A) 5.91 BM
B) 3.87 BM
C) 1.73 BM
D) 2.82 BM
Correct Answer: D
Solution :
[d] In \[{{[Cr\left( NO \right)\left( N{{H}_{3}} \right){{\left( CN \right)}_{4}}]}^{2-}},\] \[C{{r}^{2+}}({{d}^{4}})\] is given as: i.e., 2 unpaired electrons \[\mu =\sqrt{2(2+2)}=\sqrt{8}=2.82\,BM\]You need to login to perform this action.
You will be redirected in
3 sec