`Cl_(2)(g)+Ba(OH)_(2) to X(aq.)+BaCl_(2)+H_(2)O`
`X+H_(2)SO_(4) to Y+BaSO_(4)`
`Y underset(Deltagt365K)overset(Delta) Z+H_(2)O+O_(2)`
Y and Z are respectively:

Options

A. `HClO_(4),ClO_(2)`
B. `HClO_(3),ClO_(2)`
C. `HClO_(3),ClO_(6)`
D. `HClO_(4),Cl_(2)O_(7)`

Explanation

To solve the given problem, we will analyze the reactions step by step.

### Step 1: Reaction of Cl₂ with Ba(OH)₂
The first reaction is:
\[ \text{Cl}_2(g) + \text{Ba(OH)}_2 \rightarrow X(aq) + \text{BaCl}_2 + \text{H}_2\text{O} \]

In this reaction, chlorine gas (\( \text{Cl}_2 \)) reacts with barium hydroxide (\( \text{Ba(OH)}_2 \)). Chlorine undergoes a disproportionation reaction in the presence of a strong base. The products formed are barium chlorate and barium chloride along with water.

The balanced reaction is:
\[ 6 \text{Cl}_2 + 6 \text{Ba(OH)}_2 \rightarrow 5 \text{BaCl}_2 + 2 \text{Ba(ClO}_3\text{)}_2 + 6 \text{H}_2\text{O} \]

Here, \( X \) is identified as barium chlorate, which is \( \text{Ba(ClO}_3\text{)}_2 \).

### Step 2: Reaction of X with H₂SO₄
Now we consider the next reaction:
\[ X + \text{H}_2\text{SO}_4 \rightarrow Y + \text{BaSO}_4 \]

Substituting \( X \) with \( \text{Ba(ClO}_3\text{)}_2 \):
\[ \text{Ba(ClO}_3\text{)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2 \text{HClO}_3 \]

This indicates that \( Y \) is chloric acid (\( \text{HClO}_3 \)).

### Step 3: Heating of Y
Next, we analyze the heating of \( Y \):
\[ Y \overset{\Delta}{\rightarrow} Z + \text{H}_2\text{O} + \text{O}_2 \]

When chloric acid (\( \text{HClO}_3 \)) is heated, it decomposes:
\[ 2 \text{HClO}_3 \rightarrow 2 \text{ClO}_2 + \text{H}_2\text{O} + \text{O}_2 \]

From this, we can conclude that \( Z \) is chlorine dioxide (\( \text{ClO}_2 \)).

### Final Answer
Thus, the final answers are:
- \( Y \) is \( \text{HClO}_3 \) (chloric acid)
- \( Z \) is \( \text{ClO}_2 \) (chlorine dioxide)

### Summary
The correct option is:
- \( Y \) and \( Z \) are respectively \( \text{HClO}_3 \) and \( \text{ClO}_2 \).