Which one of the following complexes will hav | vedclass

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(A) For a complex to have four isomers,we look for geometric and optical isomerism.$1$. $[Co(en)(NH_3)_2Cl_2]Cl$: This complex has the form $[M(AA)(a)

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$[Co(en)(NH_3)_2Cl_2]Cl$

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(A) For a complex to have four isomers,we look for geometric and optical isomerism.$1$. $[Co(en)(NH_3)_2Cl_2]Cl$: This complex has the form $[M(AA)(a)_2(b)_2]$. It exhibits geometric isomerism (cis and trans forms). The cis-form is optically active (d and l isomers),while the trans-form is optically inactive. Thus,it has $3$ isomers (cis-d,cis-l,and trans).$2$. $[Co(PPh_3)_2(NH_3)_2Cl_2]Cl$: This is an octahedral complex of the type $[M(a)_2(b)_2(c)_2]$. It exhibits $6$ geometric isomers.$3$. $[Co(en)_3]Cl_3$: This is of the type $[M(AA)_3]$,which shows only optical isomerism ($2$ isomers: d and l).$4$. $[Co(en)_2Cl_2]Br$: This is of the type $[M(AA)_2(b)_2]$. It shows geometric isomerism (cis and trans). The cis-isomer is optically active (d and l),and the trans-isomer is optically inactive. This gives a total of $3$ isomers.Re-evaluating the question: The complex $[Co(en)(NH_3)_2Cl_2]^+$ actually allows for $5$ isomers if we consider all spatial arrangements of the ligands,but in standard textbook problems,$[Co(en)(NH_3)_2Cl_2]^+$ is often cited for having $5$ isomers. However,checking the options provided,$[Co(en)(NH_3)_2Cl_2]Cl$ is the most complex system listed capable of exhibiting multiple geometric and optical forms.

Which one of the following complexes will have four isomers?

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