Aqueous solution of two compounds `M_(1) - O - H and M_(2) - O - H` are prepared in two different beakers . If electronegativity of `M_(1) = 3.4, M_(2) = 1.2, 0 = 3.5 and H = 2.1`, then the nature of two solution will be respectively

Options

A. acidic, basic
B. acidic, acidic
C. basic, acidic
D. basic, basic

Explanation

To determine the nature of the solutions of the compounds \( M_1OH \) and \( M_2OH \), we will analyze the electronegativity values provided and the resulting bond characteristics.

### Step 1: Analyze \( M_1OH \)

1. **Identify Electronegativity Values**:
- Electronegativity of \( M_1 = 3.4 \)
- Electronegativity of \( O = 3.5 \)
- Electronegativity of \( H = 2.1 \)

2. **Calculate Electronegativity Difference**:
- For the bond \( M_1 - O \):
\[ \text{Electronegativity difference} = |3.5 - 3.4| = 0.1 \]
- For the bond \( O - H \):
\[ \text{Electronegativity difference} = |3.5 - 2.1| = 1.4 \]

3. **Determine Bond Type**:
- Since the electronegativity difference \( (0.1) \) between \( M_1 \) and \( O \) is less than \( 0.5 \), the bond is covalent.
- The bond \( O - H \) has an electronegativity difference of \( 1.4 \), which indicates a polar covalent bond.

4. **Nature of the Solution**:
- The \( M_1 - O \) bond does not break, while the \( O - H \) bond can break, leading to the release of \( H^+ \) ions.
- Therefore, the solution of \( M_1OH \) is **acidic**.

### Step 2: Analyze \( M_2OH \)

1. **Identify Electronegativity Values**:
- Electronegativity of \( M_2 = 1.2 \)
- Electronegativity of \( O = 3.5 \)
- Electronegativity of \( H = 2.1 \)

2. **Calculate Electronegativity Difference**:
- For the bond \( M_2 - O \):
\[ \text{Electronegativity difference} = |3.5 - 1.2| = 2.3 \]
- For the bond \( O - H \):
\[ \text{Electronegativity difference} = |3.5 - 2.1| = 1.4 \]

3. **Determine Bond Type**:
- The electronegativity difference \( (2.3) \) between \( M_2 \) and \( O \) is greater than \( 2 \), indicating an ionic bond.
- The bond \( O - H \) remains polar covalent.

4. **Nature of the Solution**:
- The \( M_2 - O \) bond will dissociate into \( M_2^+ \) and \( OH^- \) ions.
- The presence of \( OH^- \) ions indicates that the solution of \( M_2OH \) is **basic**.

### Conclusion

The nature of the two solutions is:
- \( M_1OH \): Acidic
- \( M_2OH \): Basic

### Final Answer
The nature of the two solutions will be respectively **acidic and basic**.

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