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 $d^2sp^3$.

Column-$I$ Complex	Column-$II$ Structure / Geometry / Property
$A$. $[Ni(CN)_4]^{2-}$	$I$. Tetrahedral / Paramagnetic
$B$. $[Ni(CO)_4]$	$II$. Tetrahedral / Diamagnetic
$C$. $[NiCl_4]^{2-}$	$III$. Square planar / Diamagnetic

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

Similar Questions

Which of the following is an outer orbital complex?

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes,respectively,are

Match the following:

Column-$I$ Complex Column-$II$ Structure / Geometry / Property

$A$. $[Ni(CN)_4]^{2-}$ $I$. Tetrahedral / Paramagnetic

$B$. $[Ni(CO)_4]$ $II$. Tetrahedral / Diamagnetic

$C$. $[NiCl_4]^{2-}$ $III$. Square planar / Diamagnetic

The correct match is:

The hybridization states of the central metal atom in the complexes $[Ni(CO)_4]$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$ are respectively:

Which of the following is true for the $[NiCl_4]^{2-}$ complex?

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