Question

Question #a36d7

Answer 1

The key to understanding the formation of the ammonium ion (`"NH"_4^(+)`) is to recognize the fact that it doesn't throw an electron away to become stable. I'll explain how and why that happens.

In a simple covalent bond, two atoms share a pair of electrons. Each of these two atoms contributes an electron to the bond, the way the resulting bonding electron pair is shared being determined by the atoms' electronegativities.

However, there are cases when one atom supplies both electrons to the bond, while the other one supplies none. This is called a dative covalent bond, or a co-ordinate covalent bond. This is what happens in the case of `"NH"_4^(+)`.

Let's analyze the reaction between ammonia (`"NH"_3`) and hydrochloric acid (`"HCl"`). Nitrogen is bonded to three hydrogen atoms, but still has 1 lone pair of electrons.

The reaction between ammonia and hydrochloric acid takes place in aqueous solution; since `"HCl"` is a strong acid, it will dissociate completely into `"H"^(+)` and `"Cl"^(-)` ions.

Here comes the interesting part. `"H"^(+)` is a proton, i.e. the nucleus of the hydrogen atom, hence the `"(+)"` sign; when the `"HCl"` molecule dissociated into ions, `"Cl"^(-)` took away hydrogen's electron, hence the `"(-)"` charge.

The lone pair on nitrogen will be attracted to the `"H"^(+)` ion. However, when the two bond, nitrogen will provide both the bonding electrons from its lone pair, since `"H"^(+)` has lost its electron to `"Cl"^(-)` `->` a co-ordinate covalent bond is formed.

The ammonium ion is formed by the protonation (the addition of a proton) of ammonia. Therefore, the ammonium ion does not throw away an electron, since it did not bond with a fourth hydrogen atom, it bonded with a proton, `"H"^(+)`.

Here's a video on dative covalent bonds and the formation of the ammonium ion: