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ALEXANDER MACKENZIE HIGH SCHOOL

SCH4U CHEMISTRY

CHAPTER 4 – CHEMICAL BONDING

 Dalton’s atom story starts even before Mendeleev’s periodic table  Frankland (1852) “each element has a fixed valence that determines its bonding capacity”  Kekulé (1858) – illustrating a bond by using a dash between bonding atoms  van’t Hoff & Le Bel (1874) extended these structures to 3-D  all this without knowing about the nucleus and bonding electrons, etc.  Abegg (1904) suggested “that the stability of the inert or noble gases was due to the number of outermost electrons in the atom”  Theorized that a Cl atom had 1 less electron than Ar so if Cl would gain 1 electron it would become more stable, also if Na having 1 electron in its outer shell would loss it to become stable i.e. Na + Cl -- Na+ + Cl-

 Lewis (1916) produced the 1st clear understanding of chemical bonding: 1. atoms are stable if they have a noble gas-like electron structure; i.e. stable octet of electrons 2. electrons are most stable when paired 3. atoms form chemical bonds to achieve a stable octet of electrons 4. a stable octet may be achieved by an exchange of electrons between metal & non-metal atoms 5. a stable octet of electrons may be achieved by the sharing of electrons between non-metal atoms 6. the sharing of electrons results in the formation of covalent bonds  to show these bonds Lewis devised Lewis Structures 2

 example of Lewis Structures:

Cl2 ·· ·· ·· ·· ·· ·· ··Cl· + ·Cl·· --- ··Cl··Cl·· ------ ··Cl-Cl·· ·· ·· ·· ·· ·· ··

NaCl ·· ·· Na· + ·Cl·· ------[Na]+ + [··Cl··]- ·· ··

Homework: Nelson 12 Review Page 225 Sample Problem

Lewis Structures and Quantum Mechanics

 putting Lewis structures and quantum mechanics together a connection between the 2 theories becomes apparent  in a Lewis structure the 4 sides of the rectangle of electrons around the core represent the s & p energy sublevels  Summerfeld added these energy sublevels in 1915 & Lewis proposed his theory in 1916  Combining the 2 theories we can see for Mg

6s 5p 5p 5p 4d 4d 4d 4d 4d 5s 4p 4p 4p 3d 3d 3d 3d 3d 4s  3p 3p 3p 3s     2p 2p 2p 2s  1s which corresponds to Mg, valence of 2, ·Mg· , 1s2 2s2 2p6 3s2 3

 For N, valence 3, 1s2 2s2 2p3 6s 5p 5p 5p 4d 4d 4d 4d 4d 5s 4p 4p 4p 3d 3d 3d 3d 3d 4s 3p 3p 3p 3s     2p 2p 2p 2s  1s ·· · N· ·

Homework: Nelson 12 Page 227, #’s 1-5

Extending the Lewis Theory of Bonding

 Lewis’ theory explained many simple molecules but some molecules including polyatomic molecules could not be explained  Work by Sidgwick showed Lewis structures can work if we don’t require that each atom contribute 1 electron to a shared pair in a covalent bond i.e. one atom could contribute both electron pairs that are shared (co-ordinate covalent bond)  Sidgwick also realized that an octet of electrons around an atom may be desirable but is not necessary in all molecules & polyatomic ions

1-- Method 1: eg. NO3 4

1) arrange atoms in most symmetrical arrangment 2) find total number of valence electrons including the charge of the ion e.g. 5+[(3) x 6] + 1 = 24 e- 3) put pairs between the central atom and each of the surrounding atoms that uses up 6 4) distribute the rest and you have this;

·· ··O·· ·· N ·· ·· ··O·· ··O·· ·· ··

a. are atoms surrounded by 8 electrons? 5) Because N only has 6 we can move one pair of electrons from O to help N have a stable octet

·· ··O·· ·· N ·· ·· ·· ··O·· ··O·· ·· 7) draw the Lewis structure with the single and double bonds and enclosing the polyatomic ion with a square bracket and including the charge

