Nature of the Bonding in C2/Ti2, N2/V2, Cr2 and related species.

Henry S. Rzepa

Department of Chemistry, Imperial college London

This short tutorial presents an analysis of the bonding in some simple diatomic species, contrasting the behaviour of these species deriving from the main group elements of the periodic table and the corresponding transition series.

Dicarbon and Dititanium

Titanium and carbon are related by virtue of their relationship in the periodic table; carbon occupies the main group and titanium the same position in the transition series. Each has four valence electrons, but when these are combined to form bonds, they achieve rather different results.

Atomic carbon is usually said to organise its valence electrons into a 2s22p12p12p0 configuration, which gives total of eight electrons (for two atoms). Symmetric and antisymmetric combinations of the (doubly occupied) pure 2s atomic orbitals (AOs) would result in a bonding and antibonding σ-combination of molecular orbitals, both doubly occupied and the resulting bond orders of which would exactly cancel. The two singly occupied 2p orbitals overlap in parallel fashion to form two bonding doubly occupied π molecular orbitals which contribute a net bond order of two. This gives us the (text book) picture of predicting a total bond order of 2 for this molecule (which is nevertheless unusual, in having only π bonds, with no underlying σ bond).

However, another way thinking about it is to start from a 2s12p12p12p1 atomic orbital configuration. Whilst this (initially) takes a little more energy, it is hoped this energy will be (more than) recovered in any eventual molecular orbitals formed by combining the AOs. Firstly a bonding orbital is formed by end-on (σ) overlap of (mostly) the (singly occupied) 2s atomic orbitals, Two equal (degenerate) molecular orbitals are formed by parallel (π) overlap of two of the 2px/y orbitals on each atom. The remaining 2pz1 orbital starts to overlap end-on. This is where things start to get less simple. Overlap of two p orbitals end on

Dicarbon, with triple bond, comprising:
Two bonding 2pπ (parallel p) Non-bonding 2pσ (end-on p) One Bonding 2sσ
c2-e6.3dmf c2-4.3dmf c2-3.3dmf
Dititanium, with quadruple bond, comprising:
One Bonding 3dδ (parallel d) Two bonding 3dπ (end-on d) One Bonding 4sσ
ti2-22.3dmf ti2-21.3dmf ti2-19.3dmf

The end-on overlap is shown in Scheme 1, in which the two 2p orbitals are shown vertically offset for clarity. In 1 the two pz orbitals approach out of phase. As the two opposite phases begin to interact (2) they eventually get close enough that the two out-of-phase lobes overlap maximally, and hence 3 would correspond to an anti-bonding overlap. But, as the two AOs get even closer (4) a new in-phase overlap begins to neutralise the original out-of-phase overlap until in 5 they are more or less balanced. You could perform this exercise starting from an initial in-phase approach, but the end result would be the same. At this point (5) we have a non-bonding orbital! If the (2s) σ bonding orbital is added to the two (2p) π bond orders, the overall bond order reaches 3! (albeit a situation rarely if ever shown in text books). Clearly, its not quite as simple as that, since one must also superimpose upon this picture some degree of s/p hybridisation (which in effect mixes the two starting atomic configurations to achieve something in between, in the trade this is known as a multi-reference configuration), and obviously the final bond length between the two carbon atoms determines the degree of overlap. Needless to say, when a full quantum-mechanical calculation is done to calculate all these effects, the net result is something close to a predicted triple rather than a double bond for C2!

Titanium plays the same game, this time starting from an atomic electronic configuration of 4s13d13d13d1. Now each of the single electrons can pair up (yes you guessed, we are forming a singlet state here, see below for a reality check!) to form only a bonding interaction. Firstly the two 4s1 electrons pair up to form a single σ bond. Next two of the 3d1 d-orbitals overlap "end-on" to form two "π" bonds formed by d-orbitals. Unlike the end-on overlap of the (non-bonding) 2p orbitals in C2, this combination is strongly bonding. The final 3d1 orbital on each atom now overlaps in parallel fashion to form a final so-called δ bond. The overall bond order is four, i.e.. a quadruple bond, a higher order than with singlet C2 due to the difference in nodal properties between the carbon 2pz and the titanium 3d atomic orbitals.

Reality Check: Of course, Ti2 as a real species does not actually have a quadruple bond (and a singlet ground state). In fact, it forms a high spin triplet state, where two of the electrons uncouple their spins, with a corresponding decrease in the bond order from four to three. The real C2 molecule is however a singlet ground state, i.e.. all the electrons are spin paired; the unpaired triplet state being somewhat higher in energy.

Dinitrogen and Divanadium

These two elements also have the same periodic table relationship. The nitrogen atom really does have an atomic configuration of 2s22p12p12p1. Here, the 2s/2p mixing is much less than with carbon (due to the greater energy difference between the two) and the (mostly) 2s orbitals combine to form bonding and anti-bonding combinations, both of which are populated with 2 electrons each. This results in a zero combined bond order (unlike dicarbon where this combination overall results in a bond order of 1). Each of the three (mostly) 2p orbitals form bonding combinations of one in-phase end-on σ overlap and two π type overlaps, and the six electrons populating these result in a total bond order of 3, but now originating in a rather different way than that of carbon.

Vanadium has an atomic configuration of 4s13d13d13d13d 1. Two of the d AOs overlap in parallel fashion to form two δ bonds. The 4s orbitals overlap to form a σ bond and the last two π bonds form by end-on d-overlap. Thus the complete absence of the 2s bonding/antibonding cancellation takes the V2 bond order to five!

