Base strength of `underset"(i)"(H_3C-oversetthetaCH_3), underset"(ii)"(H_2C=oversetthetaCH) "and" underset"(iii)"(H-C-=C^theta)` is in the order of

Options

A. `(i) gt (iii) gt (ii)`
B. `(i) gt (ii) gt (iii)`
C. `(ii) gt (i) gt (iii)`
D. `(iii) gt (ii) gt (i)`

Explanation

To determine the order of base strength of the compounds \( (i) \, H_3C-CH_3 \), \( (ii) \, H_2C=CH \), and \( (iii) \, H-C \equiv CH \), we need to analyze the hybridization of the carbon atoms in each compound and how it relates to their basic strength.

### Step-by-Step Solution:

1. **Identify the Hybridization of Each Compound:**
- For \( (i) \, H_3C-CH_3 \) (ethane):
- The carbon atoms are \( sp^3 \) hybridized because they are bonded through single bonds.
- For \( (ii) \, H_2C=CH \) (ethylene):
- The carbon atoms are \( sp^2 \) hybridized due to the presence of a double bond.
- For \( (iii) \, H-C \equiv CH \) (acetylene):
- The carbon atoms are \( sp \) hybridized because of the triple bond.

2. **Determine the S-character:**
- The S-character is the percentage of the s orbital in the hybridization:
- \( sp^3 \) has 25% s-character (75% p-character).
- \( sp^2 \) has 33% s-character (67% p-character).
- \( sp \) has 50% s-character (50% p-character).

3. **Relate S-character to Basic Strength:**
- As the S-character increases, the basic strength decreases. This is because higher S-character means more electronegativity, which makes the atom less willing to donate its lone pair of electrons (thereby acting as a base).
- Therefore, the order of basic strength will be inversely related to the S-character.

4. **Establish the Order of Basic Strength:**
- Since \( (i) \) has the lowest S-character (25%), it will be the strongest base.
- \( (ii) \) has a moderate S-character (33%), making it a weaker base than \( (i) \).
- \( (iii) \) has the highest S-character (50%), making it the weakest base.

5. **Final Order:**
- The order of base strength is:
\[ (i) \, H_3C-CH_3 > (ii) \, H_2C=CH > (iii) \, H-C \equiv CH \]