Download 1 - WordPress.com

Tutorial 4
Chapter 4
Electronic Structure of Atoms and Periodic Table
- SOLUTION1. Explain the relation between quantum numbers and atomic orbitals.
Solution:
n
l
ml
Number of
orbitals
Atomic orbital
designation
１
2
０
0
0
0
1
1
1s
2s
3
1
1
2
-1, 0, 1
-1, 0, 1
-2, -1, 0, 1, 2
3
3
5
2px, 2py, 2pz
2px, 2py, 2pz
3dxy, , 3dyz,
.
.
.
3dxz,
3dx2- y2, 3dz2
.
.
.
.
.
.
.
.
.
.
.
.
2. Explain the meaning of the symbol 4d5.
Solution:
4 denotes the principal quantum number n
d denotes the angular momentum quantum number l
5 denotes the number of electrons in the orbital or subshell
3. What is the total number of orbitals associated with the principal quantum
number n = 3?
Solution:
For n = 3, l = 0, 1, 2.
So, there is one 3s orbital (n = 3, l = 0 and m l = 0);
There are three 3p orbitals (n = 3, l = 1 and m l = -1, 0, 1);
There are five 3d orbitals (n = 3, l = 2 and m l = -2, -1, 0, 1, 2)
 The total number of orbitals is 1 + 3 + 5 = 9
4. Give the values of n, l and ml for orbitals in the 4d subshell.
Solution:
n =4
1
l
=2
m l = -2, -1, 0, 1, 2
5. What is the maximum number of electrons that can be present in the principal
level for which n = 3?
Solution:
When n = 3, then l = 0, 1 and 2. The number of orbitals for each value of l is
given by
n
Number of orbitals (2l + 1)
0
1
2
1
3
5
Total number if orbitals is nine. Because of each orbital can accommodate
two electrons, the maximum number of electrons that can reside in the
orbitals is 9 x 2 = 18. Or by formulas 2n2 = 2(3)2 = 18
6. Write electron configurations for each of the following elements.
a. Mg
b. S
c. Ga
Solution:
a. Mg
1s22 s2 2p63s2 or
[Ne] 3s2
b. S
1s22 s2 2p63s23p4
or
c. Ga
1s22 s2 2p63s23p64s23d104p1
[Ne] 3s23p4
or
[Ar] 4s23d104p1


7. Write an orbital diagram for silicon.
Solution:

1s

2s


2p

3s

3p
8. Compare the electron configuration for:
a. Mg and Mg2+
b. F and FSolution:
a. Mg
1s22s22p63s2
b. F 1s22s22p5
Mg2+
F-
2
1s22s22p6
1s22s22p6
9. Write the four quantum numbers for an electron in a 3p orbital.
Solution:
For n = 3, l = 1 (because of p orbital), m l = -1, 0, 1, ms = either +1/2 or -1/2.
The possible ways to designate the electron is:
(3, 1, -1, +1/2)
(3, 1, -1, -1/2)
(3, 1, 0, +1/2)
(3, 1, 0, -1/2)
(3, 1, 1, +1/2)
(3, 1, 1, -1/2)
10. Define the terms of paramagnetic and diamagnetic and give an example.
Solution:
Paramagnetic:
Attracted to magnet. A paramagnetic substance contains one or more
unpaired electrons.
Example: Li
Diamagnetic:
Repelled by a magnet; a diamagnetic substance contains only paired
electrons.
Example: Be
11. Give the outer electron configuration for each of the following columns in the
periodic table.
a. 1A
Solution:
a. 1A
b. 2A
c. 5A
d. 7A
b. 2A
c. 5A
ns1
ns2
ns2np3
ns2 np5
3
d. 7A