4] (potassium cuprocyanide)
SiF4 + 2HF ------> H2[SiF6] (hydrofluosilicic acid) In all these cases the two reactants get added up to form a complex molecule. The names of complex compounds given inside parenthesis are their common names. The systematic IUPAC names are not given here. (B) REDOX REACTIONS Reaction which involves changes in ON is called redox reactions. One reactant is oxidised and is called the reducing agent(RA)and other gets reduced and is called the oxidising agent(OA). We have worked with a large number of redox equations in the previous sections. You know that oxidation and reduction go together. In other words one process cannot occur without the occurrence of the other. Let us take the example to demonstrate this. Often when a student is asked to write the products of the reaction, MnO2 + HCl---->?, they answer,
MnO2 + HCl ----> MnCl2 + H2O. This is wrong. Look to the changes in the ON. +4 -1 +2 -1 MnO2 ++ HCl MnCl2 H2O
While the ON of Mn is reduced from +4 to +2, nowhere there is oxidation. This is not possible.
So the prediction is wrong. Another product is missing. It is Cl2. So that the ON of Cl is increased from -1 to 0. MnO2 is the OA and HCl is RA. +4 -1 +2 0 + MnO2 ++ HCl MnCl2 H2O Cl2 This is called studying the logic of a reaction. The cover page of this book carries this message. If you have not marked it in the cover page, see it now. Every reaction has a logic i.e we can explain why these products are obtained. Often there is a feeling that the hundreds and thousands of chemical reactions are to be memorised in an unintelligent manner. This is wrong. We shall, in this book, try to learn the logic at each step and try to disprove the above notion about chemical reactions. Now the question arises how shall you know, just by looking to the reactants, whether the reaction will be a redox type or a metathesis type? We list a few clues for the students to guess whether a reaction belongs to redox category or not.
Dr. S.S Tripathy Logic of Inorganic Reactions 69
(i) First check that the reaction does not fall into any of the eleven metathesis reactions cited before. If not, then it could be a redox reaction.
For example, if the reaction is: AlP + H2O----> ? . This cannot be a redox reaction as we know that this belongs to a metathesis type under the head, "hydrolysis of nitride, phosphide, carbide etc". (ii) If one of the reactants is not soluble in water(not assigned with aq. symbol) and hence does not produce ions, the reaction might be a redox type. In other words if you find at least one of the reactants is a gas(g), liquid(l) or solid(s), the reaction can be a redox type.
NH3(g) + CuO(s)------> This cannot be metathesis as the reactants are gas and solid and it does not belong to any category of metathesis reactions studied before. Hence it has to be a redox reaction. (ii) If one of the reactants remains in the uncombined state(elemental) state, the reaction belongs to redox type.
Cl2(g) + NaOH(aq) ---->?
In this case one of the reactants Cl2 is in the uncombined state. So it has to be a redox reaction.
CuO + H2 -----> ?, This reaction also has to be a redox reaction as one of the reactants is in uncombined state(H2). So intelligently you have to first ensure whether the reaction could be metathesis or redox. If metathesis then predict the products as per the discussion made in the previous section on metathesis reaction. But what could be the products if it is a redox reaction? How can you guess the products? We shall discuss this aspect little later. Before that we discuss some special types of redox reactions which bear special names. Note that a majority of redox reactions have no special names and simply are branded as redox reactions. But some few belong to specific classes of redox reactions. These are discussed below. (i) Displacement Reaction: In this reaction, one element displaces another from its salt. Let us take metal displacement first. In the example below, you will find that Zn can displace Cu from CuSO4, but the reverse reaction is not possible, i.e Cu cannot displace Zn from ZnSO4.
Ex. Zn + CuSO4 ------> ZnSO4 + Cu
Cu + ZnSO4 CuSO4 + Zn (not possible) This is because Zn is a more active metal than Cu. A more active metal can displace a less active metal from the salt of the less active metal. Could you say what changes in ON have taken place in the first reaction? Zn is oxidised to Zn2+(ON changes from 0 to +2) while Cu2+ is reduced to Cu(ON changes from +2 to 0). The table below gives the list of a few metals forming a series called Metal Activity Series in which metals are arraned in the order of decreasing activity. A more active metal can displace a less active metal from the salt of the latter. Hydrogen is taken as standard for comparison and is situated in the intermediate position. Metals lying before hydrogen can displace hydrogen gas from acids while metal lying below hydrogen cannot displace hydrogen gas from acids.
Mg + HCl ------> MgCl2 + H2 , Cu + HCl ------> No reaction
Dr. S. S. Tripathy 70 Concepts in Chemistry
Since Mg lies above H, it can displace hydrogen gas from HCl whereas Cu lies below H, therefore it cannot displace hydrogen gas from HCl. That is why we have written 'No Reaction'. METAL ACTIVITY SERIES Li > K >Cs > Ba > Ca > Na > Mg > Al > Mn > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > H > Sb > Bi > As > Cu > Hg > Ag > Pd > Pt > Au (NB. Many uncommon metals have not been included in the above series) The first six metals given in this series upto Na are so reactive that they liberate hydrogen gas even from cold water. Then the other members after Na i.e starting from Mg upto Cd, can displace H2 from boiled water or steam(not cold water) and sometimes the metal is to be heated to red hot conditions. The metals lying after Cd upto Pb are not active to the extent of displacing hydrogen from water in any condition, whether cold or hot. They can displace hydrogen only from dilute acids. Note that all metals lying above hydrogen can displace hydrogen from dilute acids.
SAQ 16: Give the conditions of the reaction and predict the products.
(i)Ca + H2O ----->? (ii)Mg + H2O ------>? (iii)Fe + H2O------>
(iv)Sn + H2O ------>? (v)Al + H2O------> SAQ 17: Predict the products: (i)Ni + HCl ----->? (ii)Ag + HCl ------>? (iii)Zn + HBr------> (iv)Fe + HCl ------>? (v)Hg + HCl ----->?
Halogen Activity Series:
The four halogens namely F2, Cl2, Br2 and I2 form a halogen activity series.
F2>Cl2>Br2>I2
F2 is most active while I2 is least active. In other words, F2 has the greatest tendency to be - - reduced to F while I2 has the least tendency to be reduced to I . .All haloges are good oxidising agents and are themselves reduced to their halide ions(fluoride, chloride, bromide and iodide) - easily. Among the halogens, F2 is reduced to F most easily. A more active halogen can displace a less active halogen from the salt of less active halogen. For example
F2 + NaCl ------> NaF + Cl2, - - Here F2 is reduced to F while Cl is oxidised to Cl2.This reaction occurs because F2 is more reactive than Cl2. Note that when the more active halogen is reduced, the less active halogen gets oxidised. Here - Cl is oxidised to Cl2(from -1 to 0).
Cl2 + NaF ------> No reaction
Since Cl2 is less reactive than F2, the former cannot displace the latter from the salt of the latter.
Dr. S.S Tripathy Logic of Inorganic Reactions 71
SAQ 18: Predict the products:
(i)Cl2 + KI ---->? (ii)I2+ KBr ----->? (iii)F2 + KBr ----->?
(iv)Br2 + NaCl ---->? (v) Cl2 + NaBr ------>? SAQ 19: Predict the reactions and write the products formed.
(i)Cu + HCl ------> (ii)K2O + Mg ------>
(iii)KCl + Br2 ------> (iv)FeBr3 + Cl2 ------>
(v)Al + Cr2O3 ------> (vi)Hg + H2SO4(dil) ------>
(vii)Na + ZnO -----> (viii)AgNO3 + Cu ------>
(ix)Zn + AuCl3 ------>? (x)Fe + Mn3O4------> (ii) Combustion Reactions: Case-I: Complete Combustion: Combustion or burning of a substance(mostly organic substance) in presence of sufficient quantity of air to form CO2 and H2O is called a combustion reaction. It is an exothermic reaction, that means heat energy is liberated.
CH4(g) + 2O2 ------> CO2 + 2H2O + 820 kJ
C6H12(l) + 9 O2 ------> 6CO2 + 6H2O + 4127 kJ Note that whatever may be the organic compound, on combustion or burning it will prouduce
CO2 and H2O. Case-II: Incomplete Combustion:
If the compound does not get enough oxygen, then instead of producing CO2, it will produce a poisonous gas CO. This is called incomplete combustion.
3 CH4 + 2 O2 ------> CO + 2H2O Besides CO, other vapours like formaldehyde, methyl alcohol etc. are also produced in some cases. (iii) Synthesis Reaction: When two elements directly combine to form a compound, it is called a synthesis reaction.
Mg + Cl2 ------> MgCl2 In this example, the ON of Mg of changes from 0 to +2, while the ON of Cl changes from 0 to-1.
H2 + Cl2 ------> HCl SAQ 20: Give the synthesis reaction for the formation of the following compounds. Show the changes in ON in each case.(Balancing the equation is not necessary)
(i)NH3 (ii)CO2 (iii)PCl3 (iv)H2O (iv) Disproportionation Reaction: If same substance gets oxidised as well as reduced, it is called a disproportionation reaction. It may be different elements present in one substance, out of which one element gets oxidised and the other gets reduced. It may be the same element present in the substance which is oxidised as well as reduced. See this example of first type.
Dr. S. S. Tripathy 72 Concepts in Chemistry
[O]
+5 -2 -1 0 KClO3 KCl + O2 [R]
Here the same substance KClO3 gets oxidised and reduced. O is oxidised from -2 to 0 and Cl is reduced from +5 to -1. Let us take another example of the second type.
Cl2 + NaOH ------> NaCl + NaClO + H2O Here the same substance and the same element Cl gets oxidised(from 0 to +1 in NaClO) as well as reduced (from 0 to -1 in NaCl). Hence both types of reactions shown above belong to this type of reaction called disproportionation. In such reactions, we do not specify which is oxidising agent(OA) and which reducing agent(RA) becasue the same species is OA and RA. Let us take few more examples.
HI ------> H2 + I2 (H is reduced from +1 to 0 while I oxidised from -1 to 0)
Pb(NO3)2 -----> PbO + NO2 + O2(N reduced from +5 to +4 and O oxidised from -2 to 0) SAQ 21: Write the products of the following reactions. Show the changes in ON. What type of redox reaction is this?
(i)NO2 + H2O ------>? (ii)NO2 + NaOH ----->?
(iii)ClO2 + H2O ----->? (iv)NaOH + ClO2 ------>? (v) Decomposition(Analysis) Reaction: When a compound is broken down into smaller molecule by the action of heat, light or electricity, it is called a decompositon or Analysis reaction.
Pb(NO3)2 ------heat------> PbO + NO2 + O2 Can you show the changes in ON? The ON of N changes from +5 to +4 and that of O from -2 to zero.
When lead nitrate is strongly heated we get a reddish brown gas, NO2, in addition to O2 gas. The residue obtained is PbO.
HgO ------heat------> Hg + O2 + - Electrolysis Na Cl Na + Cl2 (cathode) (Anode)
When molten(liquid) NaCl is electrolysed we get Na at cathode and Cl2 gas at anode. This is decompositon of a compound by electrical energy. Note that all nitrates except that of alkali metals(Na, K, Rb, Cs) on thermal decomposition give the reddish brown NO2 gas, O2 gas and oxide of the metal. But the alkali metal nitrates give only
O2 and the corresponding nitrite.
