Question

Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case? a. Ra b. In c. \(P\) d. \(T e\) e. \(\mathrm{Br}\) f. \(\mathrm{Rb}\)

Short answer

The most likely ions for each atom are as follows: a. \(Ra^{2+}\) b. \(In^{3+}\) c. \(P^{3-}\) d. \(Te^{2-}\) e. \(Br^{-}\) f. \(Rb^{+}\)

Step-by-step solution

Ra

: Ra (Radium) is in group 2 of the periodic table. Thus, it has 2 valence electrons. In order to achieve a stable full outer shell, it's more likely for Ra to lose these 2 electrons. When Ra loses 2 electrons, it forms a cation with a charge of +2. Therefore, the most likely ion for Ra is \(Ra^{2+}\). b.

In

: In (Indium) is in group 13 of the periodic table. Thus, it has 3 valence electrons. In order to achieve a stable full outer shell, it's more likely for In to lose these 3 electrons. When In loses 3 electrons, it forms a cation with a charge of +3. Therefore, the most likely ion for In is \(In^{3+}\). c.

P

: P (Phosphorus) is in group 15 of the periodic table. Thus, it has 5 valence electrons. In order to achieve a stable full outer shell, it's more likely for P to gain 3 electrons. When P gains 3 electrons, it forms an anion with a charge of -3. Therefore, the most likely ion for P is \(P^{3-}\). d.

Te

: Te (Tellurium) is in group 16 of the periodic table. Thus, it has 6 valence electrons. In order to achieve a stable full outer shell, it's more likely for Te to gain 2 electrons. When Te gains 2 electrons, it forms an anion with a charge of -2. Therefore, the most likely ion for Te is \(Te^{2-}\). e.

Br

: Br (Bromine) is in group 17 of the periodic table. Thus, it has 7 valence electrons. In order to achieve a stable full outer shell, it's more likely for Br to gain 1 electron. When Br gains 1 electron, it forms an anion with a charge of -1. Therefore, the most likely ion for Br is \(Br^{-}\). f.

Rb

: Rb (Rubidium) is in group 1 of the periodic table. Thus, it has 1 valence electron. In order to achieve a stable full outer shell, it's more likely for Rb to lose this 1 electron. When Rb loses 1 electron, it forms a cation with a charge of +1. Therefore, the most likely ion for Rb is \(Rb^{+}\).

Key concepts

Valence Electrons

Valence electrons are the electrons present in the outermost shell of an atom and play a crucial role in determining how an atom will interact with others to form chemical compounds. They are the electrons involved in forming bonds between atoms, either by being shared with another atom or by being transferred to or from another atom.

An atom's tendency to form a cation or an anion is largely dependent on its number of valence electrons. Atoms with fewer valence electrons than are needed to complete their outermost shell tend to lose them to form positive ions or cations, while atoms with nearly full valence shells tend to gain electrons to complete their shell and form anions, which are negatively charged ions.

For instance, Radium (Ra), with two valence electrons, readily loses them, resulting in a +2 charge, forming the cation, Ra²⁺. This principle is ubiquitous throughout the periodic table and is paramount in predicting the chemical behavior of elements.

Cation and Anion Formation

The process of forming ions from atoms involves either losing or gaining electrons to achieve a stable electron configuration, usually similar to the nearest noble gas. Cations are positively charged ions formed when an atom loses one or more valence electrons. Typically, metals which are found on the left side of the periodic table form cations. For example, Rubidium (Rb) loses its single valence electron to become Rb⁺.

Anions, on the other hand, are negatively charged ions created when an atom gains electrons. Non-metals, usually found on the right side of the periodic table, often form anions. Bromine (Br), eager to complete its octet, will gain one electron to become Br^-. The type of ion an element forms directly affects the properties of the resulting ionic compound, impacting melting points, electrical conductivity, and other physical characteristics.

Group Trends in Ionization

The periodic table exhibits clear trends in the energy required to remove an electron from an atom, known as ionization energy, which varies across groups and periods. Ionization energy generally increases across a period from left to right and decreases down a group.

As you move down a group, additional electron shells are added, and the outer electrons become further away from the positively charged nucleus. This distance reduces the force of attraction between the nucleus and the valence electrons, making it easier for the atom to lose these electrons. Therefore, lower group elements require less energy to form cations. Radium (Ra), found lower in group 2, loses its two valence electrons more readily than Magnesium (Mg) higher up in the same group.

Conversely, it becomes progressively hard to remove electrons across a period due to increased nuclear charge that 'pulls' electrons closer, thus increasing ionization energy. The varying trends in ionization energy across groups and periods are key to understanding why elements behave differently when forming ions.