Question

Consider the following complexes –
(i) $K_{2}PtC{l}_6$
(ii) $PtC{l}_4.2N{H}_3$
(iii) $PtC{l}_4.3N{H}_3$
(iv) $PtC{l}_4.5N{H}_3$
Their electrical conductance in aqueous solutions are:

• A. $256\text{S},0\text{S},97\text{S},404\text{S}$
• B. $404\text{S},0\text{S},97\text{S},256\text{S}$
• C. $256\text{S},97\text{S},0\text{S},404\text{S}$
• D. $404\text{S},97\text{S},256\text{S},0\text{S}$

Answer 1

The correct option is A $256\text{S},0\text{S},97\text{S},404\text{S}$

Coordination number of $Pt$ is $6$. Hence on aqueous solution the number of dissociated ions will be
$I{K}_2\left[PtC{l}_6\right]-\text{Three ions}$
$II\left[Pt\right(N{H}_3{)}_{2}C{l}_4]-\text{Zero ions}$
$III\left[Pt\right(N{H}_3{)}_{3}C{l}_3]Cl-\text{Two ions}$
$IV\left[Pt\right(N{H}_3{)}_{5}Cl]C{l}_3-\text{Four ions}$
More is the number of ions, more is the electrical conductance. Thus (A) is the correct option here.