Which of the following complex compounds exhibits geometrical isomerism?

  • A
    $[Fe(DMG)_2]$
  • B
    $[Be(gly)_2]$
  • C
    $[PdClBr(gly)]$
  • D
    $[Cd(NH_3)Cl(gly)]$

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On treatment of $[Ni(NH_3)_4]^{2+}$ with concentrated $HCl,$ two compounds $I$ and $II$ having the same formula,$Ni(NH_3)_2Cl_2$ are obtained. $I$ can be converted into $II$ by boiling with dilute $HCl.$ $A$ solution of $I$ reacts with oxalic acid to form $[Ni(NH_3)_2(C_2O_4)]$ whereas $II$ does not react. Point out the correct statement of the following:

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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:

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