Question:medium

Total number of geometrical isomers possible for the complexes [NiCl$_4$]$^{2-}$, [CoCl$_2$(NH$_3$)$_4$]$^+$, [Co(NH$_3$)$_3$(NO$_2$)$_3$] and [Co(NH$_3$)$_5$Cl]$^{2+}$ is

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Memorize the common cases for geometrical isomerism in octahedral complexes: - [MA$_4$B$_2$]: 2 isomers (cis, trans) - [MA$_3$B$_3$]: 2 isomers (fac, mer) - [M(AA)$_2$B$_2$]: 2 isomers (cis, trans) where AA is a symmetric bidentate ligand. - [MA$_2$B$_2$C$_2$]: 5 isomers Tetrahedral [MA$_4$] and square planar [MA$_4$] do not show geometrical isomerism.

Updated On: Mar 30, 2026

2
3
4
5

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The Correct Option is C

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