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# Magnetic moment of $K_3Fe(C_2O_4)_3.3H_2O$ is

$\begin{array}{1 1}(a)\;4.9B.M&(b)\; 5.92B.M\\(c)\;2.6B.M&(d)\;0.00B.M\end{array}$

Can you answer this question?

Magnetic moment $\mu=\sqrt{n(n+2)}$B.M
No of unpaired electrons $n=5$ in $K_3Fe(C_2O_4)_3.3H_2O$
Then $\mu=\sqrt{5(5+2)}$B.M
$\mu=\sqrt{35}$B.M
$Fe^{+3}:3d^54s^2$
$\mu=5.92$B.M
Hence (b) is the correct answer.
answered Mar 10, 2014