We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H2 molecule, which contains a purely covalent bond. Each hydrogen atom in H2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure (PageIndex1)):The electrons in the two atoms repel each other because they have the same charge ( The electrons in the two atoms repel each other because they have the same charge (E > 0).
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Similarly, the protons in adjacent atoms repel each other (E > 0). The electron in one atom is attracted to the oppositely charged proton in the other atom and vice versa (E Recall that it is impossible to specify precisely the position of the electron in either hydrogen atom. Hence the quantum mechanical probability distributions must be used.
A plot of the potential energy of the system as a function of the internuclear distance (Figure (PageIndex2)) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r0 (the observed internuclear distance in H2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figures (PageIndex1) and (PageIndex2), which described a system containing two oppositely charged ions. The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities.
At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms.
Using Lewis Dot Symbols to Describe Covalent Bonding
The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. brickandmortarphilly.comists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. brickandmortarphilly.comists usually list this central atom first in the brickandmortarphilly.comical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32−, for example, we add two electrons to the total because of the −2 charge. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
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The (H_2O) MoleculeBecause H atoms are almost always terminal, the arrangement within the molecule must be HOH. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.