Choosing between resonance and hybridization

February 11, 2015
Categories: AP | AP TOPIC 08 | Big Idea 2

A question came up on the College Board’s AP chemistry discussion group recently that I have seen many times before (including many variations of the same), and that often confuses students. It usually looks something like this;

The fact that the nitrogen-to-oxygen bond lengths in the nitrite ion are all the same, can be attributed to;

(A) ionic bonds
(B) Hybridization
(C) Resonance
(D) London dispersion forces

In these questions, whatever is offered as a potential answer other than hybridization or resonance is basically totally irrelevant, and therefore could be almost anything related to bonding. Essentially the question is testing the students’ ability to distinguish between resonance and hybridization.

So let’s consider the nitrite ion in more detail. When one draws a Lewis structure for the nitrite ion in the traditional way, it should yield the structure below (lone pairs are omitted for clarity).

Screen-Shot-2015-02-09-at-4.45.18-PMKnowing that a double bond is made up of a combination of;

a. a hybridized sigma bond of some type (in this case sp2), plus,

b. the overlap of an unhybridized p orbital on one atom (N), and another unhybridized p orbital on the other atom in the bond (O),

one can see that different bond lengths (one single bond, and one double bond) can exist in conjunction with hybridization. Logically, this means that hybridization is NOT what is creating bonds of equal length, since hybridization exists in the presence of unequal bond length.

Of course, resonance is the reason for the bond lengths being equal, but that’s a different discussion, and what is written above is really supposed to simply illustrate why the answer to the original question is not ‘hybridization’.


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