AP Unit 2 Unit 2 - Molecular and Ionic Compound Structure and Properties

2.1 Types of Chemical Bonds 2.2 Intramolecular Force and Potential Energy 2.3 Structure of Ionic Solids 2.4 Structure of Metals and Alloys 2.5 Lewis Diagrams 2.6 Resonance and Formal Charge 2.7 VSEPR and Bond Hybridisation ISPS Chemistry Aug 2020 page 1 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 This logo shows it is a Topic Question - it should only require knowledge included in this Topic and it should be giving practice in the Science Practice associated with this Topic.

ISPS Chemistry Aug 2020 page 2 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.1 Types of Chemical Bonds

ISPS Chemistry Aug 2020 page 3 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2

Electronegativity Values The key to understanding why different types of bonds exist is an understanding of the importance of - the ability to attract shared .

In essence, our 4 different labels for bonding (metallic, ionic, polar covalent & pure covalent) are really just different ways in which electrons can be shared.

ISPS Chemistry Aug 2020 page 4 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 4 Main Types of Bonding All bonding is based on the attractions (and repulsions) that exist between charged particles. The strength of bonding is governed by Coulombs Law:

F q1 q2

2 r Metallic Bonding - minimal (but equal) sharing ∝ - with similar low (metals) - leads to delocalised electrons

- Fattraction results from multiple attractions between positive cores (temporary cations) and delocalised electrons

Pure Covalent Bonding - strong equal sharing - atoms with very similar high electronegativities (non-metals) - electrons localised midway between both atoms

- Fattraction results from mutual attractions between two positive nuclei and the equally shared pair of electrons

Polar Covalent Bonding - unequal sharing - atoms with different high electronegativities (non-metals) - electrons localised closer to of higher electronegativity

- Fattraction results from uneven attractions between two positive nuclei and the unequally shared pair of electrons

Ionic Bonding - very unequal sharing - atoms with very different electronegativities (metal/non-metal) - electrons transferred to atom of higher electronegativity

- Fattraction results from multiple attractions between positive cations and negative anions

Metallic bond: electrons shared equally but delocalised

The crossover between highly polar cvalent and ionic can be diffcult: if non-metal—non-metal then best to assume highy polar covalent and if metal—non-metal then asssume ionic. ISPS Chemistry Aug 2020 page 5 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Properties Related to Bonding The properties of a substance will help determine the type of bonding but care is needed as different structures (e.g. molecular or network) and the presence of other attractions (particlarly between ) will also have a mojor effect on properties. In general,

Metallic Bonding - delocalised electrons mean they are good conductors of both heat and electricity and can reflect light (shiny).

- multiple attractions mean that they often have high MPts & high BPts - most are solids at room temperature - delocalised electrons allow bonds to break and reform easily so metals are ductile and malleable

Covalent Bonding - localised electrons mean they are poor conductors of both heat and electricity and tend to absorb light (matt)

- multiple attractions in networks result in very high MPts & BPts - solids at room temperature - localised attractions in molecules result in low MPts & BPts - solids, or gases at room temperature - localised electrons make it difficult for bonds to reform easily so tend to be soft or brittle

Ionic Bonding - localised electrons mean they are poor conductors of both heat and electricity when solid - but when are free to move (when molten or in ), then they become good conductors

- multiple attractions in networks result in very high MPts & BPts - solids at room temperature - directional nature make it difficult for ionic bonds to reform easily so tend to be brittle

During Unit 3, you will take a more detailed look at the connection between properties and types of bonding - in particular, the wide variety of properies found in covalent substances due to the existence of attractions between molecules (intermolecular) which can be more important than the attractions found within molecules (intramolecular). ISPS Chemistry Aug 2020 page 6 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.1 Practice Problems 1. On the basis of the information opposite, which of the following arranges the binary compounds in order of increasing bond polarity?

A CH < SiCl < SF B CH < SF < SiCl O 4 4 4 4 4 4

C SF4 < CH4 < SiCl4 D SiCl4 < SF4 < CH4

2. Which of the following has the bonds arranged in order of decreasing polarity? OA H−F > N−F > F−F B H−I > H−Br > H−F C O−N > O−S > O−Te D Sb−I > Sb−Te > Sb−Cl

3. Which of the following scientific claims about the bond in the molecular compound is most likely to be true?

A There is a partial negative charge on the H atom.

B Electrons are shared equally between the and atoms.

C The bond is extremely weak OD The bond is highly polar 4. Which of the following compounds contains both ionic and covalent bonds?

A SO B C H OH C MgF D H S E NH Cl 3 2 5 2 2 O 4 5. Two pure elements react to form a compound. One element is an metal, X, and the other element is a , Z. Which of the following is the most valid scientific claim that can be made about the compound?

A It has the formula XZ2. B It does not dissolve in . OC It contains ionic bonds. D It contains covalent bonds.

ISPS Chemistry Aug 2020 page 7 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 6. The elements C and Se have the same electronegativity value, 2.55. Which of the following claims about the compound that forms from C and Se is most likely to be true?

A The carbon-to-selenium bond is unstable. OB The carbon-to-selenium bond is nonpolar covalent. C The compound has the empirical formula CSe .

D A of the compound will have a partial negative charge on the carbon atom.

7. Of the following compounds, which is the most ionic?

A SiCl B BrCl C PCl D Cl O E CaCl 4 3 2 O 2 8. Of the following single bonds, which is the LEAST polar? A N—H B H—F OC O—F D I—F E O—H

ISPS Chemistry Aug 2020 page 8 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.2 Intramolecular Force & Potential Energy

ISPS Chemistry Aug 2020 page 9 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Covalent Bonding & Potential Energy All bonding is based on the attractions (and repulsions) that exist between charged particles. The strength of bonding is governed by Coulombs Law:

F q1 q2

r2 ∝ For simple covalent molecules such

as H2, the balance between attractive and repulsive forces is at an optimum when the internuculear distance, r, is 74 pm (74 x 10-12 m). This is equivalent to:- covalent radius = 37 pm H—H bond length = 74 pm

The stronger the Force of attraction, F , the more energy will be released (exothermic) when the H—H bond forms (or more energy will be needed (endothermic) to break the H—H bond). This is called Bond Energy and, for H2 : Bond Energy = 432 kJ mol-1

