AP EAMCET · Chemistry · General Organic Chemistry
The formulae of ammonium phosphomolybdate \((X)\) and the compound \((Y)\) responsible for Prussian blue colour
- A \(\begin{array}{cc}
\underline{\mathrm{X}} & \underline{\mathrm{Y}} \\
\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{XH}_2 \mathrm{O}
\end{array}\) - B \(\begin{array}{cc}
\underline{\mathrm{X}} & \underline{\mathrm{Y}} \\
\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_3 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{XH}_2 \mathrm{O}
\end{array}\) - C \(\begin{array}{cc}
\underline{\mathrm{X}} & \underline{\mathrm{Y}} \\
\left(\mathrm{NH}_4\right)_2 \mathrm{PO}_3 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2 \cdot \mathrm{XH}_2 \mathrm{O}
\end{array}\) - D \(\begin{array}{cc}
\underline{\mathrm{X}} & \underline{\mathrm{Y}} \\
\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_5\right]_2 \cdot \mathrm{XH}_2 \mathrm{O}
\end{array}\)
Answer & Solution
Correct Answer
(A) \(\begin{array}{cc}
\underline{\mathrm{X}} & \underline{\mathrm{Y}} \\
\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 & \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{XH}_2 \mathrm{O}
\end{array}\)
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