For a sparingly soluble salt $$AB_2$$, the equilibrium concentrations of $$A^{2+}$$ ions and $$B^-$$ ions are $$1.2 \times 10^{-4} M$$ and $$0.24 \times 10^{-3} M$$, respectively. The solubility product of $$AB_2$$ is :

The solubility product $$K_{sp}$$ of a sparingly soluble salt $$AB_2$$ can be determined from the equilibrium concentrations of its ions in solution. Given that $$[A^{2+}] = 1.2 \times 10^{-4}$$ M and $$[B^-] = 0.24 \times 10^{-3} = 2.4 \times 10^{-4}$$ M, the dissolution equilibrium is $$AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq).$$

By definition, $$K_{sp} = [A^{2+}][B^-]^2,$$ so substituting the values yields $$K_{sp} = (1.2 \times 10^{-4})(2.4 \times 10^{-4})^2.$$

Squaring the concentration of $$B^-$$ gives $$(2.4 \times 10^{-4})^2 = 5.76 \times 10^{-8},$$ since $$(2.4)^2 = 5.76$$ and $$(10^{-4})^2 = 10^{-8}.$$ Thus the product becomes $$K_{sp} = 1.2 \times 10^{-4} \times 5.76 \times 10^{-8} = (1.2 \times 5.76) \times 10^{-12} = 6.912 \times 10^{-12},$$ which rounds to $$6.91 \times 10^{-12}.$$

Therefore, the solubility product of $$AB_2$$ is $$6.91 \times 10^{-12}.$$

The correct answer is Option 1: $$6.91 \times 10^{-12}$$.