JEE Main 2023 · 1 February · evening · Single correct · +4 / −1
The correct order of bond enthalpy (kJ mol⁻¹) is :
- (a) C-C > Si-Si > Sn - Sn > Ge-Ge
- (b) Si-Si > C-C > Ge-Ge > Sn-Sn
- (c) C-C > Si-Si > Ge-Ge > Sn-Sn
- (d) Si-Si > C-C > Sn-Sn > Ge-Ge
Answer and solution
Answer (c) C-C > Si-Si > Ge-Ge > Sn-Sn
Solution
Chemistry ma’am · 3 steps
Bigger atoms make longer bonds and weaker overlap, so bond enthalpy falls steadily down group 14.
- Size controls strength. Going down group 14 from C to Sn, the atoms get bigger. A bigger atom forms a longer bond, and its orbitals overlap less well, so less energy is needed to break that bond.
Bond Covalent radius of atom Bond enthalpy C-C 77 pm 348 kJ mol⁻¹ Si-Si 118 pm 297 kJ mol⁻¹ Ge-Ge 122 pm 260 kJ mol⁻¹ Sn-Sn 140 pm 240 kJ mol⁻¹ - No anomaly here. Carbon is the smallest atom in the group, and C-C is the strongest bond. This strong bond is also why carbon catenates far more than Si, Ge or Sn.
- So. The order is C-C > Si-Si > Ge-Ge > Sn-Sn, which is option (c).
The idea to keep. For a homonuclear bond down a group, a bigger atom means a longer bond and a smaller bond enthalpy. The only famous exception is F₂ in group 17. Group 14 has none.
Answer from the NTA final answer key. Solution by PrepWiser.