The hybridization of the complex [CrCl_2(NO_2 | vedclass

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(C) $1$. Determine the oxidation state of $Cr$: Let the oxidation state of $Cr$ be $x$. The ligands are $Cl^-$ $(-1)$,$NO_2^-$ $(-1)$,and $NH_3$ $(0)$

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$d^2sp^3$

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(C) $1$. Determine the oxidation state of $Cr$: Let the oxidation state of $Cr$ be $x$. The ligands are $Cl^-$ $(-1)$,$NO_2^-$ $(-1)$,and $NH_3$ $(0)$. The overall charge is $-1$.$x + 2(-1) + 2(-1) + 2(0) = -1$$x - 4 = -1$$x = +3$$2$. Electronic configuration of $Cr^{3+}$: The atomic number of $Cr$ is $24$. The ground state configuration is $[Ar] 3d^5 4s^1$. Thus,$Cr^{3+}$ is $[Ar] 3d^3$.$3$. Coordination number: The complex has $6$ ligands,so the coordination number is $6$,implying an octahedral geometry.$4$. Hybridization: Since $NO_2^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals. The $3d^3$ configuration has $3$ unpaired electrons in the $t_{2g}$ orbitals. For an octahedral complex with $d^3$ configuration,the inner $d$-orbitals are available for hybridization. Therefore,the hybridization is $d^2sp^3$.

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

The hybridization of the complex $[CrCl_2(NO_2)_2(NH_3)_2]^-$ is

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