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CHEM19 01 /3 Answers to Problem Sheet 6
- (a) hydrogen cyanide,
HCN
H C N Bond order = 3 (triple bond)
(b) ethanol, CH
3
CH
2
OH
H C C O
H
H
H
H
H
Two lone pairs on oxygen.
(c) pyridine, C 5
H
5
N
C
C
N
C
C
C
C
C
C
N
C
C
H
H
H H
H
H
H
H H
H
Pyridine has delocalized ‘aromatic’ electrons like benzene with a lone pair on
nitrogen instead of a C-H bond. Note that this lone pair is pointing away from the
ring and is not delocalized.
(d) acetylene, C
2
H
2
H C C H
(e) SOCl 2
Cl S Cl
O
Cl S Cl
O
The resonance structure on the left shows the (allowed) expanded valence shell
around S with 5 electron pairs (4 bonds for the S(IV) atom and 1 lone pair). The
resonance structure on the right shows all atoms obeying the octet rule leading to
charges having to be placed on S and O.
(f) N 2
O
4
Each resonance form is equivalent – the N-O bonds are intermediate between single
and double bonds in character. There is no multiple bond between the nitrogen atoms.
N N
O O
O O
N N
O O
O O
N N
O
O
O
O
N N
O
O
O
O
(g) phosphoric acid,
H
3
PO
4
[=PO(OH)
3
]
P
O
O
O
H
H
H O P
O
O
O
H
H
H O
The resonance structure on the left shows the bonding with an expanded valence
shell around P (five bonds for the P(V) atom). The resonance structure on the right
shows all atoms obeying the octet rule but with charges.
(h) phosphate anion,
PO
4
3 –
P
O
O
O O
O P O
O
O
P
O
O
O O
O P O
O
O
The resonance structures all feature one double P=O and three single P-O bonds.
Overall, each P-O bond is equivalent with:
bond order = ¼ (3 × 1 (single)+ 1 × 2 (double)) = 1.
(i) methylsulfate anion,
CH
3
OSO
3
S
O
O
C O O
H
H
H
S
O
O
C O O
H
H
H
S
O
O
C O O
H
H
H
(f) SF 4
F S F
F
F
trigonal
bipyramid
see-saw
(g) ClF 3
F Cl F
F
trigonal
bipyramid
T-shaped
(h) XeF 2
F Xe F
trigonal
bipyramid
linear
(a) (b) (c)
H C
3
C H
O
H
3
C C OH
H
H
H C
3
C OH
O
(d) (e) (f)
H C
3
C O
O
CH
3
C
H
3
C
C
CH
3
H
3
C
H
H C
3
C C
H
C
C
C C
C
C
C C N CH
3
O H
H
H O
H H
H
O
H
H
H H
