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 $\mu = \sqrt{n(n+2)} \ BM$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

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

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