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Inorganic Chemistry 411/511
Midterm Exam # 1 KEY
60 minutes
Partial credit is given, please show all your work. A periodic table is on the last page if needed.
1.
For the following: [8 pts each]
(i) give a Lewis structure, (ii) draw the molecules indicating the correct geometry and (iii)
indicate expected deviations (if any) from ideal VSEPR coordination angles
a.
XeF4
Count valence electrons: 8 + 7(4) = 36 = 18 e pairs
4 e pairs to connect Xe and F ligands, and each F has 3 lone pairs, which is a total of 16
e pairs
therefore there are 2 lone pairs on Xe
Xe has 6 groups about it (4 bonds to F and 2 lone pairs), so geometry is based on
octahedron
the lone pairs will occupy trans positions across the octahedron (this will mitigate their
large repulsive interactions) leading to a square planar molecule
FXeF bond angles should be 90° without deviation since lone pairs act equally on all 4
ligands from both sides
(if image is not opening it should show square planar coordination with 90 deg FXeF
bond angles)
b.
PF3
count valence electrons: 5 + 3(7) = 26 or 13 e pairs
3 e pairs to connect P and F ligands, and each F has 3 lone pairs, which is a total of 12
e pairs,
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therefore there is 1 lone pair on P
P has 4 groups about it (3 bonds to F and 1 lone pair), so geometry is based on
tetrahedron
the lone pair will occupy one position, giving a pyramidal molecular geometry (as in
ammonia)
FPF bond angles should be less than the ideal 109.5 ° because the lone pair is more
repulsive than the bonds to F and will force these ligands somewhat closer together
2a.
Construct an MO diagram for N2. Show atomic and molecular orbital levels with
appropriate energies and symmetry labels, and indicate the electron filling. [10 pts]
2b.
Draw a picture of the electron distribution in the HOMO (highest-occupied molecular
orbital) in N2. [6 pts]
This is a σg orbital, involving constructive overlap of spz hybrids on each atom.
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2c.
How should the bond lengths of N2+ (cation) and N2- (anion) compare with the bond
length in N2 ? Explain. [5 pts]
The cation and anion will have longer bonds than N2. The neutral molecule has a bond
order of 3, because the 3σ and two 1π are all bonding orbitals. Removing an electron
from 3σ to make the cation reduces the bond order to 2.5, as does adding an electron to
the antibonding 2π. The reduced bond order means a weaker and longer bond between
the N nuclei.
Please note – you cannot argue this problem based on effective nuclear charges or
electron-electron repulsions. Those are issues for atomic orbital energies, but cannot be
used to predict bonds between nuclei.
3a.
Fluorine gas (F2) is a powerful oxidizer. Perhaps it can react with Xe to make the ionic salt
Xe+F-(s). Using a Born-Haber approach, and the Kapustinskii equation, calculate ∆Hf° for
XeF(s). You will need some (but not all) of the information provided below. [20 pts]
Kapustinskii equation:
ΔHL = - (1210 kJ/mol Å) (nZ+Z- / d0) (1 – 0.345 Å /d0)
r (Xe+) ≈ 1.0 Å ; r (F-) = 1.33 Å
Ionization energies: Xe (g) = +1170 kJ/mol ; F (g) = +1681 kJ/mol
Electron affinities: Xe (g) = -77 kJ/mol; F (g) = +328 kJ/mol
F-F bond energy: +155 kJ/mol
Xe (g)  Xe(g) + + e½ F2 (g)  F (g)
F (g) + e-  F- (g)
Xe(g) + + F- (g)  XeF (s)
I (Xe) = 1170 kJ / mol
½ D0 (F2) = ½ (155) = 78
- Ea (F) = - 328
- ΔHL
Summing:
Xe (g) + ½ F2 (g)
 XeF (s)
ΔHf = 1170 + 78-328 - ΔHL
Use Kapustinskii eqn to calculate ΔHL:
ΔHL = - (1210 kJ/mol Å) (nZ+Z- / d0) (1 – 0.345 Å /d0)
Z+ = +1, Z- = -1, n = 2, d0 = 2.33 Å so ΔHL = +885 kJ/mol
And therefore ΔHf = +35 kJ/mol
A positive number argues against stability, but it’s only slightly positive and our
calculation of lattice enthalpy is only approximate, so it is definitely worth trying !
Actually XeF (s) does not exist, but the XeF+ ion does, in salts such as XeF+AsF6-(s).
Also XeF2 is a stable molecular species.
3b.
Will the formation of XeF(s) from the gaseous elements be more thermodynamically
favorable if the reaction temperature is increased from 25 °C to 300 °C ? Explain. [5 pts]
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The entropy change will also contribute to the overall change in free energy (ΔG = ΔH TΔS). ΔS will be negative since we are converting gas to solid in the reaction, so the TΔS term will increase ΔG and make the reaction less thermodynamically favorable. The
higher the temperature, the higher ΔG. So increasing the temperature will tend to
destabilize XeF(s).
3c.
Why didn’t you need to know whether Xe+F-(s) forms with the rocksalt, sphalerite or
other lattice structure in part (a) above? [4 pts]
The Kap equation uses an average Madelung constant that is not structure specific.
4a.
Draw a unit cell for perovskite (CaTiO3). [10 pts]
5.
Answer every question; there is no deduction for incorrect answers. [4 pts each]
i.
If the energy of an electron in a ground-state He+ ion is –E, what is the energy of an
electron in a ground-state Be3+ ion?
(a) -16E
(g) +16E
ii.
(b) -4E
(c) -3E
(h) +5280 E
(f) +2E
(d) (AcBaCb)n
(e) (AcB)n
(f) (ABC)n
(g) [(ABC)n]m
(h) (AaBb)n
How many unpaired electrons in the ground state of Fe3+(g)?
(a) 0
(g) 6
iv.
(e) 0
What is the stacking sequence of hexagonal 2D layers in the CdI2 layered structure?
Capital letters are anion layers, lowercase are cation layers.
(a) (AaBbCc)n
(b) (ABCabc)n
(c) (AbABaB)n
iii.
(d) -3/2 E
(b) 1
(h) 7
(c) 2
(d) 3
(e) 4
(f) 5
Which of the following has the smallest lattice enthalpy (ΔHL has the smallest value)?
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(a) LiF
(g) BeO
(b) CaO
(h) LiCl
(c) NaCl
(d) Al2O3
(e) NiO
(f) CsI
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v.
How many Cs+ ions occupy second nearest neighbor locations around Cs+ in CsCl?
(a) 0
vi.
(b) 1
(c) 2
(d) 3
(e) 4
(f) 6
(g) 8
(h) 12
(i) 16
N (g) atoms in the ground-state have the electronic configuration 1s22s22p3. Which of the
following is a correct statement and explanation?
(a)
(b)
(c)
(d)
(e)
N is not spherically symmetric, because 3 is an odd number
N is not spherically symmetric, because the p orbitals are only half-filled
N is not spherically symmetric, because only molecules can have that symmetry
N is spherically symmetric, because atoms are generally spherically symmetric
N is spherically symmetric, because angular terms cancel after adding the
electron distribution functions
(f) N is spherically symmetric, because the core levels are filled