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 = 1Can 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

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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