A) 3 BM
B) 6 BM
C) 4 BM
D) 5 BM
Correct Answer: C
Solution :
The electronic configuration of Mn is \[_{25}Mn=[Ar]\,3{{d}^{5}}4{{s}^{2}}\] \[M{{n}^{4+}}=[Ar]\,3{{d}^{3}}\] Thus, three unpaired electrons are present. \[\therefore \] Spin only magnetic moment,\[\mu =\sqrt{n(n+2)}\] \[\therefore \] \[n=3\] \[\therefore \] \[\mu =\sqrt{3(3+2)}\] \[=\sqrt{15}=3.87\] = 4 BMYou need to login to perform this action.
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