Which of the following molecules or ions will exhibit delocalized bonding?

1 Answer

• Delocalized bonding…..

  SO2 and SO3 both have delocalized pi bonds. And so does SO3^2-, although if you draw the Lewis structure it looks like all the bonds are single bonds, but the bond order is greater than 1.0.

  Consider Lewis structures and resonance structures. Resonance structures are simply attempts to fit the bonding into a Lewis structure because Lewis structures have no way to represent fractional bond orders. SO2 is written with two resonance structures (alternating single and double bonds), but both bonds are exactly alike and both have bond orders of 1.80.

  The octet rule and formal charge. The octet rule isn’t all its cracked up to be. It works well enough with some of the elements in period 2, but there are many deviations from the octet rule fo periods 3 and up. Therefore, for central atoms in periods 3 and up we should consider the arrangement which minimizes formal charge.

  Formal charge is calculated for each element in a compound.

  Formal charge = valence electrons – nonbonding electrons – 1/2 bonding electrons

  For SO2, minimizing the formal charges gives two double bonds (which violates the octet rule, but that’s ok). Therefore, the bond order is predicted to be 2.0. It is actually found to be 1.80, so two double bonds is a pretty good representation. And of course, this means that there is delocalized pi bonding in SO2.

  The same is true for SO3 where formal charge indicates three double bonds. In SO3 all three bonds are alike and each has a bond order of 1.87, so we have even better agreement. And there is delocalized pi bonding in SO3.

  It looks like all three S-O bonds follow the octet rule. But there are five resonance structures for SO3^2-. Instead, all three bonds are identical. When you use formal charges you will see that the predicted bond order for each is 1.33 and and the actual bond order is closer to 1.50. So in SO3^2- we also have delocalized pi bonding.

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