The shape of `CH_3^(+) and CH_3^(-)` is respectively

Options

A. plner,planer
B. plner,pyramidal
C. pyramidal,pyramidal
D. pyramidal,planer

Answer

To determine the shapes of the ions \( \text{CH}_3^+ \) and \( \text{CH}_3^- \), we will follow these steps:

### Step 1: Determine the hybridization of \( \text{CH}_3^+ \)

1. **Count the valence electrons**: Carbon has 4 valence electrons, and each hydrogen has 1 valence electron. Thus, for \( \text{CH}_3^+ \):
- Carbon: 4 electrons
- Hydrogen: 3 × 1 = 3 electrons
- Total = 4 + 3 = 7 electrons
- Since it's a cation (\( +1 \)), we subtract 1: \( 7 - 1 = 6 \) electrons.

2. **Calculate the hybridization index (HI)**:
\[ \text{HI} = \frac{1}{2} \left( \text{Valence electrons} + \text{Number of monovalent atoms} - \text{Positive charge} + \text{Negative charge} \right) \]
\[ \text{HI} = \frac{1}{2} (4 + 3 - 1) = \frac{6}{2} = 3 \]

3. **Determine the hybridization**: A hybridization index of 3 corresponds to \( \text{sp}^2 \) hybridization.

4. **Determine the shape**: \( \text{sp}^2 \) hybridization leads to a trigonal planar shape.

### Step 2: Determine the hybridization of \( \text{CH}_3^- \)

1. **Count the valence electrons**: For \( \text{CH}_3^- \):
- Carbon: 4 electrons
- Hydrogen: 3 × 1 = 3 electrons
- Total = 4 + 3 = 7 electrons
- Since it's an anion (\( -1 \)), we add 1: \( 7 + 1 = 8 \) electrons.

2. **Calculate the hybridization index (HI)**:
\[ \text{HI} = \frac{1}{2} (4 + 3 + 1) = \frac{8}{2} = 4 \]

3. **Determine the hybridization**: A hybridization index of 4 corresponds to \( \text{sp}^3 \) hybridization.

4. **Determine the shape**: \( \text{sp}^3 \) hybridization leads to a trigonal pyramidal shape.

### Conclusion

- The shape of \( \text{CH}_3^+ \) is **planar** (trigonal planar).
- The shape of \( \text{CH}_3^- \) is **pyramidal** (trigonal pyramidal).

Thus, the answer is that \( \text{CH}_3^+ \) is planar and \( \text{CH}_3^- \) is pyramidal.