We know that species that have a lone pair of electrons are Lewis bases (like ammonia). The structue of $\ce{HCN}$ is $\ce{H-C#{N:}}$; nitrogen has a lone pair of electrons and it is still acidic. Why?

We know that species with odd electron counts such as nitrogen dioxide (having odd electrons) dimerise to acheive greater stability. The structue of $\ce{HNC}$ is $\ce{H-N#C}$. So why it does not dimerise?

  • 2
    $\begingroup$ How is HNC an odd-electron species? $\endgroup$ – orthocresol Dec 17 '16 at 5:31
  • $\begingroup$ Nitrogen has 5 electrons in its outermost orbit. Out of 5 electrons, 4 electrons are bonded. Hence N has 1 electrons left. $\endgroup$ – user399511 Dec 17 '16 at 6:31
  • $\begingroup$ You heard about dipolar bonds and compounds? en.wikipedia.org/wiki/Dipolar_compound $\endgroup$ – Mithoron Dec 17 '16 at 16:35

The usual rule is one question per post unless strongly related; while your questions are not really related I’ll still answer both of them since it’s only two.

Why is $\ce{HCN}$ not a base?

Well — it is a base. You seem to be thinking that being an acid and being a base is a dichotomy for some reason; rest assured that aside from the very simple Arrhenius theory no acid-base theory predicts that bases cannot be acids and vice-versa.

You correctly point out that there is a lone pair on nitrogen, and that lone pair can indeed act as a (Brønsted-Lowry) base. The only ‘difficulty’ is that the carbon side of the $\ce{^-C#N}$ entity is more basic than the nitrogen side; thus protonation will happen on carbon first.

$$\ce{H-C#N^+-H <=> H-C#N + H+ <=> ^-C#N + 2 H+}\tag{1}$$

Equation $(1)$ shows the complete acid-base equilibrium.

Why is $\ce{NCH}$ not an odd-electron compound?

I took the liberty of rephrasing your second question. If you count electrons as is done in equation $(2)$, you will realise that $\ce{CNH}$ comes in at ten electrons — an even number — just like $\ce{HCN}$.

$$\sum n_{\ce{e-}} = n_{\ce{e-}}(\ce{C}) + n_{\ce{e-}}(\ce{N}) + n_{\ce{e-}}(\ce{H}) = 4 + 5 + 1 = 10\tag{2}$$

When writing the Lewis structure, you should remember formal charges. Indeed, your description of $\ce{CNH}$ is bad; instead you should write its Lewis structure as $\ce{^-C#N^+-H}$. All valences are satisfied, there is no reason for the compound to dimerise. Assuming five electrons to be on nitrogen (and thus four to be on carbon) would generate a very unstable diradical which would violate the octet rule in a bad way (nine electrons on nitrogen). Lower electron count violations of the octet rule are fine but going over the electron count for a main group element does not happen.

However, there is a great tendency for the compound to de- and reprotonate as written in equation $(3)$:

$$\ce{CNH <=>> CN- + H+ <=>> HCN}\tag{3}$$


The structure of hydrogen cyanide is $\ce{H-C#N}$ not $\ce{H-N#C}$ so there is no lone electron on the carbon atom to dimerise as $\ce{H-N#C-C#N-H}$.

Hydrogen cyanide is an acid ionizing as $\ce{H^+}$ and $\ce{^-C#N}$ in aqueous solution.

  • $\begingroup$ Nitrogen has 5 electrons in its outermost orbit but in cyanide and isocyanide only 4 electrona are bonded . 1 electron is left $\endgroup$ – user399511 Dec 17 '16 at 17:22
  • 1
    $\begingroup$ This does point out some inconsistencies but does not answer the question., $\endgroup$ – Jan Dec 17 '16 at 19:21

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