Choose the molecule with the strongest bond: $\ce{F2, Cl2, Br2, I2,}$ or all are equal.

All of them are nonpolar-covalently bonded, so I had thought that the answer would be "all are equal". However, apparently, $\ce{Cl2}$ has the strongest bond, and I couldn't figure out why. Could someone help me understand this? Any help would really be appreciated.


2 Answers 2


The Bond Dissociation Energies of these molecules are in the order Cl2 > Br2 > F2 > I2

All these molecules involve a single covalent bond. Covalent bonds rely on sharing of a pair of electrons between two atoms, an electron each from both atoms involved. The better the sharing, the stronger the covalent bond formed. This sharing is reflected in the valence shell orbital overlap that occurs between the two atoms, when a more effective overlap occurs, a stronger covalent bond is formed.

Now, the radii of the halogen atoms increases in the order of writing them down the group, F < Cl < Br < I < At. As the atoms get larger, so do the valence shell orbitals. The increase in size of orbitals carrying the same number of electrons would mean that the electron density would be lower in larger orbitals. So, for the same proportional overlap between two similar orbitals of similar atoms, the sharing of electrons would be poorer and would consequently result in a weaker covalent bond.

This should explain the strength of bonds decreasing as we go down the group.

However, in this orbital overlap consideration gets forgotten another factor that becomes important when atoms are small in size, and that is of electronic repulsions.

Fluorine is extremely small in size and so the electron density is pretty high as well. This results in an unstable arrangement if the atoms come too close, since like charges repel.

The interplay of the above mentioned effects results in Chlorine having the highest bond strength.

I may also add that the same is the case for (1) N-N bond energy as against P-P bond energy and (2) O-O bond energy as against S-S bond energy. But, do keep in mind this stands true only for single bonds.


This is all to do with bond lengths.

F - F bond has the shortest bond length between them all, for its smallest atomic radius (least shielding effect). However, the length is so close, that the lone pairs from the two atoms start to repel each other, therefore the bond would be compromised and slightly weaker than that of Cl - Cl.

For Br - Br and I - I bonds, the bond lengths are longer than Cl - Cl's, resulting in a decrease in bond strength as the attraction between the bonded electron from one atom and the nuclei from the other atom decreases. (further away, smaller the attraction)

  • $\begingroup$ Could you tell me which elements other than F have bond lengths so short as to repel each other (and have lower bond energy than it would otherwise)? $\endgroup$ Nov 18, 2018 at 10:34
  • $\begingroup$ also, I'm guessing that in general, the trend is that as you go down a period, you have smaller bond energy — am I right in making this observation? What about across a period? $\endgroup$ Nov 18, 2018 at 10:35
  • $\begingroup$ There are quite a few that have bond lengths so short, for example N-N and O-O if you compare their bond enthalpies with P-P and S-S, if I am not mistaken. But do be careful with multiple bonds. Going across the group, however, to me doesn't seem to have an obvious fixed trend, especially if you compare O-O and S-S with their respected periods. I suggest you to search for some bond length and enthalpies tables. I always recommend the IB data booklet for chemistry. $\endgroup$
    – Mike
    Nov 18, 2018 at 12:15

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