A) \[~n=3,l=0,m=0,s=-\text{ }1/2\]
B) \[~n=2,l=3,m=0,s=+1/2\]
C) \[~n=3,l=1,m=0,s=-1/2\]
D) \[~n=3,l=2,m=1,s=+\,1/2\]
Correct Answer: D
Solution :
For \[3d\]orbital, \[n=3\] For \[d\]orbital,\[l=2\] and \[m=-2,-1,0,+1,+\,2\] \[s=\pm \,\frac{1}{2}\] \[\therefore \] The correct set for 3d orbital is \[n=3,l=2,m=1,s+\frac{1}{2}\]You need to login to perform this action.
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