Chemistry 121 Problem set VI solutions - 1
1.
H
2
Te a) Te is in Group VI, so Lewis structure is analogous to H
2
O (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)
Polar.
≈
0. Resultant vector comes from the two lone pair
∆
EN vectors (second structure)
H
H Te H
H
CBr
4 a) Lewis structure is first structure b) VSEPR 4 bp + 0 lp = 4 shape is tetrahedral c) Molecular shape is tetrahedral d) Hybridization is sp
3
Nonpolar.
(VSEPR 4 pairs on central atom so need 4 orbitals)
∆
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
AsCl
5
Br a) Lewis structure is first structure
Br b) VSEPR 5 bp + 0 lp = 5
Br shape is trigonal bipyramid c) Molecular shape is trigonal bipyramid d) Hybridization is sp
3
Nonpolar. d (VSEPR 5 pairs on central atom so need 5 orbitals)
∆
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 Cl
As
Cl Cl
AsCl
3 a) Lewis structure is first structure and has an extra lone pair on the central atom b) VSEPR 3 bp + 1 lp = 4 shape is tetrahedral c) Molecular shape is trigonal pyramid (second structure). d) Hybridization is sp
3
(VSEPR 4 pairs on central atom so need 4 orbitals) 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
As Cl
Cl
Cl Cl
Cl
CO
2 a) Lewis structure is first structure b) VSEPR 2 bp + 0 lp = 2 (multiple bonds count as one bp) c) Molecular shape is linear d) Hybridization is sp (VSEPR 2 pairs on central atom so need 2 orbitals) shape is linear

Chemistry 121 Problem set VI solutions - 2
Nonpolar.
∆
EN (O-C) vectors cancel.
O C O
IF
5 a) Lewis structure is first structure and has an extra lone pair on the central atom 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
F
F
I
F
F
F
F
I
F
F
SF
4 a) Lewis structure is first structure and has an extra lone pair on the central atom b) VSEPR 4 bp + 1 lp = 5 shape is trigonal bipyramidal c) Molecular shape is seesaw (second structure). d) Hybridization is sp
3 d (VSEPR 5 pairs on central atom so need 5 orbitals) 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
F
F S F
F
F
F
ICl
3 a) Lewis structure is first structure and has two extra lone pairs on the central atom b) VSEPR 3 bp + 2 lp = 5 shape is trigonal bipyramidal c) Molecular shape is T shaped (second structure). d) Hybridization is sp
∆
3 d (VSEPR 5 pairs on central atom so need 5 orbitals)
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
Cl
I
Cl Cl
Cl
XeF
2 a) Lewis structure is first structure and has three extra lone pairs on the central atom b) VSEPR 2 bp + 3 lp = 5 shape is trigonal bipyramidal c) Molecular shape is linear, lone pairs go into the horizintal trigonal plane. (second structure). d) Hybridization is sp
3
Nonpolar. d (VSEPR 5 pairs on central atom so need 5 orbitals)
∆
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.

Chemistry 121 Problem set VI solutions - 3
F
F Xe F
F
NO
2
- 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) c) Molecular shape is bent. (second structure). d) Hybridization is sp
2
(VSEPR 3 pairs on central atom so need 3 orbitals) shape is trigonal. e) Polar and the resultant of the two NO vectors will not cancel with the lone pair vector. Molecule has a net negative charge.
N
O O O O
2. C
A
VSEPR 4 bp + 0 lp = 4 pair tetrahedral sp
3
(VSEPR 4 pairs on central atom so need 4 orbitals)
C
B
VSEPR 3 bp + 0 lp = 3 pair trigonal planar (multiple bonds count as one bp) sp
2
(VSEPR 3 pairs on central atom so need 3 orbitals)
CH
3
CH
3
H H
H C C H
H H
Carbon VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp
3
2 sp
3
2s
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.

Chemistry 121 Problem set VI solutions - 4
Each of the seven bond orbitals is occupied by two valence electrons of opposite spins, accounting for the seven bonds in ethane.
H
H C O H
H
Carbon VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp
3
Oxygen VSEPR: 2 bp + 2 lp = 4 pairs; tetrahedral and hybridization is sp
3
(energy diagram as above)
2 sp
3
2s
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.
H H
C C
H H
Carbon VSEPR: 3 bp + 0 lp = 3 pairs; trigonal planar and hybridization is sp
2
2 2p sp
2
2s
Carbon

Chemistry 121 Problem set VI solutions - 5
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.
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 2p y from each carbon overlap to form a
π
bond orbital, as do the 2p z
orbitals to form a second
π
orbitals
bond orbital.
The two
π
bonds in ethyne (acetylene) constitute a barrel-shaped distribution of electron density in the bond region. e) Carbon monoxide

Chemistry 121 Problem set VI solutions - 6
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 2p y
orbitals on carbon and oxygen and the two 2p z
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 y x
*
2p z
2p
2p x
2p y
2p z
2p 2p
*
2p y
* x
*
2p z
2p x
2p y
2p z
Chemistry 121 Problem set VI solutions - 7
*
2p y x
*
2p z
2p 2p 2p
2p x
2p y
2p z
2s
*
2s 2s 2s
*
2s 2s
2s
Species N
2
+
Valence electrons 9
(9
−
4)
Bond Order
=
2
2.5
Magnetism paramagnetic
Bond strength
5a)
* x
*
2p y
*
2p z
*
2p y
* x
N
2s
2
10
(10
−
4)
2
*
2p z
=
3 diamagnetic
N
2
> N
2
+ = N
2
−
*
2p y
2p y
2p z
2p y
2p z
2p y
*
2p x
2p z
*
2p z
2p x 2p x
2p x
*
2s
*
2s
*
2s
2s
*
2s 2s
2s
N
2
-
11
(10
−
5)
=
2
2.5
paramagnetic
*
2p y
2p y x
*
2p z
2p z
2p x
*
2s
2s
*
1s
1s
Species O
2
+
Total electrons 15
(10
−
5)
Bond Order
=
2.5
2
(i) weakest bond is O
2
2 -
2s
*
1s
1s
O
2
16
(10
−
6)
=
2
2
2s
*
1s
1s
O
2
-
17
(10
−
7)
=
1.5
2
2s
*
1s
1s
O
2
2 -
18
(10
−
8)
=
1
2

5b)
(ii) shortest bond is O
2
+
σ∗
2 p
π∗
2 p
π∗
2 p
σ
2 p
π
2 p
π
2 p
σ∗
2 p
π∗
2 p
π∗
2 p
σ
2 p
π
2 p
π
2 p
σ∗
2 s
σ
2 s
σ∗
1 s
σ
1 s
Species NO
Total electrons 15
Increasing paramagnetism CN- < NO < B
2
σ∗
2 s
σ
2 s
σ∗
1 s
σ
1 s
CN-
14
Chemistry 121 Problem set VI solutions - 8
σ∗
2 p
π∗
2 p
π∗
2 p
σ
2 p
π
2 p
π
2 p
σ∗
2 s
σ
2 s
σ∗
1 s
σ
1 s
B
2
10