The spin-only magnetic moment of Ni^{2+} in a | vedclass

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(C) The atomic number of Nickel $(Ni)$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In an aqueous solution,$Ni^{2+}$ ion is fo

Vedclass · Accepted Answer

$2.84$

Vedclass · Answer

(C) The atomic number of Nickel $(Ni)$ is $28$.
The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In an aqueous solution,$Ni^{2+}$ ion is formed by losing two electrons from the $4s$ orbital.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the $3d$ subshell,there are $8$ electrons. According to Hund's rule,these are filled as: $5$ electrons singly and $3$ electrons paired,resulting in $n = 2$ unpaired electrons.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ \mu_B$.
Substituting $n = 2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ \mu_B$.

The spin-only magnetic moment of $Ni^{2+}$ in an aqueous solution in Bohr magneton $(\mu_B)$ is .......

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