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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)?

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2 Answers 2

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.

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

Seems that unpaired electrons on antibonding orbitals can only pi-overlap. In NO the electronic density is shifted thowards 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 dimere bond is weak ( enthalpy of formation is about 17 kilojoules)

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