Structure and Bonding:  The Hydrogen Peroxide Molecule

Let's examine the hydrogen peroxide molecule, H2O2, in which the oxygen has an oxidation state of -1.  Draw the Lewis dot structure of hydrogen peroxide and determine the bond order of the O-O bond.  What type of geometry does each the oxygen have?

Dioxygen, O2, has a bond order of 2 or a double bond with a bond length of 121 pm.  What would you predict the bond length in hydrogen peroxide to be?

Examine the two structures of hydrogen peroxide given below.  What is the difference between the two structures?  If hydrogen peroxide has a single oxygen-oxygen bond, could one structure be derived from the other?  If so, explain how.

O-O bond length:   _____ pm         O-O bond length:   _____ pm

                                                        H-O-O bond angle ______

Measure the O-O bond length in each image above (distance x 100 = picometers).  

The oxygen in water has two lone pairs of electrons just like each oxygen in hydrogen peroxide.  Measure the two H-O-O bond angles in the hydrogen peroxide structure on the right.  How does the angle in hydrogen peroxide compare to the H-O-H bond angle in water?  What does the difference imply about the structure of hydrogen peroxide?

The value of the O-O bond length is typical of O-O single bonds.  The decreased bond angle in hydrogen peroxide compared to water (104.5o ) indicates more distortion due to the lone pairs of electrons.  The free rotation around a single bond should allow the two structures above to be generated by rotation about the O-O single bond.  Other possiblilities could also exist.  The structure to the left above would be the thermodynamically stable one with the two hydrogen as far apart as possible or directly opposite each other.  What would be the least stable arrangement of hydrogen peroxide?

Convert the two images above to their electrostatic potential surfaces - Click here to go to the Chime Guide instructions.  Is either molecular structure polar?  If free rotation about the O-O bond occurs, would hydrogen peroxide have a permanent dipole moment?

Hydrogen peroxide has a permanent dipole moment of 2.0 D which is greater than water at 1.9 D.  Now a permanent dipole and the free rotation about the single O-O bond just do not go together.  Why?

The free rotation would cause the dipole to vary with rotation; hence, it would not be permanent.  The structure on the left above would be non-polar due to a symmetrical distribution of charge.  A planar arrangement with both oxygen on the same side of the molecule would be the most polar and least stable.

Think about the structure as each oxygen is tetrahedral in geometry with a hydrogen, an oxygen, and two lone pairs.  How would the two lone pairs of electrons on each oxygen behavior toward each other?

Measure the torsion angle in the molecule on the right.  As this is the experimental value found in the molecule.  Right click, select mouse click action, then select torsion.  To measure a torsion angle you click on four adjacent atoms in succession.  The torsion angle is the angle between the first and fourth atom while sighting down the length of the second and third or the O-O bond.  The angle is reported after clicking on the fourth atom.  A good animation of a torsion angle can be found at Colby College's Chemistry website.  You will need Shockwave installed to view it. 

torsion angle _______

The lack of free rotation is due to orbital interactions in the molecule.  This is seen in the vibration mode shown below.  Move the molecule around to look down the O-O bond.  Hence, here is restricted motion about a single bond.

Could the correct structure of hydrogen peroxide (above on the right and below on the left) be a chiral molecule?  The mirror image, formed by placing the mirror directly behind the left molecule as you look at the screen, is shown on the right below.  Decide if they are a pair of enantiomers.  Can you superimpose one structure on the other?

Hydrogen peroxide solutions are racemic mixtures, as they contain an equal mixture of both enantiomers or optical isomers.  No optical rotation occurs because the one isomer offsets the other.

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