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Question

The quantum number of four electrons are given below: I. n=4,l=2,ml=âˆ’2,ms=âˆ’1/2 II. n=3,l=2,ml=1,ms=+1/2 III. n=4,l=1,ml=0,ms=+1/2 IV. n=3,l=1,ml=1,ms=âˆ’1/2 The correct order of their increasing energies will be:

A
I<III<II<IV
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B
I<II<III<IV
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C
IV<II<III<I
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D
IV<III<II<I
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Solution

The correct option is C IV<II<III<IFor energy calculations, more is the n+l value greater is the energy. If two electrons have the same value of n+l, the orbital with higher n value will have a higher energy. So the correct order will be IV<II<III<I

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