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

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(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.$Ni^{2+}$ ion has the configuration $[Ar] 3d^8$.In an aqueous solution,$Ni^{2+}$

Vedclass · Accepted Answer

$\sqrt{8} \ BM$

Vedclass · Answer

(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.$Ni^{2+}$ ion has the configuration $[Ar] 3d^8$.In an aqueous solution,$Ni^{2+}$ forms the octahedral complex $[Ni(H_2O)_6]^{2+}$.In the $3d$ subshell,the $8$ electrons are arranged as follows: three orbitals are doubly occupied and two orbitals are singly occupied.Therefore,the number of unpaired electrons $(n)$ $= 2$.The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \ BM$.Substituting $n = 2$,we get $\mu = \sqrt{2(2 + 2)} = \sqrt{8} \ BM$.

The "spin only" magnetic moment of $Ni^{2+}$ in aqueous solution would be [At. no. of $Ni = 28$]

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