The formulae of ammonium phosphomolybdate $(X)$ and the compound $(Y)$ responsible for Prussian blue colour

The formulae of ammonium phosphomolybdate $(X)$ and the compound $(Y)$ responsible for Prussian blue colour
  1. $\begin{array}{cc} \underline{\mathrm{X}} & \underline{\mathrm{Y}} \\ \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{XH}_2 \mathrm{O} \end{array}$
  2. $\begin{array}{cc} \underline{\mathrm{X}} & \underline{\mathrm{Y}} \\ \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_3 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{XH}_2 \mathrm{O} \end{array}$
  3. $\begin{array}{cc} \underline{\mathrm{X}} & \underline{\mathrm{Y}} \\ \left(\mathrm{NH}_4\right)_2 \mathrm{PO}_3 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2 \cdot \mathrm{XH}_2 \mathrm{O} \end{array}$
  4. $\begin{array}{cc} \underline{\mathrm{X}} & \underline{\mathrm{Y}} \\ \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_5\right]_2 \cdot \mathrm{XH}_2 \mathrm{O} \end{array}$

Solution

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Asked in: AP EAMCET 2017 (24 Apr Shift 1)

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