Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. (A) $\mathrm{n}=3, \ell=2, \mathrm{~m}=+1$ (B) $n=4, \ell=0, m=0$ (C) $\mathrm{n}=6, \ell=1, \mathrm{~m}=0$ (D) $\mathrm{n}=5, \ell=1, \mathrm{~m}=+1$ (E) $\mathrm{n}=2, \ell=1, \mathrm{~m}=+1$ Choose the correct answer from the options given below :

(D) E $<$ B $<$ A $<$ D $<$ C

(A) C $<$ D $<$ B $<$ A $<$ E

(B) B $<$ A $<$ E $<$ C $<$ D

(C) E $<$ C $<$ D $<$ B $<$ A

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Atomic Structure

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
(A)

n = 3, ℓ = 2, m = + 1

(B)

n = 4, ℓ = 0, m = 0

(C)

n = 6, ℓ = 1, m = 0

(D)

n = 5, ℓ = 1, m = + 1

(E)

n = 2, ℓ = 1, m = + 1

Choose the correct answer from the options given below :

Answer: (D)

For a multi-electron atom the orbital energy rises with the sum

n + ℓ

, and when two sets share the same sum the one of smaller

n

lies lower. The given sets correspond to

3 d

for A with

n + ℓ = 5

4 s

for B with

4

6 p

for C with

7

5 p

for D with

6

and

2 p

for E with

3

. Ordering from the smallest sum upwards gives

E < B < A < D < C

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option