Among the following, total no. of planar species is :
(i) `SF_(4) " " (ii) BrF_(3) " " (iii) XeF_(2) " " (iv) IF_(5)`
(v)` SbF_(4)^(-) " " (vi) SF_(5)^(-) " " (vii) SeF_(3)^(+) " " (viii) CH_(3)^(+)`
(ix) `PCl_(4)^(+)`

Answer

To determine the total number of planar species among the given compounds and ions, we will follow these steps:

### Step 1: Identify the central atom and its valence electrons
For each compound/ion, we will identify the central atom and count its valence electrons.

1. **SF₄**: Central atom is S (Sulfur), which has 6 valence electrons.
2. **BrF₃**: Central atom is Br (Bromine), which has 7 valence electrons.
3. **XeF₂**: Central atom is Xe (Xenon), which has 8 valence electrons.
4. **IF₅**: Central atom is I (Iodine), which has 7 valence electrons.
5. **SbF₄⁻**: Central atom is Sb (Antimony), which has 5 valence electrons plus 1 from the negative charge.
6. **SF₅⁻**: Central atom is S (Sulfur), which has 6 valence electrons plus 1 from the negative charge.
7. **SeF₃⁺**: Central atom is Se (Selenium), which has 6 valence electrons minus 1 from the positive charge.
8. **CH₃⁺**: Central atom is C (Carbon), which has 4 valence electrons minus 1 from the positive charge.
9. **PCl₄⁺**: Central atom is P (Phosphorus), which has 5 valence electrons plus 1 from the positive charge.

### Step 2: Calculate the hybridization
Using the formula for hybridization:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Positive charge} + \text{Negative charge} \right) \]

1. **SF₄**:
\[ \text{Hybridization} = \frac{1}{2}(6 + 4 - 0) = 5 \rightarrow \text{sp}^3\text{d} \]

2. **BrF₃**:
\[ \text{Hybridization} = \frac{1}{2}(7 + 3 - 0) = 5 \rightarrow \text{sp}^3\text{d} \]

3. **XeF₂**:
\[ \text{Hybridization} = \frac{1}{2}(8 + 2 - 0) = 5 \rightarrow \text{sp}^3\text{d} \]

4. **IF₅**:
\[ \text{Hybridization} = \frac{1}{2}(7 + 5 - 0) = 6 \rightarrow \text{sp}^3\text{d}^2 \]

5. **SbF₄⁻**:
\[ \text{Hybridization} = \frac{1}{2}(5 + 4 + 1) = 5 \rightarrow \text{sp}^3\text{d} \]

6. **SF₅⁻**:
\[ \text{Hybridization} = \frac{1}{2}(6 + 5 + 1) = 6 \rightarrow \text{sp}^3\text{d}^2 \]

7. **SeF₃⁺**:
\[ \text{Hybridization} = \frac{1}{2}(6 + 3 - 1) = 4 \rightarrow \text{sp}^3 \]

8. **CH₃⁺**:
\[ \text{Hybridization} = \frac{1}{2}(4 + 3 - 1) = 3 \rightarrow \text{sp}^2 \]

9. **PCl₄⁺**:
\[ \text{Hybridization} = \frac{1}{2}(5 + 4 - 1) = 4 \rightarrow \text{sp}^3 \]

### Step 3: Determine the geometry and planarity
- **SF₄**: Seesaw (not planar)
- **BrF₃**: T-shaped (planar)
- **XeF₂**: Linear (planar)
- **IF₅**: Square pyramidal (not planar)
- **SbF₄⁻**: Tetrahedral (not planar)
- **SF₅⁻**: Square planar (planar)
- **SeF₃⁺**: Trigonal pyramidal (not planar)
- **CH₃⁺**: Trigonal planar (planar)
- **PCl₄⁺**: Tetrahedral (not planar)

### Step 4: Count the planar species
The planar species identified are:
1. BrF₃
2. XeF₂
3. SF₅⁻
4. CH₃⁺

Thus, the total number of planar species is **4**.

### Final Answer:
The total number of planar species is **4**.

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