NaNO3 ------>NaNO2 + O2 ( and not Na2O + NO2 + O2)
Dr. S.S Tripathy Logic of Inorganic Reactions 73
In this case the ON of N changes from +5(NaNO3) to +3(NaNO2) and that of O changes from
-2 to 0. LiNO3 is an exception to this behaviou which on heating gives Li2O, NO2 and O2 like other nitrates. Note that all thermal decomposition(analysis) reactions(under redox category)as explained above are also disproportionation reactions. The same substance gets oxidised as well as reduced.
Decomposition of NH4NO2 and NH4NO3: These two compounds when strongly heated leave no residue. All the products are gaseous and escape out of the reacting vessel. For all these, you have to remember a trick. Remove the maximum number of water molecules from them, whatever residual formula remains, that is the other gaseous product formed.
NH4NO2 ------> 2H2O + N2
In this case, after removing 2H2O molecules, we are left with N2 gas. The ON of N changes + - from -3(NH4 ) to 0(N2) and also from +3(NO2 ) to 0. It is also a disproportionation reaction.
NH4NO3------> 2H2O + N2O
After removing 2H2O molecules, we are left with the laughing gas, N2O. This is also a + disproportionation reaction, in which the ON of N changes from -3(NH4 ) to +1(N2O) and from - +5(NO3 ) to +1. SAQ 22: What happens when:
(i)A mixture of NaNO2 and NH4Cl solution is heated.
(ii)A mixture of NaNO3 and NH4Cl solution is heated. SAQ 23: Indicate which are synthesis, analysis and disproportionation reactions.Show the changes in ON in each reaction.
(i)HgO ---heat-----> Hg + O2 (ii) Hg + O2 ------> HgO
(iii)KMnO4 --heat----->K2MnO4 + MnO2 + O2 (iv)Cu + O2 ------> Cu2O SAQ 24: Show the changes in ON in the following disproportionation reactions.
(i)Pb(NO3)2 ------>PbO + NO2 + O2 (ii)NaNO3 ------> NaNO2 + O2
(iii) P4 + NaOH ------> PH3 + NaH2PO2 (iv)Cl2 + NaOH--> NaClO3 + NaCl + H2O (vi)Amphoterism: You remember that in the metathesis reactions we studied a phenomenon called amphoterism. The oxides and hydroxides of a few metals such as Al, Zn, Sn, Pb etc. reacted both with strong acids and strong bases to produce salt and water. Here we also study amphoterism, but this is redox amphoterism. In this case, in stead of metallic oxide or hydroxide, we take the metal and allow it to react with strong base like NaOH or KOH. We get the same salt(known from the table given in metathesis amphoterism reaction). But we get H2 gas instead of H2O. Thus it becomes a redox reaction.
0 +1 +2 0 Zn+ NaOH Na2ZnO2 + H2 We get sodium zincate and hydrogen gas. You can see the ON changes that take place in the
Dr. S. S. Tripathy 74 Concepts in Chemistry above reaction. Just think what we would have got if ZnO would have been taken instead of Zn.
ZnO + NaOH ------> Na2ZnO2 + H2O (metathesis)
We would have got the same sodium zincate but instead of H2 we get H2O in such case. SAQ 25: Predict the product in the following reactions. Also show the changes in ON.
(i) Al + NaOH + H2O ------> (ii)Sn + KOH -----> (iii) Pb + NaOH ------> (iv)Si + NaOH ------>
(vii) Hydrolysis of metallic hydrides:
Metallic hydrides react with water to produce H2 gas and metallic hydroxides. -1 +1 0 NaH + H2O NaOH + H2 Hydrogen is both oxidised from -1 to 0 and reduced from +1 to 0.
Some Simple Redox Reactions:
Some common OAs : MnO2, PbO2, Pb3O4, KMnO4, K2Cr2O7, F2, Cl2, Br2, O2,
HNO3, H2SO4, O3
Some common RAs : H2S, FeSO4, NH3, HI, HBr, HI, all metals, I2, P4, S, H2 , CO,
C, SnCl2
Some agents which can act both as RA and OA : H2O2, SO2, HNO2, NO, NO2,
N2O (i) MnO2 + HCl ——> MnCl2 + H2O (wrong) +4 -1 +2 -1 MnO2 ++ HCl MnCl2 H2O (there is reduction of Mn but no oxidation) +4 –1 → +2 0 Mn O2 + HCl Mn Cl2 + Cl2 + H2O (correct) ( Mn is reduced from +4 to +2 and Cl is oxidized from –1 to 0) → (ii) NH3 + Cl2 HCl + N2 (N is oxidised from -3 to 0 while Cl is reduced from 0 to -1)
If excess of NH3 is used in the above reaction, NH4Cl is formed in stead of HCl
(iii) CuO + H2 Cu + H2O
(iv) HBr + H2SO4(conc) SO2 + Br2 + H2O
(v) HNO3(conc) + C NO2 + CO2 + H2O
(vi) CO2+ C 2CO (vii) Cl2 + KI KCl + I2
(viii) KMnO4 + H2SO4 + FeSO4
K2SO4 + MnSO4 + Fe2 (SO4) 3 + H2O
(the purple colour of KMnO4 is discharged)
(ix) K2Cr2O7 + H2SO4 + Na2SO3 K2SO4 + Cr2 (SO4) 3 + Na2SO2 + H2O (the orange colour of potassium dichromate turns green due to chromic sulphate)
(x) NH3 + O2 N2 + H2O (Show the changes in ON in all the cases)
Dr. S.S Tripathy Logic of Inorganic Reactions 75
(xi) Cu + HNO3 (conc) Cu(NO3) 2 + NO2 + H2O
(xii) Cu + H2SO4 (conc.) CuSO4 + SO2 + H2O Some important hints for prediction: Usually the ON increases from -ve state to 0 and decreases from +ve state to 0. (Exception : Cl decreases from +ve state to -1, Mn from +7 to +2 in acidic medium, Cr from +6 to +3, Pb and Mn from +4 to +2, Fe from +3 to +2, Hg from +2 to +1 and then to 0, Cu from +2 to +1, Sn inreases from +2 to +4, Fe from +2 to +3, Mn from +2 to +7, Cr from +3 to +6 etc.)
STRATEGY MAKING AND PREDICTING REACTIONS: In this section we shall know how to predict the products of a reaction. If we are given the reactants only, how shall we write the correct products. Often students, particularly the beginners believe that inorganic reactions are to be simply memorised in an unintelligent manner. When the question of remembering hundreds of reactions comes, naturally the study of inorganic chemistry often appears monotonous and boring. In reality it is not so. If you know the logic of a reaction, you are not required to memorise a reaction. The following example will illustrate this point.
Predict the reaction: Al(OH)3 + NaOH ----? After seeing the reactants, we are immediately reminded of amphoterism that we studied in metathesis reaction. The produts are sodium meta-aluminate and water.
Al(OH)3 + NaOH ----> NaAlO2 + H2O So we are not supposed to remember the products. It will automatically occur to us when we know the logic of predicting reactions. Let us take another example,
H2S(g) + SO2(g) ------> ?
If you have memorised the products from before, they are S + H2O. But if you are asked why these products are formed and why not other products? Then you are in trouble. Remember that every reaction has a logic i.e why we get these products and why not some others? Let us analyse the logic of the above reaction. Just looking to the reactants you will be sure that it cannot be a metathesis reaction. Fristly because it does not fall into any of the categories of metathesis reaction. Secondly it has gaseous reactants. [O] -2 +4 0 H2S + SO2 S + H2O [R]
In this reaction the ON of S in H2S has to increase from -2 to a higher value as it cannot decrease below -2. Therefore the ON of S in SO2 has to decrease from +4 to a lower value. Since the element is same(S) in both RA and OA which undergo change in ON, the decrease and increase take place to the same value(0). So the common product formed from them is S.
Then to guess the other product from the other elements is easy. The other produt is H2O. If you have not understood this, keep patience, you will be given several examples to make your understanding clear. Dr. S. S. Tripathy 76 Concepts in Chemistry
Before prediction, you should be sure that the reaction given is either a metatheis reaction or a redox reaction. Because these are the only two types of reactions possible. When you look to the reactants, first ensure whether it belongs to any of the categories of metathesis reactions that we had studied before or not. (i) If both the reactants are soluble in water(associated with aq. symbol, otherwise you have to see the solubility rules): they will go for double replacement reaction often resulting precipitaion.
AgNO3(aq) + HCl(aq) ------> AgCl(s) + HNO3(aq) (ii) If one reactant is an acid and the other reactant is a base(oxide or hydroxide of metal), then it will be neutralisation reaction(under double replacement type) to produce salt and water, irrespective of whether one of the reactants is soluble in water or not.
MgO(s) + H2SO4(aq) ------> MgSO4 + H2O(l) Here we find that even though one of the products is insoluble(s), it forms the products, salt and water. (iii) If one of the reactants is an oxide or hydroxide of an amphoteric metal like Al, Zn, Sn etc. and the other is a strong base like NaOH/KOH, then the reaction is metathesis(Amphotersism).
ZnO + KOH ------> K2ZnO2 + H2O (iv) If one reactant is a non-metallic or a soluble metallic oxide and the other reactant is water, we get the corresponding acid or alkaline hydroxide(metathesis reaction) respectively.
N2O5 + H2O -----> HNO3
CaO + H2O ------> Ca(OH)2 (v) If one reactant is a carbonate/bicarbonate or sulphite/bisulphite/thiosulphate or sulphide
or a nitrite and the other is an acid: In such case we get CO2, SO2 ,H2S and NO2(alongwith NO) gas respectively and those are under metathesis reaction.
CaCO3 + HCl ------> CO2 + CaCl2 + H2O
Na2SO3 + H2SO4 ------> Na2SO4 + SO2 + H2O
ZnS + HCl ------> ZnCl2 + H2S (vi) If one reactant is a metallic nitride, phosphide, carbide or sulphide and the other is water:
then it is the hydrolysis reaction producing NH3, PH3, CH4(or C2H2) and H2S respectively.
Mg3N2 + H2O -----> Mg(OH)2 + NH3
AlP + H2O ------> PH3 +Al(OH)3 and so on. (vii) If one reactant is ammonium salt and the other reactant is an alkali/base; such mixture
on heating will produce NH3 gas.
NH4Cl + NaOH ---heat------> NH3 + NaCl + H2O
(some students write NH4OH + NaCl as products which is not wrong, but on heating
NH4OH produces NH3 and H2O)
Dr. S.S Tripathy Logic of Inorganic Reactions 77
(viii) If it is a case of thermal decompositon of carbonate and bicarbonate:
CaCO3 ----heat------> CaO + CO2 (ix) If it is a case of thermal decomposition of metallic hydroxide to produce its oxide or (x) If it is a case of displacement of more volatile acid by a less volatile acid or (xi) If it is a case of complexation reaction If you find that the reaction is not a metathesis reaction, you may guess it to be a redox reaction.
A redox reaction should have the following specialities:
(i) The reaction should not belong to any of the metathesis reactions cited above. (ii) One of the reactants may exist in the gaseous(g), liquid(l) or solid(s) state.
Example : NH3(g) + CuO(s) ------> ? In this case you find that the reactants are in gaseous and solid state, so it has to be a redox reaction. Once you ascertained that the reaction is a redox type then your work is to assign the ON of the elements in the reactants. Then you have to guess which element will undergo change in ON and to what values. In the above example, the ON of N is -3 in LHS and it has to increase as it cannot decrease below -3. The ON of Cu is +2 in LHS and it has to decrease. Note that the increase and decrease of ON usually takes place via zero state. Although there are many exceptions, but students always should first think of the product to come to zero ON. So the products in this case are N2 and Cu in which both remain in zero oxidation state. Note that H and O will react to form water and thus they will not undergo any change in ON.
