Chemistry · Chemical Bonding · NEET
In light molecules (B2, C2, N2) the 2s and 2p atomic orbitals are close in energy, so they mix (this is called s-p mixing). This mixing pushes the sigma2pz orbital UP in energy, above the two pi2p orbitals. So the order for these molecules is: ...sigma*2s < (pi2px = pi2py) < sigma2pz < (pi*2px = pi*2py) < sigma*2pz. In O2 and F2 the 2s-2p energy gap is large (because nuclear charge is higher), so s-p mixing is very weak. Without that push, sigma2pz stays LOW, below the pi2p orbitals: ...sigma*2s < sigma2pz < (pi2px = pi2py) < (pi*2px = pi*2py) < sigma*2pz. That is the whole crossover in one idea.
It depends on the molecule. For B2, C2, N2 (14 electrons or fewer): pi2p comes FIRST (lower energy), then sigma2pz. For O2, F2, Ne2 (more than 14 electrons): sigma2pz comes FIRST (lower energy), then pi2p. A quick rule: if the molecule has 14 or fewer valence-shell electrons in the n=2 scheme, use the pi-below-sigma order.
s-p mixing means the sigma orbitals made from 2s and the sigma orbital made from 2pz have the same symmetry, so they interact. When two orbitals of the same symmetry are close in energy, the lower one drops and the upper one rises. Here the sigma2pz is pushed UP. This only matters when 2s and 2p are close in energy, which is true for B, C, N. By oxygen and fluorine the 2s is much lower than 2p, the gap is big, mixing is tiny, and sigma2pz is no longer pushed above pi2p.
For counting bond order it often does not matter, because bond order = (bonding - antibonding)/2 and you fill the same number of electrons either way. It matters for two NEET-favourite facts: (1) magnetic behaviour (which orbital holds the LAST electrons, so whether they are paired or unpaired), and (2) statements like 'C2 has only pi bonds' or 'which orbital is the HOMO'. Get the order wrong and you get these wrong.
Count the total electrons. B2 = 10, C2 = 12, N2 = 14 all use pi-below-sigma (pi2p first). O2 = 16, F2 = 18, Ne2 = 20 all use sigma-below-pi (sigma2pz first). The switch happens right after N2. So N2 is the last molecule with the pi-first order; O2 is the first with the sigma-first order.
The correct order of energies of the molecular orbitals of the N2 molecule is:
Which of the following diatomic molecular species has only pi bonds according to Molecular Orbital Theory?
Match List-I (molecule) with List-II (bond/s between the two carbon atoms): A. ethane; B. ethene; C. carbon molecule C2; D. ethyne. List-II: I. one sigma and two pi; II. two pi; III. one sigma; IV. one sigma and one pi.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B2, C2 and N2 (10, 12 and 14 electrons). All molecules up to and including N2 use this order because their 2s and 2p orbitals are close enough in energy to mix.
O2, F2 and Ne2 (16, 18 and 20 electrons). From oxygen onwards the 2s-2p energy gap is large, s-p mixing is negligible, and sigma2pz sits below the pi2p orbitals.
CO has 14 electrons and is isoelectronic with N2, so it follows the N2 order (pi2p below sigma2pz). Species like CN- and NO+ (also 14 electrons) do the same.
No. Because you fill the same number of electrons in bonding and antibonding orbitals either way, bond order is unchanged. The crossover mainly affects the HOMO identity and magnetic properties.
Using their correct orders, N2's last electrons pair up fully (diamagnetic), while O2's last two electrons go into two degenerate pi*2p orbitals singly (Hund's rule), leaving two unpaired electrons, so O2 is paramagnetic.
