How to find the bond order in delocalized pi bonding of more than two atoms? [duplicate]








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[Bond order for carbonate ion for resonance](/questions/1302/bond-order-for-carbonate-ion-for-resonance)

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J.D Lee has written down some rules for calculating the bond order. 



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I have no difficulty following step 1 and 2. It's easier to work with an example so I'll highlight my issues with step 3 , 4 by using $\ce{CO\_3^{2-}}$ .  

It's been used by the book. 



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Here's my thought process 


1. The structure is trigonal planar
2. Number of electrons; 24


Now, the parts I have difficulty handling 


3. 3$\mathrm{\sigma}$ bonds .They've written that oxygen has four non-bonding electrons.  

Is it because oxygen's valency is 6 and after assigning the $\mathrm{\sigma}$ bonds.Each oxygen gets 2 of bonding electrons and a leftover of four which are satisfied as the non bonding electrons?
4. I follow that 4 molecular orbitals will form but how is it decided how many bonding , non-bonding and anti-bonding ?




**Point 3)**


Why some electrons are $\mathrm{\sigma}$-bonding, $\mathrm{\pi}$-bonding or non-bonding is easier to understand if you first think about the orbitals involved. I'll discuss it in terms of hybridisation.


The central $\ce{C}$ atom has an $\mathrm{sp^2}$ hybridisation (corresponding to a trigonal planar structure) so it presents three $\mathrm{sp^2}$ orbitals, one towards each of the $\ce{O}$ atoms, plus one non-hybridised $\mathrm{2p\_z}$ orbital, perpendicular to the plane of the molecule. $\ce{O}$ atoms, for their part, we can picture as aligned towards the $\ce{C}$ atom - so, for each $\ce{O}$ atom, their $\mathrm{2p\_x}$ orbital will point to the central $\ce{C}$ atom, their $\mathrm{2p\_z}$ orbital will be perpendicular to the plane of the molecule (and so parallel to the carbon $\mathrm{2p\_z}$ orbital), and the remaining $\mathrm{2p\_y}$ orbital will be in the plane of the molecule but perpendicular to the $\ce{C}$-$\ce{O}$ axis.


Each of the $\mathrm{sp^2}$ orbitals has symmetry properties suitable for $\mathrm{\sigma}$-bonding with the $\ce{O}$ atoms - they are axially symmetric around the line connecting the $\ce{C}$ and $\ce{O}$ nuclei and have no nodes. They are aligned and overlap with the $\mathrm{2p\_x}$ orbitals of the oxygen atoms, allowing for effective bonding. So we can build three $\mathrm{\sigma}$ bonds that take up 6 electrons, and we have 18 electrons left.


Two of the orbitals in each $\ce{O}$ atom cannot overlap in a bonding way: the $\mathrm{2s}$ orbitals because they are too small and localised, and the $\mathrm{2p\_y}$ orbitals because they have opposite sign lobes at each side of the $\ce{C}$-$\ce{O}$ axis - so the overlap with the carbon orbitals is zero. These two orbitals per $\ce{O}$ atom will hold two pairs of non-bonding electrons - for a total of 12. So, we have 6 electrons left.


Finally, we have four $\mathrm{2p\_z}$ orbitals (one on the carbon atom, three on the oxygen atoms) which have a symmetry suitable for $\mathrm{\pi}$-bonding (they are parallel, with the nodal plane in the plane of the molecule). Combining them will require a molecular orbitals approach and take us to point 4.


**Point 4)**


So, we have 4 suitable $\mathrm{2p\_z}$ orbitals and 6 electrons left. Combining these to form $\mathrm{\pi}$ molecular orbitals is a bit more complex, though, as symmetry plays a very important role, so outside from simple cases, the results may not be immediately obvious.


The three oxygen $\mathrm{2p\_z}$ orbitals, under the trigonal planar symmetry of the molecule, form three ***symmetry adapted linear combinations***. These three symmetry adapted versions all have the same energy as the separate $\mathrm{2p\_z}$ orbitals, but they have symmetries that are compatible with the trigonal planar molecule, which is something we need if we want to see if bonds can be formed. Since they all have the same energy as the three equivalent $\mathrm{2p\_z}$ orbitals, they will also be degenerate.


[![Symmetry adapted linear combinations](https://i.stack.imgur.com/UNbWX.gif)](https://i.stack.imgur.com/UNbWX.gif)


[Bird, P. Chemistry 241 - Inorganic Chemistry I. http://faculty.concordia.ca/bird/c241/notes\_ch3-shriver.html (accessed 19/03/2018)](http://faculty.concordia.ca/bird/c241/notes_ch3-shriver.html)


So now we wonder: how do these symmetry adapted orbitals interact with the lone carbon $\mathrm{2p\_z}$ orbital?


It turns out, one of those orbitals (identified in the image as $\mathrm{\Psi\_1}$) has a non-zero overlap with the carbon $\mathrm{2p\_z}$ orbital, while the other two ($\mathrm{\Psi\_2}$ and $\mathrm{\Psi\_3}$) have an overlap of zero (because of signs and symmetry). And since overlap is a requisite for bonding, these two linear combinations will be *non-bonding* orbitals (i.e. they will have the same energy in the $\ce{CO\_{3}^{2-}}$ molecule as they would have in the isolated oxygen $\mathrm{2p\_z}$ orbitals).


So, what about the symmetry adapted linear combination that *did* have the correct symmetry to interact with the carbon $\mathrm{2p\_z}$ orbital? They do interact, forming two combinations: one which is more stable than the independent orbitals, and therefore a *bonding* molecular orbital, and another that is less stable and therefore *anti-bonding*.


[![enter image description here](https://i.stack.imgur.com/FutN6.gif)](https://i.stack.imgur.com/FutN6.gif)


[Bird, P. Chemistry 241 - Inorganic Chemistry I. http://faculty.concordia.ca/bird/c241/notes\_ch3-shriver.html (accessed 19/03/2018)](http://faculty.concordia.ca/bird/c241/notes_ch3-shriver.html)


So, we have four molecular orbitals out of our four atomic $\mathrm{2p\_z}$ orbitals: in ascending order of energy, those would be one bonding orbital, two degenerate non-bonding orbitals, and one anti-bonding orbital. We also had 6 electrons yet to assign - which will fill the bonding and two non-bonding orbitals. Those non-bonding orbitals are the molecular equivalent of "lone pairs" - so they contribute nothing to the bond order. The fully-occupied bonding orbital does; so we have a single $\mathrm{\pi}$ bond shared by the four centres, leading to a bond order of $\mathrm{1}$ (for the $\mathrm{\sigma}$ bond) + $\mathrm{\tfrac{1}{3}}$ (for the $\mathrm{\pi}$ bond) = $\mathrm{1 \tfrac{1}{3}}$.


It is important to stress that it's three bonds of order $\mathrm{1 \tfrac{1}{3}}$, and not one bond of order $\mathrm{2}$ and two of order $\mathrm{1}$: the interaction between the central $\mathrm{2p\_z}$ orbital and those on the oxygen atoms required adapting them to the trigonal planar symmetry, and, as you can see in the images above, all three oxygen atoms contribute to the bonding and anti-bonding molecular orbitals.


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