The compound in which all carbon atoms use only `sp^(3)`-hybrid orbitals for bond formation is:

Options

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

Answer

To solve the question of identifying the compound in which all carbon atoms use only `sp^(3)` hybrid orbitals for bond formation, we will analyze the given compounds one by one and count the number of sigma bonds for each carbon atom.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we need to analyze are:
- CH3COH (which can be written as CH3-CHO)
- HCOH (which is H2C=O)
- CH3C(=O)H (which is CH3-CHO)
- Carbon double bond oxygen with NH2 groups (which is NH2-C(=O)-NH2)

2. **Draw the Structures**:
- **For CH3COH**: The structure is CH3-CHO. The carbon in CH3 is bonded to three hydrogens and one carbon (in CHO). The carbon in CHO is bonded to one hydrogen and one carbon (in CH3) and has a double bond with oxygen.
- **For HCOH**: The structure is H2C=O. The carbon is bonded to two hydrogens and has a double bond with oxygen.
- **For CH3C(=O)H**: The structure is CH3-CHO. Similar to the first compound.
- **For NH2-C(=O)-NH2**: The carbon is bonded to two NH2 groups and has a double bond with oxygen.

3. **Count Sigma Bonds**:
- **For CH3COH**:
- CH3: 3 sigma bonds with H + 1 sigma bond with C (total 4 sigma bonds)
- CHO: 1 sigma bond with H + 1 sigma bond with C + 1 sigma bond with O (total 3 sigma bonds)
- Total sigma bonds for carbon in CH3COH = 4 (from CH3) + 3 (from CHO) = 7 sigma bonds.
- **For HCOH**:
- 2 sigma bonds with H + 1 sigma bond with O (total 3 sigma bonds)
- Total sigma bonds = 3 (indicating sp2 hybridization).
- **For CH3C(=O)H**:
- Similar to HCOH, total sigma bonds = 3 (indicating sp2 hybridization).
- **For NH2-C(=O)-NH2**:
- 2 sigma bonds with NH2 + 1 sigma bond with O (total 3 sigma bonds)
- Total sigma bonds = 3 (indicating sp2 hybridization).

4. **Conclusion**: The only compound where all carbon atoms are involved in 4 sigma bonds (indicating sp3 hybridization) is **CH3COH** (or CH3-CHO). The other compounds have carbon atoms that are sp2 hybridized due to the presence of double bonds.

### Final Answer:
The compound in which all carbon atoms use only `sp^(3)`-hybrid orbitals for bond formation is **CH3COH**.