3.2: Ions - Chemistry LibreTexts

Electron Transfer

We can use electron configurations to illustrate the electron transfer process between sodium atoms and chlorine atoms.

Na: 1s22s22p63s1

As demonstrated here, a sodium atom (Na) has one valence electron in the third principal energy level. It is likely to achieve an octet in its outermost shell by losing its one valence electron. The cation produced in this way, Na+, is called the sodium ion to distinguish it from the element. The sodium ion, Na+, has the electron configuration with an octet of electrons from the second principal energy level. It is now the same as that of the noble gas neon. The term isoelectronic refers to an atom and an ion of a different atom (or two different ions) that have the same electron configuration. The sodium ion is isoelectronic with the neon atom. The equation below illustrates this process.

\[\begin{array}{lcl} \ce{Na} & \rightarrow & \ce{Na^+} + \ce{e^-} \\ 1s^2 \: 2s^2 \: 2p^6 \: 3s^1 & & 1s^2 \: 2s^2 \: 2p^6 \: \text{(octet)} \end{array}\]

Figure \(\PageIndex{1}\) is a graphical depiction of this process.

Now, let's consider chlorine atom, Cl: 1s22s22p63s23p5

Only one more electron is needed to achieve an octet in chlorine’s valence shell. When a chlorine atom gains an electron, its outermost principal energy level achieves an octet. In this case, the ion has the same outermost shell as the original atom, but now that shell has eight electrons in it. Once again, the octet rule has been satisfied. The resulting anion, Cl−, is called the chloride ion; note the slight change in the suffix (-ide instead of -ine) to create the name of this anion. This process is illustrated below. (In table salt, this electron comes from the sodium atom.)

\[\begin{array}{lcl} \ce{Cl} + \ce{e^-} & \rightarrow & \ce{Cl^-} \\ 1s^2 \: 2s^2 \:2p^6 \: 3s^2 \: 3p^5 & & 1s^2 \: 2s^2 \: 2p^6 \: 3s^2 \: 3p^6 \text{(octet)} \end{array}\]

Figure \(\PageIndex{2}\) is a graphical depiction of this process.

Figure \(\PageIndex{2}\): The Formation of a Chlorine Ion. On the left, the chlorine atom has 17 electrons. On the right, the chloride ion has 18 electrons and has a 1− charge.

With two oppositely charged ions, there is an electrostatic attraction between them because opposite charges attract. The resulting combination is the compound sodium chloride. Notice that there are no leftover electrons. The number of electrons lost by the sodium atom (one) equals the number of electrons gained by the chlorine atom (one), so the compound is electrically neutral. In macroscopic samples of sodium chloride, there are billions and billions of sodium and chloride ions, although there is always the same number of cations and anions.

Example \(\PageIndex{1}\)

Write the electron configuration of aluminum atom (Z=13). How many electrons must Al lose/gain to achieve octet? Write the formula of the resulting ion and its electron configuration.

Solution

The electron configuration of Al atom is 1s22s22p63s23p1. The second shell has octet (2s22p6) while the valence shell has 3 electrons (3s23p1). Mg can achieve octet by losing the 3 valence electrons. The resulting cation is Al3+ with electron configuration, 1s22s22p6.

Exercise \(\PageIndex{1}\)

Write the electron configuration of oxygen atom (Z=8). How many electrons must O lose/gain to achieve octet? Write the formula of the resulting ion and its electron configuration.

Answer

The electron configuration of O atom is 1s22s22p4. The second shell has six electrons (2s22p4) and needs two electrons to achieve octet. Oxygen will gain 2 electrons. The resulting anion is O2− with electron configuration, 1s22s22p6.

In many cases, elements that belong to the same group (vertical column) on the periodic table form ions with the same charge because they have the same number of valence electrons. Thus, the periodic table becomes a tool for remembering the charges on many ions. For example, all ions made from alkali metals, the first column on the periodic table, have a 1+ charge. Ions made from alkaline earth metals, the second group on the periodic table, have a 2+ charge. On the other side of the periodic table, the next-to-last column, the halogens, form ions having a 1− charge. Figure \(\PageIndex{3}\) shows how the charge on many ions can be predicted by the location of an element on the periodic table. Note the convention of first writing the number and then the sign on a multiply charged ion. The barium cation is written Ba2+, not Ba+2.

Example \(\PageIndex{2}\)

Which of these ions is not likely to form?

1. Mg+
2. K+

Solution

(a) Mg is in Group 2A and has two valence electrons. It achieves octet by losing two electrons to form Mg2+ cation. Losing only one electron to form Mg+ does not make an octet, hence, Mg+ is not likely to form.

Exercise \(\PageIndex{2}\)

Which of these ions is not likely to form?

1. S3−
2. N3−

Answer

(a) S is in Group 6A and has six valence electrons. It achieves octet by gaining two electrons to form S2− anion. Gaining three electrons to form S3−does not make it octet, hence, S3− is not likely to form.