[R]
-3 +2 0 0 NH3 + CuO N2 + Cu + H2O [O]
Let us take another example.
Example :NH3 + Cl2 ------>? In this case N is in -3 state and it has to come to 0 state, while Cl is in 0 state and it has to decrease to -1. So the products are HCl and N2 [O] -30 -1 0 NH3 + Cl2 HCl + N2 [R]
IMPORTANT: Prediction of products can be made only for simple reactions with logic. But for many other complex reactions, prediction is not possible. However when the products are known the logic of the reaction can be studied. Once the logic is known it becomes easy to remember the products. Now we shall have to predict a large number of simple reactions giving proper logic. For the first
Dr. S. S. Tripathy 78 Concepts in Chemistry few reactions the products are given, you are required to put forth the logic i.e to say whether it is metathesis or redox? If metathesis, then to which category of the metathesis reaction does it belong and if it is redox, then you have to show the oxidation and reduction processes by indicating the changes in ON of elements. The beginners are required to answer the questions again and again in regular intervals to build up more confidence. The equations are not balanced and you are not required to balance them either.
SET-I:
(i)CuO + H2SO4 ------> CuSO4 + H2O (ii)N2 + H2 ------> NH3
(iii)TiCl4 + Mg ------> Ti + MgCl2 (iv)FeS + H2SO4 ------> FeSO4 + H2S
(v)N2O3 + H2O ------> HNO2 Response: (i) Metathesis: Neutralisation(Acid +Base) (ii) Redox(synthesis):N from 0 to -3 and H from 0 to +1 (iii) Redox: Displacement; Mg is more active than Ti (iv) Metathesis: reaction of sulphide with dil. mineral acid (v) Metathesis: Reaction of nonmetallic oxide with water
SET-II:
(i)CuCO3 -----heat----> CuO + CO2 (ii)H2 + Cl2 ------> HCl
(iii)Na + HCl ------> NaCl + H2 (iv)KOH + HNO3 ------>KNO3 + H2O
(v)Na2O + H2O ------> NaOH Response: (i) Metathesis: Thermal decomposition of carbonate (ii) Redox: H from 0 to +1 adn Cl from 0 to -1 (iii)Redox: Na from 0 to +1 and H from +1 to 0. (iv)Metathesis: Neutralisation(acid + base) (v)Metathesis: Reaction of metallic oxide with water SET-III:
(i)SnO+ H3PO4 ------>Sn3(PO4)3 + H2O (ii)ZnO + NaOH -----> Na2ZnO2 +
H2O
(iii)Na2SO3 + HCl ----->NaCl + SO2 + H2O
(iv)Pb(NO3)2 ---heat-----> PbO + NO2 + O2 (v)C2H6 + O2 ------>CO2 + H2O Response: (i) Metathesis: Neutralisation (ii) Metathesis: Amphoterism (iii)Metathesis: reaction of sulphite salt with dilute acid (iv) Redox: (Analysis as well as disproportionation reaction)N from +5 to +4 and O from -2 to 0.
Dr. S.S Tripathy Logic of Inorganic Reactions 79
(v) Redox: Combustion reaction.
In the following SAQs(SAQ 26-32), the products are not given. You are required to predict the products in a logical basis and indicate the main type of reaction (metathesis or redox) and the sub type to which it belongs. You are also required to show the changes in ON for a redox reaction. Indicate the metathesis reaction by the symbol, M and redox reaction by symbol, R. SAQ 26:
(i)N2O5 + H2O ------>? (ii)Ba(OH)2 + HNO3 ------>?
(iii)Zn(OH)2 + KOH ------>? (iv)AgNO3 + NaBr----->?
(v)Na2CO3 + H2SO4 ------? Hints: (i) M, Reaction of nonmetallic oxide with water (ii) M: Neutralisation (iii) M: Amphoterism (iv) M: Double Replacement,
(v) M: Liberation of CO2 from a carbonate salt SAQ 27:
(i)BaCl2(aq) + H2SO4(aq) ------>? (ii)K2O + H2O------?
(iii)K+ H2O ------>? (iv) Zn + HCl ------>?
(v)CO2 + H2O ------>? Hints: (i) M: Double replacement(precipitation) (ii) M:reaction of metalic oxide to form hydroxide (iii) R: Displacement, K is more active than H. (iv) R:Displacement (v) M:Reaction of nonmetallic oxide with water SAQ 28:
(i)Cu + ZnSO4 ------>? (ii)Al(OH)3 + H2SO4 ------>?
(iii)SnO + NaOH ---->? (iv)H2 + I2 ------>? (v)MgS + HCl ----->? Hints: (i) Cu is less active than Zn (ii) M:Neutralisation (iii) M: Amphoterism (iv) R: synthesis (v) M: reaction of sulphide salt with dilute acid SAQ 29:
(i)Zn + KOH ----->? (ii)CaCl2(aq) + (NH4)2CO3(aq)------> ?
(iii)SO2 + H2O ------>? (iv)NH3 + O2 ------>?
(v)Ca(OH)3+H3PO4 ---->? Hints: (i)Amphoterism(redox) (ii)M:double replacement (iii)M:reaction of nonmetallic oxide with water (iv)R: N will go from -3 to 0 and O from 0 to -2 (v)M: Neutralisation SAQ 30:
(i)ZnCl2(aq) + NH4OH(aq) ------>? (ii)P2O5 + H2O -----?
(iii)Mg3N2 + H2O ------>? (iv)NaHCO3 ----heat----->?
(v)H2O2 + H2S ------>?
Dr. S. S. Tripathy 80 Concepts in Chemistry
Hints: (i)M:Double replacement(precipitation) (ii)M:reaction of non-metallic oxide with water (iii)M: hydrolysis of nitride, phosphide, carbide, sulphide (iv):M:Thermal decomposition (v)R: O will go from -1 in peroxide to -2 and S will go from -2 to 0. SAQ 31:
(i)NaHCO3 + HCl------>? (ii)Bi(NO3)3(aq) + H2S(aq) ------>?
(iii)Cu + dil. H2SO4 ------> (iv)Cl2 + KI------>?
(v)PbO2 ----heat---->?
Hint: (i) M:evolution of CO2 from bicarboante (ii)M:double replacement(ppt.) (iii)Cu lies below H in the metal activity series (iv)halogen displacement (v)R: Oxygen gas will be evolved(Pb will go from +4 to +2) SAQ 32:
(i)FeS + HCl ---->? (ii)Cl2O5 + H2O------>?
(iii)Al(OH)3 + H3PO4----->? (iv)NH3 + CuO ------>?
(v)Fe + Cl2 ------>? Hints: (i)M:reaction of sulphide with dilute acid, (ii)M:Reaction of nonmetallic oxide with water (iii)M:Neutralisation (iv)R: N will go from -3 to 0 and Cu from +2 to 0. (v)R: synthesis SAQ 33: Detect the mistakes in the equations and explain
(i)Ag + HCl ----->AgCl + H2
(ii)As2O3 + H2SO4 -----> AsSO4 + H2O
(iii)ZnO + NaOH ------> NaZnO2 + H2O (iv)N2O3 + H2O ------> HNO3
(v)MnO2 + HCl------> MnCl2 + H2O Hint: (i) Ag lies below H in the metal activity series (ii) Is it the correct formula of arsenous sulphate (iii) Is the correct formula of sodium zincate?
(iv) What acid does N2O3 produces with H2O? The ON of N is +3 in it but +5 in
HNO3. Can it be possible in a metathesis reaction? (v) The ON of Mn has been reduced from +4 to +2, but nothing has undergone oxidation. Is it possible in a redox reaction? Cl will be oxidised. SAQ 34: Predict the products giving proper logic and indicating the name of the reaction type(sub type if any). Show the ON changes in redox reactions.
(i)ZnCO3 + HBr ------>? (ii)NaHCO3 ------heat------>
(iii)P2O5 + NaOH ------>? (iv)Ca + H2O ------>?
(v)NH3 + HCl ------>? Hints: (i) M:carbonate salt + acid (ii) M:Thermal decompositon
Dr. S.S Tripathy Logic of Inorganic Reactions 81
(iii) M:the corresponding acid will be form which will react with base to form the salt and water (iv) R: Ca will displace H.
(v) M: NH3 being basic substance will react with acid to form salt SAQ 35:
(i)KOH + SnO ------> (ii)PbO + H2SO4----->? (iii)Pb + NaOH ------>? (iv)Ca(NO3)2 ------heat---->? (v)As(NO3)3(aq) + H2S(aq)------>? Hints: (i)M:Amphoterism (ii)M:Neutralisation (iii)R: Amphoterism (iv)R:Decomposition of nitrates (v)M:Double Replacement SAQ 36:
(i)Fe(OH)3 + HNO3 ------>? (ii)N2O5 + KOH ----->? (iii)Na2O + H2O ------>? (iv)K + H2SO4(dil) ------>? (v) P4 + O2------>? Hints: (i) M:Neutralisation
(ii) Acid of N2O5 will be formed which reacts with KOH to form the salt. (iii) M: Reaction of metallic oxide withe water (iv) R: Displacement (v) R: synthesis SAQ 37:
(i)Ba(NO3)2(aq) + Na3PO4(aq) ------>? (ii)NO2 + H2O ------>? (iii)NaNO3 ----heat--->? (iv)NH3 + H2SO4 ------>? (v)FeSO3 + HCl ------>? Hints: (i)M:Double Replacement (ii)R:NO2 is a mixed anhydride (iii)R:Thermal decompostion of Na,K nitrates (iv)M: salt formation (v)M:reaction of sulphite salt with acid SAQ 38: Indicate the mistakes if any with proper reasoning.
(i)Mg(NO3)2 ------> Mg(NO2)2 + O2 (ii)PbO2 + HBr ------> PbBr2 + H2O (iii)FeCl3 + H2S ------> FeS + HCl (iv)Ag + HCl ------> AgCl + H2 (v)CuSO4 + K4[Fe(CN)6] ------> Cu[Fe(CN)6]2 + K2SO4 Hints: (i) Decompostion of Na, K nitrates takes place in that manner. Other nitrates produce the reddish brown gas.
(ii) The ON of Pb has decreased from +4(PbO2) to +2(PbBr2), then where is the oxidation? So one more product is missing which is formed by the oxidation of HCl. (iii) It is a redox reaction, Fe changes from +3 to +2 and S from -2 to 0. (iv) Ag lies below H in the metal activity series (v) The formula of cupric ferrocyanide is wrong SAQ 39: Indicate the mistakes if any with proper reasoning.
(i)Na2O + H2O ------> NaOH + H2 (ii)Al2O3 + NaOH ------> Na2AlO2 + H2O (iii)N2O3 + H2O ----->HNO3 (iv)Sb(NO3)3 + H2S ------> SbS + HNO3 (v)(NH4)2SO4 + NaOH ----heat----> N2 + Na2SO4 + H2O Hints: (i)the ON of H has decreased from +1 to 0 and nowhere there is oxidation. (ii)M:Formula of sodium metaaluminate is wrong. Al should be in +3 state (iii)The ON of N is increased from +3 to +5 while there is no reduction. (iv)M:The formula of antimonous sulphide is wrong
(v)M:ammonium salt reacts with alkali to give NH3 not N2.
