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Question:

The quantum number of four electrons are given below: I. n=4, l=2, m_l=-2, m_s=-1/2 II. n=3, l=2, m_l=1, m_s=+1/2 III. n=4, l=1, m_l=0, m_s=+1/2 IV. n=3, l=1, m_l=1, m_s=-1/2 The correct order of their increasing energy will be:

IV < III < II < I

I < III < II < IV

IV < II < III < I

I < II < III < IV

Solution:

Correct option is B. IV < II < III < I
Solution:- (B) IV < II < III < I
n=4; l=2 → n+l=4+2=6 or n=4; l=2 → 4d
n=3; l=2 → n+l=3+2=5 or n=3; l=2 → 3d
n=4; l=1 → n+l=4+1=5 or n=4; l=2 → 4p
n=3; l=1 → n+l=3+1=4 or n=3; l=1 → 3p
According to (n+l) rule : 3p < 3d < 4p < 4d