Arrange the following bonds according to their average bond energies in descending order :
C–Cl, C–Br, C–F, C–I

Solution

Chemistry ma’am · 4 steps

Go down the halogen group: the atom gets bigger, the C–X bond gets longer, and a longer bond is a weaker one.

1. Size sets length. From F to Cl to Br to I, the halogen atom gets bigger because each one adds an electron shell. A bigger atom sits farther from carbon, so the C–X bond gets longer.
2. Length sets strength. A longer bond means the large, diffuse halogen orbital overlaps poorly with the small carbon orbital. Weaker overlap gives a weaker bond, so less energy is needed to break it.

   Bond	Approx. bond length	Approx. bond energy
   C–F	139 pm	484 kJ mol⁻¹, strongest
   C–Cl	178 pm	338 kJ mol⁻¹
   C–Br	193 pm	276 kJ mol⁻¹
   C–I	214 pm	238 kJ mol⁻¹, weakest

3. Extra help for C–F. Fluorine is also the most electronegative atom, so the C–F bond has a large ionic contribution. That extra attraction makes it stronger still.
4. So. The descending order is C–F > C–Cl > C–Br > C–I, which is option (d).

The idea to keep. For bonds from the same atom to members of one group, smaller partner means shorter bond, better overlap and higher bond energy. Do not mix this up with the anomalously weak F–F bond.

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