Download Assignment 11 Solutions

Chem1000A
Fall 2003
Assignment 11 - Solutions
Two or three questions will be marked from each assignment.
DUE ON Nov. 28, 2003 (Friday) 1:00 PM
To be dropped off at my office (C886)
1. What is the maximum number of electrons that can be associated with the following combinations
of quantum numbers?
(a) n = 3, l = 2
These are the 3d orbitals: a maximum of 10 electrons
(b) n = 5
This is the fifth shell. We have the following orbitals in the 5th shell (with their respective maximum
electron count):
5s, l = 0(2)
5p, l = 0, ±1 (6)
5d, l = 0, ±1, ±2 (10)
5f, l = 0, ±1, ±2, ±3 (14)
5g, l = 0, ±1, ±2, ±3, ±4 (18)
in total: a maximum of 50 electrons
(c) n = 1, l = 0, ml = 0
This is the 1s orbital: a maximum of two electrons
(d) n = 6, l = 4, ml = -4, ms = ½
This set specifies one electron (spin up) in one of the nine different 6g orbital: one electron
2. Which of the following combinations of quantum numbers are not allowed? Explain your answer.
(a) n = 6, l = 6, ml = -4, ms = ½
incorrect, since l = n. The maximum l is n-1.
incorrect, since ml > l. This is not possible, ml = -l…0…+l.
(b) n = 5, l = 4, ml = 5, ms = ½
(c) n = 2, l = 1, ml = -1, ms = -½
correct, one electron in one of the three 2p orbitals
(d) n = 6, l = 3, ml = 0, ms = 1
incorrect. Ms can only be ½ or -½.
(e) n = 4, l = -3, ml = 0, ms = -½
incorrect, since l has to be a positive number (or zero).
3. How many 3d orbitals exist? Draw all of the 3d orbitals.
Five 3d orbitals exist: 3dxy, 3dxz, 3dyz, 3dx2-y2 3dz2
x
x
z
x
y
z
y
dxy
dxz
dyz
x
y
z
dx2-y2
dz2
4. Give the electron configuration of Sb (antimony) in (a) the spectroscopic notation, (b) the orbital
box notation, and (c) the noble gas notation.
(a) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3
1
(b)
↑↓
1s
↑↓
2s
↑↓
5s
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4d
↑↓ ↑↓ ↑↓
2p
↑↓
3s
↑↓ ↑↓ ↑↓
3p
↑↓
4s
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
3d
↑↓ ↑↓ ↑↓
4p
↑ ↑ ↑
5p
(c) [Kr] 5s2 4d10 5p3
5.Give the electron configuration of the following ions and their neutral parent atoms using the
spectroscopic notation.
(a) Ti2+
Ti: 1s2 2s2 2p6 3s2 3p6 4s2 3d2
Ti2+: 1s2 2s2 2p6 3s2 3p6 3d2
(b) Pd2+
Pd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 (this is the actual electron configuration of Pd due to the
stability of the filled 4d subshell. 5s2 4d8 would have been accepted as an answer.
Pd2+: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d8
(c) Sn4+
Sn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2
Sn2+: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
6. What is an orbital?
An orbital is a wavefunction of an electron, a function that describes the electron as a standing
matter wave.
7. What is the name of the equation: HΨ=EΨ
Schrödinger equation
8. Give the electron configuration of Cr using the noble gas notation. Discuss the exceptional electron
configuration using Hund’s rule.
Chromium has a half filled 3d subshell and a half-filled 4s orbital. Both types of orbitals are that
close in energy that we have to consider Hund’s rule. Hund’s rule states that the most stable electron
configurations is that with a maximum number of unpaired electrons in degenerate orbitals, all with
the same spin direction. Strictly speaking, Hund’s rule is only valid for degenerate orbitals, i.e.,
orbitals of the same energy. However, the 4s and 3d orbitals are very close in energy and Hund’s
rule has to be applied in this case.
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