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Chem 104A, Fall 2016, Midterm 1 Key
1) Construct a microstate table for a p4 configuration. Please enumerate the ms and ml for
each electron in each microstate in the table. (Use the format mlms. That is a spin -½
electron in an s orbital would be written as 0-) Provide the term symbols representing the
states constructed from the table. Make sure to include the spin-orbit coupling quantum
number, J. Then determine which of these states is the ground state. (15 pts)
ML\MS
1
0
-1
1+ 1- 0+ 0-
2
1
1+ 1- 0+ -1+
1+ 1- 0+ -11+ 1- 0- -1+
1+ 1- 0- -1-
0
1+ 0+ 0- -1+
1+ 0+ 0- -11- 0+ 0- -1+
1+ 1- -1+ -1-
1- 0+ 0- -1-
-1
1+ 0+ -1+ -1-
1+ 0- -1+ -11- 0+ -1+ -1-
1- 0- -1+ -1-
-2
0+ 0- -1+ -1-
15 microstates total.
Term Symbols: 1D2, 3P2, 3P1, 3P0, 1S0
Ground State: 3P2
1 pt for correct ML values
1 pt for correct MS values
2 pts for correct number of microstates
3 pts for correct microstate enumerations
5 pts for correct term symbols (-1 for each wrong term symbol)
3 pts for correct ground state
2) The following series of compounds have increasing dipole moments:
NF3 = 0.23 D, PF3 = 1.03 D, AsF3 = 2.57 D. Provide at least two arguments to explain
this trend. (10 pts)
The difference in electronegativity between the two atoms increases with increasing
dipole moment.
The bond distances also increase with increasing dipole moment due larger atomic radii
of the central atoms. This allows for a stronger dipole moment.
As you go down the series there is less s-p mixing. This results in bond angles closer to
90°. The more acute bonds would then result in a higher dipole moment.
5 points for each argument.
3) Assign the point group of the molecules below. Then using Slater’s rules determine the
effective nuclear charge for a 3d electron in the following metals Mn0, Ni0, and Co2+.
(12 pts)
O
O
C
C
O
C
O
Mn
Co
Ni
N
C
O
Co
Mn0: 25 – (0.35*4 + 18) = 5.6
Ni0: 28 – (0.35*7 + 18) = 7.55
Co2+: 27 – (0.35*6 + 18) = 6.9
2 points for each point group
2 points for each Zeff
D2h
O
Co
O
O
C2v
Co
O
Td
4) Consider the eclipsed 6-coordinate molecule hexahydride tungsten, WH6, shown below.
Each of the six hydride ligands are labelled a-f.
b
H
a
H
z
c
H
W
d
H
H
e
H
x
f
y
a) List all of the symmetry elements in this molecule. (6 pts)
E, C3, C32, 3C2, σh, S3, S35, 3σv
1/2 point per symmetry element, -1/2 point per incorrect element
b) What is its point group? (2 pts)
D3h
c) Provide a symmetry element that interchanges only the given set(s) of hydride
ligands. If this is not possible, write “not possible”. (4 pts)
a. a/b and d/e
σv
b. b/e and a/f and c/d
C2
d) Apply the following symmetry operations on the structure above: C3, C2(y), S3. Draw
the resulting structure with the hydrides labelled, illustrating how they have
transformed after all of the symmetry operations have been applied. What is the net
symmetry operation? (5pts)
a
H
b
H
H
H
C3
W
d
H
H
e
b
c
H
f
c
H
a
H
H
f
H
C2 (y)
W
H
e
e
H
H
b
H
H
a
a
f
H
S3
W
d
σ
d
H
Hc
(yz)
1 point for each structure, 2 points for correct net symmetry operation
c
H
H
b
W
H
d
H
f
H
e
5) Illustrate graphically the conventional shapes and phases of the following orbitals. Then
draw its corresponding radial distribution function along the x axis. (10 pts)
3dx2-y2, 5px, 2s, 4dz2, 3dxy
1 point for each orbital diagram and each radial distribution function.
6) Elemental sulfur can be used as an electrode material for lithium-sulfur batteries. At
moderate potentials, it forms molecules with the chemical formula S8, characterized by an
S-S-S angle of 107 °.
a. Draw the Lewis structure of this molecule. (3 pts)
S
S
S S
S
S
S
S
b. Qualitatively, what is the p-character of the S-S bonds relative to that of the S
lone pairs? (4 pts)
The p character of the S-S bonds is greater than that of the S lone pairs.
c. The molecule can be oxidized by two electrons to form S82+. An additional S-S
bond is formed in this oxidation. Draw the most likely structure of this molecule.
(3 pts)
S
S
S
S
S
S
S
S
d. In lithium-sulfur batteries, S8 can be reduced by 2 electrons to form Li2S8. Draw
the structure of this molecule. (3 pts)
S
S
S
S
S
S
S
S
7) Draw the Lewis structure for thiosulfuric acid (H2S2O3). What is its molecular geometry
and point group? To which atoms are the hydrogen atoms bound? Explain. (10 pts)
S
HO
S
OH
O
The molecular geometry is tetrahedral. 3 pts
The point group is Cs. 3pts
The hydrogens are bound to oxygen atoms because O is more electronegative than S.
This results in forming a more stable conjugate base when the negative charge is place
on the O instead of the S. 4 pts
8) A molecule will have a dipole moment if the summation of all of the individual bond
moment vectors is nonzero. The presence of certain symmetry elements require that a
molecule’s dipole moment be equal to zero because for the net cancellation of the
individual bond dipoles. These symmetry elements are: an inversion center, i; a
horizontal mirror plane, σh; two or more distinct Cn axes. The dipole moment of SbF5 is
0. The dipole moment of IF5 is 2.18 D.
a. Use VSEPR to determine the molecular geometry for each molecule. Then
determine their point groups. (8 pts)
F
F
F
Sb
F
F
F
F
I
F
F
F
Trigonal bipyramidal
D3h
Square pyramidal
C4v
1 point for each structure, 1 point for each geometry, 2 points for each point
group
b. Explain how these experimental data are consistent with VSEPR concepts (and
the symmetry elements each molecule possesses)? (5 pts)
SbF5 contains σh and two distinct symmetry axes includint a C3 and a C2. That is why its
dipole moment is zero. IF5, however, has none of the necessary symmetry elements. Thus, it
has a dipole moment.