If $K_1$ and $K_2$ are the equilibrium constants of equilibria A and B respectively, then the relationship between the two constants is

(A) ${\mathrm{SO}}_{2( \mathrm{g})}+\frac{1}{2}{\mathrm{O}}_{2( \mathrm{g})}\iff {\mathrm{SO}}_{3( \mathrm{g})}\to {\mathrm{K}}_1$

(B) $2{\mathrm{SO}}_{3( \mathrm{g})}\iff 2{\mathrm{SO}}_{2( \mathrm{g})}+{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{K}}_2$

 (A) ${\mathrm{SO}}_{2( \mathrm{g})}+\frac{1}{2}{\mathrm{O}}_{2( \mathrm{g})}\iff {\mathrm{SO}}_{3( \mathrm{g})}\to {\mathrm{K}}_1$

(B) $2{\mathrm{SO}}_{3( \mathrm{g})}\iff 2{\mathrm{SO}}_{2( \mathrm{g})}+{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{K}}_2$

 For reaction (A), $K_1$ is reserved and for reaction (B), $K_2$ is reserved. Now, the reaction, (C), $S{O}_3\left(g\right)\iff S{O}_2\left(g\right)+1/2O_2\left(g\right)$, $K_3$ is obtained. The relationship between $K_1$ and $K_3$ is $K_1=\frac{1}{K_3}$

When equilibrium (C), $K_3$ is multiplied by 2. So, the relationship between $K_2$ and $K_3$ is $\frac{1}{K_2}=(\frac{1}{K_3}{)}^2$

Thus, $\frac{1}{K_2}=(1/K_3{)}^2=K_{21}$

Therefore, the correct answer is option (D)${\mathrm{K}}_1^2=\frac{1}{{ \mathrm{K}}_2}$.