Question

Magnetic momentum of an ion $M^{+3}$ is $\sqrt{35}$ BM. Then, number of electron in d orbital of m element will be _______.

Answer 1

Magnetic moment = $\sqrt{n(n+2)}$ where n is number of unpaired electrons

$\sqrt{n(n+2)}=\sqrt{35}$

$n(n+2)=35$

$n^2+2n-35=0$

$n^2+7n-5n-35=0$

$n(n+7)-5(n+7)=0$

$n=5,-7$

Thus, $n=5$

An ion $M^{+3}$ in 3d series of transition elements with 5 unpaired electrons has following electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$

Thus, number of electron in d orbital of M element = 5