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(C) $1$. $[CrCl_2(NO_2)_2(NH_3)_2]^-$ is of the type $[Ma_2b_2c_2]$. It exhibits $5$ geometrical isomers.$2$. $[Co(NO_2)_3(NH_3)_3]$ is of the type $[

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$5 - 2 - 3 - 0$

Vedclass · Answer

(C) $1$. $[CrCl_2(NO_2)_2(NH_3)_2]^-$ is of the type $[Ma_2b_2c_2]$. It exhibits $5$ geometrical isomers.$2$. $[Co(NO_2)_3(NH_3)_3]$ is of the type $[Ma_3b_3]$. It exhibits $2$ geometrical isomers (facial and meridional).$3$. $[PtCl(NO_2)(NH_3)(py)]$ is a square planar complex of the type $[Mabcd]$. It exhibits $3$ geometrical isomers.$4$. $[PtBrCl(en)]$ is a square planar complex of the type $[Mab(AA)]$. Since the bidentate ligand $(en)$ occupies adjacent positions,it cannot exhibit geometrical isomerism (number of isomers = $0$).Thus,the correct sequence is $5 - 2 - 3 - 0$.

Choose the correct code regarding the possible number of geometrical isomers exhibited by the following complexes:
$I. [CrCl_2(NO_2)_2(NH_3)_2]^-$
$II. [Co(NO_2)_3(NH_3)_3]$
$III. [PtCl(NO_2)(NH_3)(py)]$
$IV. [PtBrCl(en)]$
Select the correct sequence for $I - II - III - IV$.

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$B. [Co(NH_3)_6][Cr(CN)_6]$
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$D. [Fe(NH_3)_6][Co(CN)_6]$
$E. [Co(NH_3)_6][Fe(CN)_6]$
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Choose the correct code regarding the possible number of geometrical isomers exhibited by the following complexes:$I. [CrCl_2(NO_2)_2(NH_3)_2]^-$$II. [Co(NO_2)_3(NH_3)_3]$$III. [PtCl(NO_2)(NH_3)(py)]$$IV. [PtBrCl(en)]$Select the correct sequence for $I - II - III - IV$.