For atoms that have similar values of q1 and q2 , the size of the atoms will be the main factor that determines the Force of attraction, F, and hence the bond energy. The in Group 17 all have the same effective nuclear charge of 7+ due to the screening effect of their inner shells of electrons. Their outer shell each contain 7 electrons, 7-

Cl2 - 3 shells, covalent radius = 100 pm - bond length = 199 pm ( r = 199) - bond energy = 243 kJ mol-1

Br2 - 4 shells, covalent radius = 114 pm - bond length = 228 pm ( r = 228) - bond energy = 193 kJ mol-1

I2 - 5 shells, covalent radius = 133 pm - bond length = 266 pm ( r = 266) - bond energy = 151 kJ mol-1

F ⬃ as r ⬀

ISPS Chemistry Aug 2020 page 10 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 The Hydrogenhalides will also have

similar values of q1 and q2 . All the halogens have the same effective nuclear charge of 7+ due to the screening effect of their inner shells of electrons. Each has an effective charge of 1+. F ⬃ as r ⬀

Whilst it could be tempting to argue the other way round - that increasing attraction results in a decrease in bond length - Colombs Law only provides us with the ability to describe the Force of attraction in terms of charge and distance. Similarly with our next example - multiple Bonds. Carbon has the ability to form single bonds, C—C in which one pair of electrons are shared. For example, ethane

Carbon can also form double bonds, C = C in which two pairs of electrons are shared. For example, ethene

Carbon can also form triple bonds, C C in which three pairs of electrons are shared. For example, ethyne ≡

Again, the shorter the bond the higher the bond energy. F ⬃ as r ⬀

ISPS Chemistry Aug 2020 page 11 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Ionic Bonding & Potential Energy All bonding is based on the attractions (and repulsions) that exist between charged particles. The strength of bonding is governed by Coulombs Law:

F q1 q2

2 r ∝ + For isolated ions such as Na (g) and — Cl (g), the balance between attractive and repulsive forces is at an optimum when the internuculear distance, r, is 236 pm (236 x 10-12 m). This is equivalent (but not equal) to:-

Naionic radius + Clionic radius 102 + 181 = 283

The reason why internuclear distance ( r = 236 pm) is not the same as the sum of the ionic radia (as it was for covalent radia) is due to the different ways in which covalent radia and ionic radia are determined. Covalent radia are measured while atoms are overlapping so the radia meaured (half the distance between the two nuclei) is reduced. The determination of ionic radia does not allow for overlap so will be 'full size'. In the Potential Energy diagram above, there will be a degree of overlap so internuclear distance is less than sum of the ionic radia. However, when comparing ionic compounds, the sum of the 2 ionic radia can be used as an indiction of the value of r in F = q1 q2 / r . There is also no 'single bond' in ionic compounds so the minimum potential energy (-589 kJ mol-1) is not a bond energy. Since each is forming several attractions with several neighbouring ions of opposite charge, a better measure of the strength of the ionic bond is to look ,at Point or the Lattice Enthalpy.

Lattice Sum Ionic MPt Compound Formula Enthalpy Radii (pm) (°C) (kJ mol-1) Na+ Cl— 283 801 787 sodium Na+ F— 235 996 923 2+ 2— Ca O 240 2898 3401 F ⬀ as q ⬀ strontium oxide Sr2+ O2— 258 2531 3369 barium oxide Ba2+ O2— 275 1972 3054 F ⬃ as r ⬀ ISPS Chemistry Aug 2020 page 12 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.2 Practice Problems 1. Which of the following can be inferred from the diagram opposite that shows the dependence of potential energy on the internuclear distance between two atoms? A The atoms form a bond with a bond length of 25 pm . B The atoms form a bond with O a bond length of 75 pm. C The net force between the atoms is attractive at 25 pm. D The net force between the atoms is attractive at 75 pm

2. Which of the following graphs correctly shows the relationship between potential energy and internuclear separation for two hydrogen atoms? A B

C OD

E

ISPS Chemistry Aug 2020 page 13 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 3. The potential energy as a function of internuclear distance for three

diatomic molecules, X2, Y2, and Z2, is shown in the graph opposite. Based on the data in the graph, which of the following correctly identifies the diatomic molecules,

X2, Y2, and Z2?

X2 Y2 Z2 X2 Y2 Z2 A H N O B H O N O 2 2 2 2 2 2 C N2 O2 H2 D O2 H2 N2

4. The potential energy of a system of two atoms as a function of their internuclear distance is shown in the diagram opposite. Which of the following is true regarding the forces between the atoms when their internuclear distance is x?

A The attractive and repulsive forces are O balanced, so the atoms will maintain an average internuclear distance x. B There is a net repulsive force pushing the atoms apart, so the atoms will move further apart C There is a net attractive force pulling the atoms together, so the atoms will move closer together. D It cannot be determined whether the forces between atoms are balanced, attractive, or repulsive, because the diagram shows only the potential energy.

ISPS Chemistry Aug 2020 page 14 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 5. Based on the data in the tables opposite, which of the following statements provides the best prediction for the of NaCl ?

A NaCl will have a lower boiling point than NaF because the coulombic attractions are O weaker in NaCl than in NaF . B NaCl will have a boiling point between that of NaF and MgO because the covalent character of the bonds in NaCl is intermediate between that of MgO and NaF. C NaCl will have a higher boiling point than MgO because the ions are spaced farther apart in NaCl . D NaCl will have a higher boiling point than MgO because the energy required to transfer electrons from the anion to the cation is larger in NaCl than in MgO .

6. The of a is related to the energy required to separate the ions. For which of the following pairs of ions is the energy that is required to separate the ions largest? (Assume that the distance between the ions in each pair is equal to the sum of the ionic radii.)

+ − + −

A Na (g) and Cl (g) B Cs (g) and Br (g) C Mg2+ and O2− D Ca2+ and O2− O (g) (g) (g) (g)

7. The of MgO is higher than that of NaF. Explanations for this observation include which of the following? I. Mg2+ is more positively charged than Na+. II. O2- is more negatively charged than F-. III. The O2- ion is smaller than the F- ion.

