Why doesn't Nitrogen monoxide dimerize even though there is an odd electron present whereas nitrogen dioxide does (because of the odd electron on nitrogen)?


3 Answers 3


Highstaker's answer is almost correct. The antibonding orbitals that are occupied in NO are in fact pi symmetry, but when the dimer forms that is no longer relevant. It is a sigma bond.

The enthalpy of the newly formed sigma bond in the dimer is weak because the net gain in bond is off set by the loss of a very odd set of single-electron resonance forms available for NO monomer.

Given $\Delta G= -17 \:\mathrm{kJ/mol}$ , and that dimerization is entropically disfavored, when the total free energy is considered there is no gain since entropic effects are on the order of $10-30 \:\mathrm{kJ/mol}$. Thus any small gain in enthalphy is offset by the loss of entropy.

Always look at these things through Gibbs free energy and not just enthalpy.

  • $\begingroup$ Just to clarify, entropic effects refer to T (delta S) and not just delta S right? $\endgroup$ Aug 3, 2018 at 4:25
  • $\begingroup$ Yes, the net entropic "effect" on the direction of spontaneity is indeed T*(delS), which also explains the stability of of the dimer at lower temperatures. $\endgroup$
    – ditsuke
    Dec 25, 2018 at 17:48

Nitrogen monoxide's dimer exists, but it is very unstable. NO consists of dimeres only at low temperatures (about −163 °C). Solid NO seems to consist mostly of dimers.

Seems that unpaired electrons on antibonding orbitals can only pi-overlap. In NO the electronic density is shifted towards oxygen atom, and the "petals" of pi-orbitals don't have enough electronic density to support stable bond (some call it a half-bond). So the dimer bond is weak (enthalpy of formation is about 17 kilojoules)


One alternative rationalisation of why the dimerisation of $\ce {NO}$ is not as favourable as that of the dimerisation of $\ce {NO2}$ is that on each of the nitrogens in the $\ce {N2O2}$ dimer, there is a lone pair while for the dimer $\ce {N2O4}$, there are no lone pairs on the nitrogens. The presence of lone pairs destabilise the nitrogen-nitrogen bond in the former, thus it decreases the enthalpic favourability of the bond formation between two $\ce {NO}$ molecules, whereas this destabilisation is not present in the latter dimer.


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