Question

Which of the following coordination compounds has maximum number of isomers ?

• A. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{2+}$
• B. $\left[\mathrm{Pt}(\text { gly })_3\right]^{2-}$
• C. $\left[\mathrm{Pt}(\mathrm{en})_3\right]^{4+}$
• D. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$

Answer 1

Answer: (B)

$\left[ Pt \left( NH _{3}\right)_{4} Cl _{2}\right]^{2+}$ has 2 geometrical isomers.

$\left[ Pt ( gly )_{3}\right]^{2-}$ will have 2 geometrical isomers, both of which will be optical active, giving a total of 4 isomers.

$\left[ Pt ( en )_{3}\right]^{4+}$ will have no geometrical isomers, but the compound will be optically active, hence total two isomers.

$\left[ Pt \left( NH _{3}\right)_{2} Cl _{2}\right]$ will have two geometrical isomers.