The spin magnetic moment of cobalt in Hg[Co(S | vedclass

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Question

(C) The spin-only magnetic moment is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ B.M.$In the complex $Hg[Co(SCN)_4]$,$Hg$ is in the $+2$

Vedclass · Accepted Answer

$\sqrt{15}$

Vedclass · Answer

(C) The spin-only magnetic moment is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ B.M.$In the complex $Hg[Co(SCN)_4]$,$Hg$ is in the $+2$ oxidation state,so the complex ion is $[Co(SCN)_4]^{2-}$.Let the oxidation state of $Co$ be $x$. Then $x + 4(-1) = -2$,which gives $x = +2$.The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.In a tetrahedral field,$3d^7$ configuration results in $3$ unpaired electrons $(n=3)$.Substituting $n=3$ into the formula: $\mu_{eff} = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$Thus,option $C$ is correct.

The spin magnetic moment of cobalt in $Hg[Co(SCN)_4]$ is

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