What is the hybridization of the complex [Mn( | vedclass

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(B) $1$. In the complex $[Mn(CN)_6]^{3-}$,the central metal ion is $Mn^{3+}$.$2$. The atomic number of $Mn$ is $25$,so the electronic configuration of

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$d^2sp^3$

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(B) $1$. In the complex $[Mn(CN)_6]^{3-}$,the central metal ion is $Mn^{3+}$.$2$. The atomic number of $Mn$ is $25$,so the electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.$3$. $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.$4$. The $4$ electrons in the $3d$ orbitals pair up to occupy two $3d$ orbitals,leaving two $3d$ orbitals vacant.$5$. These two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals hybridize to form six $d^2sp^3$ hybrid orbitals.$6$. Thus,the hybridization of $[Mn(CN)_6]^{3-}$ is $d^2sp^3$.

What is the hybridization of the complex $[Mn(CN)_6]^{3-}$?

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