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Arrange the following orbitals in decreasing order of energy.
A. $n = 3, l = 0, m = 0$
B. $n = 4, l = 0, m = 0$
C. $n = 3, l = 1, m = 0$
D. $n = 3, l = 2, m = 1$
The correct option for the order is

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D > B > A > C

A > D > B > C

D > B > C > A

B > D > C > A

answer is D.

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Detailed Solution

(A) $n = 3, l = 0, m = 0; 3 s$
(B) $n = 4, l = 0, m = 0; 4 s$
(C) $n = 3, l = 1, m = 1; 3 p$
(D) $n = 3, l = 2, m = 1; 3 d$
Correct order of energy will be (D) > (B) > (C) > (A)

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