A) \[3BM\]
B) \[6BM\]
C) \[4BM\]
D) \[5BM\]
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, \[\mu =\sqrt{n(n+2)}\] \[\therefore \] \[n=3\] \[\therefore \] \[\mu =\sqrt{3(3+2)}\] \[=\sqrt{15}=3.87\] \[=4BM\]You need to login to perform this action.
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