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

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 2p_x/y orbitals on each atom. The remaining 2p_z¹ orbital starts to overlap end-on. This is where things start to get less simple.

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 p_z 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 2p_z and the titanium 3d atomic orbitals.

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

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 d_z2 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.