Which of the following is the
number of covalent bonds that can be formed by an unbound atom of carbon? Carbon can be found in group 14 of
the periodic table. (A) Three bonds, (B) five bonds,
(C) zero bonds, (D) eight bonds, or (E) four bonds.
This question is asking us to find
the number of covalent bonds that an atom of carbon forms. A rule we can keep in mind to solve
this problem is that atoms tend to form bonds that fill or empty their outermost
electron shell. For example, lithium has one
valence electron, while fluorine has seven valence electrons out of a possible
eight. When these elements combine to form
lithium fluoride, lithium donates its electron to fluorine. The end result is a lithium ion
that has emptied its outermost electron shell and a fluoride ion that now has a full
outer electron shell.
To better understand how this
concept relates to the periodic table, let’s take a look at period two. As we just saw, lithium has one
valence electron and fluorine has seven. As we move across a period, the
number of valence electrons in each element goes up by one, up to the noble gas on
the right-hand side, which has a full outer electron shell.
Carbon is four electrons away from
either a full outer shell or an empty outer shell. Rather than donating or receiving
electrons like the ionic bond in our earlier example, carbon atoms share electrons
in covalent bonds. Sharing four electrons of its own
and four electrons from other atoms gives carbon a full outer shell of eight
electrons. Each covalent bond represents a
pair of shared electrons, so we can say that carbon atoms form four covalent
bonds. These four bonds will be some
combination of single bonds, like the ones found in methane; double bonds, like the
ones found in carbon dioxide; and triple bonds, like the one found in ethyne. We can see that the carbon atoms in
each of these structures contain four covalent bonds.
So, which of the following is the
number of covalent bonds that can be formed by an unbound atom of carbon? That’s choice (E), four bonds.