A) 3.62 BM
B) 9.72 BM
C) 7.9 BM
D) 10.60 BM
Correct Answer: C
Solution :
[c] \[Gd\left( 64 \right)=\left[ Xe \right]4{{f}^{7}}5{{d}^{1}}6{{s}^{2}}\] \[G{{d}^{3+}}=[Xe]\,4{{f}^{7}}5{{d}^{0}}6{{s}^{0}}\] i.e. no. of unpaired electrons = 7 \[\mu =\sqrt{n(n+2)}=\sqrt{7(7+2)}\] \[=\sqrt{63}=7.93BM\]
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