Hybridization - People

Chemistry 121 Problem set VI solutions - 1 

Chem 121 Problem Set VI Molecular Orbital Theory, Valence Bond Theory and 

Hybridization 

1. 

H2Te 

a) Te is in Group VI, so Lewis structure is analogous to H2O (first structure) 

b) VSEPR 2 bp + 2 lp = 4 shape is tetrahedral 

c) Molecular shape is bent 

d) Hybridization is sp 3 (VSEPR 4 pairs on central atom so need 4 orbitals) 

e) Polar. ∆EN (H-Te) ≈ 0. Resultant vector comes from the two lone pair ∆EN vectors (second structure) 

H 

Te H 

H 

H 

CBr4 

a) Lewis structure is first structure 


c) Molecular shape is tetrahedral 


e) Nonpolar. ∆EN (Br-C) = 0.3. All 4 ∆EN (C-Br) vectors cancel. Second structure shows the two opposing 

resultants for two ∆EN (C-Br) vectors. 

Br 

Br 

C 

Br 

Br 

Br 

Br 

Br 

Br 

AsCl5 


b) VSEPR 5 bp + 0 lp = 5 shape is trigonal bipyramid 

c) Molecular shape is trigonal bipyramid 

d) Hybridization is sp 3 d (VSEPR 5 pairs on central atom so need 5 orbitals) 

e) Nonpolar. ∆EN (Cl-As) = 0.9. All 5 ∆EN (Cl-As) vectors cancel. Second structure shows the two vertical 

vectors will cancel and the three vectors in the triangular plane will cancel. 

Cl 

Cl 

As 

Cl Cl 

Cl 

AsCl3 

a) Lewis structure is first structure and has an extra lone pair on the central atom 


c) Molecular shape is trigonal pyramid (second structure). 


e) Weakly polar. ∆EN (Cl-As) = 0.9. The 3 ∆EN (Cl-As) vectors will give a resultant pointing down. The 

lone pair gives a vector pointing up and these two vectors will come close to cancelling. 

Cl 

As 

Cl Cl 

Cl 

As 

Cl 

Cl 

CO2 


b) VSEPR 2 bp + 0 lp = 2 (multiple bonds count as one bp) shape is linear 

c) Molecular shape is linear 

d) Hybridization is sp (VSEPR 2 pairs on central atom so need 2 orbitals)

e) Nonpolar. ∆EN (O-C) = 1. The 2 opposing ∆EN (O-C) vectors cancel. 

O C O 


IF5 


b) VSEPR 5 bp + 1 lp = 6 shape is octahedral 

c) Molecular shape is square pyramid (second structure). 

d) Hybridization is sp 3 d 2 (VSEPR 6 pairs on central atom so need 6 orbitals) 

e) Weakly polar. ∆EN (F-I) = 1.5. The 4 ∆EN (F-I) vectors in the square plane will cancel. The ∆EN (F-I) 

vector will come close to cancelling with the lone pair vector. 

F 

F 

I 

F 

F F 

F 

F 

F I F 

F 

SF4 


b) VSEPR 4 bp + 1 lp = 5 shape is trigonal bipyramidal 

c) Molecular shape is seesaw (second structure). 


e) Weakly polar. ∆EN (F-S) = 1.5. The 2 ∆EN (F-S) vectors in the vertical plane will cancel. The resultant 

of the two ∆EN(F-S) vectors will come close to cancelling with the lone pair vector in the horizontal 

triangular plane. 

F 

F 

S 

F 

F 

F 

F 

F 

F 

ICl3 

a) Lewis structure is first structure and has two extra lone pairs on the central atom 


c) Molecular shape is T shaped (second structure). 


e) Polar. ∆EN (Cl-I) = 0.5. The 2 ∆EN (F-S) vectors in the vertical plane will cancel. The resultant of the 

two lone pair vectors will be much greater than the ∆EN (F-S) vector in the horizontal triangular plane. 

Cl 

Cl 

I Cl 

Cl 

Cl 

Cl 

XeF2 

a) Lewis structure is first structure and has three extra lone pairs on the central atom 


c) Molecular shape is linear, lone pairs go into the horizintal trigonal plane. (second structure). 


e) Nonpolar. ∆EN (F-Xe) = 1.4. The 2 ∆EN (F-Xe) vectors in the vertical plane will cancel. The 3 lone pair 

vectors in the horizontal triangular plane will cancel.

