Ch 3 Lecture 2 Bonding Considerations

I. VSEPR Theory ContinuedA. Lone Pair Repulsion

1) Lone pairs are as important as atoms in VSEPR

2) Irregular structures result when central atoms have lone pairs

3) Lone pairs (lp) occupy more space on the central atom than bonding pairs (bp)

a. bp electrons are attracted to a second nucleus

b. lp electrons are localized on the central atom

c. (lp—bp) repulsion is greater than (bp—bp) repulsion

d. Ammonia H—N—H bond angle = 106.6o instead of 109.5o

e. Water H—O—H bond angle = 104.5o instead of 109.5o

B. Subtle Geometric Influences

1. Size of the central atom

a. Larger central atom leads to longer bonds and less bp—bp repulsion

b. Smaller bond angles result

c. Example: H2O, H2S, H2Se, H2Te

2. Electronegativity of outer atoms

a. More electronegative outer atom pulls bp towards it

b. Reduces bp—bp repulsion and makes smaller bond angles

c. Example: PF3, PCl3, PBr3, PI3

3. Size of outer atoms

a. Very small outer atoms (H) may reverse expected trends

b. Example: SH2, SF2, SCl2

4. Exact predictions are rarely needed, qualitative predictions are often enough.

5. Table 3-5 gives geometric data

C. Choose the structure that allows maximum lone pair separation

1. Example ClF3

a. SN = 5, so structure is based on trigonal bipyramid

b. Three possible structures

c. lp—lp most important: eliminates B as a possibility

i. 90o interactions are very unfavorable

ii. Angles > 90o give about the same interaction

d. Structure A has 6 90o lp—bp interactions

e. Structure C has 4 900 lp—bp interactions

f. C is the best structure

2. Always place lone pairs in equatorial positions in SN = 5 and SN = 7

linear

bent

bent

trigonal pyramid

see-saw T-shaped linear

square pyramid

square planar

3. Remember: lone pairs count in SN, but are left out in structure name

4. Examples: SbF4-, SF5

-, SeF3+

5. Exercise 3-2

D. Multiple Bonds

1. Double and Triple bonds have greater repulsion than

single bonds due to the large -orbitals involved

2. Multiple bonds are placed to minimize interactions, similar to lp

a. Multiple bonds are placed equatorially in SN=5, SN=7

b. lp > mp > bp

3. Examples: HCP, IOF4-, SeOCl2

4. Exercise 3-3

II. ElectronegativityA. Definitions

1. Electronegativity: power of an atom in a molecule to attract electrons

2. Polar Bond: between pure covalent (sharing e-) and ionic (transfer of e-)

a. Homonuclear Diatomics: completely nonpolar H—H

b. Heteronuclear Diatomics: polar H—Cl

B. Pauling’s Electronegativity Scale

1. Definitions

a. D = Bond Dissociation Energy (A—A A· + A·)

b. ’ = Ionic Resonance Energy (kJ)

c. A = Electronegativity of A

2. Equations

2

)()()()(

BBDAADBADBA

102.0BA

3. F set at = 4.0 and other atoms determined from it

4. Example: Find ’ of HCl using (H = 2.20, Cl = 3.16)

5. Find D of H—O. (H—H = 432kJ/mol, O—O = 213 kJ/mol)

molkJBA

BA

/6.88102.0

16.320.2

102.0

102.022

molkJBAHO /8.147

102.0

44.320.2

102.0

22

)(

molkJmolkJOHD

BBDAADBABAD

BBDAADBADBA

/3.4702

213432/8.147)(

2

)()()()(

2

)()()()(

6. Large differences give ionic bonds; small differences give polar covalent bonds

7. Actual values come from averaging several compounds of the atom to remove experimental error

8. Actual values in a given molecule may vary from the average

9. Noble gases have high values because of high effective nuclear charges

a. Small size

b. Large + nucleus

III. Polar BondsA. Any bond between atoms of different electronegativities is polar

1) Electrons concentrate on one side of the bond

2) One end of the molecule is (+) and one end is (-)

B. Calculating Dipole Moments

1) Dielectric Constant = Capacitance of Full Cell / Capacitance of Vacuum

2) The orientation of polar molecules cancels out capacitance and this difference gives the dielectric constant

3) The temperature dependence of this value lets us calculate dipole moment ()

= Qr [units = Cm or D(debye)]

Q = difference in charge

r = distance of separation

1 D = 3.38 x 10-30 Cm

4) Complex molecules: vector addition of all

a. Not very accurate (table 3-7) because of variation lengths and angles

b. Lone pairs have a large influence on dipole moment as well

A B- +vs.

C. Examples:

1. Large dipoles result from lone pair reinforcement of bond dipoles

2. Small dipoles result from lone pair cancellation of bond dipoles

3. Symmetric molecules have no net dipole

IV. Hydrogen BondingA. Hydrogens bonded to an electronegative atom (O, N, F, Cl, Br,…) can form a

weak bond to the lone pair on another electronegative atom

1) Water boils about 200 oC higher than expected without this interaction

2) Ammonia does not H-Bond with itself well, instead forming a N—N dimer

O

H

H

O

H

H

B. Unusual properties of water due to H-Bonding

1. Freezing point >> similar compounds

2. Decreases in density on freezing

a. H-Bonding network gives an open structure

b. Ice floats (think biology and geology)

C. Other H-Bonding Phenomena include Protein and Nucleic Acid Folding