Which one of the following is planar ?

Options

A. (a) `XeF_4`
B. (b) `XeO_4`
C. (c) `XeO_3F`
D. (d) `XeO_3F_2`

Explanation

To determine which of the given compounds is planar, we will analyze the hybridization and geometry of each compound step by step.

### Step 1: Analyze each compound for hybridization

**Compound A: XeF4**
- Valence electrons of Xenon (Xe) = 8
- Number of monovalent atoms (F) = 4
- Total charge (C) = 0

Using the formula for hybridization:
\[ H = \frac{1}{2} \left( V + M - C \right) \]
\[ H = \frac{1}{2} \left( 8 + 4 - 0 \right) = \frac{12}{2} = 6 \]

Hybridization: sp³d² (octahedral geometry)
- There are 4 bond pairs and 2 lone pairs, resulting in a square planar shape.

**Compound B: XeO4**
- Valence electrons of Xenon (Xe) = 8
- Number of monovalent atoms (O) = 0 (O is divalent)
- Total charge (C) = 0

Using the formula:
\[ H = \frac{1}{2} \left( 8 + 0 - 0 \right) = \frac{8}{2} = 4 \]

Hybridization: sp³ (tetrahedral geometry)
- The shape is tetrahedral.

**Compound C: XeO3F**
- Valence electrons of Xenon (Xe) = 8
- Number of monovalent atoms (F) = 1
- Total charge (C) = 0

Using the formula:
\[ H = \frac{1}{2} \left( 8 + 1 - 0 \right) = \frac{9}{2} = 4.5 \]

This indicates that XeO3F does not exist as a stable compound.

**Compound D: XeO3F2**
- Valence electrons of Xenon (Xe) = 8
- Number of monovalent atoms (F) = 2
- Total charge (C) = 0

Using the formula:
\[ H = \frac{1}{2} \left( 8 + 2 - 0 \right) = \frac{10}{2} = 5 \]

Hybridization: sp³d (trigonal bipyramidal geometry)
- The shape is trigonal bipyramidal.

### Step 2: Identify the planar compound
From the analysis:
- **XeF4** has a square planar shape.
- **XeO4** has a tetrahedral shape.
- **XeO3F** does not exist.
- **XeO3F2** has a trigonal bipyramidal shape.

Thus, the compound that is planar is **XeF4**.

### Final Answer:
**Option A: XeF4 is the planar compound.**

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