The quantum number of four electrons are given below: I. $\mathrm{n}=4, \mathrm{I}=2, \mathrm{~m}_{\mathrm{l}}=-2, \mathrm{~m}_{\mathrm{s}}=-1/2$ II. $\mathrm{n}=3, \mathrm{I}=2, \mathrm{~m}_{\mathrm{l}}=1, \mathrm{~m}_{8}=+1/2$ III. $\mathrm{n}=4, \mathrm{I}=1, \mathrm{~m}_{\mathrm{l}}=0, \mathrm{~m}_{\mathrm{s}}=+1/2$ IV. $\mathrm{n}=3, \mathrm{I}=1, \mathrm{~m}_{\mathrm{l}}=1; \mathrm{~m}_{\mathrm{s}}=-1/2$ The correct order of their increasing energies will be:

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Atomic Structure

The quantum number of four electrons are given below:
I.

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

II.

n = 3, I = 2, m_{l} = 1, m_{8} = + 1/2

III.

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

IV.

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

The correct order of their increasing energies will be:

Answer: (C)

Order the electrons by the

(n + l)

rule, breaking ties by the lower

n

. Electron I is

n = 4, l = 2

, a

4 d

state with

n + l = 6

; II is

n = 3, l = 2

, a

3 d

state with

n + l = 5

; III is

n = 4, l = 1

, a

4 p

state with

n + l = 5

; and IV is

n = 3, l = 1

, a

3 p

state with

n + l = 4

. The smallest sum, IV, lies lowest; II and III share

n + l = 5

but II has the smaller

n

and so sits below III; I has the largest sum and is highest. The increasing energy order is IV

<

<

III

<

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option