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)

${\left[Pt{\left(N{H}_3\right)}_{4}C{l}_2\right]}^{2+}$ 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.

$\left[Pt{\left(N{H}_3\right)}_{2}C{l}_2\right]$ will have two geometrical isomers.