The correct answer is $\boxed{\text{A}}$. HI has the highest percentage of ionic character among the following molecules.
Ionic character is a measure of the polarity of a bond. A bond with a high percentage of ionic character is one in which the electrons are shared unequally between the two atoms. This means that one atom has a partial negative charge and the other atom has a partial positive charge.
The percentage of ionic character in a bond can be calculated using the following equation:
$\text{Percentage of ionic character} = \frac{1}{2} \left( \frac{|E_A – E_B|}{E_A + E_B} \right)$
where $E_A$ and $E_B$ are the electronegativities of the two atoms.
The electronegativities of the atoms in the four molecules are as follows:
- Hydrogen: 2.20
- Fluorine: 4.00
- Chlorine: 3.00
- Bromine: 2.80
Substituting these values into the equation, we get the following percentages of ionic character:
- HI: 61%
- HF: 41%
- HCl: 17%
- HBr: 19%
As you can see, HI has the highest percentage of ionic character, followed by HF, HCl, and HBr. This is because fluorine has the highest electronegativity of any element, followed by chlorine, bromine, and hydrogen. Therefore, the electrons in the HI bond are shared unequally, with fluorine having a partial negative charge and hydrogen having a partial positive charge.
The other three molecules have lower percentages of ionic character because the electronegativities of the atoms are closer together. In HCl, for example, the electrons are shared more equally between chlorine and hydrogen, resulting in a lower percentage of ionic character.