We can't draw diagrams or structures on the boards because of spacing problems. But thanks for the answer because it tells me what your problem is.
You have drawn the structure and it contains 1 N-O bond and 1 N=O bond as well as a pair of unshared electrons on the N atom. Right? So you see 1 single N-O bond and 1 N=O double bond. Right? But that isn't what happens. There is a resonance structure that can be drawn. The electrons can be rearranged so that what was the N-O bond becomes the N=O bond and what was the N=O bond becomes the N-O bond. So there are two structures, each containing 1 N-O bond and 1 N=O bond BUT the single and double bonds are to different O atoms. So which one is the correct structure. Neither is. We say we can draw two equivalent resonance structures and the N-O bond is an average between a single and a double bond. If the N-O bond length is measured in the lab, we do NOT get two N-O bond lengths (which we would do IF there were two kinds of bonds); we get just one bond length and it is between a single and a double bond in strength. Avoid the idea that the electrons are flipping back and forth between a single and double bond for they don't do that. What we are seeing is a resonance hybrid. So why don't we just draw it the way it really is. Because we just don't know enough about the structure to do that. We can draw the two forms and say it is something in between and that's the best we can do.
draw the lewis structure for NO2- including any valid resonance structures. which of the following statements are true?
a. the nitrite ion contains 2 N-O bonds that are equivalent to 1 1/2 bonds.
b. the nitrite ion contains 2 N=O double bonds
c. the nitrite ion contains 1 N-O single bond and one N=O double bond
d. the nitrite ion contains 2 N-O single bonds
the answer is A, but I don't understand why its not C
1 answer
1 answer