`H-C-=Cunderset(-)aC-=Cunderset(-)bCH_3` ,
Compare the bond lengths a and b:

Options

A. a = b
B. `a gt b`
C. `b gt a`
D. `a gt gt gt b`

Explanation

To compare the bond lengths \( a \) and \( b \) in the compound \( H-C \equiv C \underset{(-)}{a} C \equiv C \underset{(-)}{b} CH_3 \), we need to analyze the hybridization and characteristics of the bonds involved.

### Step-by-Step Solution:

1. **Identify the Bonds**:
- Bond \( a \) is between two carbon atoms that are part of a triple bond (C≡C).
- Bond \( b \) is between an sp hybridized carbon (from the triple bond) and an sp³ hybridized carbon (from the methyl group, CH₃).

2. **Determine Hybridization**:
- For bond \( a \): Both carbon atoms are sp hybridized due to the triple bond. In sp hybridization, the s character is 50%.
- For bond \( b \): One carbon is sp hybridized (from the triple bond) and the other is sp³ hybridized (from the methyl group). In sp³ hybridization, the s character is 25%.

3. **Electronegativity Consideration**:
- The percentage s character is directly proportional to electronegativity. Therefore:
- For bond \( a \): Both carbons are sp hybridized, leading to higher electronegativity.
- For bond \( b \): The sp hybridized carbon has higher electronegativity than the sp³ hybridized carbon.

4. **Bond Strength and Length**:
- A bond with higher electronegativity is generally stronger and shorter.
- Thus, bond \( a \) (sp-sp) will be shorter than bond \( b \) (sp-sp³) because the electronegativity difference results in a stronger bond.

5. **Conclusion**:
- Since bond \( a \) is stronger and shorter, we conclude that:
\[ \text{Bond length } b > \text{Bond length } a \]
- Therefore, the relationship is \( b > a \).

### Final Answer:
The bond length \( b \) is greater than bond length \( a \) (i.e., \( b > a \)).