Please help Santa and Banta to decode the lock of their car XYZTU
`{:(Code,XeF_(2),XeF_(4),XeF_(6),XeO_(3),XeOF_(4),XeO_(2)F_(2)),(,1,2,3,4,5,6):}`
(a) No . Of compound having planar shape=T
(b) Code of the compound which is linear =Z
( c) Code of the compound which has distorted octahedral shape=X
(d) Code of the compound which is pyramidal=Y
(e ) Code of the compound which is square pyramidal=U

Explanation

To solve the problem, we need to analyze the compounds given and determine their shapes based on hybridization and molecular geometry. The compounds we need to consider are:

1. **XeF₂**
2. **XeF₄**
3. **XeF₆**
4. **XeO₃**
5. **XeOF₄**
6. **XeO₂F₂**

We will find the required values step by step.

### Step 1: Determine the Hybridization and Shape of Each Compound

**(a) XeF₂**
- Valence Electrons: 8 (Xe) + 2 (F) = 10
- Steric Number = (8 + 2) / 2 = 5
- Hybridization: sp³d
- Shape: Linear (due to 3 lone pairs)

**(b) XeF₄**
- Valence Electrons: 8 (Xe) + 4 (F) = 12
- Steric Number = (8 + 4) / 2 = 6
- Hybridization: sp³d²
- Shape: Square planar (due to 2 lone pairs)

**(c) XeF₆**
- Valence Electrons: 8 (Xe) + 6 (F) = 14
- Steric Number = (8 + 6) / 2 = 7
- Hybridization: sp³d³
- Shape: Distorted octahedral (due to 1 lone pair)

**(d) XeO₃**
- Valence Electrons: 8 (Xe) + 3 (O) = 11
- Steric Number = (8 + 0) / 2 = 4
- Hybridization: sp³
- Shape: Pyramidal (due to 1 lone pair)

**(e) XeOF₄**
- Valence Electrons: 8 (Xe) + 4 (F) + 6 (O) = 18
- Steric Number = (8 + 4 + 2) / 2 = 6
- Hybridization: sp³d²
- Shape: Square pyramidal (due to 1 lone pair)

**(f) XeO₂F₂**
- Valence Electrons: 8 (Xe) + 2 (O) + 4 (F) = 14
- Steric Number = (8 + 2 + 2) / 2 = 6
- Hybridization: sp³d²
- Shape: See-saw (due to 1 lone pair)

### Step 2: Count the Compounds with Planar Shapes
- Planar shapes: Linear (XeF₂), Square planar (XeF₄)
- Total planar compounds = 2

### Step 3: Identify the Codes for Each Shape
- **Linear**: XeF₂ (Code = 1)
- **Distorted Octahedral**: XeF₆ (Code = 3)
- **Pyramidal**: XeO₃ (Code = 4)
- **Square Pyramidal**: XeOF₄ (Code = 5)

### Step 4: Assign Values to T, Z, X, Y, U
- (a) No. of compounds having planar shape (T) = 2
- (b) Code of the compound which is linear (Z) = 1
- (c) Code of the compound which has distorted octahedral shape (X) = 3
- (d) Code of the compound which is pyramidal (Y) = 4
- (e) Code of the compound which is square pyramidal (U) = 5

### Final Code
Putting it all together, we have:
- XYZTU = 34125

### Final Answer
The final lock code for Santa and Banta's car is **34125**.

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