Question

Which of the following is non-ionisable?

• A. $Pt(N{H}_3{)}_{2}C{l}_4$
• B. $Pt(N{H}_3{)}_{3}C{l}_4$
• C. $Pt(N{H}_3{)}_{4}C{l}_4$
• D. $Pt(N{H}_3{)}_{6}C{l}_4$

Answer 1

The correct option is B $Pt(N{H}_3{)}_{3}C{l}_4$

$Primaryvalency$ represents the oxidation state of central metal and secondary valency represents the coordination number of the metal. Primary valency is satisfied by negative ions. This is also called principal, ionisable or ionic valency.

The secondary valency is satisfied by neutral molecules or negative ions. This valency is non-ionic or non-ionisable.

Negative ions present in the coordination sphere are not ionized.

In all the given complexes, the oxidation state of $Platinum$ is $+4$.

Ammonia $N{H}_3$ is a neutral ligand and has $zero$ charge.

Chlorine $Cl$ is an anionic ligand and has $-1$ charge.

i) $\left[Pt\right(N{H}_3{)}_{2}C{l}_4]$

Charge on complex ion = Oxidation number of metal ion + charge on ligands

= $[+4+(0\ast 2)+(-1\ast 4\left)\right]$

$Pt$ $N{H}_3$ $Cl$

Charge on the complex = $0$

$\left[Pt\right(N{H}_3{)}_{2}C{l}_4]$ is non-ionisable because $C{l}^{-}$ is present in the coordination sphere and charge on the complex is zero.

ii) $\left[Pt\right(N{H}_3{)}_{3}C{l}_4]$ or $\left[Pt\right(N{H}_3{)}_{3}C{l}_3]Cl$

Charge on the complex = +1

$\left[Pt\right(N{H}_3{)}_{3}C{l}_4]\leftrightharpoons [Pt(N{H}_3{)}_{3}C{l}_3{]}^{+}+C{l}^{-}$

$\left[Pt\right(N{H}_3{)}_{3}C{l}_4]$ dissociates to give two ions in aqueous solution.

Therefore, it is ionisable.

iii) $\left[Pt\right(N{H}_3{)}_{4}C{l}_4]$ or $\left[Pt\right(N{H}_3{)}_{4}C{l}_2]C{l}_2$

Charge on the complex = +2

$\left[Pt\right(N{H}_3{)}_{4}C{l}_4]\leftrightharpoons [Pt(N{H}_3{)}_{4}C{l}_2{]}^{+2}+2C{l}^{-}$

$\left[Pt\right(N{H}_3{)}_{4}C{l}_4]$ dissociates to give three ions in aqueous solution.

Therefore, it is ionisable.

iv) $Pt(N{H}_3{)}_{6}C{l}_4$ or $\left[Pt\right(N{H}_3{)}_6]C{l}_4$

Charge on the complex = +4

$\left[Pt\right(N{H}_3{)}_{6}C{l}_4]\leftrightharpoons [Pt(N{H}_3{)}_6{]}^{+4}+4C{l}^{-}$

$Pt(N{H}_3{)}_{6}C{l}_4$ dissociates to give five ions in aqueous solution.

Therefore, it is ionisable.

$\left[Pt\right(N{H}_3{)}_{2}C{l}_4]$ is non-ionisable.

Hence, correct option is $\left(A\right)$.