- O

N Homework: Nelson 12, Page 230 #’s 1-4 O O

1— Method 2: eg. NO3 5

1) identify the central atom as the least electronegative atom - exceptions are H and generally O . central atom is N 2) place the surrounding atoms around the central atom (on sides) 3) draw single bonds between the surrounding atoms to the central atom O

O N O 4) determine the number of bonds in the molecule a) calculate the # valence electrons for all the atoms, including electrons added or removed due to polyatomic charge # valence e-- = 23 e- added = 1 # valence e- + added e- = 24 b) calculate the # electrons if each atom had a filled valence shell by itself # e- if valence shell filled for each atom = 32 c) calculate the difference between 4a and 4b difference = 8 d) each pair of electrons in 4c is a bond # bonds = 4 5) add any double or triple bonds between the central and surrounding atoms to achieve the # bonds determined in 4d - atoms that often form multiple bonds are C, N, O, S O

O N O 6) draw the electron dots to represent the remaining unshared pair of electrons around the atoms, with charge if needed 1- O

O N O 6

7) identify whether coordinate bonds are present by determining which atom “owns” the electrons forming the bonds 1- O

O N O

Exceptions to the Octet Rule

1. Expansion of the Valence Level

 elements of the third (3rd) period or greater can form molecules where the central atom is surrounded by greater than the octet of valence electrons – due to the presence of the “d” subshell  eg. Phosphorus has 5 e—that can form as many as 5 bonds

. PF3 is common

. PF5 is less common, but does exist F F 2 e— from 3s and 3 e—from 3p are promoted to 3d subshell, F P with each 3d orbital having F an lone electron for pairing F Sulfur has 6 e—that can form as many as 6 bonds

. SF2 vs SF6 +1 other examples may include: SF4, XeF2, SeF4, TeF6, XeF5

2. Molecules with an Odd Number of Valence Electrons  the single unpaired electrons are attracted to the atom by the paramagnetic field . the strength of attraction is proportional to the # of unpaired e—  eg. NO valence e-- = 11 total e—when valence level filled = 16 5 e— # of bonds = 2 # e—left over = 1 (unpaired) 7

N = O N = O

Lewis Diagrams are consistent formulas with NO

-- NO2 valence e = total e—when valence level filled = e— # of bonds = # e—left over = (unpaired) 3. Molecules with Insufficient Valence Electrons

 Be and B do not form ionic bonds, but instead form covalent bonds . ionization energies are too high for Be and B . belong to group IIA and IIIB

. eg. BF3 - check electronegativities of B and F valence e-- = 24 total e—when valence level filled = 32 8 e— # of bonds expected = 4

suggests a Lewis structure F – B = F

F however, experimental evidence shows the structure has NO double bond B is shown to have only 6 valence e- F – B – F i.e. electron deficient exhibits partial double bond character F

+ other examples AlCl3, BeCl2, BCl2, TlCl2 , BeBr2 8

The Nature of the Chemical Bond

 Shroedinger ‘s mathematical equation to describe the standing wave for an electron works fine for an atom but when a molecule is considered, with 2 nuclei, extra electrons, etc. quantum mechanics are still being studied  Linus Pauling developed the valence bond theory when a covalent bond is formed 2 orbitals overlap and produce a new combined orbit with 2 electrons of opposite spin is produced.

 Any 2 half-filled orbitals can overlap in the same way, e.g. HF i.e. the H’s 1s orbital is thought to overlap the half-filled 2p orbital

 This also works with larger molecules, consider the O atom     _  _ 1s2 2s2 2p4 9

 If this reacted with H, it would seem reasonable that the 1s orbital would overlap the 2 half-filled 2p orbitals of O with an angle between the 2 bonds to be 90°.  In fact, what is seen is that an angle of 105° is observed which means that there is a problem with the theory or other factors are involved

Homework: Nelson 12 page #’s 1-4

Hybrid Orbitals

 Lewis bonding theory could not explain the 4 equal bonds in

compounds like CH4(g), nor its tetrahedral shape, and the existence of double & triple bonds  Pauling and others proposed that C bonds form in the shape of a tetrahedron 10

