18
$\begingroup$

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)?

$\endgroup$
14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Just to clarify, entropic effects refer to T (delta S) and not just delta S right? $\endgroup$ – Tan Yong Boon Aug 3 '18 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. $\endgroup$ – Dan Dec 25 '18 at 17:48
6
$\begingroup$

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)

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.