# Magnetic moment of $K_3Mn(C_2O_4)_3.3H_2O$ is

$\begin{array}{1 1}(a)\;\mu=5.92B.M&(b)\;\mu= 4.90B.M\\(c)\;\mu=1.38B.M&(d)\;\mu=3.6B.M\end{array}$

Magnetic moment $\mu=\sqrt{n(n+2)}$B.M
Where $n$=no of unpaired electrons
In $K_3Mn(C_2O_4)_3.3H_2O,Mn^{+3}=4s^03d^4$
$n=4$
$\mu=\sqrt{4(4+2)}$B.M
$\mu=\sqrt{24}$B.M
$\mu=4.90$B.M
Hence (b) is the correct answer.