Question

The quantum numbers of six electrons are given below. Arrange them in order of increasing energy. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = $-\frac{1}{2}$
2. n = 3, l = 2, ml = 1 , ms = $+\frac{1}{2}$
3. n = 4, l = 1, ml = 0 , ms = $+\frac{1}{2}$
4. n = 3, l = 2, ml = –2 , ms = $-\frac{1}{2}$
5. n = 3, l = 1, ml = –1 , ms = $+\frac{1}{2}$
6. n = 4, l = 1, ml = 0 , ms = $+\frac{1}{2}$

Answer 1

For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).