A I only OB I and II only C I and III only D II and III only D I, II and III

ISPS Chemistry Aug 2020 page 15 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.3 Structure of Ionic Solids

Revision of Some Properties - localised electrons mean they are poor conductors of both heat and electricity when solid - but when ions are free to move (when molten or in solution), then they become good conductors

- multiple attractions in networks result in very high MPts & BPts - solids at room temperature increase as charge on ion increases increase as size of ion decreases

- directional nature make it difficult for ionic bonds to reform easily so tend to be brittle

ISPS Chemistry Aug 2020 page 16 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Forming Ionic Compounds A lot of things happen during the formation of a typical ionic compound such as NaCl or CsF. Typically, the metal is a solid and it will be necessary to break apart the metal atoms, breaking metallic bonds in the process:

e.g. Na(s) Na(g) (sublimation or atomisation) ⇨ Typically, the non-metal is a molecule and covalent bonds will need to be broken:

e.g. Cl—Cl(g) 2Cl(g) (bond energy or ) ⇨ The metal atoms will then need to lose (s):

+ - e.g. Na(g) Na (g) + e (ionisation) ⇨ - - While the non-metal atoms gain electron(s): e.g. Cl(g) + e Cl (g) (electron affinity) ⇨ Notice that most of the steps involved are endothermic so our gaseous ions are very high energy - very unstable. The stability all comes from the final step, when the gaseous ions begin to attract/repel each other and come together to form the lattice or network.

+ - e.g. Na (g) + Cl (g NaCl(s)

(lattice energy) ⇨

ISPS Chemistry Aug 2020 page 17 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Ionic Lattice Structures The final arrangement of the ions will, as always, be governed partly by the need to maximise attractions but mainly by the need to minimise repulsions. Obvious deviations such as and will not happen and the normal arrangement will be to have alternating cations and anions in all directions;

The exact detailed arrangements of ions does not need to be known but the role of ionic radii in determining structure should be understood. In general, cations are smaller than anions but it is the relative sizes that will determine the structure. Structure 1: as seen in NaCl,

Na+ - 116 pm Na+ / Cl— = 116/167 = 0.69 Cl— - 167 pm In this structure the relatively small size of the Na+ ion means that a maximum of 6 Cl— ions can be attracted.

Structure 2: as seen in CsCl,

Cs+ - 181 pm Cs+ / Cl— = 181/167 = 1.08 Cl— - 167 pm In this structure the relatively large size of the Cs+ ion means that a maximum of 8 Cl— ions can be attracted.

As previously shown, ionic radii play a significant role in determining the strength of the ionic bond as shown by melting point and lattice energy. Difference in structure could make comparisons more difficult but the vast majority of simple ionic substances will have the sodium chloride structure. ISPS Chemistry Aug 2020 page 18 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.3 Practice Problems 1. Which of the following correctly indicates whether the solid represented by the particulate model shown opposite conducts electricity and explains why or why not? A It conducts electricity because it is made of positive and negative ions.

B It conducts electricity because it is made of particles of different sizes. OC It does not conduct electricity because its ions cannot move freely within the solid. D It does not conduct electricity because there are small spaces between the particles.

2. The particle-level diagram opposite represents the structure of solid KF.

Although the molar mass of KCl is greater than that of KF, the density of KCl is actually less than that of KF.

Which of the following representations of the structure of KCl best helps to explain this phenomenon?

A B O

C D

ISPS Chemistry Aug 2020 page 19 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 3.

The energy required to dissociate an ionic solid into gaseous ions (lattice energy) for the

compounds NaF and MgF2 is shown in the table above. On the basis of Coulomb’s law, which of the following best helps to explain the large

difference between the lattice energies of NaF and MgF2 ?

A The of MgF2 is less than that of NaF . B The electronegativity of Mg is greater than that of Na . C The mass of the Mg cation is greater than that of the Na cation. OD The charge of the Mg cation is larger than that of the Na cation.

4. Of the following diagrams, which best represents the structure of solid KF ?

A B O

C D

ISPS Chemistry Aug 2020 page 20 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 5. Which of the following diagrams best illustrates how a displacement in an ionic results in cleavage and ? OA before displacement after displacement

B before displacement after displacement

C before displacement after displacement

D before displacement after displacement

ISPS Chemistry Aug 2020 page 21 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.4 Structure of Metals and Alloys

Metallic Lattices Metallic lattices (elements and alloys) benefit from many strong interactions at short distances. 2 F = q1 q2 / r Melting points can be used to compare the strengths of metallic bonds. For example, mercury is a at room temperature, and caesium melts at 28.4°C, whereas tungsten melts at 3680°C.

ISPS Chemistry Aug 2020 page 22 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Metals can form alloys so easily because replacing one metal atom with another metal atom rarely disrupts the strength of the metallic bonding too much. For example, an alloy of sodium and potsssium can be made, but the alloy formed is a liquid at room temperature showing that the metallic bond is slightly weaker. Alloys can be divided into two main categories: substitutional alloys and interstitial alloys. These alloys are both homogeneous mixtures in which components are dispersed randomly but uniformly. The type of alloy formed is largely determined by relative atomic sizes.

Substitutional alloys are formed when the two metallic components have similar atomic radii and similar chemical-bonding characteristics. Brass is an example of a substitutional alloy. Zinc and copper atoms are of very similar size so the original structure is maintained, though strengthened. This alloy retains properties such as malleability and as metallic bonding is easily re-established.

Interstitial alloys are formed when the atoms added are much smaller than the main metal stoms. Typically, the interstitial element is a non-metal that makes covalent bonds to the neighboring metal atoms.

Steel is an interstitial alloy with much smaller carbon atoms in the spaces between the iron atoms. The presence of the carbon atoms enhances the structure so steel is harder than the original iron, but it will also be less malleable and less ductile as a result.

The wide variety of alloys makes it difficult to make accurate predictions about properties, but, in general, alloys will be poorer conductors, less malleable, less ductile but stronger/harder. Also, in general, smaller atoms that are able to form interstitial alloys will have a more disruptive effect on metallic bonding, making it harder to displace atoms resulting in stronger alloys.

ISPS Chemistry Aug 2020 page 23 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.4 Practice Problems 1. Copper atoms and zinc atoms have the same atomic radius, 135 picometers. Based on this information, which of the following diagrams best represents an alloy containing only copper and zinc atoms?

