The number of geometrical isomers for octahed | vedclass

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(B) $1$. For the octahedral complex $[CoCl_4(NH_3)_2]^-$,the general formula is $[MA_4B_2]$. It exhibits two geometrical isomers: $cis$ (where the two

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$2, 2, 0$

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(B) $1$. For the octahedral complex $[CoCl_4(NH_3)_2]^-$,the general formula is $[MA_4B_2]$. It exhibits two geometrical isomers: $cis$ (where the two $NH_3$ ligands are adjacent) and $trans$ (where the two $NH_3$ ligands are opposite).$2$. For the square planar complex $[AuBr_2Cl_2]^-$,the general formula is $[MA_2B_2]$. It exhibits two geometrical isomers: $cis$ (where the two $Cl^-$ ligands are adjacent) and $trans$ (where the two $Cl^-$ ligands are opposite).$3$. For the square planar complex $[PtCl_2(en)]$,the general formula is $[M(AA)B_2]$. Since the bidentate ligand $(en)$ must occupy adjacent positions,only one configuration is possible. Therefore,it shows $0$ geometrical isomers.

The number of geometrical isomers for octahedral $[CoCl_4(NH_3)_2]^-$,square planar $[AuBr_2Cl_2]^-$ and $[PtCl_2(en)]$ are

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