In which of the following compounds, all carbon uses `SP^(3)` hybrid orbitals only for bonding ?

Options

A. HCOOH
B. `CH_(3)CHO`
C. `(CH_(3))_(3)COH`
D. `(NH_(2))_(2)CO`

Answer

To determine which of the given compounds has all carbon atoms using only sp³ hybrid orbitals for bonding, we will analyze each option step by step.

### Step 1: Analyze Option A - HCOOH (Formic Acid)
1. **Draw the structure**: HCOOH has one carbon atom bonded to one hydrogen atom, one hydroxyl group (–OH), and one carbonyl group (C=O).
2. **Count sigma bonds**: The carbon in HCOOH forms:
- 1 sigma bond with H
- 1 sigma bond with O (in –OH)
- 1 sigma bond with O (in C=O)
- Total = 3 sigma bonds
3. **Count lone pairs**: The carbon has no lone pairs.
4. **Calculate steric number**: Steric number = number of sigma bonds + lone pairs = 3 + 0 = 3.
5. **Determine hybridization**: A steric number of 3 corresponds to sp² hybridization.

**Conclusion**: Option A is not the answer.

### Step 2: Analyze Option B - CH₃CHO (Acetaldehyde)
1. **Draw the structure**: CH₃CHO has two carbon atoms. The left carbon is part of the methyl group (–CH₃) and the right carbon is part of the carbonyl group (C=O).
2. **Count sigma bonds for the left carbon**:
- 3 sigma bonds with H (in –CH₃)
- 1 sigma bond with the right carbon
- Total = 4 sigma bonds
3. **Count lone pairs**: The left carbon has no lone pairs.
4. **Calculate steric number**: Steric number = 4 + 0 = 4.
5. **Determine hybridization**: A steric number of 4 corresponds to sp³ hybridization.

6. **Count sigma bonds for the right carbon**:
- 1 sigma bond with the left carbon
- 1 sigma bond with O (in C=O)
- Total = 2 sigma bonds
7. **Count lone pairs**: The right carbon has no lone pairs.
8. **Calculate steric number**: Steric number = 2 + 0 = 2.
9. **Determine hybridization**: A steric number of 2 corresponds to sp hybridization.

**Conclusion**: Option B is not the answer.

### Step 3: Analyze Option C - (CH₃)₃COH (Tert-Butyl Alcohol)
1. **Draw the structure**: (CH₃)₃COH has one carbon connected to three methyl groups (–CH₃) and one hydroxyl group (–OH).
2. **Count sigma bonds for the central carbon**:
- 3 sigma bonds with three CH₃ groups
- 1 sigma bond with the OH group
- Total = 4 sigma bonds
3. **Count lone pairs**: The central carbon has no lone pairs.
4. **Calculate steric number**: Steric number = 4 + 0 = 4.
5. **Determine hybridization**: A steric number of 4 corresponds to sp³ hybridization.

6. **Repeat for the other three carbons**: Each of the three CH₃ carbons also forms 4 sigma bonds and has no lone pairs, leading to a steric number of 4 and sp³ hybridization.

**Conclusion**: All carbon atoms in this compound are sp³ hybridized. Option C is a valid answer.

### Step 4: Analyze Option D - NH₂O₂CO
1. **Draw the structure**: NH₂O₂CO has one carbon connected to two oxygens (one double bond and one single bond) and two nitrogen atoms.
2. **Count sigma bonds**: The carbon forms:
- 1 sigma bond with O (double bond)
- 1 sigma bond with O (single bond)
- 1 sigma bond with N (in NH₂)
- Total = 3 sigma bonds
3. **Count lone pairs**: The carbon has no lone pairs.
4. **Calculate steric number**: Steric number = 3 + 0 = 3.
5. **Determine hybridization**: A steric number of 3 corresponds to sp² hybridization.

**Conclusion**: Option D is not the answer.

### Final Answer
The compound in which all carbon atoms use sp³ hybrid orbitals only for bonding is **Option C: (CH₃)₃COH**.

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