The number of zinc ions present in the molecular formula of Q is _____.

The correct answer is 3.

To determine the number of zinc ions present in the molecular formula of Q, let us break down the reaction steps:

Step 1: Formation of Complex P
When potassium iodide (KI) reacts with potassium ferricyanide ($K_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$) in an acidic medium, a redox reaction occurs. The ferricyanide ion oxidizes the iodide ion ($I^{-}$) to molecular iodine ($I_2$), while ferricyanide itself is reduced to ferrocyanide ($\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}^{4-}$). The reaction is represented as follows:

$2K_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]+2\mathrm{KI}\to 2K_4[\mathrm{Fe}\left(\mathrm{CN}{)}_6\right]+I_2$

Thus, the complex P is potassium ferrocyanide, $K_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$.

Step 2: Formation of the Sparingly Soluble Complex Q
When zinc chloride (${\mathrm{ZnCl}}_2$) is added to potassium ferrocyanide (complex P) in a slightly acidic medium, a sparingly soluble, neutral complex Q precipitates out. The reaction is represented as:

$2K_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]+3{\mathrm{ZnCl}}_2\to K_2{\mathrm{Zn}}_3[\mathrm{Fe}(\mathrm{CN}{)}_6{]}_2+6\mathrm{KCl}$

The molecular formula of the sparingly soluble white precipitate Q is $K_2{\mathrm{Zn}}_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}_2$.

Step 3: Finding the number of Zinc Ions in Q
Looking at the chemical formula of Q, which is $K_2{\mathrm{Zn}}_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}_2$, we can clearly see that there are 3 zinc (${\mathrm{Zn}}^{2+}$) ions present in its molecular formula.