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

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.

• 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. – Lighthart Feb 29 '16 at 19:00

$$\begin{array}{|c|c|}\hline \text{Halogen}&\text{Bond energy (kJ/mol)}\\\hline \ce{F-F}&\pu{156}\\\hline \ce{Cl-Cl}&\pu{243}\\\hline \ce{Br-Br}&\pu{193}\\\hline \ce{I-I}&\pu{151}\\\hline \end{array}$$ 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.