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It seems like the first molecule has a bond order of $1$, the second and third have a bond order of $7/6$ and the fourth molecule has a bond order of $3/2$. Why is the answer not $4 < 2 = 3 < 1$?


2 Answers 2


Well, 2 cannot be equal to three to start off, because a double bond it way shorter than a single bond.

The Answer to this is D, 2 has the shortest bond because it is a double bond, next 4 is a benzene ring, which is a resonance structure, which means that all the carbons technically have partial bonds between each other, or you can imagine the double bonds to be continuously rotating between the different carbons. Lastly 3 and 1 both have normal single bonds, so would have an equal bond length.

Therefore, since the double bond is shortest, the resonance between a single and double is a bit less short, but not as long as a single bond, and a single bond is the longest, the answer it 2(double bond)<4(between double and single bond)<3&1(which are both single bonds)

  • $\begingroup$ Why are 2 and 3 not also resonance structures? $\endgroup$
    – user99364
    Commented Dec 6, 2015 at 3:16
  • 1
    $\begingroup$ @user99364 Chemical formula for 2&3 is C6H10, but benzene's is C6H6, if you draw the two molecules, you'll realize that for benzene, if a Hydrogen is placed on each Carbon, regardless of where you place the double bonds, the molecule will always satisfy the octet rule for all it's atoms. But, if you draw 2 or 3, you will see that the double bond cannot move, but must remain where it is in the molecule to allow all its' atoms to conform to the octet rule.Therefore, benzene or (4)'s double bonds can move around freely, aka resonance, creating partial double bonds between all it's carbon atoms. $\endgroup$
    – Caelan
    Commented Dec 6, 2015 at 3:47

I can try to explain this by valence bond theory also:-
In first figure carbon is $sp^3$ hybridised having 25% s character and 75% p character.The same is for the third figure where there is carbon carbon single bond.
In second figure carbon is $sp^2$ hybridised with 3 $sp^2$ hybrid orbitals which have more s character than p unlike sp3 hybridised orbital.
Now s being more electronegative makes the the double bond shorter than the single bond since it has more s character in sp2 hybridised orbital than sp3 hybridised orbital,so bond length of 1and 3 figure is shorter than that of 2 figure.
The bond in figure 4 fall somewhat intermediate to double bond and single bond and so is its bond length(intermediate). Now the order of bond length will be:-2 less than 4 less than 1=3.


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