Microsoft Word - Memo Chem2001 Paper 1 Inorganic June 2019.docx
CHEM2001 June 2019 Page 1 of 13
Answer ALL questions in the Answer book.
Question XXXXXXXXXXmarks)
(a) Give the electron configuration for Fe(0), Fe(II) and Fe(III XXXXXXXXXX)
Fe(0) [Ar]4s
2
3d
6 √
Fe(II) [Ar]3d
6
Fe(III) [Ar]3d
5 √ XXXXXXXXXX)
(b) The decrease in the size of the ionic radius of Fe(II), compared to the atomic radius of Fe(0), is 44
pm. However, the ionic radius of Fe(III) is only 12 pm shorter than the ionic radius of Fe(II). Explain
this observation by making use of Slater’s rules for calculating the effective nuclear charge.
(Given: radius(Fe
0
) = 120 pm; radius(Fe
2+
) = 76 pm; radius(Fe
3+
) = 64 pm). (5)
For Fe(0): (1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
6
)(4s
2
)
Z* = Z – σ = 26 – [(0.35 x XXXXXXXXXXx XXXXXXXXXXx 10)] = 26 – 22.25 = 3.75 √
For Fe(II): (1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
6
)
Z* = Z – σ = 26 – [(0.35 x XXXXXXXXXXx 18)] = 26 – 19.75 = 6.25 √
For Fe(III): (1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
5
)
Z* = Z – σ = 26 – [(0.35 x XXXXXXXXXXx 18)] = 26 – 19.40 = 6.60 √
On going from Fe(0) to Fe(II) , ΔZ* = 6.25 – 3.75 = 2.5 : . effective nuclear charge increase
y 2.5
: . large decrease in ionic radius as outermost electron strongly attracted to nucleus and
overall positive charge of metal. √
Going from Fe(II) to Fe(III), ΔZ* = 6.6 – 6.25 = 0.35
: . Much smaller increase in effective nuclear charge (more shielded d-valence electron),
therefore smaller decrease in ionic radius. √ (5)
[7]
Surname (Optional): XXXXXXXXXXStudent No:
CHEM2001 June 2019 Page 2 of 13
Question XXXXXXXXXXmarks)
(a) Use the VSEPR theory to draw reasonable Lewis structures of the molecules IF4
+
and IF4
-
.
XXXXXXXXXX)
(b) Calculate the formal charges on the central atoms XXXXXXXXXX)
For IF4
+
: FC = valence e
-
- nonbonding e
-
- no. bonds = 7 – 2 – 4 = +1 √
XXXXXXXXXXFor IF4
-
: FC = valence e
-
- nonbonding e
-
- no. bonds = 7 – 4 – 4 = -1 √ (2)
(c) Give the molecular geometries for both ions, and indicate the size of the F-I-F bond angles in both
cases XXXXXXXXXX)
For IF4
+
: Saw-horse/see-saw;
XXXXXXXXXXFor IF4
-
: Square planar;
(d) For the reaction:
IF4
+
(aq)
+ F
-
(aq) → IF5 (aq)
Identify the Lewis acid and the Lewis base in the complex formation. (1)
IF4
+
= Lewis acid, F
-
= Lewis base √ (1)
(e) Can IF5 exist? Explain with reference to the valence electron count of the atoms in the molecule IF5.
XXXXXXXXXX)
Yes, Valence electron count of each F atom is 8; thus octet rule obeyed. But for I, 12
valence electrons : . expanded octet. √ √
Hypervalence possible due to presence of unoccupied 5d o
itals in the valence shell n=5. √ (3)
[12]
Question XXXXXXXXXXmarks)
(a) Use the Valence Bond (VB) Theory to describe the bonding in HCN, in terms of hy
idisation theory,
with o
ital energy diagrams XXXXXXXXXX)
CHEM2001 June 2019 Page 3 of 13
o
v
e rl ap
o
v
er la p
lo
n
e
p
ai
o
v
er la p
o
v
e rla p
(b) Demonstrate the expected molecular geometry for the HCN molecule, by making use of a simple
o
ital overlap diagram in terms of the VB Theory XXXXXXXXXX)
(c) The molecular o
ital (MO) energy level diagram of cyanide, CN
-
is given below.
(i) Re-draw the energy level diagram given below, in your answer book, and fill in (populate
with electrons) the diagram for CN
-
.
(ii) Label the molecular o
itals, indicating bonding, antibonding and/or non-bonding o
itals
in the diagram.
(iii) Indicate the HOMO (highest occupied molecular o
ital) and the LUMO (lowest unoccupied
molecular o
ital XXXXXXXXXX)
CHEM2001 June 2019 Page 4 of 13
(d) For the molecular ions CN+ and CN-, compare the relative ca
on-nitrogen bond borders AND bond
lengths. XXXXXXXXXX4)
For CN
+
: (1σ)
2
(1σ*)
2
(1Ï€)
4
: . BO = (bonding e
-
- antibonding e
-
) / = 6-2/2 =2 √
For CN
-
: (1σ)
2
(1σ*)
2
(1Ï€)
4
(2σ)
2
: . BO = (bonding e
-
- antibonding e
-
) / = 8-2/2 =3 √
: . Relative C-N bond length: CN
-
< CN
+ √√ (4)
(e) How would the relative energies of the MO energy level diagram differ for a homonuclear diatomic
atom such as N2, compared to CN
-
? XXXXXXXXXX)
More symmetric: both nitrogen atoms’ contributing to the molecule (and MO's) will have atomic
o
itals of similar energy. √ (1)
[19]
Question XXXXXXXXXXmarks)
(a) For the reaction:
CHEM2001 June 2019 Page 5 of 13
[NiHL]
+
(aq) [NiL] (aq) + H
+
(aq), where L = ligand,
estimate the pKa of the reaction by refe
ing to the species distribution diagram below. (1)
(b) Given the following equations eq. 1 and eq. 2 below, where α = mol fraction, determine the
percentage of deprotonated nickel(II) complex, α[NiL] present in solution at a pH = 6.00, for Ka1 = 4.0
x 10
-4
and Ka2 = 6.3 x 10
-6
.
(Hint: α[Ni
2+
] + α[NiHL
+
] + α[NiL] = 1 )
α[Ni
2+
] = [H
+
] / β ..….. eq. 1
β = [H
+
] + Ka1[H
+
] + Ka1Ka2 ……. eq XXXXXXXXXX)
pH = 6.00 : . [H
+
] = 1.00 x 10
-6
√
β = (1 x 10
-6
XXXXXXXXXXx 10
-4
)(1 x 10
-6
XXXXXXXXXXx 10
-4
)(6.3 x 10
-6
) = XXXXXXXXXXx 10
-6
(from eq. 2) √
α[Ni
2+
] = [H
+
] / β = 1.00 x 10
-6
1.002 x 10
-6
= XXXXXXXXXXfrom eq. 1) √
Ka1 = [NiHL
+
][H
+
] : . α [NiHL
+
] = Ka1 α[Ni
+2
] = Ka1[H
+
] = Ka1 √
[Ni
2+
] [H
+
] [H
+
]β XXXXXXXXXXβ
pKa = pH at α = 0.5, or 50% Ni
pKa ≈ 5.2 √ (1)
CHEM2001 June 2019 Page 6 of 13
Ka2 = [NiL][H
+
] : . α[NiL] = α Ka2[NiHL] = Ka1Ka2[H
+
] = Ka1Ka2 √ = (4.0 x 10
-4
)(6.3 x 10
-6
)