 Consider the energy level diagram for carbon in its ground state:

3s    2p 2p 2p 2s  1s  Initially they proposed that an electron from the 2s orbital was promoted to the third 2p orbital to create 4 orbitals with single electrons

3s     2p 2p 2p 2s  1s

 Each electron in the orbitals could then pair up with an electron from another atom, such as hydrogen - it was believed that more energy would be gained when the bond was formed than for the electron to be promoted  Unfortunately this bonding arrangement would create 3 equal bonds with the 2p orbitals, but one different bond with the 2s orbital  Evidence showed that all the bonds were equal in shape & energy so out went that thought 3  Now CH4 is explained by hybridization to 4 identical hybrid sp orbitals carbon – ground state carbon - hybridized

3s   3s  2p 2p 2p     2s sp3 sp3 sp3 sp3 11

  1s 1s

 The explanation being that the 2s electrons and the 2p electrons form 4 identical hybrid sp3 orbitals with one bonding electron in each, which explains the bonding capacity of 4 in C  These orbitals are hybridized only when bonding occurs not in an isolated atom 12

 Hybridization is used to explain the bonding of atoms in covalent molecules ONCE the shape has been determined  Conditions for bonding: . occurs along the axis of orbitals involved . covalent bond formed by the overlap of orbitals sharing two (2) electrons . bond is a result of nuclei attraction on electrons from both atoms . # of hybrid orbitals formed always equals # of atomic orbitals used

 Pauling suggested that a whole series of hybridizations could occur

to explain BF3(g) and BeH2(g) , etc (see Table 1 in text page 234) 13

Additional hybrid orbitals include:

Hybrid Orbital Geometry #Orbitals Example

sp linear 2 Be in BeF2 2 sp trigonal planar 3 B in BF3 3 sp tetrahedral 4 C in CH4 3 sp d trigonal bipyramidal 5 P in PCl5 14

3 2 sp d octahedral 6 S in SF6

 Shape of molecule and therefore hybrid orbital used is determined by the single bonds to surrounding atoms and any lone pairs of electrons (see VSEPR Theory) . questions in homework below do not involve lone pairs of electrons; as such the number of single bonds will determine the number of hybrid orbitals

Homework: Nelson 12 page 235 #’s 8,10,11,12 & 14. 15

Double & Triple Covalent Bonds

 Difficult to explain double & triple bonds i.e. explaining the

differences between C2H6(g), C2H4(g) and C2H2(g)  Lewis structures suggests that there must be sharing of 1, 2 or 3 electron pairs in order to obtain a stable octet  How could electrons in the sp3 orbitals overlap not once but two or three times with just one atom  Valence bond theory tells us that two kinds of orbital overlap are possible:

1. the end to end overlap of s orbitals, p orbitals, hybrid orbitals or a combination of these orbitals This type of the overlap produces a sigma “” bond (figure 9)

(a) represents s orbitals overlapping end to end (b) represents p orbitals overlapping end to end (c) represents hybrid orbitals overlapping end to end 16

2. two orbitals can overlap side to side to form a pi “π” bond (see Fig. 10)

Double Bonds Fig. 11 (above)

 C is the most common central atom in molecules with double or triple bonds  Orbitals of C atoms can be hybridized to form 4 sp3 orbitals (4 single bonds)  “New” theory is that we have partial hybridization of available orbitals to leave 1 or 2 p orbitals with single unpaired electrons

carbon – ground state carbon – hybridized for double bond (3, 1π)

3s   3s   2p 2p 2p    2p 2s sp2 sp2 sp2   1s 1s 17

 i.e. promote an electron from a 2s orbital to a 2p orbital and we will have 3 sp2 orbitals and 1 p orbital with a single electron (still have 4 orbitals) . the 3 sp2 orbitals gives the trigonal planar shape from the central C atom bonded to two H atoms and one C atom . the single p orbital aligns itself perpendicular to the trigonal planar shape – the “dumb bell” shape is above and below the plane . see Fig 12

2  in C2H4, 3 hybrid sp orbitals are used to form  bonds between the C and two H’s and the C and the other C (Fig. a) 18

 half-filled p orbitals overlap sideways (Fig. b) to form pi bonds, a region of electron density above & below the C-C sigma bond  a pi bond is a combined orbital which has a pair of electrons having opposite spins  the extra shared pair of electrons in the π bond makes a greater attraction between the 2 C nuclei which is why the double covalent is shorter and stronger than a single bond.

