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:

Question

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:

Quantrex Academy · Accepted Answer

Correct answer: A. D > B > C > A

In multi-electronic species, energy is decided on the basis of ( $n + l$ ) rule.

$3s = 3 + 0 + 0 = 3$

$4s = 4 + 0 + 0 = 4$

$3p = 3 + 1 + 0 = 4$

$3d = 3 + 2 + 1 = 6$

For 4s and 3p, an orbital with a higher value of n has higher energy.

So increasing the order of energy is $3d > 4s > 3p > 3s$ .

Arrange the following orbitals in decreasing order of energy. A. n...

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