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bond order of n2
Asked by visalvinod06 | 23 Jun, 2022, 07:39: AM

If you are looking for bond order of any same atoms what you do is next. You first check how many electrons the N atom has. So N atom has 7 electrons. If we have 2 N atoms then we have 14 electrons. What you do next is draw a picture like the one from google. You simply fill up the bonding orbitals first and then you go to antibonding if there is no more room in bonding.

Note to take here you need to watch at energy levels. 1s bonding has lower energy than antibonding. So the bonding orbital gets filled up first. Same principle applies later.

Now the tricky part you see 1s and 2s have same number of atoms in bonding and antibonding so they cancel out.

Let’s go to 2p orbitals. To try and simplify this problem to only bond order let us ˝dumb down˝this picture like i did at the beginning. As you know you can put 6 electrons in p orbitals. So what we do is fill up the bonding orbitals first. We see that we have used up all our 14 electrons ( 1s takes 4, 2s takes 4 and 2p takes 6). What we see now is that the antibonding orbitals are still free but damn we don’t have any spare electrons left.

Now for the calculation. You can simply cancel out 1s and 2s orbitals since they have the same amount of bonding and anti bonding electrons. So what we now get is

(6–0)/2=3 or if you want general formula including all electrons

(bonding-antibonding)/2=bond order. So translating that to our N2 problem here we get (10–4)/2=3

Answered by Ravi | 27 Jun, 2022, 08:41: PM

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