A B O

C D

2. A particle-level diagram of a metallic element is shown opposite. Typically, metals are both malleable and ductile. The best explanation for these properties is that the electrons involved in bonding among metal atoms are A unequally shared and form nondirectional bonds B unequally shared and form highly directional bonds OC equally shared and form nondirectional bonds D equally shared and form highly directional bonds 3. Steel is an alloy containing Fe atoms and C atoms. Which of the following diagrams best represents the particle-level structure of steel?

A B O

C D

ISPS Chemistry Aug 2020 page 24 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 4.

To make Au stronger and harder, it is often alloyed with other metals, such as Cu and Ag. Consider two alloys, one of Au and Cu and one of Au and Ag, each with the same mole ratio of Au. If the Au/Cu alloy is harder than the Au/Ag alloy, then which of the following is the best explanation based on the table above.

A Cu has two common oxidation states, but Ag has only one. B Cu has a higher melting point than Au has, but Ag has a lower melting point than Au has. C Cu atoms are smaller than Ag atoms, thus the interfere more with the displacement O of atoms in the alloy. D Cu atoms are less polarizable than Au or Ag atoms, thus Cu has weker interparticle forces.

5. Which of the following diagrams best depicts an alloy of Ni and B?

A B O

C D

ISPS Chemistry Aug 2020 page 25 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 11. Steel is an alloy consisting of Fe with a small amount of C. Elemental Cr can be added to steel to make the steel less likely to rust; Cr atoms react with oxygen in the air to form a nonreactive layer of chromium oxide on the surface of the steel, preventing the oxidation of underlying Fe atoms. A sample of steel-chromium alloy contains 15 percent Cr by mass. Which of the following diagrams best shows a particle-level view of a surface section and an interior section of the alloy represented below at the left? (The atomic radii of the atoms involved are given in the table below at the right.)

A

B O

C

D

ISPS Chemistry Aug 2020 page 26 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.1 - 2.4 Quick Check FRQ - some questions were used in Unit 1 1. Answer the following questions related to the chemical bonding in substances containing halogens.

a) What type of chemical bond is present in the molecules of halogens such as Cl2? The response indicates that the bond is covalent. (No explanation is required, however, the bond is covalent because the two atoms have the same electronegativity and will form a nonpolar .)

b) All of the halogens have the same molecular formula. Which molecule will have the longest bond length? Justify your choice in terms of atomic structure.

The response indicates Iodine or I2 . The response gives an explanation such as the following.

I2 has the longest bond length because the radius of the atom is greater than the radii of the other halogen atoms. Thus, the distance between the nuclei of atoms

in I2 is greater than it is in smaller halogens.

c) Cl2 reacts with the element Sr to form an ionic compound. Based on periodic

properties, identify a molecule, X2 , that is likely to to react with Sr in a way similar

to how Cl2 reacts with Sr. Justify your choice. The response meets both of the criteria below:

F2 , Br2 , I2 or At2 is written.

the element chosen is in the same group or family as Cl2 . d) A graph of potential energy versus internuclear distance for two Cl atoms is given below. On the same graph, carefully sketch a curve that corresponds to potential energy versus internuclear distance for two Br atoms. The sketched curve meets both ( 2 marks) of the criteria below: The curve has its minimum value to the right of where the given curve has its minimum value(because the length of the bond is greater than that of the bond).

The minimum value of the curve is higher than (i.e., above) the minimum value of the given curve (because the strength of the bond is less than that of the bond).

ISPS Chemistry Aug 2020 page 27 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2. White gold is a common alloy of gold and palladium that is often used in jewelry. The atomic radii of the metals are given in the table opposite A particular ring is made from an alloy that is 75 mole percent gold and 25 mole percent palladium. Using the box below, draw a particle-level diagram of the solid alloy consisting of 12 atoms with a representative proportion of atom types. Your diagram should clearly indicate whether the alloy is interstitial or substitutional. Use empty circles for gold and shaded circles for palladium.

1 point is earned for the correct number of each type of circle. 1 point is earned for showing an acceptable arrangement (i.e., regular array of circles of about the same size). Diagram should show 12 circles in a regular array, 9 empty and 3 shaded.

ISPS Chemistry Aug 2020 page 28 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.5 Lewis Diagrams

Molecular Models When it comes to representing covalent molecules we have a number of different ways of 'modelling' their structures:

CH4

molecular lewis structural molecular formula diagram formula orbital model

perspective ball-and-stick space-filling electrostatic formula model model potential map

Many of these models will already be familiar to you. In the lessons that follow, we will be progressing through Lewis Diagrams to Molecular Orbitals (Hybridisation) to allow us to better understand the shapes of molecules and will make good use of Perspective Formulae to show these 3D shapes on 2D paper. Underpinning all of this will be our Valence Shell Electron Pair Repulsion (VSEPR) theory. ISPS Chemistry Aug 2020 page 29 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Lewis Symbols Lewis symbols are a natural progression from our 'electrons in boxes' approach used earlier.

However, there is one very significant difference. These boxes represent atomic orbitals and, in the valence shell, there are two types (s and p ) at two slightly different energies. As will be explained in more detail in a later lesson, when reacting with other atoms these atomic orbitals 'mix and merge' (hybridise) to form 4 identical molecular orbitals. 'Atomic' Carbon

'Molecular' Carbon

So, in Lewis Symbols, we place our electrons N, E, S & W ( in any order) around the symbol for the atom, first singly and then pairing up.

Though we will be concentrating mainly on covalent molecules, Lewis symbols can also be used to represent ionic reactions:

Notice, that the ions end up with filled valence shells like the nearest Noble Gas - the 'stable octet'. ISPS Chemistry Aug 2020 page 30 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Lewis Structures - the Octet Rule A Lewis structure is a representation of covalent bonding in which shared electron pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms. Only valence electrons are shown in a Lewis structure. Though (as always) there are exceptions - a correct Lewis structure should result in a stable octet (or stable duet in the case of hydrogen) for each atom in the molecule. This often referred to as the Octet Rule - An atom other than hydrogen tends to form bonds until it is surrounded by eight valence electrons. Pairs of electrons shared between two atoms (bonding pairs) are usually represented by a line. Pairs of electrons that belong to a single atom (lone pairs) are represented by a pair of dots.

