Hybridisation in `CH_(3), overset(+)(C)H_(3) and overset(-)CH_(3)` are respectively

Options

A. `sp^(2), sp^(2), sp^(3)`
B. `sp^(2), sp^(3), sp^(3)`
C. `sp^(3), sp^(3), sp^(2)`
D. `sp^(3), sp^(2), sp^(2)`

Explanation

To determine the hybridization of the species \( CH_3 \cdot \), \( CH_3^+ \), and \( CH_3^- \), we can use the formula for hybridization:

\[ H = \frac{V + M - C + A}{2} \]

Where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms connected to the central atom
- \( C \) = total positive charge (cation charge)
- \( A \) = total negative charge (anion charge)

### Step 1: Hybridization of \( CH_3 \cdot \)

1. **Identify the valence electrons**:
- Carbon has 4 valence electrons (\( V = 4 \)).

2. **Count the number of monovalent atoms**:
- There are 3 hydrogen atoms connected to carbon (\( M = 3 \)).

3. **Determine the charge**:
- The species \( CH_3 \cdot \) has no charge, so \( C = 0 \) and \( A = 0 \).

4. **Plug the values into the formula**:
\[ H = \frac{4 + 3 - 0 + 0}{2} = \frac{7}{2} = 3.5 \]
Since hybridization is determined by the whole number, we round down to 3, indicating \( sp^2 \) hybridization.

### Step 2: Hybridization of \( CH_3^+ \)

1. **Identify the valence electrons**:
- Carbon still has 4 valence electrons (\( V = 4 \)).

2. **Count the number of monovalent atoms**:
- There are still 3 hydrogen atoms connected to carbon (\( M = 3 \)).

3. **Determine the charge**:
- The species \( CH_3^+ \) has a +1 charge, so \( C = 1 \) and \( A = 0 \).

4. **Plug the values into the formula**:
\[ H = \frac{4 + 3 - 1 + 0}{2} = \frac{6}{2} = 3 \]
This indicates \( sp^2 \) hybridization.

### Step 3: Hybridization of \( CH_3^- \)

1. **Identify the valence electrons**:
- Carbon still has 4 valence electrons (\( V = 4 \)).

2. **Count the number of monovalent atoms**:
- There are still 3 hydrogen atoms connected to carbon (\( M = 3 \)).

3. **Determine the charge**:
- The species \( CH_3^- \) has a -1 charge, so \( C = 0 \) and \( A = 1 \).

4. **Plug the values into the formula**:
\[ H = \frac{4 + 3 - 0 + 1}{2} = \frac{8}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

### Final Answer:
The hybridization of \( CH_3 \cdot \), \( CH_3^+ \), and \( CH_3^- \) are respectively \( sp^2 \), \( sp^2 \), and \( sp^3 \).