# Question #8e1f1

Sep 12, 2015

To determine the shape you need to start by drawing the lewis dot structure, then count the number of electronic domains, and determine the type for ${\text{NO}}_{2}$, it is $A {X}_{2} E$, memorize the shapes.

#### Explanation:

When you draw the lewis dot structure of ${\text{NO}}_{2}$, you have 17 valence electrons to work with, and you end up with $\text{N}$ is the middle double bonded to oxygen on one side.

The $\text{N}$ atom is also single bonded to the other oxygen atom, with one electron left over which is on nitrogen (this is a resonance structure ).

Basically, using $A X E$ notation, where $A$ is the center atom, there are two bonding domains represented by $X$ and one nonbonding domain given by $E$, so the short hand is $A {X}_{2} E$.

Every molecule you can break down in to a $A X E$ type, so $\text{H"_2"O}$ is $A {X}_{2} {E}_{2}$, because there are two bonds, and two lone pairs.

Then you need to just memorize the structures for the different types.

It is easiest to just start with the structures with no lone pairs.
$A {X}_{2}$ linear, $A {X}_{3}$, trigonal planar, $A {X}_{4}$ tetrahedral, etc, the lone pairs slightly distort the geometry because the repulsion is different.

Keep in mind that that structure the molecule takes, is one that will minimize the amount of electron - electron repulsion .

A table with all the possible structures is given here:

http://chemistry.ncssm.edu/labs/molgeom/