Dr. S. S. Tripathy 82 Concepts in Chemistry
100 REACTIONS WITH LOGIC Below you shall find 100 reactions whose products are given. The equations are not balanced. You are advised to balance the equations by any method you like. The main class of reaction, whether Metathesis(M) and Redox(R) has been indicated in each case. The ON changes are also given for redox reactions. The prediction of such reactions from the reactants is often not possible, however if we study the logic of the reaction it becomes easier for us to remember the products. Beginners in the secondary level should only look to logic.
1. KOH + MnO2 + O2 ------> K2MnO4 + H2O (R) (Mn: +4 to +6; O : 0 to -2)
2. C + H2O ------> CO + H2 (R) (C :0 to +2; H : +1 to 0) 3. CO + NaH ------> HCOONa + C (R) ( H: -1 to +1; C: +2 to 0)
4. (NH4)2S2O8 + H2O ------> NH4HSO4 + H2O2 (M): (the peroxy linkage -O-O- present
in peroxidisulphate ion gets hydrolysed to H2O2)
5. Zn + H2SO4 + NaClO3 ------> NaCl + Na2ZnO2 + Na2SO4 + H2O (R): (Zn: 0 to +2; Cl: +5 to -1)
6. Al + NaNO3 + NaOH ------> NH3 + NaAlO2 + H2O (R): (Al: 0 to +3; N: +5 to -3)
7. PbSO4 + CaH2 ------> PbS + Ca(OH)2 (R): (S: +6 to -2; H: -1 to +1)
8. SOCl2 + H2O ------> H2SO3 + HCl (M)
9. S + NaOH ------> Na2S2O3 + Na2S + H2O (R) (S: 0 to +2; S: 0 to -2)
10. Si + NaOH + H2O -----> Na2SiO3 + H2 (R) ( Si: 0 to +4; H: +1 to 0)
11. CaNCN + NaCl + C ------> NaCN + CaCl2 (R) (C: +4 to +2; C: 0 to +2)
12. NaNH2 + H2O ------> NaOH + NH3 (M)
13. NaNH2 + CO ------> NaCN + NaOH + NH3 (M)
14. NaNH2 + N2O ------> NaN3 + H2O (R) : (N: -3 to -1/3; N:+1 to -1/3)
15. KBrO3 + C ------> KBr + CO (R) (Br:+5 to -1; C: 0 to +2)
16. KOH(hot and conc.) + I2 ------> KIO3 + KI +H2O (R) : (I: 0 to -1; I: 0 to +5)
17. CuSO4 + KI -----> Cu2I2 + K2SO4 + I2 (R)(Cu: +2 to +1; I: -1 to 0)
18. K4[Fe(CN)6] + K ------> KCN + Fe (R) ( Fe : +2 to 0; K: 0 to +1)
19. Ca(ClO3)2 + KCl ------> KClO3 + CaCl2 (M)
20. KClO3 + HCl ------> KCl + Cl2 + ClO2 + H2O (R) ( Cl: +5 to +4; Cl: -1 to 0)
21. BeO + NaOH ------> Na2BeO2 + H2O(M) Amphoterism
22. Be2C + H2O ------> CH4 + Be(OH)2 (M) 23. BaO + Al ------> No reaction (Ba lies before Al in the metal activity series)
24. Ca(HCO3)2 ----heat----> CaCO3 + CO2 + H2O (M)
25. CaC2 + N2 ---heat------> CaNCN + C (R) (C:-1 to +4 in CaNCN also to 0 in C; N:0 to -3)
26. CaNCN + H2O ------> CaCO3 + NH3 (M)
27. CaO + SiO2 ------> CaSiO3 (M)
28. BaCO3 + C ------> BaO +CO (R) (C: +4 to +2; C: 0 to +2)
29. BaSO4 + C + CaCl2 ------> BaCl2 + CaS + CO (R): (S: +6 to -2; C: 0 to +2)
30. B2O3 + PCl5 ------> BCl3 + POCl3 (M)
Dr. S.S Tripathy Logic of Inorganic Reactions 83
31. AlCl3 + H2O -----> Al(OH)3 + HCl (M)
32. Al + Na2CO3 + H2O ------> NaAlO2 + H2 +CO2 (R) : (Al: 0 to +3; H: +1 to 0)
33. SnCl4 + Sn -----> SnCl2 (R) (Sn: +4 to +2; Sn: 0 to +2)
34. I2O5 + CO ------>I2 + CO2 (R) (I:+5 to 0; C: +2 to +4)
35. SiO2 + Mg ------> Si + MgO (R) Displacement
36. PbO + PbS ------> Pb + SO2 (R) (Pb: +2 to 0; S: -2 to +4)
37. Pb3O4 + HCl ------> PbCl2 + Cl2 + H2O (R): (Pb: +8/3 to +2; Cl: -1 to 0)
38. NH3 + NaOCl -----> N2 + NaCl + H2O (R) (N: -3 to 0; Cl: +1 to -1)
39. KNO3 + H2SO4 + FeSO4 ------> Fe2(SO4)3 + K2SO4 + NO + H2O (R) (N: +5 to +2; Fe: +2 to +3)
40. KMnO4 + H2SO4 + NO ------> K2SO4 + MnSO4 + HNO3 + H2O (R): (Mn: +7 to +2; N: +2 to +5)
41. NO + Cl2 ------> NOCl (R) (N: +2 to +3; Cl: 0 to -1)
42. NO + NO2 + KOH ------> KNO2 + H2O (R) ( N: +2 to +3; N: +4 to +3)
43. SO2 + NO2 + H2O ------> H2SO4 + NO (R) ( S: +4 to +6; N: +4 to +2)
44. HNO3 + P4O10 ------> N2O5 + HPO3 (M) Dehydration of HNO3 by P4O10
45. N2O5 + I2 ------> NO2 + I2O5 (R) (N: +5 to +4; I: 0 to +5)
46. HNO2 ---heat-----> HNO3 + NO + H2O (R) Dispror: (N:+3 to +5; N: +3 to +2)
47. HNO2 + HI ------> I2 + H2O + NO (R) (N: +3 to +2; I: -1 to 0)
48. FeSO4 + H2SO4 + HNO2 ------> Fe2(SO4)3 + NO + H2O (R) (Fe: +2 to +3; N: +3 to +2)
49. HNO3 ---sunlight -----> NO2 + H2O + O2 (R) (N:+5 to +4; O:-2 to 0)
50. Sn + HNO3(conc) -----> H2SnO3 + NO2 + H2O (R) (Sn: 0 to +4; N: +5 to +4)
51. Zn + HNO3 (v.dil) ------> Zn(NO3)2 + NH4NO3 + H2O (R) (Zn: 0 to +2; N: +5 to -3)
52. PH3 + Cl2 -----> PCl5 + HCl (R) (P: -3 to +5; Cl: 0 to -1)
53. CuSO4 + PH3 ------> Cu3P2 + H2SO4 (M)
54. PCl3 + 3H2O ------> H3PO3 + 3HCl (M)
55. P4O10 + C ------> P4 + CO (R) (P: +5 to 0; C: 0 to +2)
56. H3PO2 ---heat-----> PH3 + H3PO3 (R) Dispropr. (P: +1 to +3; P: +1 to -3)
57. H3PO2 + HgCl2 + H2O ------> Hg2Cl2 + H3PO4 + HCl (R) (P: +1 to +5; Hg: +2 to +1)
58. H3PO3 + CuSO4 + H2O------> H3OP4 + Cu + H2SO4 (R) (P: +3 to +5; Cu: +2 to 0)
59. As + NaOH ------> Na3AsO3 + H2 (R- Ampho.) (As: 0 to +3; H: +1 to 0)
60. As + HNO3(conc.) ------> H3AsO4 + NO2 + H2O (R) (As: 0 to +5; N: +5 to +4)
61. Na3AsO4 + HCl + H2S ------> As2S5 + NaCl + H2O (M)
62. H3AsO4 + H2S ------> H3AsO3 + S + H2O (R) ( As: +5 to +3; S: -2 to 0)
63. Sb + HNO3(conc.) ------> H3SbO4 + NO2 + H2O (R) (Sb: 0 to +5; N: +5 to +4)
64. Sb2O3 + NaOH ------> NaSbO2 + H2O (M- Ampho.)
65. Sb2O5 + HCl + KI ------> SbCl3 + KCl + I2 + H2O (R) ( Sb: +5 to +3; I: -1 to 0)
66. SbCl3 + H2O ------> SbOCl(antimony oxycloride or antimonyl chloride)+ HCl (M)
67. Bi(NO3)3 ----heat------>: Bi2O3 + NO2 + O2 (R)
68. Bi2O5 + H2SO4 ------> Bi2(SO4)3 + H2O + O2 (R) (Bi: +5 to +3; O: -2 to 0)
Dr. S. S. Tripathy 84 Concepts in Chemistry
69. MnO2 + H2SO4 ------> MnSO4 + O2 + H2O (R) (Mn: +4 to +2; O:-2 to 0)
70. KMnO4 +H2SO4 ------> K2SO4 + MnSO4 + O2 + H2O (R) ( Mn: +7 to +2; O: -2 to 0)
71. Na2O2 + H2O ------>NaOH + O2 (R): O:-2 to 0; O: -1 to -2) +2 72. Fe3O4 + HCl ------> FeCl2 + FeCl3 + H2O (M): Fe3O4 (FeO.Fe2O3)contains both Fe and Fe+3)
73. PbS + O3 ------> PbSO4 + O2 (R) (S: -2 ro +6; O: 0 to -2: note that O2 is merly an additional product)
74. O3 + S + H2O ------> H2SO4 + O2(R) (S: 0 to +6; O: 0 to -2)
75. K2S + 4S ------> K2S5 (R) (S:-2 to -2/5; S: 0 to -2/5)
76. CuS + HNO3(dil.) ------> Cu(NO3)2 + NO + S + H2O (R) (S: -2 to 0; N:+5 to +2)
77. Na2S5 + HCl ------> NaCl + H2S + S (R) (opposite of eq. 75)
78. CaSO4 + C ------> CaO + SO2 + CO2 (R) (S: +6 to +4; C: 0 to +4)
79. SO2 + H2O + FeCl3 ------> FeCl2 + H2SO4 + HCl (R) (S: +4 to +6; Fe: +3 to +2)
80. KIO3 + H2SO3 ------> KHSO4 + H2SO4 + I2 + H2O (R) (I: +5 to 0; S: +4 to +6)
81. KClO3 + H2SO4(conc.) ------> KHSO4 + HClO4 + ClO2 + H2O (R-Dispropr. Cl:+5 to +7; Cl:+5 to +4)
82. K2S2O8 + KI ------> K2SO4 + I2 ( R) (O: -1 to -2; I: -1 to 0) peroxydisulphate has a peroxy group)
83. Na2S2O3 + I2 ------> Na2S4O6 + NaI (R) (S:+2 to +2.5; I: 0 to -1)
84. FeSO4 ----heat-----> Fe2O3 + SO2 + SO3 (R) (Fe: +2 to +3; S:+6 to +4)
85. Na2S + Na2SO3 + I2 ------> Na2S2O3 + NaI (R) (S: -2 to +2; S: +4 to +2 and I: 0 to -1)
86. F2 + NaOH -----> NaF + OF2 + H2O (R) (F: 0 to -1; O: -2 to +2)
87. ClO2 + NaOH ------> NaClO2 + NaClO3 + H2O (R- dispropr.) Cl: +4 to +3; Cl: +4 to +5)
88. Ca(OCl)Cl + HCl ------> CaCl2 + Cl2 + H2O (R) (Cl: +1 to 0; Cl: -1 to 0)
89. HClO3 ------> HClO4 + O2 + Cl2 + H2O (R-dispropr. Cl: +5 to +7; O: -2 to 0; Cl: +5 to 0)
90. F2 + H2O ------> HF + O2 + O3 (R) (F: 0 to -1; O: -2 to 0)
91. NaBiO3 + MnSO4 + HNO3 ------> HMnO4 + Bi(NO3)2 + H2SO4 + H2O (R) (Bi: +5 to +3; Mn: +2 to +7)
92. Bi(OH)3 + Na2SnO3------> Bi + Na2SnO3 + H2O (R) (Bi: +3 to 0; Sn: +2 to +4)
93. Zn + NaNO2 + NaOH ------> Na2ZnO2 + NH3 + H2O (R) (Zn: 0 to +2; N: +3 to -3)
94. K2CrO4 + H2SO4 ------> K2Cr2O7 + K2SO4 + H2O (M)
95. HgI2 + 2KI ------> K2[HgI4] (M- Complexation)
96. CuSO4 + K4[Fe(CN)6] ------> Cu2[Fe(CN)6] + K2SO4 (M)
97. K2Cr2O7 + HCl ------> KCl + CrCl3 + Cl2 + H2O (R) (Cr:+6 to =3; C: -1 to 0)
98. K2Cr2O7 + H2SO4 + H2O2 ------> K2SO4 + CrO5 + H2O (M)
99. K2Cr2O7 + KCl + H2SO4 ------> K2SO4 + CrO2Cl2 +H2O (M)
100. CrO2Cl2 + NaOH ------> Na2CrO4 + NaCl + H2O (M)
Dr. S.S Tripathy Logic of Inorganic Reactions 85
PRACTICE QUESTIONS 50 reactions are given below in each set. You are required to predict the products with proper reasoning, giving its type(sub-type if any) and for redox reaction, you have to show the changes in ON. LEVEL-I SET-I
