Which among the following complexes has the m | vedclass

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(D) To determine the number of isomers,we analyze each complex:$1$. $K_3[Co(OX)_2Br_2]$: This is an $M(AA)_2b_2$ type complex. It shows geometric isom

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$[Pt(py)(NH_3)BrCl]$

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(D) To determine the number of isomers,we analyze each complex:$1$. $K_3[Co(OX)_2Br_2]$: This is an $M(AA)_2b_2$ type complex. It shows geometric isomerism (cis and trans). The cis-isomer is optically active,existing as a pair of enantiomers. Total isomers = $2$ (geometric) + $1$ (enantiomer of cis) = $3$ isomers.$2$. $[Cr(NH_3)_6][Co(CN)_6]$: This is a coordination isomer. It can form $[Cr(NH_3)_5(CN)][Co(CN)_5(NH_3)]$,$[Cr(NH_3)_4(CN)_2][Co(CN)_4(NH_3)_2]$,etc. However,these are distinct coordination isomers. The complex itself does not show stereoisomerism.$3$. $[Pt(H_2O)_3Cl_3]^+$: This is an $Ma_3b_3$ type complex. It shows facial $(fac)$ and meridional $(mer)$ isomers. Total = $2$ isomers.$4$. $[Pt(py)(NH_3)BrCl]$: This is a square planar complex of the type $Mabcd$. It exhibits $3$ geometric isomers (cis-trans arrangements of ligands).Comparing the stereoisomeric potential,$[Pt(py)(NH_3)BrCl]$ is a classic example of $Mabcd$ square planar geometry which yields $3$ geometric isomers,while $K_3[Co(OX)_2Br_2]$ also yields $3$ (cis-trans + enantiomer). However,in standard textbook problems of this type,the $Mabcd$ square planar complex is often cited for its distinct geometric isomer count.

Which among the following complexes has the maximum number of isomers?

Similar Questions

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