F 

Xe F 

F 

F 


NO2 - 

a) Lewis structure is first structure and and has a formal charge on one oxygen to give a net charge of -1. 

Nitrogen forms one double bond and there are two resonance hybrids. 

b) VSEPR 2 bp + 1 lp = 3 (multiple bonds count as one bp) shape is trigonal. 

c) Molecular shape is bent. (second structure). 


e) Polar and the resultant of the two NO vectors will not cancel with the lone pair vector. Molecule has a 

net negative charge. 

O 

N 

O 

O 

O 

2. CA VSEPR 4 bp + 0 lp = 4 pair tetrahedral 

Hybridization sp 3 (VSEPR 4 pairs on central atom so need 4 orbitals) 

CB VSEPR 3 bp + 0 lp = 3 pair trigonal planar (multiple bonds count as one bp) 

Hybridization sp 2 (VSEPR 3 pairs on central atom so need 3 orbitals) 

3. a) CH3CH3 

H 

H 

C 

H 

H 

C H 

H 

Carbon VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp 3 

2 

2s 

sp 3 

Carbon 

The six carbon-hydrogen bond orbitals in ethane result from the over lap of sp 3 orbitals on the carbon 

atoms and 1s orbitals on the hydrogen atoms. The carbon-carbon bond orbital results from the overlap of 

two sp 3 orbitals, one from each carbon atom. There are (6 x 1) + (2 x 4) = 14 valence electrons in ethane.


Each of the seven bond orbitals is occupied by two valence electrons of opposite spins, accounting for 

the seven bonds in ethane. 

b) CH3OH 

H 

H 

C 

H 

O H 

Carbon VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp 3 (energy diagram as above) 

Oxygen VSEPR: 2 bp + 2 lp = 4 pairs; tetrahedral and hybridization is sp 3 

2 

2s 

sp 3 

Oxygen 

The three carbon-hydrogen bond orbitals and the oxygen-hydrogen bond orbital in methanol result from the 

overlap of sp 3 orbitals and 1s orbitals on the hydrogen atoms. The carbon-oxygen bond orbital results from 

the overlap of a carbon sp 3 orbital and an oxygen sp 3 orbital. Ten of the valence electrons (4 x 1) + 4 + 6 = 

14 in methanol occupy five bond orbitals and the other four valence electrons occupy the two remaining sp 3 

orbitals as lone pairs on the oxygen atom. 

c) Ethene 

H 

H 

C 

C 

H 

H 

Carbon VSEPR: 3 bp + 0 lp = 3 pairs; trigonal planar and hybridization is sp 2 

2 2p 

2s 

sp 2 

Carbon


All six atoms of the ethene molecule lie in one plane. The first diagram shows the two sp 2 hybridized 

carbon atoms, each with three sp 2 orbitals and one remaining 2p orbital. Each orbital contains one electron 

as shown in the energy diagram. The second diagram shows the σ framework. The four carbon-hydrogen 

bond orbitals result from the overlap of carbon sp 2 orbitals and hydrogen 1s orbitals. The two carbon atoms 

are joined by the overlap of an sp 2 orbital from each. The resulting bond orbital is cylindrically symmetric 

around the carbon-carbon axis and is therefore a σ bond orbital., as are the carbon-hydrogen bond orbitals. 

Finally the last diagram shows the overlap of the two 2p orbitals to form a π bond, the second bond in the 

double bond. There are (2 x 4) + 4 x 1) = 12 valence electrons in ethene, 10 of these occupy the five σ 

bonds and the last two occupy the π bond. 

d) Ethyne 

H C C H 

Carbon VSEPR: 2 bp + 0 lp = 2 pairs; linear and hybridization is sp 

2 2p 

2s 

sp 

Carbon 

The first diagram shows the σ bond framework for ethyne. The carbon-carbon σ bond orbital results from 

overlapping two sp orbitals, one from each carbon atom. Each of the two carbon-hydrogen σ bond orbitals 

results from overlapping a carbon sp orbital and a hydrogen 1s orbital. When carbon is sp hybridized, there 

are two 2p orbitals, each with one electron, unused (energy diagram). The second diagram shows the 2p 

orbitals from both carbons (one set along the y-axis and the other set along the z-axis). The 2py orbitals 

from each carbon overlap to form a π bond orbital, as do the 2pz orbitals to form a second π bond orbital. 

