# Question #77567

May 28, 2015

The molecules you listed don't match that description. Here's why.

Right from the get-go, ozone, ammonia and water are eliminated because they are polar molecules. This leaves you with nitrogen, ${N}_{2}$, and methane, $C {H}_{4}$.

Nitrogen is a homonuclear diatomic molecule formed when two nitrogen atoms form a covalent bond.

Since the bond is formed between two identical nitrogen atoms, the bonding electrons will not be more attracted to one of the two atoms,. This means that the bond will be nonpolar covalent, i.e. the bonding electrons will be shared equally between the two atoms.

Methane's Lewis structure looks like this

Now, a difference in electronegativity between carbon and hydrogen does exist, but it's too small for the $\text{C-H}$ bond to be considered polar covalent.

${\Delta}_{E N} = E {N}_{C} - E {N}_{H} = 2.55 - 2.20 = 0.35$

As a rule of thumb, if the difference in electronegativity between two atoms is smaller than 0.5, the bond is said to be nonpolar.

So, if you were to consider the $\text{C-H}$ as polar, despite the fact that you have no reason to do so, the molecule would indeed be nonpolar because of its tetrahedral molecular geometry.