# How are the hydrogens attached to the nitrogen atom in the ammonium ion?

Naively attaching hydrogens to a nitrogen atom's electron orbitals results in a picture like this:

However, there are two problems with this drawing that I can see.

First, I have heard that the ammonium ion has a tetrahedral geometry so the hydrogens would have to be moved around and the orbitals would have to be stretched around or distorted for this picture to make sense.

Second, while I have drawn each of the hydrogens attaching to different external orbital I don't actually know which orbitals the hydrogens would attach to or why.

How are the hydrogens actually attached to the nitrogen atom in the ammonium ion?

I suppose one possibility might be to have the molecule be a hybrid of various $\ce{NH3 + H^+}$ configurations. I guess this could be tested by analyzing the charge distribution of ammonium and to see if the positive charge is more distributed around the outside or the inside.

• This is because atomic orbital methods like you've described here don't accurately reflect how bonding takes place in real molecules. You can google "VSEPR theory" to learn more, or someone should come along in time and do a proper explanation in the answers section. Great question, I wish I saw more of this in chemistry students. Feb 4, 2016 at 1:16
• You heard about hybridisation? Feb 4, 2016 at 14:55

The nitrogen atom, on both ammonium ion ($\ce{NH4+}$) and ammonia ($\ce{NH3}$), is $\ce{sp^3}$ hybridized. That's because we have 4 sigma bonds on ammonium ion and 3 sigma bonds + 1 electron pair in ammonia. The most stable orbital configuration for this set up is a tetrahedral shape.
Nitrogen can have a $\ce{sp^2}$ configuration if it has 3 sigma bonds and a pi bond, like in $\ce{H2C=NH2}$. The $\ce{sp^2}$ orbitals are "used" in the sigma bonds and the remainder $2p_z$ orbital is "used" in the pi bond. This gives a triangular plane shape to the molecule.
Finally, nitrogen can also have a $\ce{sp}$ configuration if it has 1 sigma bond and 2 pi bonds, in $\ce{N2}$ for example. In this case, the $\ce{sp}$ orbital is "used" in the sigma bond and the remainder $2p_y$ and $2p_z$ orbitals are "used" in the pi bonds. This gives $\ce{N2}$ a linear geometry.