Chemistry 9701/12 — October/November 2010

Background Concept

Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself. It is a periodic property: within a period it increases from left to right, and within a group it decreases going down. Fluorine is the most electronegative element known (Pauling value 3.98). A covalent bond between two atoms of different electronegativity is polar because the shared electrons are pulled more strongly towards the more electronegative atom, creating a permanent dipole. The greater the electronegativity difference, the more polar (more ionic in character) the bond becomes.

Understanding the Question

The question states the definition of electronegativity and tells us that a larger difference in electronegativity between two bonded atoms produces a more polar bond. It then asks us to choose, from four pairs of halogens, the pair that will form the most polar covalent bond. This is a direct application of the electronegativity trend — no calculation is required, only a comparison of the electronegativity differences across the four candidate pairs.

Approach

Identify the most electronegative element among those listed and the least electronegative element among those listed. The pair combining the most electronegative with the least electronegative element will have the greatest electronegativity difference and therefore the most polar bond. Fluorine is the most electronegative element in the periodic table; among chlorine, bromine and iodine, iodine is the least electronegative. Hence the fluorine–iodine pair is the correct choice.

Step-by-Step Reasoning

1. Recall the electronegativity trend: electronegativity increases across a period (left to right) and decreases down a group. Among the halogens, the order of decreasing electronegativity is F > Cl > Br > I.
2. For each pair, consider the electronegativity difference:
   • Cl and Br: a small difference (both are relatively electronegative halogens close in the group).
   • Cl and I: a larger difference than Cl–Br, but chlorine is not the most electronegative element.
   • F and Cl: a fairly large difference, but chlorine is still quite electronegative.
   • F and I: the largest difference, because fluorine is the most electronegative element and iodine is the least electronegative of the four elements listed.
3. Since bond polarity increases with electronegativity difference, the F–I bond is the most polar. The correct option is D.

Key Takeaways

• Bond polarity is governed by the difference in electronegativity between the two bonded atoms.
• The most polar bond between two elements is formed by pairing the most electronegative element with the least electronegative element.
• For the halogens, the electronegativity order F > Cl > Br > I is essential for comparing bond polarities.

Common Mistakes

• Choosing F–Cl (option C) because it involves fluorine, without noticing that iodine is much less electronegative than chlorine, so the F–I difference is actually larger.
• Assuming that "more reactive" halogens form more polar bonds; reactivity and electronegativity are different concepts.
• Confusing bond polarity with bond strength or bond length; the question is specifically about the electronegativity difference.

Things to Be Careful About

• The question asks for the most polar bond, not the strongest or most reactive bond.
• Remember that electronegativity decreases down a group, so iodine is the least electronegative of the elements listed.
• The definition given in the stem (attraction for electrons in a covalent bond) points directly to electronegativity difference as the deciding factor.