The d-electronic configuration of [CoCl_4]^{2 | vedclass

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(C) The central metal ion is $Co^{2+}$,which has a $3d^7$ electronic configuration.$Cl^-$ is a weak field ligand $(WFL)$,so pairing does not occur.In

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The central metal ion is $Co^{2+}$,which has a $3d^7$ electronic configuration.$Cl^-$ is a weak field ligand $(WFL)$,so pairing does not occur.In a tetrahedral crystal field,the $d$-orbitals split into $e$ (lower energy) and $t_2$ (higher energy) sets.According to Hund's rule,the $7$ electrons are filled as follows: $e^4 t_2^3$.Here,$m = 4$ and $n = 3$.The number of unpaired electrons is $3$.Sum of '$m$' and the number of unpaired electrons $= 4 + 3 = 7$.

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 $d$-electronic configuration of $[CoCl_4]^{2-}$ in a tetrahedral crystal field is $e^m t_2^n$. The sum of '$m$' and the number of unpaired electrons is $.............$.

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In the extraction of silver,zinc metal is used as a reducing agent. What is the molecular structure of the zinc complex formed in this reaction?

Which of the following ions forms complexes with a coordination number of only $4$?

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:

Chromium hexacarbonyl,$Cr(CO)_6$,is an octahedral compound involving which type of hybridization?

Considering $H_2O$ as a weak field ligand,the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be (Atomic number of $Mn = 25$).

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