A) \[\sqrt{6}B.M\]
B) \[\sqrt{15}B.M\]
C) \[\sqrt{2}B.M\]
D) \[\sqrt{8}B.M\]
Correct Answer: D
Solution :
: spin only magnetic moment \[(\mu )\] \[=\sqrt{n(n+2)}BM\] where n = number of unpaired electrons. \[N{{i}^{2+}}:3{{d}^{8}}4{{s}^{0}}\] \[\therefore \] \[n=2\] \[\Rightarrow \]\[\mu =\sqrt{2(2+2)}BM\] \[\mu =\sqrt{8}BM\]
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