Homework: Nelson 12 Page 238 #’s 18-21 19

Triple Bonds

 ethyne (acetylene) is a linear molecule of C2H2 so that a tetrahedral sp3 hybridization is not likely

 the energy level diagram that permit the formation of the triple bond of ethyne is:

carbon – ground state carbon – hybridized for triple bond (2, 2π)

3s   3s    2p 2p 2p   2p 2p 2s sp sp   1s 1s

 the linear shape is achieved by the overlap of one sp orbital from the C atom with the s orbital of the H atom and the overlap of the other sp orbital from the C atom with the sp orbital of the second C atom (see Fig. 16a) - 2 identical sp hybrid orbitals orient at 180° resulting in a linear molecule  according to the valence bond theory, the unpaired electrons in the 2 p orbitals of adjacent C atoms form 2 π bonds (see Fig. 16b)  In the C-C triple bond, the C’s are bonded with 1 sigma bond and 2 π bonds (see Fig. 16c) 20

Homework: Nelson 12 Page 239 #’s 25-27

As indicated before the section on Double and Triple Bonds: . the shape of the molecule and therefore hybrid orbital used is determined by the single bonds to surrounding atoms and any lone pairs of electrons (see VSEPR Theory) . in the case of molecules with double and triple bonds the shape surrounding each “central” atom (involved with a double or triple bond) is determined by the single bonds and lone electron pairs - the double or triple bond is treated as a single bond in determining the shape . identifying the shape allows one to identify the hybrid orbital involved in the single  bonds and the  bond of the double or triple bond . the double or triple π bond will result from the overlap of one or two p orbitals of the atoms involved in the double or triple bond . as you will see in the next section on VSEPR Theory the starting point for the molecular shape and therefore the hybrid arrangement is the Lewis Diagram (or modified structure) 21

VSEPR Theory

“Valence Shell Electron Pair Repulsion Theory”

 theory based on the electrical repulsion of bonded and unbonded electron pairs in a molecule or polyatomic ion . like charges repel, unlike charges attract  the number of electron pairs are determined by adding the number of bonded atoms plus the number of lone pairs of electrons . the 3-D shape can be determined once this number is known by arranging all the pairs of electrons as far apart as possible  Basic Shapes: (see below) 1. Linear . 2 electron pairs on central atom . 180° angle between pairs 2. Trigonal Planar . 3 electron pairs on central atom . 120° angles between pairs 3. Tetrahedral . 4 electron pairs on central atom . 109.5° angles between pairs 4. Trigonal Bipyramidal . 5 electron pairs on central atom 5. Octahedral . 6 electron pairs on central atom 22

Trigonal Bipyramidal

Octahedral

 All of the shapes of molecules are derived from one of the above five basic shapes . the molecular shape that results is dictated by the electron pairs that form bonds to other atoms versus lone electron pairs 23 24 25

Steps to predict the shape of a molecule are: 1. Draw the Lewis diagram for the molecule, including the electron pairs around the central atom 2. Count the total number of bonding pairs (bonded atoms) and lone pairs of electrons around the central atom. Count a multiple bond as one pair of electrons. 3. Arrange the pairs of electrons according to the appropriate diagram – use basic shapes 4. Obtain the molecular geometry from the directions of bonding pairs . Refer to Table 1 (p 245) / Table 2 (p 247) or Hand-out (from class) and draw the shape of the molecule using a structural formula as per text drawings

