Chemistry · Chemical Bonding · NEET
Bond Order = ½ × (number of electrons in bonding molecular orbitals − number of electrons in antibonding molecular orbitals). We write it as B.O. = ½ (Nb − Na). First fill all the electrons of the molecule into the MO energy-level diagram (bonding orbitals like sigma and pi, antibonding orbitals marked with a star *). Then count Nb and Na and put them in the formula. A higher bond order means a stronger and shorter bond. This is a very common NEET question, so learn the filling order by heart.
This is the biggest reason MOT is on the NEET syllabus. In O2 (16 electrons), the last two electrons go one each into the two equal-energy pi-star (pi*2p) antibonding orbitals following Hund's rule. So O2 has TWO unpaired electrons and is paramagnetic. In N2 (14 electrons), every orbital is completely filled with paired electrons, so there are no unpaired electrons and N2 is diamagnetic. Simple Lewis dot structures cannot explain O2's magnetism, but MOT can. That is why NEET loves this example.
Fill the electrons into the MO diagram, then look for unpaired electrons. If even ONE electron is unpaired (alone in an orbital), the molecule is paramagnetic (attracted by a magnetic field). If ALL electrons are paired up, it is diamagnetic (slightly pushed away by a magnet). Quick trick for NEET: if the total number of electrons is odd (like NO with 15), it must have an unpaired electron, so it is paramagnetic. O2, O2+, O2-, B2 and NO are all paramagnetic; N2, CO, CN-, F2 and O2^2- are diamagnetic.
For light molecules up to N2 (total electrons 14 or fewer, i.e. B2, C2, N2), the order is: sigma1s < sigma*1s < sigma2s < sigma*2s < (pi2px = pi2py) < sigma2pz < (pi*2px = pi*2py) < sigma*2pz. Here the two pi2p orbitals come BEFORE sigma2pz. For O2 and heavier (more than 14 electrons), sigma2pz drops below the pi2p orbitals. Mixing this up changes your bond order and magnetism answer, so remember: 'up to N2, pi is below sigma.'
Start from O2 which has bond order 2 (16 electrons, with 2 unpaired electrons in pi*2p). Removing an electron gives O2+ (dioxygenyl): you remove it from the antibonding pi*2p, so bond order rises to 2.5. Adding one electron gives O2- (superoxide): antibonding electrons increase, bond order falls to 1.5. Adding two gives O2^2- (peroxide): bond order 1. So the order of bond strength is O2+ (2.5) > O2 (2) > O2- (1.5) > O2^2- (1). Note O2+ still has one unpaired electron, so it is paramagnetic, not diamagnetic.
Any diatomic with 14 electrons that fills up to sigma2pz has bond order 3. This isoelectronic family includes N2, CO, CN- and NO+ (all 14 electrons, all diamagnetic, all bond order 3). CN has only 13 electrons and bond order 2.5, CN+ has 12 electrons and bond order 2. So among CN+, CN-, NO and CN, the highest bond order is CN- because it reaches 14 electrons. NEET repeats this 14-electron rule many times.
Which of the following is paramagnetic?
Consider the following species: CN+, CN-, NO and CN. Which of these will have the highest bond order?
Which amongst the following is an incorrect statement?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A bond order of zero means the molecule does not exist as a stable molecule. For example, He2 has 2 bonding and 2 antibonding electrons, so B.O. = ½(2−2) = 0, and He2 is not formed. A positive bond order is needed for the molecule to be stable.
Yes. When there is an odd number of electrons in bonding versus antibonding orbitals, bond order comes out as a fraction. NO has bond order 2.5, O2+ has 2.5, and O2- has 1.5. Fractional bond order is normal and correct in MOT.
Yes. As bond order increases, bond enthalpy (strength) increases and bond length decreases. This is why N2 (bond order 3) has a very short, very strong triple bond, while O2^2- (bond order 1) has a longer, weaker bond.
A paramagnetic substance is attracted into a magnetic field because it has unpaired electrons. A diamagnetic substance, with all electrons paired, is weakly pushed out of the field. Liquid O2 sticking between magnet poles is the famous demonstration of O2's paramagnetism.
Not always. A fast NEET shortcut: if the molecule has an ODD total number of electrons (like NO = 15, O2+ = 15), it must have at least one unpaired electron and is paramagnetic. For even-electron species you should still draw or recall the MO filling, because O2 is even yet paramagnetic.
