Chem1000a - Set #2 - Answers
Fall 2003
Study Questions:
1. What is the electron configuration of Sc and Cl? Write these configurations using both
the spectroscopic notation and orbital box diagrams. Describe the position of the
elements in the periodic table.
Sc: 1s2 2s2 2p6 3s2 3p6 4s2 3d1
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑
1s
2s
2p
3s
3p
4s
3d
Scandium is a transition metal element. Sc is in group 3, row 4. Alternatively, you can
say it is a frist row transition metal element (since it is in the first row of the d-block).
Cl: 1s2 2s2 2p6 3s2 3p5
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑
1s
2s
2p
3s
3p
Chlorine is a main-group element. Cl is a halogen (group 17 element), second period
(row).
2. Which of the following combination of quantum numbers is wrong? Explain your
answer.
a) n = 3, l = 3, ml = -2, ms = -1/2
b) n = 4, l = 2, ml = -2, ms = -1
c) n = 1, l = 0, ml = 1, ms = 1/2
d) n = 5, l = 0, ml = 0, ms = -1/2
e) n = 2, l = 1, ml = -1/2, ms = +1/2
f) n = 0, l = 0, ml = 0, ms = -1/2
a) Impossible combination: n = 3, l = 3. A 3f orbital does not exist.
b) Incorrect quantum number: ms can only be +1/2 or -1/2
c) Impossible combination: A 1s orbital (n = 1, l = 0) cannot have ml = 1.
d) This is a 5s orbital with spin down. Correct.
e) Incorrect quantum number: ml cannot be -1/2, ml can only have integer values.
f) Incorrect quantum: n cannot be 0, n = 1 is the first shell.
3. Which of the following statements are correct?
(a) Compounds with unpaired electrons are diamagnetic.
(b) Paramagnetic compounds are drawn into a strong magnetic field.
(c) Ferromagnetic compounds are repelled by a magnetic field.
(d) Electrons have a magnetic moment.
a) incorrect. Compounds with unpaired electrons are paramagnetic.
b) correct
c) incorrect. Ferromagnetic compounds are drawn into a magnetic field.
d) correct
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4. What is the maximum number of electrons that can be identified with each of the
following sets of quantum numbers?
a)
b)
c)
d)
e)
n= 2, l = 1
n=3
n = 3, l = 2
n = 4, l = 1, ml = -1, ms = -1/2
n = 5, l = 0, ml = +1
a) 2p orbitals: six electrons
b) third shell: two s electrons, 6 p electrons and 10 d electrons = in total: 18 electrons
c) 3d orbitals: ten electrons
d) one specific 4p orbital with an electron spin down: this set specifies a single electron.
e) this is an incorrect combination. A 5s orbital cannot have ml = 1.
5. Depict electron configurations for V2+, V3+, and Co3+. Use orbital box diagrams and
the noble gas notation. Are any of the ions paramagnetic? If so, give the number of
unpaired electrons.
V2+: [Ar] 4s0 3d3
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
↑↓ ↑↓ ↑↓
3p
4s
↑ ↑ ↑
3d
V3+: [Ar] 4s0 3d2
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
↑↓ ↑↓ ↑↓
3p
4s
↑ ↑
3d
Co3+: [Ar] 4s0 3d6
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
↑↓ ↑↓ ↑↓
3p
4s
↑↓ ↑ ↑ ↑ ↑
3d
All of the three cations (V2+, V3+, and Co3+) are paramagnetic, with three, two and four
unpaired electrons, respectively.
6. Name two elements for each of the following categories.
(a) main-group elements
(b) alkaline earth elements
(c) transition metal elements
(d) halogens
(e) row 5 elements
a) Si and P
b) Mg and Ca
c) Mo and Ti
d) F and Br
e) Rb and Sn
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7. A possible excited state for the H atom has an electron in a 5d orbital. List all possible
sets of quantum numbers, n, l, ml for this electron.
n = 5, l = 2, ml = -2
n = 5, l = 2, ml = -1
n = 5, l = 2, ml = 0
n = 5, l = 2, ml = 1
n = 5, l = 2, ml = 2
8. State which of the following orbitals cannot exist according to the quantum theory: 2s,
2d, 3p, 3f, 4f, and 5s. Briefly explain your answer.