 Students are responsible for knowing the shapes (and their modification based on lone pairs of electrons) from Table 1 (p 245) and the basic shapes trigonal pyramidal and octahedral – Table 2 (p 247)

Hybridization and VSEPR Theory

 As mentioned during the discussion on Hybridization the shape of the molecule and therefore hybrid orbital used is determined by the single bonds to surrounding atoms and any lone pairs of electrons

 The steps to follow when determining the hybrid orbital used are: 1. Draw the Lewis electron-dot diagram for the molecule, including the electron pairs around the central atom 2. Count the total number of bonding pairs (bonded atoms – remember that multiple bonds are treated as one bonding pair) and lone pairs of electrons around the central atom 3. Use the VSEPR model to obtain the arrangement of electron pairs about the central atom 4. From the geometric arrangement of the electron pairs, deduce the type of hybrid orbitals on the central atom required for the bonding description (see Table C3 above) 26

5. Assign electrons to the hybrid orbitals of the central atom one at a time, pairing them only when necessary 6. Form  bonds to this central atom by overlapping single occupied orbitals of other atoms with the singly occupied hybrid orbitals of the central atom 7. If a double or triple bond exists, then pair up the p orbitals from the two atoms that are involved to form a π bond 27

Polar Molecules

Electronegativity & Polarity of Bonds

 Pauling realized that electron pairs could be shared evenly or unevenly  Created “electronegativity” to explain & predict the polarity of molecules based on

 F has highest electronegativity (EN) at 4.0  When you compare the electronegativities between 2 atoms, the greater the ∆EN (difference in electronegativity) the greater the polarity of the chemical bond

 The smaller the ∆EN the more non-polar the bond  A very polar bond is an ionic bond while a non-polar bond is a 28

covalent bond  Pauling’s polar bond resulted when 2 different kinds of atoms (non-metals) formed a bond  If the bond is polar it meant that the electrons were “spending more time closer to one of the atoms than to the other  The end where the electrons stayed, was slightly more negative, or partially negative and represented as  - , and the end where the electron stayed away from was  +

 General rule is that when the ∆EN is greater than 1.7, the percent ionic character exceeds 50%

Examples: H – H 2.1- 2.1=0, non-polar bond

P – Cl 2.1-3.0=0.9, indicates a polar covalent bond

Na - Br 0.9 – 2.8=1.9, indicates an ionic bond 29

Polar Molecules

 Polar bonds in a molecule do not mean that you have a polar molecule

 CO2 is considered to be a non-polar molecule even though the C=O bonds are considered to be polar

 According to Lewis structures & VSEPR theory, CO2 is a linear molecule

 The bond dipole – proportional to the ∆EN – is shown by an arrow pointing from the lower EN ( + ) to the higher EN ( - )  The bond dipole is a vector quantity where direction is important – two dipoles of equal magnitude, but opposite direction will cancel each other out

. shown in the CO2 molecule, i.e. the CO2 molecule exhibits no polarity and the molecule is said to non-polar

 If we look at H2O, water is polar and the Lewis structure & VSEPR theory predict that the molecule would be V-shaped (based on the tetrahedral basic shape)

. The dipoles do not cancel and instead the vertical components add together to produce a non-zero molecular dipole (red resultant arrow in middle) . The molecule has a definite overall polarity and is said to be a polar molecule . i.e. it is partially negative at the O end and partially positive at the H end 30

. In this case the shape helps produce the 2 oppositely charged ends on the molecule  From these 2 examples one can see why shape is as important as the polarity of the molecule  “Both the shape of the molecule and the polarity of the bond are necessary to determine if a molecule is polar or non-polar”

 check out CH4

. The outer part of the molecule is positive on all sides and none of the ends are charged differently

. This is because the CH4 molecule is symmetrical

 “In all symmetrical molecules, the sum of the bond dipoles is 0 and the molecule is non-polar”

 other examples of non-polar molecules are CCl4 and BF3 ,etc. 31

Intermolecular Forces

 Intermolecular forces are the forces of interaction that may exist between molecules.  Intermolecular forces are much weaker than covalent bonds: if covalent bonds are assigned a strength of about 100, then intermolecular forces are generally 0.001 to 15  types of intermolecular forces can be classified as: dipole-dipole forces, London forces, hydrogen bonding

