Chem 110                                      Resonance Hybrids

 

At this point you should be feeling fairly comfortable with writing Lewis structures for species that have resonance.  Chang Sec 9.6 discusses resonance.  Be sure to go through the examples, and in-text problems in the chapter.

 

The topic here is beyond writing Lewis structures, and focuses on drawing resonance hybrids.

 

Whenever it’s possible to write more than one valid electron-dot structure for a molecule, the actual electornic structure is an average of the different possibilities, called a resonance hybrid.

 

Example 1

Taking the example of the nitrite ion, the resonance structures are

 

      

 

Neither one is a “true” picture of the nitrite ion.  A better representation is an average of the two, the resonance hybrid.  To draw the average, first note which bonds are always there, and which bonds are only sometimes there.  Also note which lone pairs are always there and which are only sometimes there.

 

For NO2–, each oxygen is bonded to N by at least one single bond.  It is the second bond that is only there half the time.  In the hybrid we show the single bond as a solid line, but the second bond is shown as a dotted line.  The dotted line represents a bond that is only sometimes there. 

 

As for the lone pairs, each oxygen in the nitrite ion always has 2 lone pairs.  It is the third lone pair that is sometimes there.  Since we cannot show only part of a dot, this third lone pair is not shown at all.  Thus, the resonance hybrid for the nitrite ion is as shown below:

 

                    

 

What is the shape of this resonance hybrid?  We follow the same steps used for regular Lewis structures.  Examine the central atom (N).  How many charge clouds (lone pairs and atoms) are on it?  (The number of charge clouds is determined by counting the number of lone pairs on the central atom, and the number of atoms directly attached to the central atom.)  The N atom in the resonance hybrid shown above has 3 charge clouds (2 atoms and 1 lone pair).  You would predict an sp2 hybridization, and classify it as AX2E.  The shape would be expected to be bent, with bond angles of around 120°.  The nitrite ion is NOT linear as shown above, but is expected to be thus:

                     

 

 

Example 2:

 

Now, let’s go through another example:

Draw the resonance hybrid of the formate ion (HCO2–) .

Step 1:  Work out the resonance structures of HCO2–.

(This is Problem 7.12c on p. 260 in McMurry & Fay 3rd edition.  The solution is given at the back of the book.)

Step 2:  Which bond are always there?  Which lone pairs are always there?  Bonds that are only sometimes there are represented as dotted lines.

 

The resonance hybrid for HCO2– should look like this:

 

     

 

What is the hybridization of the central atom?  Write it down.

What is the molecular geometry?  Write it down.

What is the bond angle?  Write it down.

Don’t peek.  Write down your answers before scrolling down.

 

 

 

 

 

 

 

 

 

 

 

 

Ans. The central atom is C, and it has an sp2 hybridization (with 3 atoms and no lone pairs attached)  It belongs in the AX3 classification. The molecular geometry is trigonal planar.

 

 

 

Example 3:

Draw the resonance hybrid of “laughing gas”, N2O.  (This is problem 7.11 on p. 260.  Draw all the resonance structures yourself and then check the answer given at the back of the book..)

 

Note that your textbook did not include N-N=O as a resonance structure, the reason being that it is very unlikely for O to have three bonds (it prefers two bonds) when there are the other two arrangements.  Oxygen has a very high electronegativity.  It would not accommodate a triple bond (which is electron-deficient) unless there is no other choice (as in carbon monoxide.)

 

Taking the two resonance structures shown for N2O, now draw the resonance hybrid.  (No peeking!  Work on it yourself!)

 

 

 

 

 

 

 

 

 

 

 

 

Answer:

 

What is the hybridization, molecular type, molecular geometry?

 

 

 

 

 

 

 

 

 

Answer:

sp hybridization, AX2, linear

 

Example 4:

 

The organic compound, benzene has the following resonance structures. 

 

  

Each corner of the hexagon is a carbon atom.  Also not shown, are the H atoms (one attached to each C).

What does its resonance hybrid look like?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ans.

 

 

In fact, benzene is commonly written

 

 

 

 


The circle signifies alternating double bonds that shift positions.  We say that the electrons are “delocalized” (not localized in one position).  This delocalization makes benzene much more stable than one would have predicted by considering only one or the other of the two resonance structures.  There is a discussion of this on p.1002 of McMurry & Fay, 3rd edition.

 

What is the hybridization of the C atoms?  What are the bond angles?

 

 

 

 

 

 

 

 

Ans.

sp2 hybridization.  Each C can be classified as AX3, trigonal planar with bond angles are 120°.  In fact the whole molecule is planar.

 

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