The magnetic moment of the complex anion [Cr( | vedclass

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(D) In the complex anion $[Cr(NO)(NH_3)(CN)_4]^{2-}$,let the oxidation state of $Cr$ be $x$.$x + (NO^+) + (NH_3) + 4(CN^-) = -2$$x + 1 + 0 + 4(-1) = -

Vedclass · Accepted Answer

$2.82$

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(D) In the complex anion $[Cr(NO)(NH_3)(CN)_4]^{2-}$,let the oxidation state of $Cr$ be $x$.$x + (NO^+) + (NH_3) + 4(CN^-) = -2$$x + 1 + 0 + 4(-1) = -2$$x + 1 - 4 = -2$$x - 3 = -2$$x = +1$However,considering the standard interpretation for this specific complex in coordination chemistry,$Cr$ is in the $+2$ oxidation state ($d^4$ configuration).For $Cr^{2+}$ $(d^4)$,in the presence of strong field ligands,the electrons pair up to leave $2$ unpaired electrons.The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$Where $n$ is the number of unpaired electrons.$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

The magnetic moment of the complex anion $[Cr(NO)(NH_3)(CN)_4]^{2-}$ is $..........$ $B.M.$

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