The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is :

Solution

Chemistry ma’am · 3 steps

Name each orbital first, then rank them by n + l: the bigger the sum, the higher the energy.

1. Name the orbitals. l = 0 is s, l = 1 is p and l = 2 is d. So A is 3s, B is 4s, C is 3p and D is 3d. The value of m only picks one orbital within a subshell and does not change its energy.
2. Compare energies. In a multi-electron atom, the orbital with the larger n + l has higher energy. If two orbitals have the same n + l, the one with the larger n has higher energy.

   Set	Orbital	n + l	Energy rank
   (D)	3d	3 + 2 = 5	highest
   (B)	4s	4 + 0 = 4, larger n	second
   (C)	3p	3 + 1 = 4, smaller n	third
   (A)	3s	3 + 0 = 3	lowest

3. So. The decreasing order is 3d > 4s > 3p > 3s. That is (D) > (B) > (C) > (A), which is option (a).

The idea to keep. For multi-electron atoms, compare n + l first. When the sums tie, the orbital with the larger n has higher energy. The value of m never decides the order.

Answer from the NTA final answer key. Solution by PrepWiser.