As nitrogen has 1 lone pair and 3 electrons, either it should have maximum covalency of 5 or 3. But why does it have a maximum covalency of 4 instead?
Why did it leave 1 electron? Why did it have to break a lone pair?


Recall that covalency is the number of shared electron pairs formed by an atom of that element. Nitrogen's maximum covalency is indeed $4$. And no, it does not break up its lone pair.

I'll give you a simple example. Have a look at the Lewis structure of the ammonium ion:

Ammonium ion (source)

Notice that nitrogen's octet is complete as soon it bonds with three $\ce{H}$ atoms (aka forms ammonia). The fourth covalent bond is actually a coordinate covalent bond, formed when that nitrogen atom's lone pair gets donated to a proton.

This is also the maximum covalency for the nitrogen atom, since it has no more unpaired electrons that could be paired up with other atoms, to form more covalent bonds.

Homework exercise: Can you now deduce the maximum covalency of nitrogen's elder brother, oxygen?

Caution: extra stuff ahead this will help those with knowledge beyond high school level. If you're at or below high school level, go back home and play with your cat.

After-thought: Valency is a useless term that doesn't help you do any chemistry. Different sources define it differently (two definitions on Wikipedia). It just helps fill high school textbooks with more pages, but it becomes irrelevant with the introduction of better terms like coordination number, which actually tells you something about the structure of the molecule. While "valency" maybe useful in basic beginners course at getting a grip, there are limitations to this point of view.

Here's an example of such a contradiction. You may expect elements of period $3$ and above to display higher covalencies. One such example is phosphorus. Though it belongs to the same group as nitrogen, it can form compounds like $\ce{PCl5}$, (apparently) increasing its maximum covalency to $5$ instead. The reasons for these are usually attributed to hypervalency/octet-expansion, but they are wrong and obsolete concepts, having been superseded by newer concepts. In fact, $\ce{P}$ still has a covalency of four in $\ce{PCl5}$, since there are only four shared pair of electrons (the non-bonding electrons don't count). This reinforces the idea that coordination numbers are a better and more useful term than valency. ($\ce{P}$ now has a coordination number of $5$ in $\ce{PCl5}$)

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    $\begingroup$ It must be noted that as generally only valence electrons participate in bonding, the 1s electrons cannot be donated to form 5 bonds from 5 available orbitals. $\endgroup$ – Apoorv Potnis Feb 13 '18 at 11:13
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    $\begingroup$ Gaurang, your "homework exercise" is rather difficult. Certainly oxygen can share 3 pairs as H3O+, but can it share 4 pairs? sciencedirect.com/science/article/abs/pii/0009261487804906 ; pubs.acs.org/doi/pdf/10.1021/ja00265a031 $\endgroup$ – DavePhD Mar 30 '18 at 12:39
  • $\begingroup$ @DavePhD Yes, you're right, thanks for bringing it here. I had later also come across this Zinc acetate. So, tetravalent oxygen does exist. Though, I was only thinking from an elementary point of view. $\endgroup$ – Gaurang Tandon Mar 30 '18 at 13:16
  • $\begingroup$ Hi Gaurang, does Oxygen shows a maximum covalency of four (2 cb, 2 coord-cb) or three (as 2 coord-cb isn't stable due to high EN of O) ? $\endgroup$ – rv7 Mar 11 at 6:25

The nitrogen has an electronic configuration $\ce{1s^2 2s^2 2p^3}$, therefore it could be expected to make 3 covalent bonds accommodating 3 electrons in orbitals $\ce{p}$. However, there is less repulsion of the electron domains when the $\ce{2s}$ orbital hybridizes with the $\ce{p}$ orbitals to form 4 $\ce{sp^3}$ orbitals. In one of the orbitals there will be a pair of paired electrons that can make a coordinate bond. The other 3 orbitals will make 3 covalent bonds accommodating 3 electrons.

Note: In the ammonia molecule the electron pair alone is in an orbital $\ce{sp^3}$, different from the isolated atom. There is no distinction between the bonds (covalent and coordinate) in the ammonium ion.

  • $\begingroup$ [...] there is less repulsion of the electron domains when the 2s orbital hybridizes [...] Hybridisation is a mathematical concept, what you are probably trying to say is that s and p orbitals mix, which is similar, but different. (You can correctly describe the molecule without hybrid orbitals, you cannot describe it correctly without sp mixing.) Also, nitrogen does not form four sp³ orbitals, that is an approximation, as the lone pair will still have a slightly larger s contribution. Nevertheless, this answer is correct in its essence. $\endgroup$ – Martin - マーチン Feb 14 '18 at 5:13

Nitrogen has maximum covalency 4, because we know that maximum electrons a shell can accommodate is $2n^2$, and hence it can have only 8 electrons. So covalency cannot exceed more than 4. `

  • $\begingroup$ This doesn't seem to make sense. By your reasoning, all elements with the same n should have the same covalency. $\endgroup$ – Dmitry Grigoryev Feb 13 '18 at 12:03
  • $\begingroup$ @Dmitry You omitted maximum, and that is the key word, therefore the statement is correct (in the confines of the definition of [co-]valency). $\endgroup$ – Martin - マーチン Feb 14 '18 at 5:16

Simple answer would be to say that nitrogen bonds with any element with only 4 orbitals,1s and orbital and 3p orbital.So its covalency is limited to 4.


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