The valuse of `EN` of atoms `A` and `B` are 1.80 and 4.0 respectively The percentage of ionic character of `A-B` bond is .

Options

A. `43%`
B. `50%`
C. `55.3%`
D. `65%`

Explanation

To find the percentage of ionic character of the A-B bond given the electronegativity values of atoms A and B, we can follow these steps:

### Step 1: Identify the electronegativity values
- The electronegativity (EN) of atom A is 1.80.
- The electronegativity (EN) of atom B is 4.0.

### Step 2: Determine the correct formula to use
Since the electronegativity of atom B (4.0) is greater than that of atom A (1.80), we will use the formula for calculating the percentage of ionic character:
\[ \text{Percentage of ionic character} = 16 \times (\text{EN}_B - \text{EN}_A) + 3.5 \times (\text{EN}_B - \text{EN}_A)^2 \]

### Step 3: Substitute the values into the formula
Substituting the values of EN into the formula:
\[ \text{Percentage of ionic character} = 16 \times (4.0 - 1.80) + 3.5 \times (4.0 - 1.80)^2 \]

### Step 4: Calculate the difference in electronegativity
Calculate \(4.0 - 1.80\):
\[ 4.0 - 1.80 = 2.2 \]

### Step 5: Substitute the difference back into the formula
Now substituting back into the formula:
\[ \text{Percentage of ionic character} = 16 \times 2.2 + 3.5 \times (2.2)^2 \]

### Step 6: Calculate the square of the difference
Calculate \((2.2)^2\):
\[ (2.2)^2 = 4.84 \]

### Step 7: Substitute and calculate
Now substituting these values:
\[ \text{Percentage of ionic character} = 16 \times 2.2 + 3.5 \times 4.84 \]
\[ = 35.2 + 16.94 \]

### Step 8: Add the results together
Now add the two results:
\[ 35.2 + 16.94 = 52.14 \]

### Step 9: Final result
Thus, the percentage of ionic character of the A-B bond is approximately:
\[ \text{Percentage of ionic character} \approx 52.14\% \]

### Conclusion
The final answer is that the percentage of ionic character of the A-B bond is approximately 52.14%, which can be rounded to 50%.

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