The $\ce{CN}$ molecule has been observed spectroscopically in comets where pressure has been estimated to be as low as $10^{-10}\ \mathrm{mmHg}$. Draw a Lewis structure for $\ce{CN}$. Suggest an explanation as to why this species can persist in a comet but is unstable under normal terrestrial conditions?

Having drawn the Lewis structure, the molecule is a free radical, hence it is very unstable. Would the species be unstable under terrestrial conditions due to a higher temperature (hence pressure) there? Or more specifically, how does a change in pressure effect a free radical molecule?

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    $\begingroup$ Temperature is unrelated to pressure. As for the radical, it is stable by itself, but very reactive (=unstable) when it meets another radical (or just about anything else, for that matter), which happens more often when there is more stuff around. $\endgroup$ – Ivan Neretin Jan 21 '16 at 20:29
  • $\begingroup$ @IvanNeretin I'm not a native english speaker, but reactive not means unstable. For example the complex of cobalt with EDTA is very very stable in water but if you have cobalt in water as a complex like $[Co(H_2O)_6]^{3+}$ it is absolutely unstable but to totally inert, then if you add a bit of EDTA you will never get the complexe of Co with EDTA if you don't heat up your solution... $\endgroup$ – ParaH2 Jan 21 '16 at 20:52
  • $\begingroup$ Neither am I. Sure, these two words are not synonyms, but in this particular context they can be used this way. OK, I should have said "extremely reactive to everything around (including itself), and hence unstable". $\endgroup$ – Ivan Neretin Jan 21 '16 at 21:04
  • $\begingroup$ Some molecules are inherently unstable, CN, CN- or even CN+ aren't. $\endgroup$ – Mithoron Jan 21 '16 at 21:16
  • $\begingroup$ Is there a direct relationship of the stability of a molecule to the pressure it is undergoing? @IvanNeretin $\endgroup$ – mnmakrets Jan 22 '16 at 2:28

$\ce{NC-CN <=> 2CN}$

Which side is favored at low pressure?

  • $\begingroup$ Would the right side be favoured at low pressure because there is less force exerted on the molecules to drive them together, hence will remain separate? $\endgroup$ – mnmakrets Jan 29 '16 at 1:27
  • $\begingroup$ @mnmakrets The right side will be favored, but not for the reason you have given. The concentration (and partial pressure) of CN will be squared in the equilibrium expression. So for example if K=1, then the system will be at equilibrium when, for example, p(CN) = p(NCCN) = 1. But if the pressure is decrease such that p(CN) = p(NCCN) = 0.01, then the system will not be at equilibrium and will need to shift far to the right to achieve equilibrium. Read more about pressure effects on equilibrium generally. $\endgroup$ – DavePhD Jan 29 '16 at 11:52

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