[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.

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

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