2 Added more precise language edited Nov 1 '17 at 11:12 Martin - マーチン♦ 34.7k99 gold badges117117 silver badges245245 bronze badges Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding. Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms roughlyhaving three $$\ce{sp^2}$$ hybridised(roughly) sp2 hybrid orbitals, giving rise to two $$\sigma$$σ bonds to the neighbouring carbons and one $$\sigma$$σ bond to hydrogen. The remaining 6 perpendicular $$\ce{p}$$p orbitals form the delocalised $$\pi$$π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. $$D_\mathrm{6h}$$D6h, three of them are bonding and three of them are antibonding. In the $$\pi$$π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.) The nitrate ion $$\ce{NO3-}$$ is what I like to call a Y-aromatY-aromat since itsit has some similarity in the electronic structure. The bonding situation is very similarcomparable to benzene. The resonance structures, however, look much more complicated. In In quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals $$\ce{sp^2}$$ hybridised(and one p orbital) and the oxygen as $$\ce{sp}$$(roughly) having two sp (and 2 p) orbitals. You follow the same scheme, sotherefore you obtain four equivalent nitrogen oxygen $$\sigma$$σ bonds. Each oxygen has one lone pair in the in-plane $$\ce{p}$$ orbitalp orbital and one lone pair in the $$\ce{sp}$$ hybridsp hybrid. The remaining out-of-plane $$\ce{p}$$ orbitalsp orbitals of nitrogen and oxygen build up the delocalised $$\pi$$π system. This time there are only four orbitals. Because of symmetry constraints, i.e. $$D_\mathrm{3h}$$D3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene. Since Since the carbonate anion $$\ce{CO3^{2-}}$$ is isoelectronic to nitrate, approximately the same orbital picture will result. * Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another. For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $$\ce{N-O}$$ bonds and antibonding towards the other. (It is also bonding with respect to a $$\ce{O\bond{~}O}$$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that. Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding. Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms roughly $$\ce{sp^2}$$ hybridised, giving rise to two $$\sigma$$ bonds to the neighbouring carbons and one $$\sigma$$ bond to hydrogen. The remaining 6 perpendicular $$\ce{p}$$ orbitals form the delocalised $$\pi$$ system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. $$D_\mathrm{6h}$$, three of them are bonding and three of them are antibonding. In the $$\pi$$ system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. The nitrate ion $$\ce{NO3-}$$ is what I like to call a Y-aromat since its similarity in the electronic structure. The bonding situation is very similar to benzene. The resonance structures, however, look much more complicated. In quite good approximation, you can regard the nitrogen as $$\ce{sp^2}$$ hybridised and the oxygen as $$\ce{sp}$$. You follow the same scheme, so you obtain four equivalent nitrogen oxygen $$\sigma$$ bonds. Each oxygen has one lone pair in the in-plane $$\ce{p}$$ orbital and one lone pair in the $$\ce{sp}$$ hybrid. The remaining out-of-plane $$\ce{p}$$ orbitals of nitrogen and oxygen build up the delocalised $$\pi$$ system. This time there are only four orbitals. Because of symmetry constraints, i.e. $$D_\mathrm{3h}$$, one is bonding, two are (close enough to) non-bonding, and one is antibonding. With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene. Since the carbonate anion $$\ce{CO3^{2-}}$$ is isoelectronic to nitrate, the same orbital picture will result. Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding. Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp2 hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. D6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.) The nitrate ion $$\ce{NO3-}$$ is what I like to call a Y-aromat since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene. The resonance structures, however, look much more complicated. In quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. D3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene. Since the carbonate anion $$\ce{CO3^{2-}}$$ is isoelectronic to nitrate, approximately the same orbital picture will result. * Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another. For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $$\ce{N-O}$$ bonds and antibonding towards the other. (It is also bonding with respect to a $$\ce{O\bond{~}O}$$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that. 1 answered Dec 18 '14 at 9:49 Martin - マーチン♦ 34.7k99 gold badges117117 silver badges245245 bronze badges Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding. Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms roughly $$\ce{sp^2}$$ hybridised, giving rise to two $$\sigma$$ bonds to the neighbouring carbons and one $$\sigma$$ bond to hydrogen. The remaining 6 perpendicular $$\ce{p}$$ orbitals form the delocalised $$\pi$$ system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. $$D_\mathrm{6h}$$, three of them are bonding and three of them are antibonding. In the $$\pi$$ system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. The nitrate ion $$\ce{NO3-}$$ is what I like to call a Y-aromat since its similarity in the electronic structure. The bonding situation is very similar to benzene. The resonance structures, however, look much more complicated. In quite good approximation, you can regard the nitrogen as $$\ce{sp^2}$$ hybridised and the oxygen as $$\ce{sp}$$. You follow the same scheme, so you obtain four equivalent nitrogen oxygen $$\sigma$$ bonds. Each oxygen has one lone pair in the in-plane $$\ce{p}$$ orbital and one lone pair in the $$\ce{sp}$$ hybrid. The remaining out-of-plane $$\ce{p}$$ orbitals of nitrogen and oxygen build up the delocalised $$\pi$$ system. This time there are only four orbitals. Because of symmetry constraints, i.e. $$D_\mathrm{3h}$$, one is bonding, two are (close enough to) non-bonding, and one is antibonding. With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene. Since the carbonate anion $$\ce{CO3^{2-}}$$ is isoelectronic to nitrate, the same orbital picture will result.