# Bond dissociation energy order for C−C, N−N, O−O and F−F single bonds

Data from Chemistry LibreTexts — Strength of Covalent Bonds, Table 8.8.1: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K:

$$\begin{array}{cr} \hline \text{Bond} & D/\pu{kJ mol-1} \\ \hline \ce{C-C} & 346 \\ \ce{N-N} & \approx 167 \\ \ce{O-O} & \approx 142 \\ \ce{F-F} & 155 \\ \hline \end{array}$$

I couldn't understand the trend of bond dissociation energy of various single bonds. Why

$$D(\ce{C-C}) > D(\ce{N-N}) > D(\ce{F-F}) > D(\ce{O-O})?$$

I had mugged up the trend the now, and according to me BDE should continuously increase while going from left to right in a period due to decrease in size and increase in ENC. Why this irregular trend?

• A simple explanation would be using the idea of lone pair-lone pair repulsions. As we go from C-C to F-F, there is an increasing number of lone pairs on the bonding atoms. Thus, this increasing destabilising interaction results in the bond energies following that particular trend. Aug 1 '18 at 5:11

1. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., $$\ce{N, O, F}$$), which are unusually weak compared with single bonds between their larger congeners. It is likely that the $$\ce{N–N, O–O,}$$ and $$\ce{F–F}$$ single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms.
2. $$\ce{N-N}gt \ce{F-F}$$, in $$\ce{F}$$ there are greater number of lone pairs than in $$\ce{N}$$ so it experiences strong repulsion between its electrons that make it vulnerable to be broken.
3. $$\ce{F-F} \gt \ce{O-O}$$ it's so because $$\ce{F}$$ atom is very small that enables strong electrostatic attraction between nucleus and electrons For further information on 3) you can follow this link.