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

Naive Ammonium Ion Drawing

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.

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    $\begingroup$ 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. $\endgroup$ – chipbuster Feb 4 '16 at 1:16
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    $\begingroup$ You heard about hybridisation? $\endgroup$ – Mithoron Feb 4 '16 at 14:55
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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.

Sp3 nitrogen

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.

Sp2 nitrogen

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.

Sp nitrogen

Molecular orbital theory and hybridization are a little hard to process (at least for me it was and it's still being), so take your time to read some more about it, but at least with those examples you should have an idea of what to expect from some molecules geometries.

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