Identify the complex ion with spin only magne | vedclass

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(D) The spin-only magnetic moment $(\mu_{eff})$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For

Vedclass · Accepted Answer

$[MnCl_6]^{3-}$

Vedclass · Answer

(D) The spin-only magnetic moment $(\mu_{eff})$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $\mu = 4.90 \ BM$,$n$ must be $4$ (since $\sqrt{4(4+2)} = \sqrt{24} \approx 4.90$).
$A$: $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $d^6$. $NH_3$ is a strong field ligand,causing pairing. $n = 0$.
$B$: $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $d^3$. $n = 3$.
$C$: $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $d^4$. $CN^{-}$ is a strong field ligand,causing pairing. $n = 2$.
$D$: $[MnCl_6]^{3-}$: $Mn^{3+}$ is $d^4$. $Cl^{-}$ is a weak field ligand,no pairing occurs. The configuration is $t_{2g}^3 e_g^1$,resulting in $n = 4$ unpaired electrons.
Therefore,$[MnCl_6]^{3-}$ has a magnetic moment of $4.90 \ BM$.

Identify the complex ion with spin only magnetic moment of $4.90 \ BM$.

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