A) 1.73 and 1.73 BM
B) 1.73 and 5.92 BM
C) 0.0 and 1.73 BM
D) 0.0 and 5.92 BM
Correct Answer: D
Solution :
\[\because \] No. of unpaired electrons in [Fe(CN)6]4 = 0 \[\therefore \] Spin-only magnetic moment \[=\sqrt{n(n+2)}BM\]= 0 Again, no. of unpaired electrons in \[{{[Fe{{F}_{6}}]}^{3-}}=5\] \[\therefore \] Spin-only magnetic moment \[=\sqrt{n(n+2)}BM\] \[=\sqrt{5(5+2)}BM\] \[=\sqrt{35}BM\] = 5.916 BM
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