Arrange the following orbitals in decreasing order of energy ? (A) $\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0$ (B) $\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0$ (C) $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0$ (D) $\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1$ The correct option for the order is :

(A) $\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}$

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Atomic Structure

Arrange the following orbitals in decreasing order of energy ?
(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

The correct option for the order is :

Answer: (B)

Orbital energy is compared through the

(n + l)

value, and when two orbitals share the same

(n + l)

the one with the larger

n

lies higher. The four orbitals are

n = 3, l = 0

(3s,

n + l = 3

n = 4, l = 0

(4s,

n + l = 4

n = 3, l = 1

(3p,

n + l = 4

) and

n = 3, l = 2

(3d,

n + l = 5

). The 3d orbital has the largest sum and sits highest; 4s and 3p both give 4, but 4s has the larger

n

and so lies above 3p; 3s is lowest. The orbitals in decreasing order of energy are

3 d > 4 s > 3 p > 3 s

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option