1. FeS + H2SO4(dil)------> 2. CO2 + H2O------>?
3. Cr(OH)3 + H2SO4 ------>? 4. Na + H2O ------>?
5. CaO + H2O ------>? 6. BaCO3 + HCl ------>?
7. Ca3N2 + H2O ----->? 8. NH4Cl + CaO ---heat---->?
9. Cu + AgNO3 ------>? 10. Cl2 + KI ------>?
11. NH3 + HCl ----->? 12. N2O5 + H2O ------>?
13. Pb(CH3COO)2(aq) + K2CrO4(aq)------>?
14. CaCO3 ----heat---->? 15. Na2SO3 + dil. H2SO4 ------>?
16. Al + HCl ------>? 17. AgNO3(aq) + NaCl(aq)------>?
18. Al4C3 + H2O ------>? 19. CrCl3(aq) + NH4OH(aq) ------>?
20. PbO2 ----heat ----->? 21. SO3 + NaOH------>?
22. Cu + Zn(NO3)2 ------>? 23. Mg + HCl ------>?
24. MnCl2(aq) + H2S(aq)---->? 25. H2 + I2 ------>?
26. C2H6 + O2 ------>? 27. Ca + H2O------>?
28. Al + Cl2 ------>? 29. Pb(NO3)2-----heat----->?
30. Ca(OH)2 + H2SO4 ------>? 31. NH4NO2 ---heat---->?
32. Mg + SO2 ------>? 33. K2CO3 + H2SO4(dil)------>?
34. PbO + HBr ------>? 35. AlP + H2O ------>?
36. KClO3 ----heat---->? 37. (NH4)2SO4 + KOH ----heat----->?
38. Al(OH)3 + NaOH------>? 39. Hg2(NO3)2(aq) + HCl(aq) ----->
40. Fe2O3 + H2SO4 ------>? 41. KNO3 -----heat----->?
42. Fe(red hot) + H2O(steam)----->? 43. Hg(NO3)2(aq) + H2S(aq)------>?
44. NaOH + SnO ------>? 45. Al + Fe2O3 ------>?
46. SO2 + H2O------>? 47. Br2 + KCl ------>?
48. CaCl2(aq) + (NH4)2C2O4(aq)------>?
49. MnO2 + HBr ----heat--->? 50. H2O2 + H2S ------>? SET-II
1. N2O3 + H2O------>? 2. Fe2(SO4)3(aq) + NH4OH(aq)------>?
3. CO2 + Ca(OH)2------>? 4. NaNO3(aq) + H2SO4(conc.)----->?
5. Sn(OH)2 + KOH------>? 6. P2O5 + NaOH------>?
7. Ca3N2 + H2O------>? 8. Ca(NO3)2------>?
9. K2O + H2O------>? 10. K + H2O------>?
11. FeSO4 + H2SO4 + H2O2------>? 12. MnO + HNO3 ------>?
13. BiCl3(aq) + H2S(aq)------>? 14. Zn + KOH------>?
15. NH4Br + KOH ----heat---->? 16. Hg + CuSO4 ------>?
Dr. S. S. Tripathy 86 Concepts in Chemistry
17. F2 + NaBr ------>? 18. MnO2 + HI ------>?
19. Cu + H2SO4(dil)------>? 20. Cu + H2SO4(conc.)------>?
21. NH3 + H2SO4 ------>? 22. NH4NO3------heat------>?
23. Ca(OH)2 -----heat---->? 24. H2S + O2 ------>?
25. Pb(NO3)2(aq) + K2CrO4(aq)--->? 26. Mg + PbO ------>?
27. MgSO3 + HBr------>? 28. Hg2(NO3)2(aq)+ HCl(aq)----->? 29. Ca + HCl------>? 30. Ag + HCl ------>?
31. PbO+ NaOH----->? 32. NaNO3 ----heat------>?
33. SO3 + KOH----->? 34. Ba(NO3)2 + Na2CO3------>?
35. Mn3O4 + Al----->? 36. NH3 + Cl2 ------>?
37. NH3 + H2O----->? 38. KI + H2O2 ------>?
39. Na + NH3 ------>? 40. AlCl3(aq) + NH4OH(aq)------>?
41. NiCl2(aq) + H2S(aq)----->? 42. Cl2 +KF------>?
43. FeSO4(aq) + NaOH(aq)------>? 44 FeSO4+ H2SO4 + Cl2------>
45. SnCl2 + HgCl2 ------>? 46. SrCO3 + H2SO4(dil)----->?
47. HgS + O2 ------>? 48. Pb + NaOH------>?
49. Fe(OH)3 + HCl------>? 50. HNO3(conc) + H2S------>?
LEVEL-II SET-I:
1. S+O2------>? 2. BaCO3------heat--->?
3. PbS+O2------>? 4. Pb(NO3)2(aq)+HCl(aq)------>?
5. HI +H2O2------>? 6. N2O3 + NaOH---->?
7. Al(OH)3+NaOH------>? 8. NaOH+HNO3----->?
9. ZnO + C------>? 10. CaO + H2O----->?
11. BaO2 + H2SO4 ------>? 12. TiCl4+Mg------>?
13. Cr(OH)3+ H2SO4------>? 14. Ba(OH)2+CH3COOH------>? 15. ZnO+KOH------>? 16. Sn+ KOH------>?
17. PbO2-----heat---->? 18. HBr + O3 ------>?
19. Cu + H2SO4(conc.)------>? 20. Cu + HNO3(conc.)-----?
21. Sr(NO3)2(aq) + CaSO4(aq)---->? 22. HgO + H2------>?
23. HNO3(conc) + C ------>? 24. P2O5 + KOH----->?
25. PbO2 + HCl----->? 26. MgO + H3PO4------>?
27. AlP + H2O----->? 28. CaC2 + H2O----->?
29. H2S + SO2 ------>? 30. SnO2 + HCl------>?
31. Al2S3 + H2SO4 ------>? 32. BaO + O2 ------>?
33. NaHCO3 ------heat----->? 34. KNO3 -----heat----->?
35. (NH4)2CO3 + NaOH----heat---? 36. NH4NO3----heat---->?
