Identification of Iron Salt Reagent Y Forming Turnbull Blue

To determine the identity of the compound $\mathbf{Y}$, we analyze the reactions described in the passage:

1. Formation of Methylene Blue:
   The synthesis of methylene blue is a standard qualitative test for sulfide ions ($\text{S}^{2-}$). When $p$-amino-$N,N$-dimethylaniline reacts with sulfide ions (from $\mathbf{X} = \text{Na}_2\text{S}$) in strongly acidic medium in the presence of an oxidizing agent $\mathbf{Y}$, methylene blue is formed. The common laboratory oxidizing agent used for this reaction is ferric chloride, $\text{FeCl}_3$.

2. Reaction with Potassium Hexacyanoferrate(II):
   When an aqueous solution of $\text{Fe}^{3+}$ ions is treated with potassium hexacyanoferrate(II), $\text{K}_4[\text{Fe}(\text{CN})_6]$, an intense blue precipitate known as Prussian blue is formed:
   $4\text{Fe}^{3+} + 3[\text{Fe}(\text{CN})_6]^{4-} \longrightarrow \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \downarrow \quad (\text{Prussian blue})$
   or, in colloidal/soluble form in the presence of potassium ions:
   $\text{K}^+ + \text{Fe}^{3+} + [\text{Fe}(\text{CN})_6]^{4-} \longrightarrow \text{KFe}[\text{Fe}(\text{CN})_6] \downarrow$

3. Reaction with Potassium Hexacyanoferrate(III):
   When $\text{Fe}^{3+}$ is treated with potassium hexacyanoferrate(III), $\text{K}_3[\text{Fe}(\text{CN})_6]$, no precipitate forms; instead, a brown solution of $\text{Fe}[\text{Fe}(\text{CN})_6]$ ($\mathbf{Z}$) is obtained:
   $\text{Fe}^{3+} + [\text{Fe}(\text{CN})_6]^{3-} \longrightarrow \text{Fe}[\text{Fe}(\text{CN})_6] \quad (\text{brown coloration})$

These reactions confirm that the metal cation present in $\mathbf{Y}$ is $\text{Fe}^{3+}$. Thus, the compound $\mathbf{Y}$ is iron(III) chloride, $\text{FeCl}_3$.

Therefore, the correct option is C.