The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is nearly

Answer

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
- Given:
- Electronegativity of atom A (E_A) = 1.20
- Electronegativity of atom B (E_B) = 4.0

### Step 2: Determine which atom has higher electronegativity
- Since E_B (4.0) > E_A (1.20), we will consider atom B as atom 1 and atom A as atom 2 for our calculations.

### Step 3: Use the formula for percentage of ionic character
The formula for calculating the percentage of ionic character (IC) is:
\[ \text{Percentage of Ionic Character} = 16 \times (E_1 - E_2) + 3.5 \times (E_1 - E_2)^2 \]
Where:
- \(E_1\) = Electronegativity of the more electronegative atom (B)
- \(E_2\) = Electronegativity of the less electronegative atom (A)

### Step 4: Substitute the values into the formula
- Calculate \(E_1 - E_2\):
\[ E_1 - E_2 = E_B - E_A = 4.0 - 1.20 = 2.8 \]
- Now substitute into the formula:
\[ \text{IC} = 16 \times (2.8) + 3.5 \times (2.8)^2 \]

### Step 5: Calculate each term
1. Calculate the first term:
\[ 16 \times 2.8 = 44.8 \]
2. Calculate the second term:
\[ (2.8)^2 = 7.84 \]
\[ 3.5 \times 7.84 = 27.44 \]

### Step 6: Add the two terms together
\[ \text{IC} = 44.8 + 27.44 = 72.24 \]

### Step 7: Round the result
- The percentage of ionic character is approximately:
\[ \text{Percentage of Ionic Character} \approx 72.2\% \]

### Final Answer:
The percentage of ionic character of the A-B bond is nearly **72.2%**.

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