[CoCl_2(NH_3)_4]^+ + Cl^- to [CoCl_3(NH_3)_3] | vedclass

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(B) The reactant is $[CoCl_2(NH_3)_4]^+$. If the reactant is in the $cis$-form,the substitution of $NH_3$ by $Cl^-$ can occur at different positions,l

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compound is in cis-form

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(B) The reactant is $[CoCl_2(NH_3)_4]^+$. If the reactant is in the $cis$-form,the substitution of $NH_3$ by $Cl^-$ can occur at different positions,leading to both $cis$ and $trans$ isomers of the product $[CoCl_3(NH_3)_3]$. If the reactant were in the $trans$-form,it would typically yield only one isomer due to symmetry. Since two isomers of the product are obtained,the initial reactant must be the $cis$-isomer.

List-$I$ (Coordination compounds)	List-$II$ (Number of unpaired electrons)
$A$. $[Cr(CN)_6]^{3-}$	$I$. $0$
$B$. $[Fe(H_2O)_6]^{2+}$	$II$. $3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $2$
$D$. $[Ni(NH_3)_6]^{2+}$	$IV$. $4$

$[CoCl_2(NH_3)_4]^+ + Cl^- \to [CoCl_3(NH_3)_3] + NH_3$. If in this reaction two isomers of the product are obtained,which is true for the initial (reactant) complex?

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