Which isomeric form of complex compound ${\left[{\mathrm{Ma}}_4\mathrm{bc}\right]}^{\mathrm{n}\pm }$ gives three isomers as product by replacing $\left(\mathrm{a}\right)$by $\left(\mathrm{b}\right)$ [ where $\mathrm{a} , \mathrm{b} , \mathrm{c}$ are monodentate simple ligands]

${\left[{\mathrm{Ma}}_4\mathrm{bc}\right]}^{\mathrm{n}\pm } : \mathrm{Geometrical} \mathrm{Isomerism}.$

On replacing $\text{'}\mathrm{a}\text{'}$by $\text{'}\mathrm{b}\text{'}$ in cis form :

On replacing $\text{'}\mathrm{a}\text{'}$ by $\text{'}\mathrm{b}\text{'}$ in Trans form :

$\Rightarrow$They are same hence only one isomer product is obtained by replacement