Question

The "Spin-only" magnetic moment [in units of Bohr magneton,$\left({\mu }_B\right)]$ of ${Ni}^{2+}$ in aqueous solution would be (At. no. $Ni=28$):

• A. $0$
• B. $1.73$
• C. $2.84$
• D. $4.90$

Answer 1

The correct option is C $2.84$

The electronic configuration of $N{i}^{+2}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$

Number of unpaired electrons in $N{i}^{+2}$ is 2 formula used for the spin only magnetic moment

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

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

$\mu =\sqrt{8}$

$\mu =2.84BM$

Hence, the correct option is $\text{C}$