On occassions, a lone pair can then be used to form a bond. The hydrogen has lost an electron (so H+) to the chlorine. It then forms a covalent bond (shares a pair of electrons) with the nitrogen. This type of covalent bond (both shared electrons donated by one atom) is often called a Dative or Coordinate covalent bond. Notice that the positive charge that was part of the (H+) is + now shared between all 5 atoms in the ion (NH4 ). The ammonium ion is both ionic and covalent. Many compounds are held together by multiple bonds, that is, bonds formed when two atoms share two or more pairs of electrons. If two atoms share two pairs of electrons, the covalent bond is called a double bond. Double bonds are found in molecules of carbon dioxide (CO2) and ethylene (ethene - C2H4). A triple bond arises when two atoms share three pairs of electrons, as in the

nitrogen molecule (N2):

The acetylene molecule (ethyne - C2H2) also contains a triple bond, in this case between two carbon atoms: ISPS Chemistry Aug 2020 page 31 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Applying the Octet Rule - Rules for Drawing Simple Lewis Structures Drawing a Lewis structure is a skill that is easier to show than to describe on paper. Written rules and worked examples can take up a lot of pages and can make simple things look more difficult than they are. Therefore, I'm going to make use of some excellent videos and would encourage you to watch each one before working through each section of Notes.

Lewis Structures for Polyatomic Ions Drawing Lewis structures for polyatomic ions - covalent molecules with charges - is very similar except that the number of valence electrons increases for anions and decreases for cations. ISPS Chemistry Aug 2020 page 32 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2

.. —— — —.. .. —— ——...... — — .. — —

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Exceptions to the Octet Rule Hypervalent molecules - As already mentioned, atoms convert atomic orbitals into molecular orbitals when bonding together. Smaller atoms are usually limited to using just their s and p orbitals to create 4 hybrid molecular orbitals - 4 pairs of electrons - stable octet. Larger atoms also have the option of using some of the empty d orbitals. This can produce 5, 6, 7, 8 or more orbitals resulting in central atoms that will often have electrons in excess of a stable octet.

ISPS Chemistry Aug 2020 page 33 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 ..

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— ...... — ...... phosphorus trichloride sulfur tetrachloride sulfur hexachloride

PCl3 PCl5 SF4 SCl6 number of V.E. = 26 number of V.E. = 40 number of V.E. = 34 number of V.E. = 48 number of B.E. = 6 number of B.E. = 10 number of B.E. = 8 number of B.E. = 12 number of lone pairs = number of lone pairs = number of lone pairs = number of lone pairs = 20 / 2 = 10 30 / 2 = 15 26 / 2 = 13 36 / 2 = 18

stable octet? OY / N stable octet? Y / O N stable octet? Y / O N stable octet? Y / O N

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— — ...... — ...... xenon tetrachloride xenon hexafluoride xenon pentafluoride ion thionyl chloride + XeF4 XeF6 XeF5 SOCl2 number of V.E. = 36 number of V.E. = 50 number of V.E. = 42 number of V.E. = 26 number of B.E. = 8 number of B.E. = 12 number of B.E. = 10 number of B.E. = 8 number of lone pairs = number of lone pairs = number of lone pairs = number of lone pairs = 28 / 2 = 14 38 / 2 = 19 32 / 2 = 16 18 / 2 = 9 stable octet? Y / O N stable octet? Y / O N stable octet? Y / O N stable octet? Y / O N Electron Deficient molecules -

Generally, these are molecules with central atoms from groups 2 and 3 and outer atoms that are hydrogen or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH2 , and boron trifluoride, BF3 , the beryllium and boron atoms each have only four and six electrons, respectively - less than the stable octet expected. An atom like the boron atom in BF , which does not have 8 electrons, is very reactive. It readily combines with a molecule containing an atom

with a lone pair of electrons. For example, NH3 reacts with BF because the lone pair on nitrogen can be shared with the boron atom:

ISPS Chemistry Aug 2020 page 34 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Organic molecules -

The Lewis structures for many organic molecules can be written quickly and easily as they often just require the drawing of lines to represent pairs of bonding electrons:

However, the presence of halogens, oxygen or nitrogen atoms will require lone pairs to be added to the traditional structural formula:

Since all these atoms (except hydrogen) will normally form a stable octet, it can be easier to simply draw a structural formula as normal and then add the lone pairs needed to achive the stable octet:

1 bond 3 lone pairs

2 bonds 2 lone pairs

3 bonds 1 lone pair

4 bonds 0 lone pairs

ISPS Chemistry Aug 2020 page 35 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.5 Practice Problems 1. Which of the following has as a central atom with less than an octet of electrons?

A H2O B NH3 C BH3 D CH4 E SiH4 O 2. Which of the following has two lone pairs of electrons?

A H O B NH C BH D CH E SiH O 2 3 3 4 4

3. Which of the following is isomeric with CH3CH2CHO ?

A CH CH CH CH B CH CH CH OH C CH COCH 3 2 2 3 3 2 2 O 3 3 D CH3COOH E CH3CH2CH2NH2

4. The electron-dot structure (Lewis structure) for which of the following molecules would have two unshared pairs of electrons on the central atom?

A H S B NH C CH D HCN E CO O 2 3 4 2 5. Which of the following molecules contains only single bonds?

A CH COOH B CH CH COOCH C C H 3 3 2 3 O 2 6 D C6H6 E HCN

6. Which of the following is an isomer of CH3OCH3 ?

A CH CH B CH COOH C CH CH OH 3 3 3 O 3 2 D CH3CH2CH3 E CH3CH2OCH2CH3

ISPS Chemistry Aug 2020 page 36 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.6 Resonance and Formal Charge

ISPS Chemistry Aug 2020 page 37 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Limitations with the Lewis Model Though extremely useful, there are a few limitations with Lewis Structures: They tend to present a 'black & white' versions of a molecule whereas the reality can sometimes involve 'shades of grey': ①

For example, Lewis structures for ozone, O3 , would show one single and one double bond, whereas the reality is 2 identical bonds that are 'in between' - shorter than single bonds but longer than double bonds. Similarly, Lewis structures show negative charges on either of the outside oxygens whereas the reality is that the charge is shared equally over both outside oxygens.

Less serious is the fact that they encourage a 2D view (N,S,E & W) of molecules, whereas we often need to consider the 3D shape (tetrahedral) of a molecule. ②

CH4

Even less serious is the fact that they can struggle with odd numbers of electrons that don't fit into our electron pairs/stable octet Model. ③

For example, the free radical NO2• has 17 valence electrons.

