# The spin only magnetic moment of $Cr^{x+}$ is $4.9\;BM$ . Find $x$

$(a)\;x=4\\(b)\;x=2 \\(c)\;x=1 \\(d)\;x=3$

$\mu=4.9\;BM=\sqrt {n(n+2)}$
unpaired $e^-=n=4$
Cont :- $3d^44s^0$
$Cr^{2+}$ ion poseses 4 uparide : x=2
Hence b is the correct answer.