37. Al + NaOH +H2O----->? 38. Pb3O4 + HCl------>?
39. Pb + HBr------>? 40. Br2 + NaCl----->?
41. Cr(NO3)2(aq) + NaOH(aq)----->? 42. I2O5+H2O------>?
43. MnCl2 + H2S------>? 44. NH3 + H2SO4 ------>?
45. HNO3(conc)+H2S------>? 46. Zn + CuSO4 ------>?
47. NH3(excess)+ Cl2------>? 48. Fe(red hot)+H2O(steam)------>?
Dr. S.S Tripathy Logic of Inorganic Reactions 87
49. Cl2O+H2O------>? 50. Cl2O3 +NaOH------>? SET-II
1. NO2 +NaOH----->? 2. Fe+Cl2----->?
3. SnCl2+HCl+HNO3----->? 4. Al4C3+H2O----->?
5. I2+HNO3(conc.)------>? 6. PI3+H2O----->?
7. HNO3+P2O5------>? 8. Cl2+SO2+H2O----->?
9. K2O+H2O----->? 10. HIO3+HI------>?
11. Cl2O7+H2O------>? 12. NaNO2 +HCl------>?
13. KMnO4+H2SO4+H2S----->? 14. C3H6+O2------>?
15. Cu+AgNO3------>? 16. SO2+KOH------>?
17. Na2S2O3 +HCl------>? 18. Li+MgO----->?
19. K2Cr2O7+H2SO4+FeSO4----->? 20. Cr(OH)3+NaOH------>?
21. Cr(OH)3+NaOH + H2O2------>? 22. Si+NaOH------>?
23. HI +H2SO4(conc)----->? 24. Pb +KOH------>?
25. AgNO3 ----heat---->? 26. P4+HNO3------>?
27. K2CO3-----heat---->? 28. KMnO4-----heat---->?
29. O3+K4[Fe(CN)6]+H2O--->? 30. ZnSO4(aq)+K4[Fe(CN)6](aq)----->?
31. CaH2 + H2O ----->? 32. Ca + H2 ----->?
33. Fe3O4 + H2 ------>? 34. SOCl2 + H2O ------>?
35. BaO2 + H2O + CO2 ----->? 36. Al(OH)3----heat--->?
37. N2O + Cu ------>? 38. CuI2------>?
39. AgCl + Hg---->? 40. XeF4 + H2 ------>?
41. AsH3 + AgNO3 + H2O------>? 42. ZnSO4 ----heat---->?
43. NH4NO3 + NaOH---heat------>? 44. NaNO2 + Al + NaOH----heat---->?
45. NH4NO2 + HCl------>? 46. FeSO4 + K3[Fe(CN)6]------>?
47. K4[(Fe(CN)6] + Cl2------>? 48. K3[Fe(CN)6] + SO2 + KOH------>?
49. CrO3 + NaOH------>? 50. Cu2(CN)2 + 6KCN------>?
SET-III:
1. Na2O2 + H2SO4 ------> 2. (NH4)2S2O8 + 2H2O ------>
3. FeSO4 + H2SO4 + H2O2 -----> 4. Ag2O + H2O2 ------>
5. Cl2 + H2O2 ------> 6. PbS + H2O2 ------>
7. BeCl2 + Na ------> 8. B + NaOH ------> 9. Si + NaOH -----> 10. S + NaOH ------>
11. Zn(OH)2 + KOH -----> 12. NaOH + NaHCO3 ------>
13. NaNH2 + C ------>? 14. NaNH2 + H2O------>
15. I2 + NaOH(hot and conc.)------> 16. Ca3N2 + H2O ----->
17. BeO + NaOH-----> 18. Be2C + H2O ------>
19. CaCO3 + CO2 + H2O -----> 20. Ca(HCO3)2(aq) ---heat---->
21. MgH2 + H2O -----> 22. CaF2 + H2SO4 ----->
23. Al4C3 + H2O ------> 24. B2O3 + KOH ----->
25. SiCl4 + H2O ------> 26. BF3 + H2O ------>
27. H3BO3 ----heat---> 28. B2O3 + K ------>
29. P4O6 + KOH ------> 30. Al2O3+C + N2 ----->
Dr. S. S. Tripathy 88 Concepts in Chemistry
31. AgCl + 2NH3 ------> 32. KNO3 +KOH + Zn ------>
33. Fe + Al2O3 ------> 34. Al + H2SO4(conc.)------>
35. Al2O3 + C ------> 36. C + HNO3(conc) ------>
37. CO2 + C ----> 38. PbO +C ------>
39. H2C2O4 ----conc.H2SO4----> 40. HCOOH----conc.H2SO4 ----->
41. Fe + 5CO -----> 42. Fe2O3 + CO ----->
43. I2O5 + CO -----> 44. K2CO3 + H2O + CO2 ------>
45. Si + KOH + H2O ------> 46. SiO2 + Mg ----->
47. Sn + HNO3(v.dilute)-----> 48. SnCl4 + Sn ----->
49. FeCl3 + SnCl2 -----> 49. Sn(OH)2 + KOH ----->
50. Pb3O4 + HCl ----->
RESPONSE TO SAQs SAQ 1:
(i) NaOH + H2SO4 -----> Na2SO4 + H2O (The formula of sodium sulphate is Na2SO4)
(ii) Ba(OH)2 + H3PO4 ------> Ba3(PO4)2 + H2O(The formula of barium phosphate is not
BaPO4)
(iii) CrCl3 + NaOH ------> Cr(OH)3 + NaCl (The formula of chromic hydroxide and sodium chloride are wrong in the question)
(iv) Bi(NO3)2 + H2S ------> Bi2S3 + HNO3(The formula of Bismuth sulphide and nitric acid are wrongly written in the question). SAQ 2: The symbol (s) is given for the product which is formed as insoluble precipitate.
(i)Pb(NO3)2 + NaCl ------> PbCl2(s) + NaNO3 (P)
(ii)BaCl2 + K2CrO4 ------> BaCrO4(s) + KCl (P)
(iii)KF + CaCl2 ------> CaF2(s) + KCl (P)
(iv)Pb(CH3COO)2 + H2S ---->PbS(s) + HCH3COO( more correctly CH3COOH) (P)
(v)Ba(NO3)2 + K2CO3 ------>BaCO3(s) + KNO3 (P)
(vi)FeCl3 + NH4OH ----->Fe(OH)3(s)+ NH4Cl (P)
(vii)NaOH + H3PO4 ----->Na3PO4 + H2O (N)
(viii)HNO3 + Al2O3 ----->Al(NO3)3 + H2O (N)
(ix) MgO + H2SO4 ------> MgSO4 + H2O (N)
(x) Cr2O3 + H3PO4 ------>CrPO4(s) + H2O(B)
SAQ 3:These are all neutralisation reactions of a base NH3 or NH4OH with an acid to give salt.
When we take NH4OH we get water besides the salt but when we take NH3 we get only the salt.
(i)(NH4)2SO4 + H2O (ii)(NH4)2SO4 (iii)NH4Cl + H2O
(iv)NH4Cl (v)NH4NO3 SAQ 4: All these are metathesis reactions and ON does not change.
(i)H3PO4 (ON=+5) (ii)HNO3( ON=+5) (iii)P2O3(ON=+3)
(iv)Cl2O (ON=+1) (v)HNO2(ON=+3) (vi)HClO3(ON=+5)
(vii)Mn2O7(ON=+7) (viii)HClO4(ON=+7) (ix)H2CrO4(ON=+6) SAQ 5:
(i) NO2 + H2O ----->HNO2 + HNO3 (The ON of N goes from +4 in NO2 to +3 in HNO2 and again the ON of the same N goes from +4 in NO2 to +5 in HNO3). This is a redox reaction Dr. S.S Tripathy Logic of Inorganic Reactions 89 and hence not a metathesis reaction. Note that this type of redox reaction in which the same substance(NO2) gets oxidised as well as reduced has a separte name, which we shall discuss later.
(ii) ClO2 + H2O ------> HClO2 + HClO3 (The ON of Cl changes from +4 in ClO2 to +3 in HClO2 and again the ON of same Cl changes from +4 in ClO2 to +5 in HClO3).
SAQ 6:SO3, N2O3, Cl2O, CO2, Cl2O7
SAQ 7:(i)KOH (ii)Ba(OH)2 (iii)SrO (iv)No reaction; since Al2O3 is insoluble in water
SAQ 8: (i) KAlO2 + H2O(Al is in +3 ON) (ii)Na2SnO2 + H2O(Sn is in +2 state)
(iii) Na2PbO2 + H2O(Pb in +2 state) (iv)K2SnO3 + H2O (Sn is in +4 state) Write the names of these salts in language from the table given in the text. The ON of the metal is same on both the sides because these are metathesis reactions. SAQ 9:
(i) Na2ZnO2 + H2O(zincate has a valency of 2 while in the question it has been taken as 1)
(ii) K2CoO2 + H2O(cobaltate has a valency of 2, not 3)
(iii) NaAlO2 + H2O(meta-aluminate has a valency of 1, not 2) (iv) Na2SnO2 + H2O(The ON of Sn in LHS is +2 and in the RHS it cannot be +4 as it is a metathesis reaction). SAQ 10:
(i)K2SO4 + CO2 + H2O (ii)NaCl + SO2 + H2O (iii)CaCl2 + SO2 + H2O
(iv)ZnBr2 + CO2 +H2O (v)BiCl3 + H2S (vi)FeSO4 + H2S
(vii)CaSO4 + SO2+ H2O+ S (viii)K2SO4 + NO + NO2 + H2O SAQ 11:
(i)AlN + H2O ------> Al(OH)3 + NH3 (ii)Ca3P2 + H2O ----->Ca(OH)2 + PH3
(iii)CaC2 + H2O ------> C2H2 + Ca(OH)2 (iv)ZnS + H2O------>Zn(OH)2 + H2S SAQ 12:
(i)MgO+CO2 (ii)K2CO3 + CO2 + H2O
(iii)Li2O +CO2(Li2CO3 is the only alkali metal carbonate which decomposes on heating). SAQ 13: (i)NH3 + CaCl2 + H2O (ii)NH3 + K2CO3 + H2O
(iii)NH3 + BaBr2 + H2O (iv)NH3 + Na2SO4 + H2O
SAQ 14: (i)BeO +H2O (ii)MgO+H2O (iii)Cr2O3 + H2O(iv)No change
SAQ 15: (i)Na2SO4 + HCl (ii)K2SO4 + HNO3
SAQ 16: (i)(Cold water)Ca(OH)2 + H2 (ii)(boiled water)MgO + H2
(iii)Fe3O4(magnetic oxide or ferrosoferric oxide) + H2 (the reaction takes place with red hot Fe and superheated steam)
(iv)No reaction with any form of water (v)(steam)Al(OH)3 + H2
SAQ 17: (i)NiCl2 + H2 (ii)No reaction (iii)ZnBr2 + H2 (iv)FeCl2 + H2 (v)No reaction
SAQ 18: (i)I2 + KCl (ii)No reaction (iii)Br2 + KF
(iv)No reaction (v)Br2 + NaCl
SAQ 19: (i)No reaction (ii)No reaction (iii)No reaction (iv)FeCl3 + Br2
(v)Al2O3 + Cr (vi)No reaction (vii)Na2O + Zn
Dr. S. S. Tripathy 90 Concepts in Chemistry
(viii)Cu(NO3)2 + Ag (ix)ZnCl2 + Au (x)No reaction
SAQ 20: (i)N2 + H2 -----> NH3 (ii)C + O2 ------> CO2
(iii)P4 + Cl2 -----> PCl3 (iv)H2 + O2 ------> H2O (Student is advised to find the changes in ON in all the above cases) SAQ 21:
(i) NO2 + H2O -----> HNO3 + HNO2 [ N goes from +4 to +5(HNO3) and from +4 to
+3(HNO2)] Hence it is a disproportionation reaction. You know that NO2 is a mixed anhydride.
(ii) NO2 + NaOH ----> NaNO3 + NaNO2 + H2O[N goes from +4 to +5(NaNO3) and from
+4 to +3(NaNO2)]. Note that NaOH reacts with the two acids produced by NO2(HNO3
and HNO2) to form the corresponding salts(NaNO3 and NaNO2) and water.
(iii) ClO2 + H2O -----> HClO2(chlorous acid) + HClO3(chloric acid) Cl goes from +4 to
+5(HClO3) and to from +4 to +3(HClO2). Hence it is a disproportionation reaction.
(iv) ClO2 + NaOH -----> NaClO3 + NaClO2 + H2O
NaOH reacts with the two acids produced by ClO2 to produce the corresponding salts and water. SAQ 22: (i)First there will be double replacment reaction.
NaNO2 + NH4Cl -----> NH4NO2 +NaCl (i)
Then NH4NO2 on heating will give N2 and 2H2O.
NH4NO2 ----heat ----> N2 + 2H2O (ii) Adding reaction (i) and (ii), we get the overall reaction.
NaNO2 + NH4Cl ------> N2 + H2O + NaCl
(ii) Like (i)NH4NO3 will be formed first which will decompose on heating to give N2O.
NaNO3 + NH4Cl -----> N2O + H2O + NaCl SAQ 23: (i) Ana. and also Dispr.; Hg changes from +2 to 0 and O changes from -2 to 0. (ii) Synth.; Hg changes from 0 to +2 and O changes from 0 to -2
(iii) Ana. and also Dispr.; Mn changes from +7 to +6(K2MnO4) and +4(MnO2) and O changes from -2 to 0. Note that in this reaction there are two reductions and one oxidation.
(iv) Syn.; Cu changes from 0 to +1(Cu2O) while O changes from 0 to -2. SAQ 24: (i) N from +5 to +4 and O from -2 to 0 (ii)N from +5 to +3 and O from -2 to 0
(iii) P from 0 to -3 and P(PH3) from 0 to +1(NaH2PO2): disproportionation
(iv) Cl from 0 to +5(NaClO3) and Cl from 0 to -1(NaCl): disproportionation SAQ 25:
(i)NaAlO2 +H2 (ii)K2SnO3(potassium stannate) + H2
Note that in metathesis, we get K2SnO2(potassium stannite)
(iii)Na2PbO2 + H2 (iv)Na2SiO3 + H2 The student is advised to show the changes in ONs. SAQ 26: All these are metathesis reactions indicated in the hints in the SAQ.