The two π bonds in ethyne (acetylene) constitute a barrel-shaped distribution of electron density in the 

bond region. 

e) Carbon monoxide

C O 

Carbon VSEPR: 1 bp + 1 lp = 2 pairs; linear and hybridization is sp 

2 2p 

2s 

sp 

Carbon 

Oxygen VSEPR: 1 bp + 1 lp = 2 pairs; linear and hybridization is sp 

2 2p 

2s 

sp 

Oxygen 


The hybridization for carbon is sp and one of the sp orbitals contains a lp and the other a bp. There are two 

remaining 2p orbitals each occupied by one electron. Thus carbon has five valence electrons; note the 

negative formal charge on carbon in the Lewis structure. The hybridization for oxygen is sp and one of the 

sp orbitals contains a lp and the other a bp. There are two remaining 2p orbitals each occupied by one 

electron. Thus oxygen has five valence electrons; note the positive formal charge on carbon in the Lewis 

structure. The carbon-oxygen σ bond orbital results from overlapping the two sp orbitals, each contributing 

one electron. The two π bonds (a triple bond results from a σ bond and two π bonds) result from the 

overlap of the two 2py orbitals on carbon and oxygen and the two 2pz orbitals as in ethyne. Carbon 

monoxide has (4 + 6) = 10 valence electrons, two occupy the σ orbital, four occupy the two π orbitals and 

the other four occupy the two remaining sp orbitals on carbon and oxygen as lone pairs.

4. 

2p 

2s 

* 

2py 2p y 

2p x 

* 

2px 

2s 

* 

2s 

* 

2pz 2p z 

2p 

2s 

2p 

2s 

* 

2py 2p y 

2p x 

* 

2px 

2s 

* 

2s 

* 

2pz 2p z 


Species N2 + N2 N2 - 

Valence electrons 9 10 11 

Bond Order 

(9 − 4) 

= 2.5 

2 

(10 − 4) 

= 3 

2 

(10 − 5) 

= 2.5 

2 

Magnetism paramagnetic diamagnetic paramagnetic 

Bond strength N2 > N2 + = N2 − 

5a) 

* 

2py 2p y 

2px 

* 

2px 

2s 

* 

2p z 

2s 

* 

2s 2s 2s 2s 

1s 

* 

* 

2pz * 

2py 2p y 

2px 

* 

2px 

1s 

* 

* 

2pz 2p z 

1s 1s 1s 1s 

* 

2py 2p y 

2p 

2s 

2px 

* 

2s 

* 

1s 

* 

* 

2pz 2p z 

2p 

2s 

* 

2py * 

2py 2p y 

2py 

2px 

* 

2px 

2s 

* 

2px 2px 

Species O2 + O2 O2 - O2 2- 

Total electrons 15 16 17 18 

Bond Order 

(10 − 5) 

= 2.5 

2 

(i) weakest bond is 

2- 

O2 

(10 − 6) 

= 2 

2 

(10 − 7) 

= 1.5 

2 

2s 

2px 

* 

2s 

* 

1s 

* 

* 

2pz 2p z 

* 

2pz 2pz 

(10 − 8) 

= 1 

2 

2p 

2s

5b) 

(ii) shortest bond is O2 + 

σ ∗ 2p 

π∗ 2p π∗ 2p 

π 2p 

σ 2p 

σ ∗ 2s 

σ 2s 

σ ∗ 1s 

σ 1s 

π 2p 

σ ∗ 2p 

π∗ 2p π∗ 2p 

π 2p 

σ 2p 

σ ∗ 2s 

σ 2s 

σ ∗ 1s 

σ 1s 

π 2p 


σ ∗ 2p 

π∗ 2p π∗ 2p 

Species NO CN - B2 

Total electrons 15 14 10 

Unpaired electrons 1 0 2 

Increasing paramagnetism CN - < NO < B2 

π 2p 

σ 2p 

σ ∗ 2s 

σ 2s 

σ ∗ 1s 

σ 1s 

π 2p

Hybridization - People

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