Which of the following pairs of atoms is most likely to form a polar covalent bond due to a moderate difference in electronegativity while both remain non metals?

Introduction / Context:
Chemical bonds can be broadly classified as ionic, covalent or metallic, with covalent bonds further divided into polar and non polar types. Polarity depends on the difference in electronegativity between bonded atoms. Understanding which combinations of atoms are likely to form polar covalent bonds is essential for predicting molecular properties such as solubility and dipole moments. This question asks you to select the pair that forms a polar covalent bond.

Given Data / Assumptions:

• Options include H with Br, H with H, Na with Br, N with N and Na with Cl.
• We assume Pauling scale electronegativity trends.
• Elements on the left of the periodic table like sodium tend to form ionic bonds with halogens.
• Pairs of identical non metals usually form non polar covalent bonds.

Concept / Approach:
A polar covalent bond arises when two non metals share electrons but one is more electronegative, pulling the shared electron pair closer and creating partial charges. If the electronegativity difference is very large and one atom is a metal while the other is a non metal, the bond tends to be ionic. If there is no difference, as in identical atoms, the bond is non polar covalent. Hydrogen and bromine are both non metals with a moderate electronegativity difference, so the H–Br bond is polar covalent. Sodium with bromine or chlorine involves a metal and a non metal and leads to ionic bonds. Hydrogen with hydrogen and nitrogen with nitrogen form non polar covalent bonds because the atoms are identical.

Step-by-Step Solution:
Step 1: Identify which pairs involve two non metals. H with Br, H with H and N with N are non metal pairs, while Na with Br and Na with Cl involve a metal and a non metal.Step 2: Recognise that metal non metal pairs like NaBr and NaCl are typically ionic, not covalent, due to large electronegativity differences.Step 3: Consider H–H and N–N bonds. In each case, the two atoms are identical, so electronegativity difference is zero.Step 4: Zero electronegativity difference leads to non polar covalent bonds.Step 5: Consider H–Br. Both are non metals, but bromine is significantly more electronegative than hydrogen.Step 6: The electronegativity difference is moderate, so the shared pair is pulled closer to bromine, giving partial charges and a polar covalent bond.Step 7: Therefore, the best example of a polar covalent bond among the options is H and Br.

Verification / Alternative check:
Bond classification guidelines often state that an electronegativity difference between about 0.4 and 1.7 typically results in a polar covalent bond, while values above this range often correspond to ionic bonds and values near zero to non polar covalent bonds. The H–Br electronegativity difference falls into the polar covalent range. For Na–Cl or Na–Br, the difference is much larger, which is characteristic of ionic bonds in salts like sodium chloride and sodium bromide.

Why Other Options Are Wrong:
Option b, H and H, consists of identical atoms, so the H–H bond is non polar covalent. Option c, Na and Br, tends to form NaBr, an ionic compound. Option d, N and N, gives an N≡N bond in nitrogen gas, which is non polar covalent. Option e, Na and Cl, forms ionic NaCl, not a polar covalent molecule.

Common Pitfalls:
Students sometimes assume that any difference in electronegativity, even between metal and non metal, means polar covalent, forgetting that very large differences lead to ionic bonds. Others forget that identical atoms must share electrons equally, giving non polar bonds. To avoid these mistakes, remember the rough rule of thumb: identical atoms give non polar covalent bonds, moderate differences between non metals give polar covalent bonds, and large differences between metals and non metals give ionic bonds.

Final Answer:
The pair most likely to form a polar covalent bond is H and Br, which form a polar H–Br bond.