Which one of the following correctly represents the order of stability of oxides, X₂O; (X = halogen)?

Solution

Chemistry ma’am · 4 steps

Iodine wins on bond polarity, chlorine on multiple bonding, and bromine gets neither, so it comes last.

1. Iodine oxides. Iodine is much less electronegative than oxygen, so the I–O bond is strongly polar. That extra ionic character makes it strong. Iodine oxides are solids that decompose only on heating.
2. Chlorine oxides. Chlorine is small, and its Cl–O bonds gain strength from multiple bonding with oxygen. Chlorine oxides still exist at room temperature, but they are reactive oxidising agents that tend to explode.
3. Bromine oxides. Bromine gets neither advantage. Its Br–O bond is less polar than I–O, and it does not form multiple bonds as well as Cl–O does. Bromine oxides are the least stable and exist only at low temperatures. This is the middle-row anomaly.
4. So. The order of stability is I > Cl > Br, which is option (d).

The idea to keep. For halogen oxides, remember I > Cl > Br. Iodine is helped by a polar bond, chlorine by multiple bonding, and bromine, stuck in the middle, is the least stable.

Answer from the NTA final answer key. Solution by PrepWiser.