CHEM1101 Answers to Problem Sheet 5
1.
(a)
The electron configuration is
shown on the diagram,
including the parallel spins of
the electrons in 1π*:
(1σ)2 (1σ*)2 (1π)4 (2σ)2 (1π*)2
0.4
0.2
σ*
0
bond order
bond
length / Å
-0.2
-0.4
O2
½ (8 – 4) = 2
1.21
O2 +
½ (8 – 3) = 5/2
1.12
O2
½ (8 – 5) = 3/2
1.34
½ (8 – 6) = 1
1.49
−
O2 2
−
The bond order is clearly related to
the bond length – the larger the bond
order, the shorter the bond. The bond
order reflects the relative number of
bonding and antibonding electrons.
π*
energy / eV
(b)
σ
-0.6
-0.8
π
σ*
-1
-1.2
σ
-1.4
-1.6
Bonding electrons act to bind the atoms more strongly and reduce the bond
length. Antibonding electrons act to weaken and hence lengthen the bond.
2. Lobe representations of the 1σ and 1σ * orbitals of H2:
+
1σ
−
1σ∗
The nuclei are represented by dots – note that for the 1σ* orbital, the nuclei do
not lie at the centres of the two lobes.
3.
(a) The longest electronic absorbance wavelength corresponds to the
smallest energy transition. This arises from excitation of an electron
from the HOMO to the LUMO. The transition energy is therefore:
ΔE = E(LUMO) – E(HOMO) = (-8.23 eV) – (-13.85 eV) = 5.62 eV
As 1 eV = 1.602 × 10–19 J,
ΔE = 5.62 × 1.602 × 10-19 J = 9.00 × 10-19 J
From Planck’s equation, Δ E =
hc
so
λ
hc (6.626 × 10−34 J s) × (2.998 × 108 ms−1 )
λ=
=
= 2.21× 10−7 m = 221nm
ΔE
(9.00 ×10−19 J)
(b)
UV-A runs from 400 nm - 320 nm, UV-B runs from 320 nm - 290 nm and
UV-C runs from 290 nm - 100 nm. The transition in ozone is therefore
excited by UV-C radiation.
The ionic radii are: Na+ (0.102nm), Br (0.196 nm), Cl (0.181 nm). The radius
ratio for NaBr, r+/r = 0.520, is smaller than for NaCl (0.563), suggesting that Na+
can more easily fit into the octahedral interstices of the NaCl structure. There is
insufficient space to fit 8 bromide ions around sodium so the CsCl structure is not
formed.
4.
−
−
−
5.
Using Planck’s relationship:
hc (6.626 × 10−34 J s) × (2.998 × 108 ms −1 )
ΔE =
=
= 4.23 × 10−19 J
−
9
λ
(470 × 10 m)
(As 1 eV = 1.602 × 10-19 J, ΔE =
4.23 × 10 −19
= 2.60 eV )
1.602 × 10 −19
A wavelength of 470 nm corresponds to
blue light. This is the light that is
absorbed by CdS. The colour observed is
the complementary colour – white light
with the absorbed colour removed as
shown on the colour wheel opposite. The
complementary colour of blue is orange
and hence CdS is expected to be orange a sample is shown on the right.
6.
(a)
hydrogen cyanide,
HCN
H
(b) ethanol, CH3CH2OH
H
C
N
H
H
C
C
H
H
Bond order = 3 (triple bond)
O
H
Two lone pairs on oxygen.
(c)
pyridine, C5H5N
H
H
H
C
H
H
C
H
C
C
C
C
C
C
C
C
H
N
H
H
N
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, C2H2
(e)
H
C
SOCl2
C
H
O
Cl
S
O
Cl
Cl
S
Cl
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)
N2 O4
O
O
N
O
N
O
N
N
O
O
O
O
O
O
O
O
N
O
N
N
O
O
N
O
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.
The resonance structures all feature two double N=O and two single N-O bonds.
Overall, each N-O bond is equivalent with:
bond order = ¼ (2 × 1 (single)+ 1 × 2 (double)) = 1.5
(g) phosphoric acid,
H3PO4 [=PO(OH)3]
H
O
H
H
O
O
P
O
H
O
P
O
O
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, PO43–.
3O
O
P
O
O
O
O
O
P
O
O
P
O
O
O
O
O
P
O
O
P
methylsulfate anion,
CH3OSO3–
H
H
C
H
H
O
O
O
O
O
(i)
O
S
H
O
C
H
O
H
H
C
H
O
O
S
O
O
O
O
S
O
O