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 $NH_3$ $(0)$

Vedclass · Accepted Answer

$d^2sp^3$

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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-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

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

Similar Questions

An aqueous solution of $NiCl_{2}$ was heated with excess sodium cyanide in the presence of a strong oxidizing agent to form $[Ni(CN)_{6}]^{2-}$. The total change in the number of unpaired electrons on the metal centre is $.....$

Which of the following statements is correct?

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

According to $V.B.T.$,in which of the following complexes does the transition of an electron occur from one shell to another shell of the central metal atom?

The incorrect statement about $Ni(CO)_4$ is

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