Dipole-Dipole Forces

 molecules may be classified as polar or non-polar: a polar molecule occurs when the net result of the bond dipoles (proportional to the difference in electronegativities of the two atoms involved in the bond) in a molecule are NOT zero  the dipole-dipole force is an attractive intermolecular force resulting from the tendency of polar molecules to aligh themselves such that the positive end of one molecule is near the negative end of another . the strength of the dipole-dipole force is dependent on the polarity of the molecule 32

London (Dispersion) Forces

 Fritz London (1930) accounted for the weak attraction between any two molecules (specifically non-polar molecules) by recognizing that electrons orbiting the nucleus of an atom may be on any one side at any point in time . as a result there may be a small, instantaneous dipole, with one side having a partial negative charge and the other side having a partial positive charge . if another atom is present nearby, the partial negative charge of the first atom will repel the electrons of the second atom creating a partial positive charge of the second atom; the partial negative charge and the partial positive charge of each atom will result in an attractive force between the two atoms . this weak attraction occurs instantaneously because the electrons are in constant motion; nonetheless the motion of the electrons in one atom will influence the motion of the electrons of the other atom 33

 London forces (or dispersion forces) are the weak attractive forces between molecules resulting from the small, instantaneous dipoles that occur because of the varying positions of the electrons during their motion about the nuclei  London forces increase as the molecular weight increases due to: . the increased number of electrons in motion . the increased size of the molecule, which permits electrons to move further from the nucleus – more polarizable

Van der Waals Forces

 Van der Waals forces is a general term for those intermolecular forces that include dipole-dipole and London forces  these forces will affect the ease or difficulty with which a molecule leaves a liquid  can observe the effect of intermolecular forces by looking at molecular weight - increase in MW generally results in increase in London forces  can also observe effect of intermolecular forces by looking at boiling points of liquids  surface tension is also dependent on intermolecular forces 34

. surface tension is the energy needed to increase the surface area of the liquid . to increase the surface area, it is necessary to pull molecules apart against the intermolecular forces of attraction  viscosity of a liquid depends in part on the intermolecular forces . increasing attractive forces between molecules increases the resistance to flow . viscosity is also dependent on other factors, such as the possibility of molecules tangling together (long molecules) 35

Hydrogen Bonding

 comparison of fluoromethane (CH3F) and methanol (CH3OH) will show that their boiling points are significantly differently . fluoromethane: -78°C (gas under normal conditions) . methanol: 65°C (liquid under normal conditions) . both molecules have the same molecular weight and polarity (dipole) . suggests that another intermolecular force is at work, not just Van der Waals forces . chemical structure shows that fluoromethane has a C-F bond and methanol has a C-OH bond (similar to water H-OH) – the –OH group creates additional attractive forces between molecules than only the Van der Waals forces  Hydrogen Bonding is a weak to moderate attractive force that exists between a hydrogen atom covalently bonded to a very electronegative atom, X, and a lone pair of electrons on another small, electronegative atom, Y

X H- - -Y

. generally X and Y are the atoms F, O or N  compare boiling points of hydrides with Group 16 (VIA)

elements: H2O (100°C), H2S (-60°C), H2Se (-40°C), H2Te (0°C) . if London forces were the intermolecular forces present one would expect the boiling points to increase from

H2O to H2Te, based on molecular weight – obviously not the

only factor as H2O has a much higher boiling point . consistent with the view that hydrogen bonding

exists in H2O, but is virtually non-existent in the others

. observe hydrogen bonding in H2O – relates to partial charges on O and H 36

 can also see the same pattern when looking at other hydrides of: . group 17: HF, HCl, HBr, Hi

. group 15: NH3, PH3, AsH3, SbH3 . however, pattern is not evident with

group 14 hydrides: CH4, SiH4, GeH4, SnH4