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The electrons identified by quantum numbers n and l
(1) n = 4, l = 1 (2) n = 4, l = 0 (3) n = 3, l = 2 (4) n = 3, l = 1
Can be placed in the order of increasing energy as

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(3) < (4) < (2) < (1)

(4) < (2) < (3) < (1)

(2) < (4) < (1) < (3)

(1) < (3) < (2) < (4)

answer is B.

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Detailed Solution

	n	l	nl(suborbit)	n + l
(a)	4	1	4p	5
(b)	4	0	4s	4
(c)	3	2	3d	5
(d)	3	1	3p	4

Higher the value of (n + l), higher is the energy. If (n +l) are same, then sub orbit with lower value of n has lower energy. Thus,

3p < 4s < 3d < 4p

(4) < (2) < (3) < (1)

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