Consider the complexes:I. [Pd(NH_3)_2 ClBr]II | vedclass

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(A) $Pd(II)$ complexes are square planar. $I$. $[Pd(NH_3)_2 ClBr]$ is of the type $[M(a)_2bc]$,which shows $2$ geometrical isomers (cis and trans). $I

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$II$

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(A) $Pd(II)$ complexes are square planar. $I$. $[Pd(NH_3)_2 ClBr]$ is of the type $[M(a)_2bc]$,which shows $2$ geometrical isomers (cis and trans). $II$. $[Pd(NH_3)_2 Cl_2]$ is of the type $[M(a)_2b_2]$,which shows $2$ geometrical isomers (cis and trans). $III$. $[Pd(en) Cl_2]$ is of the type $[M(AA)b_2]$,which shows $0$ geometrical isomers. $IV$. $[Pd(en) ClBr]$ is of the type $[M(AA)bc]$,which shows $0$ geometrical isomers. $V$. $[Pd(en)_2 Cl_2]$ is of the type $[M(AA)_2b_2]$,which shows $0$ geometrical isomers. Thus,the number of geometrical isomers of $(I)$ is the same as that of $(II)$.

Consider the complexes:
$I$. $[Pd(NH_3)_2 ClBr]$
$II$. $[Pd(NH_3)_2 Cl_2]$
$III$. $[Pd(en) Cl_2]$
$IV$. $[Pd(en) ClBr]$
$V$. $[Pd(en)_2 Cl_2]$
(en = ethylenediamine)
The total number of geometrical isomers of $(I)$ is the same as the total number of geometrical isomers of:

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Consider the complexes:$I$. $[Pd(NH_3)_2 ClBr]$$II$. $[Pd(NH_3)_2 Cl_2]$$III$. $[Pd(en) Cl_2]$$IV$. $[Pd(en) ClBr]$$V$. $[Pd(en)_2 Cl_2]$(en = ethylenediamine)The total number of geometrical isomers of $(I)$ is the same as the total number of geometrical isomers of: