# What is the hybridization of the central atom in NO_3^-?

Aug 21, 2017

See below.

#### Explanation:

To get the answer to this question, you can take the approach of drawing the Lewis structure of $N {O}_{3}^{-}$. It looks like this:

(For clarification, you need to calculate the number of electrons in $N {O}_{3}^{-}$ (without forgetting the $-$). Then, you position the bonds with the right number of electrons, and fill in lone pairs around the atoms to get the right number of total electrons (the dots are electrons))

Now, to find hybridization, use the following table:

All we have to do now is count the number of bonds between atoms on the Lewis structure, and check the table. In the Lewis structure, you can see $3$ bonds between atoms (the dots between atoms are the bonds).

If we refer to the table under "type of hybrid orbital," this is where the hybridization should be. For "number of bonds" = $3$, the hybridization is $s {p}^{2}$. So, your answer is $s {p}^{2}$.

I hope that helps!

P.S. If you still have trouble understanding how to draw the Lewis structure, just search up "Lewis structure no3-" on YouTube. There are great videos which explain it well. I am sorry if I didn't go in-depth enough.

Aug 21, 2017

Nitrogen is $s {p}^{2}$ hybridized and a formal cation.........

#### Explanation:

Given $1 \times N + 3 \times O$ atom, we have $5 + 18$ valence electrons to distribute, PLUS one electron to make 24 electrons ACROSS four centres......

The common Lewis representation gives a quaternized nitrogen, and thus a formal cationic centre. Two of the oxygen atoms have formal negative charges, whereas one is neutral, and thus a FORMAL negative charge is associated with the ion. And thus we gots.......

$O = \stackrel{+}{N} {\left\{- {O}^{-}\right\}}_{2}$.

The oxygen-nitrogen-oxygen bond angle, $\angle O - N - O = {120}^{\circ}$ to a first approximation.