The electrons identified by quantum numbers n and I : (a) $n=4, I=1$ (b) $\mathrm{n}=4, \mathrm{I}=0$ (c) $n=3$, I $=2$ (d) $n=3, I=1$ Can be placed in order of increasing energy as :

(B) (d) $<$ (b) $<$ (c) $<$ (a)

(A) (c) $<$ (d) $<$ (b) $<$ (a)

(C) (b) $<$ (d) $<$ (a) $<$ (c)

(D) (a) $<$ (c) $<$ (b) $<$ (d)

JEE Main · 2012 · 29 Apr 2012 · Offline · Atomic Structure

The electrons identified by quantum numbers n and I :
(a)

n = 4, I = 1

(b)

n = 4, I = 0

(c)

n = 3

, I

= 2

(d)

n = 3, I = 1

Can be placed in order of increasing energy as :

Answer: (B)

Orbital energies follow the

(n + l)

rule: the smaller

(n + l)

lies lower, and for a tie the one with the smaller

n

lies lower. The four sets give

(n + l)

values of

5

for

(a) n = 4, l = 1

;

4

for

(b) n = 4, l = 0

;

5

for

(c) n = 3, l = 2

; and

4

for

(d) n = 3, l = 1

. Of the two tying at

4

(d)

with the smaller

n

is lower; of the two tying at

5

(c)

with the smaller

n

is lower. The increasing order of energy is therefore

(d) < (b) < (c) < (a)

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option