Given below are the quantum numbers for $4$ electrons.
A. $n = 3, l = 2, {m}_{1} = 1, {m}_{s} = + \dfrac{1}{2}$
B. $n = 4, l = 1, {m}_{1} = 0, {m}_{s} = + \dfrac{1}{2}$
C. $n = 4, l = 2, {m}_{1} = - 2, {m}_{s} = - \dfrac{1}{2}$
D. $n = 3, l = 1, {m}_{1} = - 1, {m}_{s} = + \dfrac{1}{2}$
The correct order of increasing energy is

Question

Given below are the quantum numbers for $4$ electrons.

A. $n = 3, l = 2, {m}_{1} = 1, {m}_{s} = + \dfrac{1}{2}$

B. $n = 4, l = 1, {m}_{1} = 0, {m}_{s} = + \dfrac{1}{2}$

C. $n = 4, l = 2, {m}_{1} = - 2, {m}_{s} = - \dfrac{1}{2}$

D. $n = 3, l = 1, {m}_{1} = - 1, {m}_{s} = + \dfrac{1}{2}$

The correct order of increasing energy is

Quantrex Academy · Accepted Answer

Correct answer: B. D < A < B < C

Energy order of sub-shell decided by (n + l) value. Greater the (n + l) value of a sub-shell greater is its energy. If two sub-shells have same (n + l) values the orbital with greater principal quantum number has greater energy A ⇒ 3d ⇒ (n + l) = 3 + 2 = 5 B ⇒ 4p ⇒ n + l = 4 + 1 = 5 C ⇒ 4d ⇒ n + l = 4 + 2 = 6 D ⇒ 3s ⇒ (n + l) = 3 + 0 = 3 D

Given below are the quantum numbers for 4 electrons. A. n = 3, l =...

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