If equal volumes of $$AB_2$$ and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $$AY_2$$ at 300 K?

(Given $$K_{sp}$$ (at 300 K) for $$AY_2 = 5.2 \times 10^{-7}$$)

When the salt $$AB_2$$ dissolves, it produces one divalent cation $$A^{2+}$$ and two anions $$B^-$$:
$$AB_2 \rightarrow A^{2+} + 2B^-$$

The salt $$XY$$ produces one monovalent cation $$X^{+}$$ and one anion $$Y^-$$:
$$XY \rightarrow X^{+} + Y^-$$

The required precipitate is $$AY_2$$, which dissociates as
$$AY_2 \rightleftharpoons A^{2+} + 2Y^-$$

Hence, the solubility-product expression is
$$K_{sp} = [A^{2+}][Y^-]^{2}$$

Let the initial molarities of $$AB_2$$ and $$XY$$ be $$C_1$$ and $$C_2$$ respectively (as given in each option). The two solutions are mixed in equal volumes, so the total volume becomes double. Every concentration therefore becomes half.

After mixing:
$$[A^{2+}] = \dfrac{C_1}{2}$$ (comes only from $$AB_2$$)
$$[Y^-] = \dfrac{C_2}{2}$$ (comes only from $$XY$$)

The ionic product (IP) for $$AY_2$$ just after mixing is
$$\text{IP} = \left(\dfrac{C_1}{2}\right)\left(\dfrac{C_2}{2}\right)^{2} = \dfrac{C_1\,C_2^{2}}{8}$$ $$-(1)$$

Precipitation of $$AY_2$$ occurs when $$\text{IP} \gt K_{sp}$$.

Substitute in $$(1)$$:
$$\text{IP} = \dfrac{(3.6\times10^{-3})(5.0\times10^{-4})^{2}}{8} = \dfrac{3.6\times10^{-3}\times25\times10^{-8}}{8} = \dfrac{9.0\times10^{-10}}{8} = 1.1\times10^{-10}$$
Since $$1.1\times10^{-10} \lt 5.2\times10^{-7}$$, no precipitate.

$$\text{IP} = \dfrac{(2.0\times10^{-4})(8.0\times10^{-4})^{2}}{8} = \dfrac{2.0\times10^{-4}\times64\times10^{-8}}{8} = \dfrac{1.28\times10^{-10}}{8} = 1.6\times10^{-11}$$
Again $$1.6\times10^{-11} \lt 5.2\times10^{-7}$$, so no precipitate.

$$\text{IP} = \dfrac{(2.0\times10^{-2})(2.0\times10^{-2})^{2}}{8} = \dfrac{2.0\times10^{-2}\times4.0\times10^{-4}}{8} = \dfrac{8.0\times10^{-6}}{8} = 1.0\times10^{-6}$$
Now $$1.0\times10^{-6} \gt 5.2\times10^{-7}$$, so $$\text{IP} \gt K_{sp}$$ and precipitate of $$AY_2$$ will form.

$$\text{IP} = \dfrac{(1.5\times10^{-4})(1.5\times10^{-3})^{2}}{8} = \dfrac{1.5\times10^{-4}\times2.25\times10^{-6}}{8} = \dfrac{3.375\times10^{-10}}{8} = 4.2\times10^{-11}$$
Since $$4.2\times10^{-11} \lt 5.2\times10^{-7}$$, no precipitate.

The only combination that satisfies $$\text{IP} \gt K_{sp}$$ is Option C.

Answer: Option C