OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 2

✅ Correct Answer: A

Option A shows δ⁻ N — I δ⁺ .

Looking up the Pauling values from the table:

• N = 3.04
• I = 2.66

Since nitrogen has a higher electronegativity value than iodine (3.04 > 2.66), nitrogen attracts the bonding electrons more strongly and carries the partial negative charge ( δ⁻ ), making iodine δ⁺ . This polarity is correct.

💡 Key Knowledge

• Electronegativity is the ability of an atom to attract the bonding electrons in a covalent bond.
• The atom with the higher electronegativity value becomes delta negative ( δ⁻ ).
• The atom with the lower electronegativity value becomes delta positive ( δ⁺ ).

🧠 Exam Technique

• Always look up both elements given in the options from the provided data table before making your choice.
• Double-check the positions of the δ⁻ and δ⁺ symbols in the question options against your calculated differences.

❌ Common Errors & Why Others Are Incorrect

• B ( δ⁻ C — F δ⁺ ): Incorrect. Fluorine is the most electronegative element (3.98) compared to Carbon (2.55). Fluorine must be δ⁻ , but the option incorrectly labels F as δ⁺ .
• C ( δ⁻ B — Cl δ⁺ ): Incorrect. Chlorine (3.16) is more electronegative than Boron (2.04), so Cl should be δ⁻ , not B.
• D ( δ⁻ Br — Cl δ⁺ ): Incorrect. Chlorine (3.16) is more electronegative than Bromine (2.96), so Cl should be δ⁻ , not Br.