The "spin-only" magnetic moment [ i | vedclass

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Question

The number of unpaired electrons in $N i^{2+}(a q)=2$

Water is weak ligand hence no pairing will take place spin magnetic moment

$=\sqrt{n(n+2)}

Vedclass · Accepted Answer

$2.84$

Vedclass · Answer

The number of unpaired electrons in $N i^{2+}(a q)=2$

Water is weak ligand hence no pairing will take place spin magnetic moment

$=\sqrt{n(n+2)}=\sqrt{2(2+2)}$ $=\sqrt{8}=2.82$

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

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