.. Putting bonds between the atoms uses up 4 of these electrons. O.. — N — O

..

Allocating lone pairs to the oxygen atoms uses up 12 electrons. .. O — N — O ......

..

Leaving only 1 electron to be allocated to the nitrogen atom. .. ..O — N — ..O

. The oxygen atoms have a stable octet but nitrogen only has .. .. 5 electrons. We can improve this by creating a double bond. .. ..O — N = O .. Again, the reality can be somewhat different:

ISPS Chemistry Aug 2020 page 38 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Assigning Formal Charges

In some of the previous examples, charges have been included into the Lewis Diagrams. Having

. applied our rules to NO2•, we ended up with: ...... O — N = O.. If we formally share out the electrons (including those in inner shells) and compare with the

number of protons, we see that some of our atoms are no longer balanced and will, as a result,

. have overall charges associated with them: ...... O — N = O inner shell electrons belong to only that atom .. .. lone pairs belong to only that atom 8 p 7 p 8 p single electrons belong to only that atom 9 e 6 e 8 e each atom owns one electron from each bonding pair 1- 1+ 0

In a similar way, we can identify the location of charge within polyatomic ions: 8 p 8 p 8 e 0 8 e 0 Most polyatomic ions involve oxygen atoms so it can be useful to note that ...... single bonded oxygen atoms will often —— — —

7 p 8 p ...... 8 p

.. — carry a negative charge while double — 7 e 0 —— .. 9 e 1- .. .. 9 e 1-

.. — .. —— ...... bonded oxygens are usually neutral. 8 p 8 p 8 p 9 e 1- 8 e 0 8 e 0

3- Often we can predict the structure of an ion, such as the phosphate ion, PO4 , using these observations rather than following the rules to a formal Lewis Structure. Resonance Structures In many of our previous examples, several 'identical but different' structures could have been drawn depending on which oxygen atom we decided to allocate lone pairs and which oxygen atoms would form the double bond. These equivalent Resonance Structures are usually combined to produce more realistic structures like those shown on the right. An analogy would be to mix a red pigment and a blue pigment together in a paint. The molecules of red and blue would remain, but the paint would not alternate between red and blue - instead we would observe a purple colour. To better understand these Resonance Structures we need the Orbital Model and to learn more about Hybridisation. ISPS Chemistry Aug 2020 page 39 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Benzene is a common organic that was previously used in ; it is no longer used for this purpose, however, because it is now known to be a carcinogen.

The benzene molecule (C6H6) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Using our rules we can arrive at 2 equivalent resonance structures:

Both structures involve alternating single bonds (longer,weaker) and double bonds (shorter, stronger). In reality, all the bonds are the same length and of equal strength. They are between single and double bonds - intermediate bonds. Lewis Structures cannot really explain how the final structure is formed. We need to consider the orbitals involved:

As previously mentioned, the atomic s and p orbitals will mix and merge to form four sp3 hybrid molecular orbitals when atoms such as carbon form molecules. In a benzene molecule, each carbon is only going to bond to 3 atoms so three sp2 hybrid molecular orbitals form, leaving one of the sp2 sp2 sp2 original p orbitals intact.

Thethree sp2 orbitals will minimise repulsions by taking up positions 120° away from each other. The p orbital will lie perpendicular to these orbitals.

This lead to ahexagonal ring of carbon atoms with sp2 orbitals overlapping to share electrons as normal - these are called sigma (σ) bonds. More σ-bonds form between C and H atoms.

Thep orbitals can overlap sideways to form three weaker π-bonds. However, instead of overlapping to form 3 double bonds, they all merge together to provide extra π-bonds equally between all the C atoms. ISPS Chemistry Aug 2020 page 40 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2

The first bond that forms between atoms is always a sigma (σ) bond. The shared electrons are directly between the two nuclei, maximising attractions and making the bond stronger. Second or third bonds are weaker π-bonds, due to the fact that the shared electrons are off to the side.

Ozone (O3) and sulfur dioxide (SO2) also have structures that are best understood by considering orbitals along side the Lewis structure:

2- Similarly for the ion, CO3 , and the nitrate ion:

The evidence for these resonance hybrid structures is usually some of the following: bonds are intermediate in length between single bond and double bond ① bonds are intermediate in strength between single bond and double bond ② molecules do not display the polarity/charges predicted by individual structures ③ molecules, such as benzene, do not undergo the addition reactions expected of ④ double bonds.

ISPS Chemistry Aug 2020 page 41 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.6 Practice Problems 1.

The diagram above shows two resonance structures for a molecule of 6C H6. The phenomenon shown in the diagram best supports which of the following claims about

the bonding in C6H6 ?

A In the C6H6 molecule, all the bonds between the carbon atoms have the same O length.

B Because of variable bonding between its carbon atoms, C6H6 is a good conductor of electricity.

C The bonds between carbon atoms in C6H6 are unstable, and the compound decomposes quickly.

D The C6H6 molecule contains three single bonds between carbon atoms and three double bonds between carbon atoms.

2. Which of the following Lewis diagrams best represents the bonding in the N2O molecule, considering formal charges? A B O C D

3. For which of the following molecules are reso­nance structures necessary to describe the bond­ing satisfactorily?­ A H S B SO C CO D OF E PF 2 O 2 2 2 3

ISPS Chemistry Aug 2020 page 42 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 4.

− Which of the following statements, if true, would support the claim that the NO3 ion, represented above, has three resonance structures? A The NO − ion is not a polar species. O 3 B The oxygen-to-nitrogen-to-oxygen bond angles are 90°

− C One of the bonds in NO3 is longer than the other two.

− D One of the bonds in NO3 is shorter than the other two.

5. Which of the following molecules has an angular (bent) geometry that is commonly represented as a resonance hybrid of two or more electron-dot structures? A CO B O C CH D BeF E OF 2 O 3 4 2 2

6. Resonance is most commonly used to describe the bonding in molecules of which of the following? A CO B O C H O D CH E SF 2 O 3 2 4 6

ISPS Chemistry Aug 2020 page 43 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.7 VSEPR and Bond Hybridisation

ISPS Chemistry Aug 2020 page 44 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Molecular Geometry (shapes) One of the most important things we gain from Lewis Structures and Orbital Models is the ability to understand and predict Molecular Geometry (Shapes). From shapes we can predict some attributes like polarity which have a big influence on properties.

