Chemistry 211
Fall 2013
Test 4
Name: __________________________
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Multiple Choice: (4 points each. Put answers in left margin as capital letters.)
h = 6.626 x 10-34 J•s, R = 0.0821 L•atm/mol•K
1. Which of the following proportionalities is false for a gas?
A) V  n
B) V  1/P
C) V  T
D) P  T
E) P  1/n
2. Chlorine and oxygen gases can react to form Cl2O7 gas as follows:
2 Cl2 (g) + 7 O2 (g) → 2 Cl2O7 (g)
If 1.2 L of chlorine reacts with 8.4 L of oxygen, what volume of dichlorine heptoxide will be
formed, assuming pressure and temperature remain constant.
A) 1.2 L
B) 2.4 L
C) 4.2 L
D) 8.4 L
E) 15.4 L
3. Which of the following should have the highest first ionization energy?
A) Al
B) Br
C) F
D) Mn
E) Na
D) 3, 1
E) 3, 2
4. How many bonds and lone pairs are present in CN-?
A) 2, 1
5.
B) 2, 2
C) 2, 3
Which of the following is isoelectronic to sulfide ion?
A) Ar
B) Na+
E) All are isoelectronic to sulfide
C) O2-
D) Se2-
6. Using the methodology we went over in class; predict place the following 3 bonds in order of
increasing bond polarity: C-S, B-F, N-O.
A) C-S < B-F < N-O
B) B-F < N-O < C-S
C) N-O < C-S < B-F
D) N-O < B-F < C-S
E) C-S < N-O < B-F
7. Which atom is most electronegative?
A) Al
8.
B) As
C) I
D) Li
Which of the following is not an odd electron molecule?
A) NO
B) NO2
E) All are odd electron molecules
C) NO3
D) N2O4
E) S
Discussion Questions: (You must show your work to receive credit.)
1. Write out the electron configurations for each of the following and provide the number of
unpaired electrons on each. (10 points)
Sb: [Kr] 5s2 4d10 5p3
Ni2+: [Ar] 3d8
2. What are the wavelength and frequency of a photon of light with an energy of 4.14 x 10-17 J.
Visible light is typically in the 400-700 nm range. Is this photon of higher or lower energy
than this and why? (10 points)
 = (4.14 x 10-17 J)/(6.626 x 10-34 J•s) = 6.25 x 1016 s-1
 = (3.00 x 108 m/s) /(6.25 x 1016 s-1) = 4.80 x 10-9 m
This 4.80 nm, which is much short wavelength than 400-700. Short wavelength light is
higher in energy than long wavelength light so this is higher energy.
3. For a given shell, place the 4 orbital types in order of increasing energy and explain your
ordering. (10 points)
s < p < d < f. In the same shell all have the same principle quantum number, n. As such,
they differ in that as one moves from s to f, the number of nodes increases. This has the
effect of pushing the electrons away from the nucleus. Separating oppositely charged
particles costs energy so the energy of the orbitals increases in the direction shown.
4. Consider atoms of the elements Cl, In, and Sb. Place them in order of increasing ionization
energy and explain the ordering. (10 points)
In < Sb < Cl. Indium and antimony are in the same period. Moving right in a period, each
new element adds one more proton and electron. There is a net greater attraction because the
new electron is not shielded by the other electrons in the same shell effectively so there is
more positive charge pulling on it. This holds the electron more and raises the energy
required to remove it. Thus, antimony has a higher ionization energy than indium.
Atomic size increases down a group because each new shell has a higher principal q.n. and so
the electron is more energetic and lies further from the nucleus. Thus, chlorine is smaller
than either indium or antimony. Again, electrons that are closer to the nucleus are harder to
remove so chlorine has the highest ionization energy.
5. Draw the Lewis structures of CO and CO2. (8 points)
..
..
:C
O:
:O
C
O:
6. The resonance Lewis structures of XeO4 can be put into three unique groups. Draw a
representative molecule of each group and show formal charges on each atom. (12 points)
..
:O:
.. Xe
:O
..
:O:
..
:O:
..
O:
..
FCXe = (8 - 4 - 0) = +4
FCO = (6 - 1 - 6) = -1
.. Xe
:O
..
:O:
..
:O:
..
O:
..
FCXe = (8 - 5 - 0) = +3
FCOsb = (6 - 1 - 6) = -1
FCOdb = (6 - 2 - 4) = 0
.. Xe
:O
..
:O:
..
..
O:
..
FCXe = (8 - 6 - 0) = +2
FCOsb = (6 - 1 - 6) = -1
FCOdb = (6 - 2 - 4) = 0
7. For the following reaction, estimate the heat of reaction, Hrxnº. (8 points)
4 NH3 (g) + 7 O2 (g)  4NO2 (g) + 6 H2O(g)
Bond
N-H
O-O
O=O
N-O
N=O
H-O
Enthalpy (kJ/mol) 388
146
495
201
607
463
H  [(12 mol)(388 kJ/mol) + (7 mol)(495 kJ/mol)]
– [(4 mol)(201 kJ/mol) + (4 mol)(607 kJ/mol) + (12 mol)(463 kJ/mol)]
 -667 kJ