The compounds in which `C` uses its `sp^(3)`- hybrid orbitals for bond formation are:

Options

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

Explanation

To determine the compounds in which carbon uses its `sp^(3)` hybrid orbitals for bond formation, we can follow these steps:

### Step 1: Understand Hybridization
- Carbon can undergo hybridization to form different types of bonds. The hybridization can be classified as:
- **sp³**: When carbon forms four single (sigma) bonds.
- **sp²**: When carbon forms three bonds (two single and one double bond).
- **sp**: When carbon forms two bonds (one triple bond or two double bonds).

### Step 2: Analyze the Given Compounds
We will analyze each compound to determine the hybridization of the carbon atoms present.

1. **Compound A: HCOH (Formaldehyde)**
- Structure: H₂C=O
- Carbon forms three sigma bonds (two with hydrogen and one with oxygen) and one pi bond (with oxygen).
- **Hybridization**: sp²

2. **Compound B: C(NH₂)₂=O (Urea)**
- Structure: H₂N-C(=O)-NH₂
- Carbon forms three sigma bonds (two with nitrogen and one with oxygen) and one pi bond (with oxygen).
- **Hybridization**: sp²

3. **Compound C: CH₃-CH₃-CH₃OH (Propan-1-ol)**
- Structure: CH₃-CH₂-CH₂-OH
- The carbon atoms in the CH₃ (methyl) groups and the CH₂ (methylene) group each form four sigma bonds.
- **Hybridization**: sp³ for all carbon atoms.

4. **Compound D: CH₃-C(=O)-H (Acetaldehyde)**
- Structure: CH₃-CHO
- The carbon in the CH₃ group forms four sigma bonds, while the carbon in the CHO group forms three sigma bonds and one pi bond.
- **Hybridization**: sp² for the carbon in the CHO group and sp³ for the carbon in the CH₃ group.

### Step 3: Identify Compounds with sp³ Hybridization
From the analysis:
- **Compound C** has carbon atoms that are sp³ hybridized.
- **Compound D** has at least one carbon (the one in the CH₃ group) that is sp³ hybridized.

### Conclusion
The compounds in which carbon uses its `sp^(3)` hybrid orbitals for bond formation are **C and D**.

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