Hydrolysis of `XeF_(4)` and CaNCN gives respectively :

Options

A. `XeO_(3)` and `CaCO_(3)`
B. `XeO_(2)` and `CaCN_(2)`
C. `XeOF_(3)` and `CaCN_(2)`
D. `XeOF_(2)` and `CaCO_(3)`

Answer

To solve the question regarding the hydrolysis of `XeF_(4)` and `CaNCN`, we will go through the reactions step by step.

### Step 1: Hydrolysis of `XeF_(4)`
1. **Write the reaction**: When `XeF_(4)` reacts with water, it undergoes hydrolysis.
\[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{HF} + \text{O}_2 \]

2. **Balance the reaction**: To balance the equation, we find that 6 moles of `XeF_(4)` react with 12 moles of water to produce 4 moles of `Xe`, 2 moles of `XeO_(3)`, 24 moles of `HF`, and 3 moles of `O_(2)`.
\[ 6 \text{XeF}_4 + 12 \text{H}_2\text{O} \rightarrow 4 \text{Xe} + 2 \text{XeO}_3 + 24 \text{HF} + 3 \text{O}_2 \]

### Step 2: Hydrolysis of `CaNCN`
1. **Write the reaction**: The hydrolysis of `CaNCN` (calcium cyanamide) with water can be represented as follows:
\[ \text{CaNCN} + 3 \text{H}_2\text{O} \rightarrow \text{CaCO}_3 + 2 \text{NH}_3 \]

### Step 3: Identify the products
From the above reactions, we can summarize the products of the hydrolysis:
- The hydrolysis of `XeF_(4)` yields `XeO_(3)` and `HF`.
- The hydrolysis of `CaNCN` yields `CaCO_(3)` and `NH_(3)`.

### Final Answer
Thus, the hydrolysis of `XeF_(4)` and `CaNCN` gives respectively:
- `XeO_(3)` and `CaCO_(3)`.