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Question
The quantum number of four electrons are given below:
I. n=4,l=2,ml=−2,ms=−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=−1/2
The correct order of their increasing energies will be:
The quantum number of four electrons are given below:
I. n=4,l=2,ml=−2,ms=−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=−1/2
The correct order of their increasing energies will be:
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Solution
The correct option is C IV<II<III<I
For 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
For 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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