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(C) In $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$. Since $Cl^-$ is a weak field lig

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$4; 1; 0$

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(C) In $[MnCl_6]^{3-}$,the oxidation state of $Mn$ is $+3$. The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$. Since $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $4$ unpaired electrons.In $[Fe(CN)_6]^{3-}$,the oxidation state of $Fe$ is $+3$. The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. Since $CN^-$ is a strong field ligand,pairing occurs,leaving $1$ unpaired electron.In $[Co(C_2O_4)_3]^{3-}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$. Since $C_2O_4^{2-}$ acts as a strong field ligand with $Co^{3+}$,all electrons are paired,resulting in $0$ unpaired electrons.

According to Valence Bond Theory,the number of unpaired electrons present in $[MnCl_6]^{3-}$,$[Fe(CN)_6]^{3-}$ and $[Co(C_2O_4)_3]^{3-}$,respectively,are

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The number of species from the following that are involved in $sp^3 d^2$ hybridization is $[Co(NH_3)_6]^{3+}, SF_6, [CrF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$ and $[MnCl_6]^{3-}$.

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