The 'Spin only' magnetic moment for [Ni(NH_3) | vedclass

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(B) $NH_3$ acts as a weak field ligand with $Ni^{2+}$.The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.For $Ni^{2+}$,the configuratio

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$28$

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(B) $NH_3$ acts as a weak field ligand with $Ni^{2+}$.The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.For $Ni^{2+}$,the configuration is $3d^8$.The $3d$ orbitals are filled as follows: two orbitals are doubly occupied,and two orbitals are singly occupied (total $2$ unpaired electrons).The spin-only magnetic moment is calculated as:$\mu = \sqrt{n(n+2)} \ BM$Where $n = 2$ (number of unpaired electrons).$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.To express this as $x 	imes 10^{-1} \ BM$,we have $2.82 = 28.2 	imes 10^{-1}$.Rounding to the nearest integer,$x = 28$.

The 'Spin only' magnetic moment for $[Ni(NH_3)_6]^{2+}$ is $ . . . . . . \times 10^{-1} \ BM$.
(Given: Atomic number of $Ni = 28$)

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The 'Spin only' magnetic moment for $[Ni(NH_3)_6]^{2+}$ is $ . . . . . . \times 10^{-1} \ BM$.(Given: Atomic number of $Ni = 28$)