As known, nitrogen could form 3 bonds based on octet rule, because it has 5 valence electrons. That means it needs 3 bonds.

On the other hand, why sometimes nitrogen forms 4 bonds?

enter image description here

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    $\begingroup$ Because lone pairs can make dipolar bonds... $\endgroup$
    – Mithoron
    Aug 11 '17 at 17:25
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    $\begingroup$ The title and the question are somewhat incoherent. The question is rather trivial, whereas the title is quite interesting. If this is about the maximum number of bonds, I would answer I saw a structure with 7. $\endgroup$
    – andselisk
    Aug 11 '17 at 17:31
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    $\begingroup$ @andselisk True, but the original title (prior to the edit) was even more interesting ;) $\endgroup$ Aug 11 '17 at 17:36
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    $\begingroup$ @paracetamol Ah, the mighty sodium. Probably OP should clarify what exactly is expected from the answer. $\endgroup$
    – andselisk
    Aug 11 '17 at 17:37
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    $\begingroup$ Well, for nitrogen the richest coordination environment I know is a capped trigonal prism, C.N. 7 (Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288). But this is a rather non-standard case as the $\ce{N}$ atom is trapped inside a metal framework. $\endgroup$
    – andselisk
    Aug 11 '17 at 17:58

I'd like to try answer the question from the title regarding the maximum number of atoms nitrogen is capable of bonding with, and also slightly expand my comment.

Metal nitrido complexes are commonly known to have up to 6 metal centers bound with a single bridging $\ce{N3−}$ ion, located in an octahedral cavity. An interstitial nitrogen can contribute 5 electrons, and the rest is provided by the group 9 and 10 metals which are electron-rich (typically, $\ce{Rh}$, $\ce{Ir}$).

There are few more exceptions where formal C.N. for nitrogen is 7: complexes of lithium amides based on $\ce{\{Li14N10\}^{6-}}$ cluster framework [1,2] and an inclusion nitrido-cluster $\ce{\{Co2RhN2\}^{3-}}$ [3]. Unfortunately in both crystal structures [1,2] with 6- and 7-fold coordinated nitrogens of $\ce{N-Ar}$ groups those are heavily disordered. Structure [3] is more suitable for the representation.

tris(Tetramethylammonium)($\mu_7$-nitrido)-($\mu_6$-nitrido)-decakis($\mu_2$-carbonyl)-undecacarbonyl-deca-cobalt-rhodium(I) $\ce{[Co10RhN2(CO)21]^3-}$ [3] contains two non-equivalent 6- and 7-fold coordinated nitrogen atoms ($\mathrm{N2}$ and $\mathrm{N1}$, respectively), sharing a triangular face:

$\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#F090A0}{\Large\bullet}~\ce{Co}$; $\color{#0A7D8C}{\Large\bullet}~\ce{Rh}$;

enter image description here

Cluster core wireframe model without carbonyl ligands:

enter image description here

Atom $\mathrm{N1}$ with C.N. 7 is coordinated with 6 cobalts and 1 rhodium, forming a capped trigonal prism. Interestingly enough, $\mathrm{Co1}$ is a capping atom, not rhodium:

enter image description here

     N1   SYMM Co5   Co4   Co6   Co3   Co2  Rh1  Co1
Co5  1.90 I    -     -     -     -     -    -    -  
Co4  1.91 I    135.0 -     -     -     -    -    -  
Co6  1.92 I    79.5  80.2  -     -     -    -    -  
Co3  1.98 I    129.7 85.1  80.2  -     -    -    -  
Co2  2.00 I    82.3  140.2 128.4 75.4  -    -    -  
Rh1  2.18 I    80.7  81.1  128.3 144.9 95.3 -    -  
Co1  2.43 I    143.5 70.3  136.5 66.4  70.1 78.6 -  

Both interstitial nitrogens play the role of internal ligands, which provide cluster valence electrons (CVE), but don't contribute to steric hindrance between external ligands such as carbonyls, making the cluster more stable [4, ch. 1.18]


  1. Armstrong, D. R.; Barr, D.; Clegg, W.; Drake, S. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wright, D. S. Angew. Chem. Int. Ed. Engl. 1991, 30 (12), 1707–1709. DOI 10.1002/anie.199117071.
  2. Armstrong, D. R.; Ball, S. C.; Barr, D.; Clegg, W.; Linton, D. J.; Kerr, L. C.; Moncrieff, D.; Raithby, P. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wheatley, A. E. H.; Wright, D. S. J. Chem. Soc., Dalton Trans. 2002, 0 (12), 2505–2511. DOI 10.1039/B107970K.
  3. Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288.
  4. Metal clusters in chemistry; Oro, L. A., Braunstein, P., Raithby, P. R., Eds.; Wiley-VCH: Weinheim; New York, 1999. ISBN 978-3-527-29549-4.
  • $\begingroup$ while esoteric examples with high numbers of "bonds" are fascinating, the OP was asking about the far more basic case of why nitrogen can form 4 covalent bonds. For that question, this answer probably just confuses (though it would be a good one to a different question). $\endgroup$
    – matt_black
    Aug 23 '21 at 19:49

Nitrogen has three electrons in its 2p orbital. Therefore, it can form three bonds by sharing its three electrons. It cannot accept any more electrons but here's how it forms the fourth bond.

Nitrogen has one lone pair of electrons in its 2s orbital. It can donate this electron pair to form a coordinate bond. This coordinate bond that nitrogen forms by donating its electron pair to the vacant orbital of other atom is how it can form 4 bonds.


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