The stepwise formation

Solution:

$\mathrm{Cu}^{2+}+\mathrm{NH}_{3} \stackrel{\mathrm{K}_{1}}{\rightleftharpoons}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)\right]^{2+}$

${\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)\right]^{2+}+\mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{K}_{3} \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{2+}\right.}$

${\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{2+}+\mathrm{NH}_{3} \stackrel{\mathrm{K}_{1}}{\rightleftharpoons}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}\right]^{2+} }$

${\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}\right]^{2+}+\mathrm{NH}_{3} \stackrel{\mathrm{K}_{4}}{\rightleftharpoons}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} }$

$\mathrm{Cu}^{2+}+4 \mathrm{NH}_{3} \underset{\rightleftharpoons}{\stackrel{\mathrm{K}}{\rightleftharpoons}}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$

So

$\mathrm{K}=\mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \times \mathrm{K}_{4}$

$=10^{4} \times 1.58 \times 10^{3} \times 5 \times 10^{2} \times 10^{2}$

$\mathrm{K}=7.9 \times 10^{11}$

Where $\mathrm{K} \rightarrow$ Equilibrium constant for formation of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$

So equilibrium constant $\left(\mathrm{K}^{\prime}\right)$ for dissociation

of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ is $\frac{1}{\mathrm{~K}}$

$\mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}}$

$\mathrm{K}^{\prime}=\frac{1}{7.9 \times 10^{11}}$

$=1.26 \times 10^{-12}=\left(\mathrm{x} \times 10^{-12}\right)$

So the value of $x=1.26$

OMR Ans $=1$ (After rounded off to th nearest integer)