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So in the same period oxygen has a higher electronegativity and carbon with a lower electronegativity participate readily in combustion, but N2 with an electronegativity of 3.04 is inert. Does anyone know why?
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Why is N2 inert?
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#2
18 January 2012 - 05:50 AM
Suxamethonium 
Baryon
Um, ok- I'm not sure where you may be at in bonding education, so i'll to try and keep it simple.
N2 needs to share 3 pairs of electrons between the nitrogen atoms to be 'happy', which results in a tripple bond. Looking at bond energies (the energy required to break a bond in 1 mole of gaseous compound) for a carbon-cabron tripple bond (812kJ/mol), and nitrogen-nitrogen tripple bond (941kJ/mol) we see that nitrogen is only stronger by 130kJ/mol. This is expected because of nitrogens higher electron affinity.
A carbon-carbon single bond (347kJ/mol) and carbon-hydrogen single bond (414kJ/mol) are much easier to break, which helps explain why combustion of hydrocarbons is 'easy' (and why the reaction yeilds so much energy).
**Side thought- If you are wondering why a tripple bond is more reactive than a single bond** It is because the bonds break in steps- it only takes 200kJ/mol to break the tripple bond to a double bond which is approx. half of the energy required to break a single bond.***
Ok, so now diatomic oxygen. We often draw O2 as O=O. But more accurately it is a di-radical looking somthing like this .O-O. . The radical exists because electrons are in what are called "anti-bonding orbitals"- i.e. not going to form a bond- which leaves the oxygen "unhappy" (but still STABLE compared to singlet (O=O) oxygen which has bonding orbitals much higher in energy). Also, the presence of the anti bonding orbital weakens the bond energy of the O-O single bond between the radicals. As such O2 is more reactive than expected by electron affinity alone.
Finaly, there are the bond energies in water and carbon dioxide (products of combustion) compared to nitrogen dioxide (combustion of nitrogen). The release of energy is significant in regular combustion, but not so in nitrogen combustion. Combustion of nitrogen does occur- but at high energy costs.
N2 needs to share 3 pairs of electrons between the nitrogen atoms to be 'happy', which results in a tripple bond. Looking at bond energies (the energy required to break a bond in 1 mole of gaseous compound) for a carbon-cabron tripple bond (812kJ/mol), and nitrogen-nitrogen tripple bond (941kJ/mol) we see that nitrogen is only stronger by 130kJ/mol. This is expected because of nitrogens higher electron affinity.
A carbon-carbon single bond (347kJ/mol) and carbon-hydrogen single bond (414kJ/mol) are much easier to break, which helps explain why combustion of hydrocarbons is 'easy' (and why the reaction yeilds so much energy).
**Side thought- If you are wondering why a tripple bond is more reactive than a single bond** It is because the bonds break in steps- it only takes 200kJ/mol to break the tripple bond to a double bond which is approx. half of the energy required to break a single bond.***
Ok, so now diatomic oxygen. We often draw O2 as O=O. But more accurately it is a di-radical looking somthing like this .O-O. . The radical exists because electrons are in what are called "anti-bonding orbitals"- i.e. not going to form a bond- which leaves the oxygen "unhappy" (but still STABLE compared to singlet (O=O) oxygen which has bonding orbitals much higher in energy). Also, the presence of the anti bonding orbital weakens the bond energy of the O-O single bond between the radicals. As such O2 is more reactive than expected by electron affinity alone.
Finaly, there are the bond energies in water and carbon dioxide (products of combustion) compared to nitrogen dioxide (combustion of nitrogen). The release of energy is significant in regular combustion, but not so in nitrogen combustion. Combustion of nitrogen does occur- but at high energy costs.
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