Are there atoms that can form covalent and ionic bonds?

Dec 27, 2015

If a metal atom's charge is great enough, combined with a small atomic radius, that metal may form covalent bonds with non-metals.

Explanation:

${\text{BeCl}}_{2}$, despite being composed of a metal and a non-metal, is a covalent compound. This seems confusing, since it seems more logical for ${\text{BeCl}}_{2}$ to be an ionic compound. Let's run through the reasons why in an ordered list:

1. A beryllium atom has $2$ electrons in its highest main energy level / outermost shell, whilst each chlorine atom has $7$.
2. A single $\text{Be}$ atom can achieve a noble gas configuration by losing two electrons. In doing so, it has become a ${\text{Be}}^{2 +}$ ion.
3. A single $\text{Cl}$ atom can achieve a noble gas configuration by gaining one electron. In doing so, it has become a ${\text{Cl}}^{-}$ (chloride) ion. Since the $\text{Be}$ atom lost two electrons, two chloride ions can be formed in this way.
4. The ${\text{Be}}^{2 +}$ ion is very small / has a small atomic radius, and has quite a positive, $\text{2+}$ charge. Because of this, it distorts the electron clouds on the two local chloride ions, and draws electrons back towards it.
5. This effectively pulls electrons from the highest main energy level of each chloride ions, but since these ions are stable they will not transfer the electrons back (this would result in them becoming less stable chlorine atoms). The result is the formation of a co-ordinate / dative covalent bond, in which one species contributes a lone pair of electrons from its highest main energy level to form a covalent bond between itself and another species.
6. Each chloride ion has now halved ownership of two electrons, i.e. overall it has 'lost' one of the two electrons that were contributed to the covalent bond. Observe that this negates the $\text{1-}$ charge of the chloride ion, so the result is a chlorine atom with no overall charge.
7. The single beryllium ion has accepted a share in two covalent bonds, and because of this it has almost 'gained back' the equivalent of two electrons. Observe now that this negates the $\text{2+}$ charge of the beryllium ion, meaning that the final result is a beryllium atom covalently bonded to two chlorine atoms:
Feb 2, 2016

Consider $G r o u p$ $V I$ and $G r o u p$ $V I I$ elements. Oxygen exists as ${O}_{2} \left(g\right)$; ${F}_{2} \left(g\right)$ also exists as a diatomic molecule; there also exists $O {F}_{2} \left(g\right)$, another covalently bound molecule. Both fluorine and oxygen atoms, as their negatively charged ions, can make ionic compounds with metals, e.g. $N {a}_{2} O$, $L i F$ etc.