Predict the correct shape of `BH_(4)^(-)`

Options

A. Pyramidal, `sp^(3)`
B. octahedral, `sp^(3)d^(2)`
C. tetrahedral, `sp^(3)`
D. none of the above

Explanation

To predict the correct shape of the ion \( BH_4^{-} \), we can follow these steps:

### Step 1: Determine the number of valence electrons
- Boron (B) has 3 valence electrons.
- Each hydrogen (H) has 1 valence electron, and there are 4 hydrogen atoms, contributing 4 electrons.
- The negative charge on the ion indicates that there is an additional electron.

**Total valence electrons:**
\[ 3 \, (from \, B) + 4 \, (from \, H) + 1 \, (for \, the \, negative \, charge) = 8 \, electrons \]

### Step 2: Determine the hybridization
- The total number of electron pairs can be calculated by dividing the total number of valence electrons by 2.
\[ \frac{8 \, electrons}{2} = 4 \, electron \, pairs \]
- Since there are 4 bonding pairs (one for each hydrogen atom) and no lone pairs, the hybridization is \( sp^3 \).

### Step 3: Determine the molecular geometry
- The \( sp^3 \) hybridization corresponds to a tetrahedral geometry.
- In a tetrahedral arrangement, the bond angles are approximately 109.5 degrees.

### Step 4: Draw the structure
- The structure can be represented as:
```
H
|
H - B - H
|
H
```
- The negative charge is typically represented outside the structure, indicating that the entire ion is negatively charged.

### Conclusion
The shape of \( BH_4^{-} \) is tetrahedral due to its \( sp^3 \) hybridization.

### Summary
- **Hybridization:** \( sp^3 \)
- **Shape:** Tetrahedral

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