00:03
In this problem we have three species, cn plus, cn and cn minus.
00:12
The number of balanced electrons in each case is what we will start our problem with.
00:20
So here it's 9, 4 from carbon and 5 from nitrogen.
00:25
Here we lose an electron so it's 8 and here we gain another electron so it's 10 electrons.
00:32
Let's fill this up in.
00:34
Our electronic configuration.
00:38
Sigma 2s 2, sigma star 2s 2, pi 2px equals pi 2, 2pi 2p.
00:48
So this accounts for all the electrons, all the 8 electrons we have.
00:54
Now for cn 2px equals 2pi.
01:02
There's one more electron so it fills up the 2p z.
01:07
So we have one electron here and in this case we have another electron so in cn your sigma 2p z was partially filled and in this case it'll be completely filled so the bond order is number of bonding electrons minus anti -bonding electrons so 1 2 3 4 5 6 6 minus 2 is 4 divided by 2 2 this case 1 2 3 4 5 6 7 to 7 minus 2 is 5 so it will be 2 .5 and in the last case 1 2 4 5 6 7 8 8 minus 2 is 6 divided by 2 will be 3 so the bond order is inversely proportional to the bond length because the higher the bond order, it means that there are more bonds that are holding these two atoms together, are bonding these two atoms together.
02:33
So they will be held closer together.
02:38
And so the bond length obviously will decrease...