Compare the energies of following sets of quantum numbers for multielectron system.
(A) n=4,1=1
(B) n=4,1=2
(C) n=3, l=1
(D) n=3,1=2
(E) n=4,1=0
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Name each orbital, add n + l, and when the sums tie, the smaller n comes first.

1. Energy rule. In a multielectron atom, energy rises with n + l. If two orbitals have the same n + l, the one with the smaller n is lower in energy.

   Set	Orbital	n + l	Tie-break
   (C)	3p	3 + 1 = 4	n = 3, lower than 4s
   (E)	4s	4 + 0 = 4	n = 4, above 3p
   (D)	3d	3 + 2 = 5	n = 3, lower than 4p
   (A)	4p	4 + 1 = 5	n = 4, above 3d
   (B)	4d	4 + 2 = 6	highest sum, no tie

2. Order them. Sum 4 gives 3p below 4s, sum 5 gives 3d below 4p, and 4d alone has sum 6. This is the same order in which these orbitals fill: 3p, 4s, 3d, 4p, 4d.
3. So. The energy order is (C) < (E) < (D) < (A) < (B), which is option (c).

The idea to keep. For multielectron atoms, rank orbitals by n + l first. Only when two sums are equal do you look at n, and the smaller n wins the lower energy.

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