For a tetrahedral complex [MCl_{4}]^{2-},the | vedclass

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(A) Given,$\mu = 3.83 \ BM$. The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.

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$Co$

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(A) Given,$\mu = 3.83 \ BM$. The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. $3.83 = \sqrt{n(n+2)}$ $\Rightarrow n(n+2) \approx 14.67$ $\Rightarrow n \approx 3$. Thus,the complex $[MCl_{4}]^{2-}$ has $3$ unpaired electrons. In $[MCl_{4}]^{2-}$,the oxidation state of $M$ is $+2$. For $Co^{2+}$ $(Z=27)$,the configuration is $[Ar] 3d^7$. In a tetrahedral field,the $3d^7$ configuration has $3$ unpaired electrons $(t_2^4 e^3)$. $(a)$ $Co = [Ar] 3d^7 4s^2$; $Co^{2+} = [Ar] 3d^7$ ($3$ unpaired electrons). $(b)$ $Cu = [Ar] 3d^{10} 4s^1$; $Cu^{2+} = [Ar] 3d^9$ ($1$ unpaired electron). $(c)$ $Mn = [Ar] 3d^5 4s^2$; $Mn^{2+} = [Ar] 3d^5$ ($5$ unpaired electrons). $(d)$ $Fe = [Ar] 3d^6 4s^2$; $Fe^{2+} = [Ar] 3d^6$ ($4$ unpaired electrons). Therefore,the element $M$ is $Co$.

For a tetrahedral complex $[MCl_{4}]^{2-}$,the spin-only magnetic moment is $3.83 \ BM$. The element $M$ is

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