The “spin only” magnetic moment of Ni2+ in aqueous solution would be [At.No.of Ni=28]

Ni : [Ar] 3d8 4s2Ni2+ : [Ar] 3d8 4s0Number of unpaired electrons, n = 2μ=n(n+2)=2(2+2)=8 B.M

The “spin only” magnetic moment of Ni2+ in aqueous solution would be [At.No.of Ni=28]

Ni : [Ar] 3d8 4s2Ni2+ : [Ar] 3d8 4s0Number of unpaired electrons, n = 2μ=n(n+2)=2(2+2)=8 B.M

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The “spin only” magnetic moment of Ni²⁺ in aqueous solution would be [At.No.of Ni=28]

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$\sqrt{6}$ B.M

$\sqrt{15}$ B.M

$\sqrt{2}$ B.M

$\sqrt{8}$ B.M

answer is D.

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Detailed Solution

Ni : [Ar] 3d⁸ 4s²
Ni²⁺ : [Ar] 3d⁸ 4s⁰
Number of unpaired electrons, n = 2
$μ = \sqrt{n (n + 2)} = \sqrt{2 (2 + 2)} = \sqrt{8} B.M$

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The “spin only” magnetic moment of Ni2+ in aqueous solution would be [At.No.of Ni=28]

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The “spin only” magnetic moment of Ni²⁺ in aqueous solution would be [At.No.of Ni=28]