`Xe+F_(2) overset(1:20) to X overset(H_(2)O) to Y overset(H_(2)O ) to Z overset(H_(2)O) to XeO_(3)`
Select correct option(s) for X, Y, Z and given chemical change:

Options

A. X, Y and Z are in same oxidation state
B. All have equal number of lone pair on central atom
C. All are non-planar
D. All have equal number of covalent bonds

Explanation

To solve the given question, we need to identify the compounds X, Y, and Z formed during the reactions involving xenon (Xe) and fluorine (F₂), and then analyze their properties based on the provided options.

### Step-by-Step Solution:

1. **Identify Compound X:**
- The reaction begins with xenon reacting with fluorine in a 1:20 ratio.
- The product formed is xenon hexafluoride (XeF₆).
- **X = XeF₆**

2. **Identify Compound Y:**
- Next, XeF₆ reacts with water (H₂O).
- This reaction leads to the formation of xenon oxyfluoride (XeOF₄), which is a partial hydrolysis of XeF₆.
- **Y = XeOF₄**

3. **Identify Compound Z:**
- The compound Y (XeOF₄) then reacts with water again.
- This results in the formation of xenon difluoride oxide (XeO₂F₂).
- **Z = XeO₂F₂**

4. **Final Product:**
- The last reaction involves Z (XeO₂F₂) reacting with water.
- This leads to the formation of xenon trioxide (XeO₃).
- **Final Product = XeO₃**

5. **Analyzing the Properties of X, Y, and Z:**
- **Oxidation State:**
- For X (XeF₆), the oxidation state of Xe is +6.
- For Y (XeOF₄), the oxidation state of Xe is also +6.
- For Z (XeO₂F₂), the oxidation state remains +6.
- All three compounds (X, Y, Z) have the same oxidation state of +6.
- **Lone Pairs:**
- XeF₆ has one lone pair.
- XeOF₄ has one lone pair.
- XeO₂F₂ has one lone pair.
- Each compound has one lone pair on the central atom.
- **Planarity:**
- XeF₆ has a distorted octahedral structure (non-planar).
- XeOF₄ has a square pyramidal structure (non-planar).
- XeO₂F₂ has a seesaw shape (non-planar).
- All three compounds are non-planar.
- **Covalent Bonds:**
- Each of the compounds has six covalent bonds.

### Conclusion:
Based on the analysis, we can conclude that:
- **X = XeF₆**
- **Y = XeOF₄**
- **Z = XeO₂F₂**

All the options provided are correct:
1. All three have the same oxidation state.
2. All have equal number of lone pairs on the central atom.
3. All are non-planar.
4. All have equal number of covalent bonds.