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

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Atomic Structure

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

n = 3, l = 2, m_{1} = 1, m_{s} = + 1/2

n = 4, l = 1, m_{1} = 0, m_{s} = + 1/2

n = 4, l = 2, m_{1} = - 2, m_{s} = - 1/2

n = 3, l = 1, m_{1} = - 1, m_{s} = + 1/2

The correct order of increasing energy is :

(A)D $<$ B $<$ A $<$ C
(B)D $<$ A $<$ B $<$ C✓ Correct
(C)B $<$ D $<$ A $<$ C
(D)B $<$ D $<$ C $<$ A

Answer: (B)

Solution

Energy of orbitals is determined by the (n + l) rule. For equal (n + l) values, the orbital with lower n has lower energy.

A: n=3, l=2 → 3d, (n+l)=5
B: n=4, l=1 → 4p, (n+l)=5
C: n=4, l=2 → 4d, (n+l)=6
D: n=3, l=1 → 3p, (n+l)=4

Since D has the lowest (n+l)=4, it has the lowest energy. C has the highest (n+l)=6, so highest energy. For A and B, both have (n+l)=5, but A has lower n=3, so A < B.

Order: D < A < B < C.

Answer: (B)

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Atomic Structure

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