Arrange the following compound in decreasing order of their basicity.
`(a)H_(2)C = CHNa " "(b)CH_(3)CH_(2)Na`
`(c)CH_(3)CH_(2)ONa" "(d)HC -=CNa`

Answer

To arrange the compounds in decreasing order of their basicity, we need to analyze the stability of the anions formed when these compounds act as bases. Basicity is inversely related to the stability of the anion; the less stable the anion, the stronger the base.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (a) H₂C=CHNa
- (b) CH₃CH₂Na
- (c) CH₃CH₂ONa
- (d) HC≡CNa

2. **Determine the Anions Formed**:
- (a) H₂C=CHNa → H₂C=CH⁻ (vinyl anion)
- (b) CH₃CH₂Na → CH₃CH₂⁻ (ethyl anion)
- (c) CH₃CH₂ONa → CH₃CH₂O⁻ (alkoxide anion)
- (d) HC≡CNa → HC≡C⁻ (acetylide anion)

3. **Analyze the Stability of the Anions**:
- **Anion (c) CH₃CH₂O⁻**: The negative charge is on oxygen, which is highly electronegative and stabilizes the negative charge effectively. Therefore, this anion is the most stable and thus the least basic.
- **Anion (d) HC≡C⁻**: The negative charge is on a carbon that is sp hybridized. While sp hybridized carbons hold the negative charge more tightly, they are less stable than oxygen anions. However, it is more basic than the alkoxide.
- **Anion (a) H₂C=CH⁻**: The negative charge is on a carbon that is sp² hybridized, making it less stable than the sp anion but more stable than the sp³ anion.
- **Anion (b) CH₃CH₂⁻**: The negative charge is on a carbon that is sp³ hybridized, which is the least stable of all the anions considered. Therefore, this is the most basic.

4. **Rank the Compounds Based on Basicity**:
- Since the stability of the anions decreases in the order: CH₃CH₂O⁻ < HC≡C⁻ < H₂C=CH⁻ < CH₃CH₂⁻, the basicity increases in the opposite order.
- Thus, the order of basicity is:
- (b) CH₃CH₂Na > (a) H₂C=CHNa > (d) HC≡CNa > (c) CH₃CH₂ONa

### Final Answer:
The decreasing order of basicity is:
**(b) > (a) > (d) > (c)**