Which set of quantum numbers cannot specify an orbital?
a. n = 2, l = 1, m1 = – 1
b. n = 3, l = 2, m1 = 0,
c. n = 3, l = 3, m1 = 2
d. n = 4, l = 3, m1 = 0
Answer: c. n = 3, l = 3, m1 = 2
Explanation
Basing from the electronic configuration:
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
Azimuthal quantum numbers describe the subshell and give the magnitude of the orbital angular momentum through the relation. It is represented by the symbol l. The s, p, d, and f orbitals have an l value of 0, 1, 2, and 3 respectively.
Magnetic quantum numbers are the projection of the orbital angular momentum along a given axis is obtained by describing the energy levels available within a subshell. It is represented by the symbol ml. The values of ml range from -l to l.
Given that:
a. n = 2, l = 1, ml = -1. This would be 2p and the ml = -1 is possible and since 2p exist in the electronic configuration. Therefore, it can occur together to specify an orbital.
b. n = 3, l = 2, ml = 0. This would be 3d and the ml = 0 is possible and since 3d exists in the electronic configuration. Therefore, it can occur together to specify an orbital.
c. n = 3, l = 3, ml = 2. This would be 3f and the ml = 2 is possible and since 3f does not exist in the electronic configuration. Therefore, it cannot occur together to specify an orbital.
d. n = 4, l = 3, ml = 0. This would be 4s and the ml = 0 is possible and since 4s exist in the electronic configuration.
Therefore, it can occur together to specify an orbital.
