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

(A) (D) $>$ (B) $>$ (C) $>$ (A)

(B) (B) $>$ (D) $>$ (C) $>$ (A)

(C) (C) $>$ (B) $>$ (D) $>$ (A)

(D) (B) $>$ (C) $>$ (D) $>$ (A)

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Atomic Structure

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, 1 = 0, m = 0

(C)

n = 3, l = 1, m = 0

(D)

n = 3, l = 2, m = 1

Answer: (A)

The energy of orbitals is determined by the

n + ℓ

rule: higher

n + ℓ

means higher energy. If

n + ℓ

values are equal, the orbital with higher

n

has higher energy.

For the given orbitals:

Since (B) and (C) both have

n + ℓ = 4

, we compare their

n

values: (B) has

n = 4

while (C) has

n = 3

, so (B) > (C).

Therefore, the decreasing order of energy is: (D) > (B) > (C) > (A).

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option