Question

A 0.001 molal solution of $[Pt{(N{H_3})_4}C{l_4}]$ in water has a freezing point depression of
${0.0054^\circ }C$. If ${K_f}$ for water is 1.80, the correct formulation of the above molecule is:
A. $[Pt{(N{H_3})_4}C{l_3}]Cl$
B. $[Pt{(N{H_3})_4}C{l_2}]C{l_2}$
C. $[Pt{(N{H_3})_4}C{l_2}]C{l_3}$
D. $[Pt{(N{H_3})_4}C{l_4}]$

Answer 1

We will calculate the value of van't hoff factor, i from the Raoult's Law equation $\Delta {T_f} = i\times {K_f} \times m$​. Then, we will see which complex from the option dissociates into the same number of ions in solution.

Complete step by step answer:
According to Raoul’s Law,
$\Delta {T_f} = i \times {K_f} \times m$
$\Rightarrow$ $0.0054 = i \times 1.80 \times 0.001$
On solving,
$\Rightarrow$ $i = \dfrac{{0.0054}}{{1.80 - 0.001}}$
$\Rightarrow$ $i = \dfrac{{54 \times {{10}^{ - 4}}}}{{18 \times {{10}^{ - 1}} \times {{10}^{ - 3}}}}$
$\Rightarrow$ $i = 3$
Van't hoff factor, $i = 3$
In the first option, when $[Pt{(N{H_3})_4}C{l_2}]C{l_2}$ dissociates in solution to produce 2 ions.
$[Pt{(N{H_3})_4}C{l_3}]Cl \to {[Pt{(N{H_3})_4}C{l_3}]^ + } + C{l^ - }$
Total number of moles = 2, so this option is incorrect.
In the second option, 1 molecule of $[Pt{(N{H_3})_4}C{l_2}]C{l_2}$ dissociates in solution to produce
3 ions.
$[Pt{(N{H_3})_4}C{l_2}]C{l_2} \to {[Pt{(N{H_3})_4}C{l_3}]^ + } + 2C{l^ - }$
So, total number of moles = 3, so this option is correct
In the third option, $[Pt{(N{H_3})_4}C{l_2}]C{l_3}$ will dissociate into,
$[Pt{(N{H_3})_4}C{l_2}]C{l_3} \to {[Pt{(N{H_3})_4}C{l_3}]^ + } + 3C{l^ - }$
So, total number of moles = 4, so this option is also incorrect
In the last option, $n = 0$ because $[Pt{(N{H_3})_4}C{l_4}]$ is a coordination compounds that is chemically neutral complex in which only at least one ion is present as a complex.
Hence, the correct formula of the complex is $[Pt{(N{H_3})_4}C{l_2}]C{l_2}$

Therefore, the correct answer is option (B).

Note: The given electrolyte, $[Pt{(N{H_3})_4}C{l_4}]$ is strong. The relation of i and $\alpha$ is given by, $\left[ {i = 1 + \left( {n - 1} \right)\alpha } \right]$ So, for strong electrolyte $\alpha = 1$. All ammonia is within the bracket because it is a strong field ligand.