The shape of carbanion `[CH_(3)]^(-) ` is:
linear
bent
pyramidal
tetrahedral

Options

A. linear
B. bent
C. pyramidal
D. tetrahedral

Explanation

To determine the shape of the carbanion \([CH_3]^-\), we can follow these steps:

### Step 1: Determine the Hybridization
1. **Identify the central atom**: In this case, the central atom is carbon.
2. **Count the valence electrons**: Carbon has 4 valence electrons. Each hydrogen atom contributes 1 valence electron, and there are 3 hydrogen atoms, contributing a total of 3 electrons. The negative charge adds 1 more electron.
\[ \text{Total valence electrons} = 4 \text{ (from C)} + 3 \text{ (from H)} + 1 \text{ (negative charge)} = 8 \]

3. **Calculate the steric number**: The steric number is calculated by dividing the total number of valence electrons by 2.
\[ \text{Steric number} = \frac{8}{2} = 4 \]

### Step 2: Determine the Geometry
1. **Geometry based on steric number**: A steric number of 4 corresponds to a tetrahedral geometry. This means that the arrangement of the atoms around the carbon atom is tetrahedral.

### Step 3: Consider the Shape
1. **Identify lone pairs**: In the case of \([CH_3]^-\), there are no lone pairs on the carbon atom, but the negative charge indicates that there are two electrons associated with it.
2. **Shape vs. Geometry**: While the geometry is tetrahedral, the shape considers only the positions of the atoms and not the lone pairs. Since there are no lone pairs affecting the shape, we consider the arrangement of the three hydrogen atoms around the carbon atom.
3. **Final shape determination**: The arrangement of the three hydrogen atoms around the carbon leads to a pyramidal shape due to the presence of the negative charge.

### Conclusion
The shape of the carbanion \([CH_3]^-\) is **pyramidal**.

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