A) \[n=4,\,\ell =1,\,m=0,\,s=+1/2\]
B) \[n=4,\,\ell =0,\,m=0,\,s=+1/2\]
C) \[n=4,\,\ell =2,\,m=1,\,s=+1/2\]
D) \[n=3,\,\ell =1,\,m=0,\,s=+1/2\]
Correct Answer: D
Solution :
[d] The lower the value of \[(n+l)\] for an orbital, the lower is its energy. If two orbitals have the same \[(n+l)\] value, the orbital with lower value of n has the lower energy i.e., highest stability. Set [d] has the lower value of n, thus has highest stability.
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