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

Potassium ferricyanide is $$K_3[Fe(CN)_6]$$. In the anion $$[Fe(CN)_6]^{3-}$$ iron is in the +3 oxidation state and all six ligands are $$CN^-$$.

Step 1 - Formation of the intermediate complex P
Iodide ion is a good entering ligand for the octahedral ferricyanide ion. The substitution that actually occurs is

$$[Fe(CN)_6]^{3-} + I^- \;\rightleftharpoons\; [Fe(CN)_5I]^{3-} + CN^-$$

The reaction is reversible because the liberated $$CN^-$$ can again replace $$I^-$$. However, in a strongly acidic medium the free $$CN^-$$ is protonated to volatile HCN:

$$CN^- + H^+ \rightarrow HCN(g)$$

Removal of $$HCN$$ (by escape of the gas) drives the equilibrium completely to the right, so the only iron-containing species present is

$$P = [Fe(CN)_5I]^{3-}$$

Step 2 - Precipitation of Q with $$ZnCl_2$$
In a slightly acidic solution, the divalent zinc ion combines with the trivalent anion $$[Fe(CN)_5I]^{3-}$$ to give a neutral, sparingly soluble salt. Let the formula of this salt be $$Zn_x[Fe(CN)_5I]_y$$.

For electrical neutrality: $$2x - 3y = 0 \quad\Longrightarrow\quad 2x = 3y$$

The smallest integer solution is $$x = 3,\; y = 2$$. Hence the precipitate is

$$Q = Zn_3[Fe(CN)_5I]_2$$

Step 3 - Number of zinc ions in Q
From the formula of Q, the number of $$Zn^{2+}$$ ions present is $$3$$.

Therefore, the required number is 3.