Question

A complex $PtC{l}_4.5N{H}_3$ shows a molar conductance of $402{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$ in water and precipitates 3 moles of $AgCl$ with $AgN{O}_3$ solution. The formula of the complex is:

• A. $\left[Pt\right(N{H}_3{)}_6]C{l}_4$
• B. $\left[Pt\right(N{H}_3{)}_{4}C{l}_2]C{l}_2$
• C. $\left[Pt\right(N{H}_3{)}_{5}Cl]C{l}_3$
• D. $\left[Pt\right(N{H}_3{)}_{3}C{l}_3]Cl$

Answer 1

The correct option is C $\left[Pt\right(N{H}_3{)}_{5}Cl]C{l}_3$

The complex $PtC{l}_4.5N{H}_3$ can be designated as $\left[Pt\right(N{H}_3{)}_{5}Cl]C{l}_3$ which ionizes to give $\left[Pt\right(N{H}_3{)}_{5}Cl{]}^{3+}+3C{l}^{-}$ ions. Thus, the total number of ions produced are four and 1 mole of complex gives 3 mole of $AgCl$ when reacted with $AgN{O}_3$.