(i)HNO3 (ii)Ba(NO3)2 + H2O (iii)K2ZnO2 + H2O
(iv)AgBr(s)+ NaNO3 (v)Na2SO4 + CO2 + H2O
SAQ 27: (i)BaSO4(s) + HCl(aq) (ii)KOH (iii)KOH + H2
(iv)ZnCl2 + H2 (v)H2CO3(carbonic acid)
Dr. S.S Tripathy Logic of Inorganic Reactions 91
SAQ 28: (i)No reaction (ii)Al2(SO4)3 + H2O (iii)Na2SnO2 + H2O
(iv)HI(H goes from 0 to +1 and I from 0 to -1) (v)MgCl2 + H2S
SAQ 29: (i)K2ZnO2 + H2 (Zn goes from 0 to +2 and H from +1 to 0)
(ii)CaCO3(s) + NH4Cl(aq) (iii)H2SO3 (iv)N2 + H2O(N from -3 to 0
and O from 0 to -2) (v)Ca3(PO4)2 +H2O
SAQ 30: (i)Zn(OH)2(s) + NH4Cl (ii)H3PO4 (iii)Mg(OH)2 + NH3
(iv)Na2CO3 + CO2 + H2O (v)H2O + S
SAQ 31: (i)NaCl + CO2 + H2O (ii)Bi2S3(s) + HNO3 (iii)No reaction
(iv)I2 +KCl (v)PbO + O2(Pb from +4 to +2 and O from -2 to0)
SAQ 32: (i)FeCl2 + H2S (ii)HClO3 (iii)AlPO4 + H2O
(iv)N2 + Cu +H2O (v)FeCl3(Fe from 0 to +3 and Cl from 0 to -1)
SAQ 33: (i)No reaction (ii)As2(SO4)3 (iii)Na2ZnO2
(iv)HNO2 (v)MnCl2 + Cl2 + H2O (Cl is oxidised from -1 to 0)
SAQ 34: (i)ZnBr2 + CO2 + H2O (ii)Na2CO3 + CO2 + H2O
(iii)Na3PO4 + H2O
(iv)Ca(OH)2 + H2(the ON of Ca changes from 0 to +2 and H from +1 to 0)
(v)NH4Cl
SAQ 35: (i)K2SnO2 + H2O (ii)PbSO4 + H2O
(iii)Na2PbO2 + H2(The ON of Pb changes from 0 to +2 and H from +1 to 0)
(iv)CaO+NO2 +O2(The ON of N changes from +5 to +4 and O from -2 to 0)
(v)As2S3(s) +HNO3(aq)
SAQ 36: (i)Fe(NO3)3 + H2O (ii)KNO3 + H2O
(iii)NaOH (iv)K2SO4 + H2
(v)P4 + O2 ------>P2O3(limited oxygen) and also P4 + O2 ------> P2O5(excess oxygen) (The ON of P changes from 0 to +3 or +5 and of O from 0 to -2)
SAQ 37: (i)Ba3(PO4)2(s) +NaNO3(aq)
(ii)HNO2 + HNO3 (Redox: Disproportionation, the ON of N changes from +4 to +5 and +4 to +3)
(iii)NaNO2 + O2(the ON of N changes from +5 to +3 and O from -2 to 0) (iv)(NH4)2SO4 (v)FeCl2 + SO2 + H2O SAQ 38: (i)MgO+ NO2 + O2 (ii)PbBr2 + Br2 + H2O(ON of Br changes from -1 to 0) (iii)FeCl2 + HCl + S (iv)No reaction (v)Cu2[Fe(CN)6] SAQ 39: (i)NaOH (ii)NaAlO2 + H2O (iii)HNO2 (iv)Sb2S3 + HNO3 (v)NH3 + Na2SO4 + H2O
ANSWERS TO PRACTICE QUESTIONS LEVEL-I SET-I
1. FeSO4 + H2S (M: suphide salt with acid)
2. H2CO3 (M:non-metallic oxide with H2O)
3. Cr2(SO4)3 + H2O (M:Neutralisation)
Dr. S. S. Tripathy 92 Concepts in Chemistry
4. NaOH + H2 (R: Displ.; Na from 0 to +1 and H from +1 to 0)
5. Ca(OH)2 (M: metallic oxide with H2O)
6. BaCl2 + CO2 + H2O (M: carbonate salt with acid)
7. Ca(OH)2 + NH3 (M:nitride salt with water)
8. CaCl2 + NH3 + H2O (M: ammonium salt with a base)
9. Ag + Cu(NO3)2 (R: Displ.; Ag from +1 to 0 and Cu from 0 to +2)
10. I2 + KCl (R: halogen displ.; Cl from 0 to -1 and I from -1 to 0)
11. NH4Cl (M: Neu., H2O is not formed because it is NH3 not NH4OH)
12. HNO3 (M: nonmetallic oxide with water)
13. PbCrO4(s) + KCH3COO(CH3COOK) (M: Double. Repl./ Preci.)
14. CaO + CO2 (M: decomp. of carbonate salt)
15. Na2SO4 + SO2 + H2O (M: sulphite salt with acid)
16. AlCl3 + H2 (R: Displ.;Al from 0 to +3 and H from +1 to 0)
17. AgCl(s) + NaNO3 (M:Double Repl./ Preci.)
18. Al(OH)3 + CH4 (M: carbide with water)
19. Cr(OH)3(s)+NH4Cl (M:Double Repl./ Preci.)
20. PbO + O2 (R: Pb from +4 to +2 and O from -2 to 0)
21. Na2SO4 + H2O (M:SO3 gives H2SO4 which reacts with NaOH) 22. No reaction (Cu lies below Zn in the metal activity series)
23. MgCl2 + H2 (R: Mg from 0 to +2, H from +1 to 0) 24. MnS(s) + HCl (M: Double Repl./ Preci.) 25. HI (R:syn.; H from 0 to +1, I from 0 to -1)
26. CO2 + H2O (R: combustion)
27. Ca(OH)2 + H2 (R: Displ.; Ca from 0 to +2, H from +1 to 0)
28. AlCl3 (R: syn.; Al from 0 to +3, Cl from 0 to -1)
29. PbO + NO2 + O2 (R: Decomp., Dispropr./ N from +5 to +4 and O from -2 to 0)
30. CaSO4 + H2O(M: Neu) + - 31. N2 + 2H2O [Decomp. of NH4NO2/ N from -3(NH4 ) to 0 and N from +3(NO2 ) to 0] 32. MgO + S (R: Mg from 0 to +2, S from +4 to 0)
33. K2SO4 + CO2 + H2O (M: carbonate salt with acid)
34. PbBr2 + H2O (M: Neu.)
35. Al(OH)3+PH3 (M:phoshphide salt with H2O)
36. KCl + O2 (R: Decomp., Dispropr./Cl from +5 to -1 and O from -2 to 0)
37. K2SO4 + NH3 + H2O (M:ammonium salt with a base)
38. NaAlO2 + H2O (M: Amphoterism)
39. Hg2Cl2(s) + HNO3 (M:Double repl./Preci.)
40. Fe2(SO4)3 + H2O (M: Neu.)
41. KNO2 + O2 (R: Decomp. of Na and K nitrate/ N from +5 to +3 and O from -2 to 0)
42. Fe3O4 (Ferrosoferric oxide or magnetic oxide) + H2 (R: Displ./Fe from 0 to +8/3 and H from +1 to 0)
43. HgS(s) + HNO3 (M:Double Repl./Preci.)
44. Na2SnO2 + H2O (M: Amphoterism)
45. Al2O3 + Fe (R: Displ./Al from 0 to +3 and Fe from +3 to 0)
46. H2SO3 (M: nonmetallic oxide with H2O)
Dr. S.S Tripathy Logic of Inorganic Reactions 93
47. No reaction (Br is below Cl in the halogen activity series)
48. CaC2O4(s)(calcium oxalate) +NH4Cl (M: Double Repl./Preci.)
49. MnBr2 + Br2 + H2O (R: Mn from +4 to +2 and Br from -1 to 0)
50. S + H2O [R: S from -2 to 0 and O from -1(peroxide) to -2(water)] SET-II:
1. HNO2(M)
2. Fe(OH)3(s)+(NH4)2SO4 (M:Double Repl)
3. CaCO3(s)+H2O(M: CO2 forms carbonic acid salt i.e carbonate)
4. HNO3 + Na2SO4 (M:Double Repl.)
5. K2SnO2+H2O (M: Ampho.)
6. Na3PO4 + H2O(P2O5 forms the H3PO4 and with NaOH forms phoshpate salt)
7. Ca(OH)2 + NH3 (M:hydrolysis of nitride)
8. CaO+NO2+O2 (R: N goes from +5 to +4 and O from -2 to 0)
9. KOH (M: metallic oxide with H2O)
10. KOH + H2(R:Displacement)
11. Fe2(SO4)3 + H2O[R: Fe goes from +2 to +3 and O from -1 in peroxide to -2 in water.
Note : that in every redox reaction FeSO4 takes the help of H2SO4 to make it Fe2(SO4)3]
12. Mn(NO3)2 +H2O(M: Neu.)
13. Bi2S3(s)+HNO3(M: DR/preci)
14. K2ZnO2 + H2(R:Amphoterism/ Zn goes from 0 to +2 and H from +1 to 0)
15. KBr + NH3 + H2O(M: decomposition of ammonium salt by a base) 16. No reaction: Hg lies below Cu
17. NaF + Br2(R: Displacement of halogen)
18. MnI2 + I2 + H2O(R: Mn from +4 to +2 and I from -1 to 0) 19. No reaction: Cu lies below H
20. CuSO4 + SO2 + H2O(R:Cu from 0 to +2 and S from +6 to +4. Note that conc. H2SO4 is
a strong oxidising agent and it is always reduced to SO2 gas. In no case H2 gas is displaced from the acid)
21. (NH4)2SO4(M: salt formation. NH3 is a base which reacts with acid to form the salt). + - 22. N2O + 2H2O(R: N goes from -3 in NH4 to +1 and also N goes from +5 in NO3 to +1)
23. CaO + H2O(M: decomp. of hydroxides)
24. S +H2O(R: S from -2 to 0 and O from 0 to -2. Note that in every redox reaction, H2S is oxidised to S)
25. PbCrO4(s) +KNO3(M:DR/preci.) 26. MgO + Pb(R: Displacement)
27. MgBr2 + SO2 + H2O(M: sulphite salt with acid)
28. Hg2Cl2(s) + HNO3(M:DR/preci.)
29. CaCl2 + H2(R: Ca lies above H) 30. No reaction(Ag lies below H)
31. Na2PbO2 +H2O(M:Ampho.)
32. NaNO2 +O2(R: N from +5 to +3 and O from -2 to 0. Note that NaNO3 and KNO3 do not
give NO2 gas like other nitrates)
33. K2SO4 +H2O(M: Neu./ SO3 forms H2SO4 which forms sulphate salt with KOH)
34. BaCO3(s) + NaNO3(M:DR/preci.)
Dr. S. S. Tripathy 94 Concepts in Chemistry
35. Al2O3 + Mn(R: Displacement)
36. N2 +HCl(R: N from -3 to 0 and Cl from 0 to -1. Note that often students commit mistake
by writing the product NH4Cl only. In such case only Cl is reduced from 0 to -1 but
there is no oxidation. So N2 is formed)
37. NH4OH(M: reaction of a base with water forming its hydroxide)
38. I2 + KOH(R: I from -1 to 0 and O from -1 in peroxide to -2 in KOH)
39. NaNH2(sodium amide) + H2(R: Na from 0 to +1 and H from +1 to 0. Note that active
metals like Na, K etc. can displace H from even NH3 forming the metal amides)
