Assignment 4 - SOLUTIONS
DUE ON OCT 7, 2002 (MONDAY) 1:00 PM
To be dropped off at my office (C884)
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.
(will be marked)
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)
b)
c)
d)
e)
f)
Impossible combination: n = 3, l = 3. A 3f orbital does not exist.
Incorrect quantum number: ms can only be +1/2 or -1/2
Impossible combination: A 1s orbital (n = 1, l = 0) cannot have ml = 1.
This is a 5s orbital with spin down. Correct.
Incorrect quantum number: ml cannot be -1/2, ml can only have integer values.
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.
(will be marked)
V2+: [Ar] 4s0 3d3
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
↑↓ ↑↓ ↑↓
3p
V3+: [Ar] 4s0 3d2
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
Co3+: [Ar] 4s0 3d6
↑↓
↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
↑↓
3s
4s
↑ ↑ ↑
3d
↑↓ ↑↓ ↑↓
3p
4s
↑ ↑
3d
↑↓ ↑↓ ↑↓
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. (will be marked)
(a) main-group elements
(b) alkaline earth elements
(c) transition metal elements
(d) halogens
(e) row 5 elements
a)
b)
c)
d)
e)
Si and P
Mg and Ca
Mo and Ti
F and Br
Rb and Sn
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