2d and 3f cannot exist. The angular momentum quantum number l can only have integer
numbers from 0 to n-l.
For n= 2, l = 1 (p orbital) is the highest number for l.
For n = 3, l = 2 (d orbital) is the highest number for l.
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9. (a) Draw the molecular orbital (MO) diagram for the nitrosyl ion, NO+. (b) Is NO+
diamagnetic or paramagnetic? If it is paramagnetic how many unpaired electrons does it
have? (c) Determine the bond order of the NO bond in NO+ from the MO diagram. (d)
When NO is ionized to from NO+, is the nitrogen-oxygen bond becoming stronger or
weaker than in NO?
a)
N+
NO+
O
E
σ2p*
π2p*
↑ ↑
2p
↑↓ ↑ ↑
2p
↑↓ σ2p
↑↓ ↑↓ π2p
↑↓
σ2s*
↑↓
2s
↑↓
2s
↑↓
σ2s
Alternatively, N can be combined with O+. The core orbitals are omitted in the diagram.
Electron configuration of NO+: (σ1s)2 (σ1s*)2 (σ2s)2 (σ2s*)2 (π2p)4 (σ2p)2
b) The NO+ cation is diamagnetic, since all electrons are paired.
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c) bond order
= ½ (number of electrons in bonding MO’s – number of electrons in antibonding
MO’s
= ½(8-2) = ½(6) = 3
The bond order of the nitrogen-oxygen bond in NO+ is three (triple bond).
d) The additional electron in NO is located in an antibonding MO. Therefore, the bond
order in NO is 2.5 (as discussed in class). Upon ionization of NO, the nitrong-oxygen
bond becomes stronger, i.e., the bond order increases from 2.5 (in NO) to 3 (in NO+).
10. Draw the Lewis structure of NO+, include all resonance structures (only paired
electrons). Discuss which resonance structure describes the cation the best considering
the information that you obtained from the MO diagram. Discuss whether this finding is
in agreement with your expectation without the knowledge of the MO theoretical results.
+ ..
:N O
:N O +
:
..
These two resonance structure are the only structures that contain one formal charge.
The first resonance structure provides the best description of the NO+ structure since it
contains a triple bond, which is also found in the MO diagram. The second resonance
structure has no significant contribution to the real structure.
The unfavorable aspect of the first resonance structure is the localization of the positive
formal charge on oxygen, the more electronegative element in the molecule. In the
second resonance structure the positive formal charge is located on nitrogen, which is
less electronegative than oxygen, however, nitrogen has an electron sextet, which is not
favorable. Therefore, the first resonance structure with electron octets around nitrogen
and oxygen is far more stable. In conclusion, the resonance structure description of the
NO+ cation is in agreement with the MO diagram.
11. Describe the bonding situation in the NO+ cation using VB theory: (a) What is the
hybridization of nitrogen and oxygen in this molecule, (b) describe the orbitals involved
in the formation of the multiple bonds in this ion. (c) How many σ and how many π
bonds do you have in the NO+ cation. (d) Draw an energy level diagram indicating the
hybridization of the nitrogen, starting from the atomic orbitals yielding the hybrid
orbitals.
(a) Nitrogen and oxygen in NO+ are sp hybridized.
(b) A σ bond is formed by overlap of an sp hydrid orbital on nitrogen with one sp hybrid
orbital on oxygen. One π bond is formed by overlap of the px orbital on nitrogen and the
px orbital on oxygen (the molecular axis is taken as the z-axis). The second π bond is
formed by overlap of the py orbital on nitrogen and the py orbital on oxygen.
(c) One σ bond and two π bonds are present between N and O in the NO+ cation.
(d)
E
↑ ↑ ↑
↑ ↑
2p
px py
↓↑ ↑
sp
↓↑
2s
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12. How many valence d electron do the following ion have:
(a) Fe2+
six d electrons
(b) Co3+
six d electrons
3+
three d electrons
(c) Cr
(d) Zn2+
ten d electrons
3+
(e) In
ten d electrons
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