Which of the following pair of molecules have identical bond dissociation energies (approx)?

  1. $\ce{H2}$, $\ce{F2}$
  2. $\ce{F2}$, $\ce{I2}$
  3. $\ce{N2}$, $\ce{CO}$
  4. $\ce{HF}$, $\ce{O2}$

I think it's either (2) or (3). (Single correct type).

Bot $\ce{F2}$ and $\ce{I2}$ bonds are weak because of inter electronic repulsions and due to large sizes respectively.

In case (3), both molecules have bond order 3.

  • 1
    $\begingroup$ Your large sizes argument is weak. I is much bigger than F. F2 bond length: 142pm. I2 bond length:267pm. Both 2 and 3 are very close in bond dissociation energies. $\endgroup$
    – Lighthart
    Feb 29, 2016 at 19:00

1 Answer 1


The answer here has nothing to do with how the molecules relate to each other, e.g. similar bond order or length. This is actually a question that highlights an unexpected break in a trend.

The bond enthalpies for the halogens are:

$$ \begin{array}{|c|c|}\hline \text{Halogen}&\text{Bond energy (kJ/mol)}\\\hline \ce{F-F}&\pu{156}\\\hline \ce{Cl-Cl}&\pu{243}\\\hline \ce{Br-Br}&\pu{193}\\\hline \ce{I-I}&\pu{151}\\\hline \end{array} $$ Plotting these data we would expect the bond enthalpy of $\ce{F2}$ to be the strongest of them all, something around $\pu{300 kJ/mol}$, but in fact it much weaker than for chlorine. Indeed it is so weak, it as almost as weak as for iodine.

The reason for this is that fluorine is so small that it breaks the trend. The atomic radius is so small that the electrostatic repulsion between the nuclei is significant, and the non-bonding electrons surrounding each atom also repel each other due to the small amount of space available. These forces counteract the strong attraction between the nuclei and the bonding pair of electrons and weakens the bond compared to what might be expected.


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