For example, at first glance we might expect CO2 and SO2 to be very similar molecules, however,

CO2 is a linear molecule (180°), contains two double bonds, and is non-polar, despite the bonds being polar, wheras

SO2 is a bent molecule (120°) with intermediate bonds, and is a polar molecule as the shape allows for a molecule with ends of opposite charge.

To understand or predict molecular geometry (shape) we usually use a combination of ideas that form the Valence Shell Electron Pair Repulsion (VSEPR) theory : identification of the type of hybrid molecular orbitals forming ① allocation of electron pairs between bonding pairs and lone pairs ② consideration of Coulombs Law to ensure minimal repulsion ③ For example, we might expect XeF4 to be a tetrahedral shaped molecule but earlier we worked out the arrangement of electrons in XeF4 using the Rules for drawing Lewis Structures:

...... From this we see that there are .... 6 electron pairs - 4 bonding and —— ...... 2 lone pairs. .. —— ...... This would require 6 orbitals so .. would need to be sp3d2 hybrids. xenon tetrachloride 6 orbitals will arrange themselves octahedrally to minimise repulsions, XeF4 so XeF4 will be based on this shape. number of V.E. = 36 number of B.E. = 8 However, the lone pairs are more number of lone pairs = strongly repulsive so they will occupy 28 / 2 = 14 positions as far away from each other as possible. This leaves the 4 bonding pairs to occupy the remaining 4 positions that forms a square planar shaped molecule. ISPS Chemistry Aug 2020 page 45 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Types of Hybrid Orbitals Most of the atoms that get involved in covalent molecules are from the second period of the Periodic Table - elements such as C , N and O. They will normally make use of the s and p orbitals to form sp, sp2 or sp3 hybrid orbitals. These will then establish the initial arrangement of electron pairs. sp3 hybrid orbitals

4 bonding pairs - molecule has same geometry as the orbitals - tetrahedral Equal repulsions ensure bond angle of 109.5° - maximum distance. 3 bonding pairs - molecule has a pyramidal shape but stronger repulsion of the lone pair reduces bond angle to 107°. 2 bonding pairs - molecule has a bent shape but stronger repulsions of the two lone pairs reduces bond angle to 104.5°. sp2 hybrid orbitals 3 orbitals will take up positions 120° away from each other to minimise repulsions. This forms a trigonal planar orbital geometry. 3 bonding pairs will result in a molecule with the same trigonal planar molecular geometry.

For example, BF3

ISPS Chemistry Aug 2020 page 46 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 The unused empty p orbital can still have a part to play. For example, an electron rich fluoride ion can use one of its lone pairs to form a fourth bond (dative or coordinate bond) with the boron.

trigonal planar tetrahedral

— The resulting BF4 ion will be tetrahedral in shape as there are now 4 bonding pairs and the orbitals will change from sp2 to sp3. Sometimes the formation of an extra dative or coordinate bond does not require a change in hybridisation (orbital geometry) but will result in a change in molecular geometry.

orbital geometry = tetrahedral orbital geometry = tetrahedral molecular geometry = pyramidal molecular geometry = tetrahedral

At other times, a half-filled p orbital can be used to form a π-bond as the second bond in a double bond:

Notice that the orbital geometry (trigonal planar) and the molecular geometry (trigonal planar) are unaffected by the formation of the π-bond - only bonding pairs in σ-bonds and lone pairs determine the orbital geometry. 3 bonding pairs orbital geometry molecular geometry in σ-bonds is trigonal planar is trigonal planar

1 bonding pair the greater repulsion of the lone pairs on in π-bond the O atom pushes the H atoms closer together ISPS Chemistry Aug 2020 page 47 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2 bonding pairs orbital geometry molecular geometry in σ-bonds & is trigonal planar is bent 1 lone pair

1 bonding pair in π-bond sp hybrid orbitals

There are not many examples of atoms that only have 2 valency electrons (group 2 metal) but form covalent bonds but BeCl2 is a good example of where both the orbital geometry and the molecular geometry are linear (180°). In acetylene (ethyne), each carbon only needs to form 2 σ-bonds so forms sp hybrid orbitals and, again both the the orbital geometry and the molecular geometry are linear (180°);

Again, the formation of the second and third π-bonds in the triple bond has no effect on the geometry - only bonding pairs in σ-bonds and lone pairs determine the orbital geometry.

TheLewis diagram for CO2 shows us that the carbon will form 2 σ-bonds so 2 sp hybrid orbitals needed, leaving 2 unused p-orbitals to form π-bonds. Each oxygen will form 1 σ-bond and will have 2 lone pairs so 3 sp2 hybrid orbitals needed leaving 1 unused p-orbitals to form a π-bond.

Thesp hybrid orbitals make the geometry linear and the sideways overlapping of p-orbitals to form the double bonds has no effect.

ISPS Chemistry Aug 2020 page 48 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2

Number Number of Number of Molecular of hybrid Example σ-bonds lone pairs Geometry orbitals

sp

sp2

sp3

ISPS Chemistry Aug 2020 page 49 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Number Number of Number of Molecular of hybrid Example σ-bonds lone pairs Geometry orbitals

sp3d

sp3d2

ISPS Chemistry Aug 2020 page 50 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 Summary of Bond Properties

Bond Lengths - will decrease as we move from single bonds double bonds triple bonds

⇨ ⇨

F2 1.43 Å O2 1.21 Å N2 1.10 Å

will increase as we increase the size (atomic radii) of atoms

F2 1.43 Å Cl2 1.99 Å Br2 2.28 Å Bond Energies - will increase as we move from single bonds double bonds triple bonds

⇨ ⇨

F2 159 kJ O2 498 kJ N2 945 kJ

347 kJ 614 kJ 839 kJ

Bond Angles - will increase as we move from sp3 hybrid sp2 hybrid sp hybrid orbitals

⇨ ⇨

will decrease as the number of lone pairs increases

Bond Dipoles/Molecular Dipoles -

ISPS Chemistry Aug 2020 page 51 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 1.7 Practice Problems 1. Which of the following complete Lewis diagrams represents a molecule containing a bond angle that is closest to 120°

A B C O D

2.

Lewis diagrams of molecules of three different hydrocarbons are shown above. Which of the following claims about the molecules is best supported by the diagrams?

