The E.N. of H,X,O are `2.1, 0.8` and `3.5` respectively comment on the nature of the compound `H-O-X`, that is :

Options

A. acidic
B. basic
C. neutal
D. amphoteric

Explanation

To analyze the nature of the compound \( H-O-X \) based on the given electronegativity values, we will follow these steps:

### Step 1: Identify the Electronegativity Values
The electronegativity (E.N.) values given are:
- Hydrogen (H): 2.1
- Oxygen (O): 3.5
- Halogen (X): 0.8

### Step 2: Determine the Electronegativity Differences
To understand the nature of the bonds in the compound \( H-O-X \), we need to calculate the differences in electronegativity between the atoms involved.

1. **Between Oxygen and Hydrogen:**
\[ \Delta EN_{O-H} = E.N._O - E.N._H = 3.5 - 2.1 = 1.4 \]

2. **Between Oxygen and Halogen:**
\[ \Delta EN_{O-X} = E.N._O - E.N._X = 3.5 - 0.8 = 2.7 \]

3. **Between Hydrogen and Halogen:**
\[ \Delta EN_{H-X} = E.N._H - E.N._X = 2.1 - 0.8 = 1.3 \]

### Step 3: Analyze the Electronegativity Differences
The differences calculated indicate how polar the bonds are:
- The bond between \( O \) and \( H \) has a moderate polarity (1.4).
- The bond between \( O \) and \( X \) has a high polarity (2.7).
- The bond between \( H \) and \( X \) has a moderate polarity (1.3).

### Step 4: Determine the Nature of the Compound
The bond with the highest electronegativity difference is between oxygen and halogen (\( O-X \)). A large difference in electronegativity suggests that the bond is ionic or highly polar, which can lead to the formation of ions upon dissociation.

### Step 5: Predict the Behavior of the Compound in Solution
When \( H-O-X \) dissociates, it is likely to form:
- Hydroxide ion (\( OH^- \))
- Halide ion (\( X^- \))

The presence of the hydroxide ion indicates that the solution will exhibit basic properties.

### Conclusion
Based on the analysis, the compound \( H-O-X \) is expected to exhibit basic nature when dissolved in water.

### Final Answer
The nature of the compound \( H-O-X \) is **basic**.

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