Question

Among the following molecules, the one with the highest percentage of ionic character is:

• A. HI
• B. HF
• C. HCl
• D. HBr

Hint:

Higher electronegativity difference with H means higher ionic character, F is most electronegative.

Answer 1

The Correct Option is B

Step 1: Recall what decides percentage ionic character.
Percentage ionic character in a covalent bond depends on the electronegativity difference between the two bonded atoms. A larger difference pulls the shared electron pair more strongly towards one atom, making the bond behave more like an ionic bond.

Step 2: Compare electronegativity of the halogens with hydrogen.
Electronegativity order of halogens is F greater than Cl greater than Br greater than I. Hydrogen has a fixed electronegativity of about 2.2. So the electronegativity difference with hydrogen is largest for F and smallest for I.

Step 3: Apply this to the four molecules.
HF has the largest electronegativity difference among the four, so it has the highest percentage ionic character. HI has the smallest difference, so it is the most covalent, essentially close to a pure covalent bond.

So option B, HF, is correct.

Why the others are wrong: HCl and HBr have moderate electronegativity differences, giving them a moderate but lower ionic character than HF. HI has the smallest electronegativity difference of all four, making it the least ionic, so it cannot be the answer.