The magnetic moment of the complex [Cr(CN)_6] | vedclass

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Question

(C) In the complex $[Cr(CN)_6]^{3-}$, the oxidation state of $Cr$ is $+3$. $Cr$ $(Z=24)$ has the electronic configuration $[Ar] 3d^5 4s^1$. $Cr^{3+}$

Vedclass · Accepted Answer

$3.87$

Vedclass · Answer

(C) In the complex $[Cr(CN)_6]^{3-}$, the oxidation state of $Cr$ is $+3$. $Cr$ $(Z=24)$ has the electronic configuration $[Ar] 3d^5 4s^1$. $Cr^{3+}$ has the configuration $[Ar] 3d^3$. Since $CN^-$ is a strong field ligand, the $3$ electrons remain in the $t_{2g}$ orbitals. The number of unpaired electrons $(n)$ is $3$. The magnetic moment $(\mu)$ is calculated as $\mu = \sqrt{n(n+2)} \ BM$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

List-$I$ (Complex ion)	List-$II$ (Calculated spin only magnetic moment $(BM)$)
$A$. $[Cr(H_2O)_6]^{2+}$	$I$. $3.87$
$B$. $[Co(H_2O)_6]^{2+}$	$II$. $5.92$
$C$. $[Cu(H_2O)_6]^{2+}$	$III$. $4.90$
$D$. $[Mn(H_2O)_6]^{2+}$	$IV$. $1.73$

The magnetic moment of the complex $[Cr(CN)_6]^{3-}$ is ..... (in $BM$)

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