Question

Calculate the percentage covalent character in the H-Br bond.

Given the electronegativity of Hydrogen is 2.2 and Bromine is 3.0.

Enter the value rounded off to the nearest integer. ___

• A. 15
• B. 85
• C. 84
• D. 16

Answer 1

The correct option is B 85

We can apply the Hanny-Smith equation here.

% ionic character = $16(|x_A-x_B|)+3.5(x_A-x_B{)}^2$

where '$x_A$ ' is the electronegativity of one element and $x_B$ is the electronegativity of the other element.

Here, our A is hydrogen and B is Bromine, so $x_A$ is 2.2 and $x_B$ is 3

So, $16(|2.2-3|)+3.5(2.2-3{)}^2$
= $(16\times 0.8)+(3.5\times (0.8{)}^2)$
= $12.8+2.24$
= $15.04$

But this is the % ionic character.

% covalent character will be 1 - 15.04 = 84.96 % ≃ 85 %