Question

Spin only magnetic moment of ni2+ in aqueous solution

Answer 1

In Ni2+ , electronic conf. will be 3d8 4s0 so in d-orbital there are 2 unpaired electrons so ,
n=2 and spin only magnetic moment =√n(n+2) =√2(2+2) =√2×4 =2√2
=2.8BM

Answer 2

Answer : The Spin only magnetic moment is  2.83 BM

Solution :

The electronic configuration of $Ni^{2+}$ is, $1s^22s^22p^63s^23p^63d^8$

The number of unpaired electrons in  $Ni^{2+}$ is 2

Formula used for spin only magnetic moment :

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

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the magnetic moment.

$\mu=\sqrt{2(2+2)}=2.83BM$

Therefore, the spin only magnetic moment is 2.83 BM