$$Al_2O_3$$ was leached with alkali to get X. The solution of X on passing of gas Y, forms Z. X, Y and Z respectively are

We are told that $$Al_2O_3$$ is leached with alkali (NaOH). This is the Bayer's process for extraction of aluminium.

When $$Al_2O_3$$ is treated with NaOH solution, it dissolves to form sodium aluminate: $$Al_2O_3 + 2NaOH + 3H_2O \to 2NaAl(OH)_4$$. So X is $$NaAl(OH)_4$$ (sodium aluminate).

Now, when $$CO_2$$ gas is passed through the sodium aluminate solution, the weakly acidic $$CO_2$$ neutralises the solution and precipitates aluminium hydroxide: $$NaAl(OH)_4 + CO_2 \to Al(OH)_3 + NaHCO_3$$. So Y is $$CO_2$$.

The precipitate $$Al(OH)_3$$ is the hydrated form of alumina, written as $$Al_2O_3 \cdot xH_2O$$. So Z is $$Al_2O_3 \cdot xH_2O$$.

Hence, X = $$NaAl(OH)_4$$, Y = $$CO_2$$, Z = $$Al_2O_3 \cdot xH_2O$$.

Hence, the correct answer is Option 4.