Reality Check

V2 also does not actually have a pentuple bond, since like Ti2, the ground state is a triplet species and not a singlet.
Nitrogen, with triple bond, comprising:
One Bonding 2pσ (end-on p) Two bonding 2pπ (parallel p)
n2-7.3dmf n2-e6.3dmf
Divanadium, with pentuple bond, comprising:
Two Bonding 3dδ (parallel d) One Bonding 4sσ Two bonding 3dπ (end-on d)
v2-e23.3dmf v2-a21.3dmf v2-e20.3dmf

Onwards to Dichromium

Dichromium, with hextuple bond, comprising:
Two Bonding 3dπ (end-on d) One bonding 3dσ (end-on dz2) Two bonding 3dδ (parallel d) One Bonding 4sσ
Cr2-homo-4e Cr2-homo-3a Cr2-homo-1e Cr2-homo-a

Yes, you guessed it, Cr2 continues the trend of Ti2 and V2 in having a hextuple bond. If atomic chromium is considered as having the configuration 4s13d13d13d13d 13d1, the resulting bonds comprise one σ bond formed from the 4s orbitals, two π and two δ bonds formed by respectively end-on and parallel overlap of the d-orbitals, and (additional to V) a σ bond formed by end-on overlap of the dz2 orbital. This time, the ground state really is a closed shell low spin singlet. Despite the remarkable bond order, it has a relatively weak bond strength of about 33 kcal/mol! In effect, all those twelve electrons crammed into the small diatomic region do not result in strong bonds.

A Closer and perhaps different look at C2, N2 and other species.

Another way of looking at the bonding in C2 is to transform the molecular orbitals to so-called localised orbitals. The following diagrams illustrate such orbitals transformed according to a scheme proposed by Pipek and Mezey. Two localised orbitals for dicarbon appear essentially identical to the molecular orbitals and correspond to the orthogonal π orbitals. The two end-on molecular orbitals jointly transform to two (identical) localised orbitals (unlike the MOs which are different), each having significant density in the C-C region (i.e.. each of these localised orbitals contributes about half a bond order). This is another way of achieving the quantum mechanical sp hybridisation referred to above.

Pipek-Mezey Localised orbitals
Carbon Nitrogen
Two π Two σ Two π Two σ One σ
c2-pm-pi c2-pm-pi c2-pm-pi c2-pm-pi c2-pm-pi
Boron B22- Nitrogen N22+
Two π Two σ Two π Two σ
c2-pm-pi c2-pm-pi c2-pm-pi c2-pm-pi

Contrast this with nitrogen. The two π orbitals are the same as before, but now there are two end-on (again identical) σ localised orbitals which consist mostly of 2s character, neither of which shows any bonding in the N-N region (this is equivalent to saying that doubly occupied bonding and anti-bonding 2s combinations cancel). This situation corresponds to the 2s22p12p12p0 atomic configuration discussed for carbon; it is favoured because the greater electronegativity of N means an electron prefers the 2s to a 2p orbital, i.e.. The greater nuclear charge on nitrogen differentiates more effectively between 2s and 2p atomic orbitals. Unlike carbon, nitrogen has a third occupied 2pz1 atomic orbital, which forms a localised and clearly bonding orbital of largely 2p end-on character. Because of the greater nuclear charge on N, the 2p atomic orbital is also rather "smaller" than that for carbon, and so the onset of non-bonding character which could arise when two of them overlap end on (scheme 1 above) is less advanced for N. The effect however can be detected, since the resulting N-N bond length is actually rather longer than it ought to be (despite often being cited as the strongest bond). If N2 is protonated to form a diazonium ion, HN2+, the N-N bond length actually shrinks because the H+ polarises the 2pz orbital, making the σ-bond stronger (this now is the record holder for bond strength!).

Going in the opposite direction, boron (as B- to make it isoelectronic with C) would be expected to increase the 2s/2p mixing because of the smaller nuclear charge (i.e.. favouring even more the 2s12p12p12p1 configuration). Thus the corresponding 2spz end on orbital for the molecule B22- (which is isoelectronic with C2) is now bonding rather than non-bonding, and the two resulting localised orbitals each contribute almost a whole single bond order rather than around half. This little molecule has a total predicted bond order of four (two π, two σ), although perversely the B-B bond length is actually longer than the C-C length.

Moving back to N22+, the positive charge has the effect of further contracting the size of the valence atomic orbitals. Perhaps the 2pz orbital is so small that it too avoids the non-bonding region (Scheme 1). At any rate, this molecule too has a total predicted bond order of around 4 (and that too has not made it into the text books!).


  1. Gutsev, Gennady L.; Bauschlicher, Charles W. Chemical Bonding, Electron Affinity, and Ionization Energies of the Homonuclear 3d Metal Dimers. Journal of Physical Chemistry A (2003), 107(23), 4755-4767.
  2. Barden, Christopher J.; Rienstra-Kiracofe, Jonathan C.; Schaefer, Henry F., III. Homonuclear 3d transition-metal diatomics: A systematic density functional theory study. Journal of Chemical Physics (2000), 113(2), 690-700.

Viewing the orbitals

To demonstrate the 3D nature of these orbitals, each thumbnail image above is linked to a 3DMF file. To view these orbital models, you will need a 3DMF viewer such as 3DMFPlugin (a browser plugin), 3DMF Optimizer or Geo3D (Macintosh applications) or 3DMF Viewer for Windows.

Windows Users

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  1. install the QuickDraw3D libraries from Apple
  2. Download this plugin
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(C) H. S. Rzepa, 2004.