A All the atoms in molecule 1 lie in one plane.

B All the molecules have the same empirical formula.

C The C-C-C bond angle in molecule 2 is close to 180°. OD The strongest carbon-to-carbon bond occurs in molecule 3.

3. In the following diagrams, elements are represented by X and Z, which form molecular compounds with one another. Which diagram represents a molecule that has a bent molecular geometry? A B O C D

4. The geometry of the SO3 molecule is best de­scribed as

OA trigonal planar B trigonal pyramidal C square pyramidal D bent E tetrahedral

ISPS Chemistry Aug 2020 page 52 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 5.

Lewis electron-dot diagrams for CO2 and SO2 are given above. The molecular geometry and polarity of the two substances are

A the same because the molecular formulas are similar

B the same because C and S have similar electronegativity values

C different because the lone pair of electrons on the S atom make it the negative end of a dipole

D different because S has a greater number of electron domains (regions of electron O density) surrounding it than C has

6. Which of the following Lewis electron-dot diagrams represents the molecule that contains the smallest bond angle?

A OB C D

7.

The hybridization of the carbon atoms in the molecule represented above can be described as

2 3 2 2 A sp B sp OC sp D dsp E d sp

8.

What is the hybridization of the carbon atoms in a molecule of ethyne, represented above?

2 3 2 2 OA sp B sp C sp D dsp E d sp ISPS Chemistry Aug 2020 page 53 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 9. Which of the following molecules contains exactly three sigma (σ) bonds and two pi (π) bonds?

A C H B CO CN HC D SO E N O 2 2 2 3 2 10. Of the following molecules, which has the largest dipole moment?

A CO B CO C O D HF E F 2 2 O 2 11. Which of the following has molecules with a pyramidal shape

A NH B BH C H D H S E HBr O 3 3 2 2 12. Which of the following molecules contains polar covalent bonds but is a nonpolar molecule?

A CH Cl B CH Cl CH N D CCl E N 3 2 2 3 O 4 2

13. Which of the following arranges the molecules N2, O2, and F2 in order of their bond enthalpies, from least to greatest?

A F2 < O2 < N2 B O2 < N2 < F2 O

C N2 < O2 < F2 D N2 < F2 < O2

14. Pi (π) bonding occurs in each of the following species EXCEPT

— A CO B C H C CN D C H E CH 2 2 4 6 6 O 4 15. Which of the following molecules has the shortest bond length?

A N B O C Cl D Br E I O 2 2 2 2 2 16. According to the VSEPR model, the progressive decrease in the bond angles in the series

of molecules CH4, NH3, and H2O is best accounted for by the A increasing strength of the bonds

B decreasing size of the central atom

C increasing electronegativity of the central atom OD increasing number of unshared pairs of electrons E decreasing repulsion between hydrogen atoms

ISPS Chemistry Aug 2020 page 54 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 17. Which of the following is a nonpolar molecule that contains polar bonds?

A F B CHF C CO Dl HC E NH 2 3 O 2 3 18. Which of the molecules represented below contains carbon with sp2 hybridization?

A CH B CH Cl C C H D C H Cl E C H Cl 4 2 2 2 6 O 2 2 2 2 4 2 19. Which of the following has a zero dipole mo­ment?

A HCN B NH C SO D NO E PF 3 2 2 O 5 20. Which of the following species is NOT planar?

2— — A CO B NO C ClF D BF E PCl 3 3 3 3 O 3

21. The BF3 molecule is nonpolar, whereas the NF3 molecule is polar. Which of the following statements accounts for the difference in polarity of the two molecules?

A In NF3, each F is joined to N with multiple bonds, whereas in BF3 , each F is joined to B with single bonds.

B N — F bonds are polar, whereas B — F bonds are nonpolar.

C NF3 is an ionic compound, whereas BF3 is a molecular compound

D Unlike BF3, NF3 has a nonplanar geometry due to an unshared pair of electrons on O the N atom.

ISPS Chemistry Aug 2020 page 55 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2 2.5 - 2.7 Quick Check FRQ 1. Answer the following questions using principles of chemical bonding and molecular structure.

2- a) Consider the carbon dioxide molecule, CO2 , and the carbonate ion, CO3 . i) Draw the complete Lewis electron-dot structure for each species. 1 point is earned for each Lewis electron-dot structure Indication of lone pairs of electrons are required on each structure

2- Resonance forms of CO3 are not required

2- ii) Account for the fact that the carbon-oxygen bond length in CO3 is greater

than the carbon-oxygen bond length in CO2.

1st point is earned for indicating double bonds are present in CO2 OR resonance 2- occurs in CO3 . 2nd point earned for BOTH of the above AND indicating the relative lengths of the bond type

b) Consider the molecules CF4 and SF4. i) Draw the complete Lewis electron-dot structure for each species. 1 point is earned for each Lewis electron-dot structure

Lone pairs of electrons are required on each structure

ii) In terms of molecular geometry, account for the fact that the CF4 molecule is

nonpolar, whereas the SF4 molecule is polar. 1st point is earned for each molecule for proper geometry and explanation

CF4 has a tetrahedral geometry, so the bond dipoles cancel, leading to a nonpolar molecule.

With five pairs of electrons around the central S atom, SF4 exhibits a trigonal bipyramidal electronic geometry, with the lone pair of electrons. In this configuration, the bond dipoles do not cancel, and the molecule is polar.

ISPS Chemistry Aug 2020 page 56 Molecular and Ionic Compound Structure and Properties AP Chemistry Unit 2

2. Propanoic , C2H5COOH, is an organic acid that is a liquid at room temperature. a) An incomplete Lewis diagram for the propanoic acid molecule is provided in the box below. Complete the diagram, showing how the remaining atoms in the molecule are arranged around the carbon atom marked with an asterisk (*). Your structure should minimize formal charge and include any lone pairs of electrons.

1 point is earned for a correct structure. There should be two O atoms attached to the C atom, one with a double bond and one with a single bond. The O atom attached with a single bond should have an H atom attached to it.

Each O atom has two lone pairs of electrons. (Figure is required.)

b) Identify the hybridization of the carbon atom marked with the asterisk 1st point is earned for the correct hybridization. sp2

ISPS Chemistry Aug 2020 page 57 Molecular and Ionic Compound Structure and Properties