The "spin-only" magnetic moment [ in units of | vedclass

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(C) The electronic configuration of $Ni$ $(Z = 28)$ is $[Ar] 3d^8 4s^2$.For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.In $3d^8$,there are $2$ unpaire

Vedclass · Accepted Answer

$2.84$

Vedclass · Answer

(C) The electronic configuration of $Ni$ $(Z = 28)$ is $[Ar] 3d^8 4s^2$.For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.In $3d^8$,there are $2$ unpaired electrons.The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ \mu_B$,where $n$ is the number of unpaired electrons.Substituting $n = 2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ \mu_B$.Thus,the closest value is $2.84 \ \mu_B$.

The "spin-only" magnetic moment [ in units of Bohr magneton,$(\mu_B)$ ] of $Ni^{2+}$ in aqueous solution would be (At. No. $Ni = 28$).

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