Amongst the species given below which have shape similar to `XeOF_4`.
(I) `XeO_3`
(II) `IF_5`
(III) `PCI_5`
(IV) `SF_4`.

Options

A. Only II
B. Only I
C. II and III
D. II and IV

Answer

To determine which of the given species have a shape similar to `XeOF_4`, we first need to analyze the molecular geometry of `XeOF_4` itself.

### Step 1: Determine the molecular geometry of `XeOF_4`

1. **Identify the central atom**: The central atom in `XeOF_4` is xenon (Xe).
2. **Count valence electrons**: Xenon has 8 valence electrons, oxygen has 6, and each fluorine has 7. Therefore, the total number of valence electrons is:
\[ 8 (Xe) + 6 (O) + 4 \times 7 (F) = 8 + 6 + 28 = 42 \text{ valence electrons} \]
3. **Draw the Lewis structure**: In `XeOF_4`, xenon forms one double bond with oxygen and four single bonds with fluorine. This uses up 4 (for the single bonds) + 2 (for the double bond) = 6 electrons for bonding. The remaining electrons will be lone pairs on the xenon.
4. **Count the bonding pairs and lone pairs**:
- 4 single bonds (to F) + 1 double bond (to O) = 5 bonding pairs
- 1 lone pair remains on xenon.
5. **Calculate the steric number**: The steric number is the sum of the number of bonding pairs and lone pairs:
\[ \text{Steric number} = 5 \text{ (bonding pairs)} + 1 \text{ (lone pair)} = 6 \]
6. **Determine hybridization**: The hybridization corresponding to a steric number of 6 is `sp^3d^2`.
7. **Identify the shape**: The molecular geometry with 5 bonding pairs and 1 lone pair is square pyramidal.

### Step 2: Analyze the given species

Now, we will analyze the shapes of the given species to see which one has a similar shape (square pyramidal).

**(I) `XeO_3`**
- Central atom: Xe
- Valence electrons: 8 (Xe) + 3 × 6 (O) = 26
- Lewis structure: 3 single bonds with O and 1 lone pair.
- Steric number = 4 (3 bonding pairs + 1 lone pair).
- Hybridization: `sp^3`.
- Shape: Trigonal pyramidal.

**(II) `IF_5`**
- Central atom: I
- Valence electrons: 7 (I) + 5 × 7 (F) = 42
- Lewis structure: 5 single bonds with F and 1 lone pair.
- Steric number = 6 (5 bonding pairs + 1 lone pair).
- Hybridization: `sp^3d^2`.
- Shape: Square pyramidal.

**(III) `PCl_5`**
- Central atom: P
- Valence electrons: 5 (P) + 5 × 7 (Cl) = 40
- Lewis structure: 5 single bonds with Cl and no lone pairs.
- Steric number = 5 (5 bonding pairs).
- Hybridization: `sp^3d`.
- Shape: Trigonal bipyramidal.

**(IV) `SF_4`**
- Central atom: S
- Valence electrons: 6 (S) + 4 × 7 (F) = 34
- Lewis structure: 4 single bonds with F and 1 lone pair.
- Steric number = 5 (4 bonding pairs + 1 lone pair).
- Hybridization: `sp^3d`.
- Shape: See-saw.

### Conclusion

From the analysis, the species that have a shape similar to `XeOF_4` (square pyramidal) is:
- **(II) `IF_5`**.

### Final Answer
**The species that have a shape similar to `XeOF_4` is `IF_5`.**

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