Our discussion of ionic compounds was confined to monatomic ions. However, more complex ions, containing several atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. covalently bonded to one another, but having a positive or negative charge, occur quite frequently in chemistry. The charge arises because the total number of valence electrons from the atoms cannot produce a stable structure. With one or more electrons added or removed, a stable structure results. Well-known examples of such polyatomic ions are the sulfate ion (SO42–),
the hydroxide ion (OH–),
the hydronium ion (H3O+),
and the ammonium ion (NH4+).
The atoms in these ions are joined together by covalent electron-pair bonds, and we can draw Lewis structures for the ions just as we can for molecules. The only difference is that the number of electrons in the ion does not exactly balance the sum of the nuclear charges. Either there are too many electrons, in which case we have an anionA negatively charged ion. An ion that is attracted toward the anode in an electrolytic cell., or too few, in which case we have a cationA positively charged ion, attracted toward the cathode in an electrolytic cell..
Consider, for example, the hydroxide ion (OH–) for which the Lewis structure is
A neutral molecule containing one O and one H atom would contain only seven electrons, six from O and one from H. The hydroxide ion, though, contains an octet of electrons, one more than the neutral molecule. The hydroxide ion must thus carry a single negative charge. In order to draw the Lewis structure for a given ion, we must first determine how many valence electrons are involved. Suppose the structure of H3O+ is required. The total number of electrons is obtained by adding the valence electrons for each atom, 6 + 1 + 1 + 1 = 9 electrons. We must now subtract 1 electron since the species under consideration is not H3O but H3O+. The total number of electrons is thus 9 – 1 = 8. Since this is an octet of electrons, we can place them all around the O atom. The final structure then follows very easily:
In more complicated cases it is often useful to calculate the number of shared electron pairs before drawing a Lewis structure. This is particularly true when the ion in question is an oxyanionAn anion containing oxygen and another element. Also called oxoanion. Examples are nitrate, sulfate, and phosphate ions. (i.e., a central atomIn a small molecule or ion, the atom to which the other atoms are bonded. The central atom is surrounded by the other atoms. is surrounded by several O atoms). A well-known oxyanion is the carbonate ion, which has the formula CO32–. (Note that the central atom C is written first, as was done earlier for molecules.) The total number of valence electrons available in CO32– is
4(for C) + 3 × 6(for O) + 2(for the –2 charge) = 24
We must distribute these electrons over 4 atoms, giving each an octet, a requirement of 4 × 8 = 32 electrons. This means that 32 – 24 = 8 electrons need to he counted twice for octet purposes; i.e., 8 electrons are shared. The a ion thus contains four electron-pair bonds. Presumably the C atom is double-bonded to one of the O’s and singly bonded to the other two:
In this diagram the 4C electrons have been represented by dots, the 18 O electrons by ×’s, and the 2 extra electrons by colored dots, for purposes of easy reference. Real electrons do not carry labels like this; they are all the same.
There is a serious objection to the Lewis structure just drawn. How do the electrons know which oxygen atom to single out and form a double bondAttraction between two atoms (nuclei and core electrons) that results from sharing two pairs of electrons between the atoms; a bond with bond order = 2. with, since there is otherwise nothing to differentiate the oxygens? The answer is that they do not. To explain the bonding in the CO32– ion and some other molecules requires an extension of the Lewis theory. We pursue this matterAnything that occupies space and has mass; contrasted with energy. further when we discuss resonance. Now we end with an example.
EXAMPLE Draw a Lewis structure for the sulfite ion, SO32–.
SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. The safest method here is to count electrons. The total number of valence electrons available is
6(for S) + 3 × 6(for O) + 2(for the charge) = 26
To make four octets for the four atoms would require 32 electrons, and so the difference, 32 – 26 = 6, gives the number of shared electrons. There are thus only three electron-pair bonds in the ion. The central S atom must be linked by a single bondAttraction between two atoms (nuclei and core electrons) that results from sharing a single pair of electrons between the atoms; a bond with bond order = 1. to each O atom.
Note that each of the S—O bonds is coordinate covalent.