40. Al(OH)3(s) +NH4Cl(M:DR/preci.) 41. NiS(s) +HCl(M:DR/preci.)
42. No reaction(Cl2 is below F2)
43. Fe(OH)2(s)+Na2SO4(M:DR/preci.)
44. Fe2(SO4)3+HCl(R: Fe from +2 to +3 and Cl from 0 to -1)
45. SnCl4 + Hg2Cl2(R: Sn from +2 to +4 and Hg from +2 to +1. Note that SnCl2 is good
reducing agent and HgCl2 is a good oxidising agent)
46. SrSO4 +CO2 +H2O(M: carbonate salt with water)
47. HgO + SO2(R: O from 0 to -2 and S from -2 to +4)
48. Na2PbO2 + H2(R: Amphoterism/ Pb from 0 to +2 and H from +1 to 0)
49. FeCl3 + H2O(M:Neu.)
50. NO2 + S + H2O(R: N from +5 to +4 and S from -2 to 0. Note that conc. HNO3 is always
reduced to reddish brown gas NO2 and H2S is always oxidised to S)
LEVEL - II SET-I:
1. SO2(R: show the changes in ON)
2. BaO+CO2(M)
3. PbO +SO2(R: S from -2 to +4 and O from 0 to -2, this is called the roasting of an ore)
4. PbCl2(s)+HNO3(M/DR/preci.)
5. I2 + H2O(R:I from -1 to 0 and O from -1 to -2)
6. NaNO2 + H2O(M: N2O3 is the anhydride of HNO2 and hence the nitrite salt is formed)
7. NaAlO2 + H2O(M:Ampho.)
8. NaNO3 +H2O(M/Neu.) 9. Zn + CO(R: Zn from +2 to 0 and C from 0 to +2)
10. Ca(OH)2(M)
11. BaSO4 + H2O2(M/DR)
12. Ti +MgCl2(R:Displ.)
13. Cr2(SO4)3+H2O(M:Neu.)
14. Ba(CH3COO)2 + H2O(M/Neu.)
15. K2ZnO2+H2O(M:Ampho.)
16. K2SnO3 +H2(R:Ampho./Sn from 0 to +4 and H from +1 to 0)
17. PbO +O2(R:Pb from +4 to +2 and O from -2 to 0)
18. O2 +Br2+H2O(R: Br from -1 to 0 and O from 0 in O3 to -2 in H2O. Note that in all
reaction involving ozone, O2 is formed alongwith other products)
Dr. S.S Tripathy Logic of Inorganic Reactions 95
19. CuSO4 +SO2+H2O(R:Cu from 0 to +2 and S from +6 to +4. Note that conc. H2SO4
reacts with most metals to produce SO2)
20. Cu(NO3)2+NO2+H2O(R: Cu from 0 to +2 and N from +5 to +4. Note that conc. HNO3
reacts with most metals to produce the reddish brown NO2 gas)
21. SrSO4(s)+Ca(NO3)2 (M:D.R/preci.)
22. Hg +H2O(R: Hg from +2 to 0 and H from 0 to +1)
23. NO2 + CO2 +H2O(R: N from +5 to +4 and C from 0 to +4).
24. K3PO4 + H2O(M:Neu./ P2O5 is the anhydride of H3PO4)
25. PbCl2 +Cl2+H2O(R:Pb from +4 to +2 and Cl from -1 to 0)
26. Mg3(PO4)2 +H2O(M:Neu.)
27. PH3+Al(OH)3(M:hydrolysis of phosphide salt)
28. Ca(OH)2 +C2H2(M:hydrolysis of carbide salt)
29. S +H2O(R: S goes from +4 to 0 and also S from -2 to 0)
30. SnCl4 + H2O(M: Sn remains at +4)
31. Al2(SO4)3 +H2S(M)
32. BaO2(R: O from -2 to -1 in peroxide and also O from 0 to -1)
33. Na2CO3 +CO2 +H2O(M:thermal decomposition of bicarbonate)
34. KNO2 + O2 (R: N from +5 to +3 and O from -2 to 0)
35. NH3 + Na2CO3 + H2O(M)
36. N2O+2H2O(R: N from -3 to +1 and N from +5 to +1)
37. NaAlO2 +H2(R:Ampho.: Al from 0 to +3 and H from +1 to 0)
38. PbCl2+Cl2+H2O(R: Pb from+8/3 to +2 and Cl from -1 to 0. Note that Pb3O4 called the
red lead reacts in the same manner as PbO2)
39. PbBr2+H2(R: displ.)
40. No reaction, Br2 is below Cl2
41. Cr(OH)2(s)+NaNO3(M/DR/preci)
42. HIO3(M: I2O5 is the anhydride of HIO3) 43. MnS(s)+HCl(M:DR/preci)
44. (NH4)2SO4 (M: salt formation)
45. NO2 +S + H2O(R: N from +5 to +4 and S from -2 to 0)
46. ZnSO4 +Cu(R:Displacement)
47. NH4Cl +N2(N from -3 to 0 and Cl from 0 to -1. Note that due to excess of ammonia,
NH4Cl is formed by the reaction HCl with ammonia)
48. Fe3O4(magnetic oxide)+H2(R: Fe from 0 to +8/3 and H from +1 to 0)
49. HOCl or HClO(Cl2O is the anhydride of hypochlorous acid)
50. NaClO2 +H2O(Cl2O3 is the anhydride of chlorous acid, HNO2 from which chlorite salt is produces in the above reaction). SET-II The reader is advised to indicate the changes of ON in the following equations.
1. NaNO2 + NaNO3 +H2O(R: NO2 is the mixed anhydride of HNO2 and HNO3)
2. FeCl3(R: syn.)
Dr. S. S. Tripathy 96 Concepts in Chemistry
3. SnCl4 + NO2 + H2O(R)
4. Al(OH)3 + CH4 (M)
5. HIO3 + NO2 + H2O (R)
6. HI + H3PO3 (M)
7. N2O5 + H3PO4(M: dehydration)
8. H2SO4 + HCl(R) 9. KOH(M)
10. I2 + H2O(R)
11. HClO4(M)
12. NaCl + NO + NO2 + H2O (R:Dispr.; N from +3 to +4 and +3 to +2. Note that when any
nitrite salt reacts with dilute acid, it gives HNO2 which gives NO and NO2 and H2O).
13. K2SO4 + MnSO4 + S + H2O(R)
14. CO2 + H2O (R: combustion)
15. Ag + Cu(NO3)2(R: Metal displ.)
16. K2SO3 + H2O(M)
17. NaCl + SO2 + S + H2O (R: note thiosulphate reacts with dilute acids to produce SO2 gas like sulphite salt and besides that sulphur is formed)
18. Mg + Li2O(R: Metal displ.)
19. K2SO4 + Cr2(SO4)3 + Fe2(SO4)3 + H2O(R)
20. NaCrO2 + H2O(M: Ampho.)
21. Na2CrO4 + H2O(R: Cr from +3 to +6 and O from -1 to -2)
22. Na2SiO3 + H2(R: Ampho.)
23. I2 + SO2 + H2O(R)
24. K2PbO2 + H2(R: Ampho.)
25. Ag + NO2 + O2(R: note that silver and mercuric nitrates decompose to their respective
metals alongwith NO2 and O2)
26. H3PO4 + NO2 +H2O(R) 27. No reaction: stable to heat
28. K2MnO4 + MnO2 + O2(R: Mn from +7 to +6 and +4 while O from 0 to -2)
29. O2 + K3[Fe(CN)6] + KOH(R: Fe from +2 to +3 while O from 0 to -2)
30. Zn2[Fe(CN)6] + K2SO4
31. Ca(OH)2 + H2 (R: H:-1 to 0; H:+1 to 0)
32. CaH2(R: Synthesis)
33. Fe + H2O (R:Fe:+8/3 to 0; H:0 to +1)
34. H2SO3 +2HCl (M)
35. H2O2+BaCO3 (M)
36. Al2O3 + H2O (M)
37. N2O + Cu ------>CuO + N2 (R)
38. Cu2I2 + I2(R:Cu:+2 to +2; I:-1to 0) 39. HgCl + Ag(R: Displacement) 40. Xe +HF(R: Xe:+4 to 0; H:0to +1)
41. As2O3 + Ag + HNO3 (R: Ag: +1 to 0; H: -1 to +1) Dr. S.S Tripathy Logic of Inorganic Reactions 97
42. ZnO + SO2 +O2 (R: S: +6 to +4; O:-2 to 0)
43. NaNO3+NH3+H2O(M)
44. NaAlO2 + NH3 + H2O(R: Al: 0 to +3; N:+3 to -3)
45. NH4Cl + NO + NO2 +H2O (R: dispropr.: N:+3 to +2; N:+3 to +4)
46. Fe3[Fe(CN)6]2 + K2SO4 (M)
47. K3[Fe(CN)6] + KCl (R: Fe:+2 to +3; Cl:0 to -1)
48. K4[Fe(CN)6] + K2SO4 (R: Fe: +3 to +2; S: +4 to +6)
49. Na2CrO4 + H2O (M)
50. 2K3[Cu(CN)4] (M: Complexation reaction) SET-III
1. Na2SO4 +H2O2 (M) 2. NH4HSO4 + H2O2(M)
3. Fe2(SO4)3 + H2O (R) 4. Ag +O2 + H2O (R)
5. O2 + 2HCl (R) 6. PbSO4 + H2O (R)
7. NaCl + Be(R) 8. Na3BO3 + H2 (R)
9. Na2SiO3 + H2 (R) 10. S + NaOH ------>Na2S + Na2S2O3 +H2O(R)
11. K2ZnO2 +H2O(M) 12. Na2CO3 + H2O (M)
13. NaCN + H2(R: C:0 to +2; H:+1 to 0) 14. NaOH + NH3(M)
15. NaI + NaIO3 +H2O(R: dispropr.) 16. NH3 + Ca(OH)2 (M)
17. Na2BeO2 + H2O (M) 18. CH4 + Be(OH)2 (M)
19. Ca(HCO3)2 (M) 20. CaCO3 + CO2 + H2O (M)
21. Mg(OH)2 + H2 (R) 22. CaSO4 + HF (M)
23. CH4 + Al(OH)3 24. KBO2 + H2O (M)
25. Si(OH)4 + HCl (M) 26. H3BO3 + HF (M)
27. B2O3 + H2O (M) 28. B + K2O (R: displacement)
29. K2HPO3(M) 30. AlN + CO(R)
31. [Ag(NH3)2]Cl(M: complexation)
32. K2ZnO2 + NH3 +H2O(M: Zn: 0 to +2; N:+5 to -3)
33. No reaction(Al is above Fe) 34. Al2(SO4)3 + SO2 + H2O (R)
35. Al4C3 +CO(R: disproprl) 36. CO2 + NO2 + H2O (R) 37. 2CO (R) 38. Pb + CO (R)
38. CO + CO2 + H2O (R:C:+3 to +2; C:+3 to +4: Dehydration)
40. CO + H2O (M: Dehydration) 41. Fe(CO)5 (M: Complexation)
42. Fe + CO2(R) 43. I2 + CO2(R)
44. KHCO3(M) 45. K2SiO3 + H2 (R) 46. MgO + Si(R)
47. Sn(NO3)2 + NH4NO3 + H2O(R) 48. 2SnCl2 (R)
49. FeCl2 + SnCl4 (R) 49. K2SnO2 + H2O (M)
50. PbCl2 +Cl2 + H2O(R)
Dr. S. S. Tripathy