Consider the following carbanions
`(i) CH_(3)- overset(Ө)CH_(2)" "(ii) CH_(2) = overset(Ө)CH" "(iii) CH -= overset(Ө)C `
Correct order of stabilityof these carboanions in decreasing order is

Options

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

Answer

To determine the correct order of stability for the given carbanions, we will analyze the hybridization and s-character of each carbanion. The stability of carbanions is influenced by the s-character of the orbital holding the negative charge; the higher the s-character, the more stable the carbanion.

### Given Carbanions:
1. **(i)** CH₃-CH₂⁻ (Carbanion 1)
2. **(ii)** CH₂=CH⁻ (Carbanion 2)
3. **(iii)** CH≡C⁻ (Carbanion 3)

### Step-by-Step Solution:

**Step 1: Determine the Hybridization of Each Carbanion**
- **Carbanion 1 (CH₃-CH₂⁻)**: The carbon with the negative charge is sp³ hybridized.
- Hybridization: sp³
- s-character: 25% (1 s orbital out of 4 total orbitals)

- **Carbanion 2 (CH₂=CH⁻)**: The carbon with the negative charge is sp² hybridized.
- Hybridization: sp²
- s-character: 33% (1 s orbital out of 3 total orbitals)

- **Carbanion 3 (CH≡C⁻)**: The carbon with the negative charge is sp hybridized.
- Hybridization: sp
- s-character: 50% (1 s orbital out of 2 total orbitals)

**Step 2: Analyze the Stability Based on s-character**
- The stability of carbanions increases with increasing s-character because higher s-character means the negative charge is held closer to the nucleus, increasing stability.
- Carbanion 1 (sp³): 25% s-character
- Carbanion 2 (sp²): 33% s-character
- Carbanion 3 (sp): 50% s-character

**Step 3: Order the Carbanions by Stability**
- Based on the s-character:
- Carbanion 3 (sp, 50% s-character) is the most stable.
- Carbanion 2 (sp², 33% s-character) is the next stable.
- Carbanion 1 (sp³, 25% s-character) is the least stable.

### Final Order of Stability:
**CH≡C⁻ > CH₂=CH⁻ > CH₃-CH₂⁻**

### Answer:
**(iii) > (ii) > (i)**