Question

On the basis of their electron configurations, predict the formula of the simple binary ionic compound likely to form when the following pairs of elements react with each other. a. sodium, \(\mathrm{Na}\), and selenium, Se b. rubidium, Rb, and fluorine, F c. potassium, \(\mathrm{K}\), and tellurium, Te d. barium, Ba, and selenium, Se e. potassium, \(\mathrm{K}\), and astatine, \(\mathrm{At}\) f. francium, Fr, and chlorine, \(\mathrm{Cl}\)

Short answer

The simple binary ionic compounds formed by the given pairs of elements are: a. Sodium and selenium: \(\mathrm{Na_2Se}\) b. Rubidium and fluorine: \(\mathrm{RbF}\) c. Potassium and tellurium: \(\mathrm{K_2Te}\) d. Barium and selenium: \(\mathrm{BaSe}\) e. Potassium and astatine: \(\mathrm{KAt}\) f. Francium and chlorine: \(\mathrm{FrCl}\)

Step-by-step solution

Identify the electron configurations

For each pair of elements, we need to find their electron configurations. To do this, we can look up their atomic numbers on the periodic table and fill in their electron orbitals. a. Sodium, Na (Z = 11), Se (Z = 34) b. Rubidium, Rb (Z = 37), Fluorine, F (Z = 9) c. Potassium, K (Z = 19), Tellurium, Te (Z = 52) d. Barium, Ba (Z = 56), Selenium, Se (Z = 34) e. Potassium, K (Z = 19), Astatine, At (Z = 85) f. Francium, Fr (Z = 87), Chlorine, Cl (Z = 17)

Determine cation and anion for each pair

Metals tend to lose electrons to form cations, whereas nonmetals tend to gain electrons to form anions. a. Na (cation), Se (anion) b. Rb (cation), F (anion) c. K (cation), Te (anion) d. Ba (cation), Se (anion) e. K (cation), At (anion) f. Fr (cation), Cl (anion)

Combine cation and anion to form a neutral compound

For each pair of elements, we need to find the ratio of cations to anions that will result in a neutral compound. This can be done by finding the lowest common multiple of the charges on the cation and anion. a. Na^+ and Se^2-: \(\mathrm{Na_2Se}\) b. Rb^+ and F^-: \(\mathrm{RbF}\) c. K^+ and Te^2-: \(\mathrm{K_2Te}\) d. Ba^2+ and Se^2-: \(\mathrm{BaSe}\) e. K^+ and At^-: \(\mathrm{KAt}\) f. Fr^+ and Cl^-: \(\mathrm{FrCl}\) In conclusion, the simple binary ionic compounds formed by the given pairs of elements are: a. Sodium and selenium: \(\mathrm{Na_2Se}\) b. Rubidium and fluorine: \(\mathrm{RbF}\) c. Potassium and tellurium: \(\mathrm{K_2Te}\) d. Barium and selenium: \(\mathrm{BaSe}\) e. Potassium and astatine: \(\mathrm{KAt}\) f. Francium and chlorine: \(\mathrm{FrCl}\)

Key concepts

Electron Configuration

Understanding electron configuration is crucial in predicting how elements will interact to form compounds. Every element on the periodic table has a specific arrangement of electrons across different energy levels, or shells, surrounding the nucleus. This arrangement is known as its electron configuration. The electron configuration helps determine the chemical properties and reactivity of an element.
When writing the electron configuration, we fill the orbitals in a specific order, following the rules of quantum mechanics. We use the Aufbau principle which says electrons occupy the lowest energy orbital available. Hund's rule and the Pauli exclusion principle further guide us in filling these orbitals correctly. For example, sodium (\(Na\)) with an atomic number of 11, has the electron configuration: \([\text{Ne}]\,3s^1\). This indicates that sodium has filled up to the neon core and has one electron in the 3s orbital. By identifying the configuration of each element, we can better understand how they might bond with other elements.

Cation and Anion Formation

In binary ionic compounds, we often deal with the formation of cations and anions. Metals generally lose electrons to form cations with a positive charge, while non-metals gain electrons to become anions with a negative charge. This transfer of electrons leads to the formation of ionic bonds.
For example, when sodium (\(Na\)) reacts, it loses one electron to form \(Na^+\), a cation. Selenium (\(Se\)), on the other hand, needs two electrons to fulfill its valence shell and forms \(Se^{2-}\), an anion. Understanding these tendencies is essential in predicting the formation of binary ionic compounds.
By determining which atom forms a cation and which forms an anion, you can predict the type of ionic bond that will occur and the resulting compound's structure.

Periodic Table

The periodic table is an invaluable tool for predicting the behavior of elements, including their electron configurations and tendencies to form cations or anions. Elements in the same group (vertical column) typically exhibit similar chemical behavior, as they have the same number of electrons in their outermost shell.
Alkali metals like sodium (\(Na\)) and potassium (\(K\)), found in Group 1, are excellent candidates for forming cations as they naturally have one electron in their outer shell. On the other hand, halogens such as fluorine (\(F\)) and chlorine (\(Cl\)), found in Group 17, are highly reactive non-metals that readily form anions by gaining an electron.
Using the periodic table, you can quickly deduce important electronic and reactive tendencies, helping predict the compounds they might form during chemical reactions.

Chemical Formulas

Chemical formulas represent the types and numbers of atoms within a compound, reflecting how elements combine to form substances. For binary ionic compounds, the formula is determined by balancing the charges of the ions involved to create a neutral compound.
Consider the combination of potassium (\(K^+\)) and tellurium (\(Te^{2-}\)). Potassium forms a \(1+\) ion while tellurium forms a \(2-\) ion. To balance these charges in a compound, two potassium ions will combine with one tellurium ion, represented by the formula \(K_2Te\).
By knowing the typical ionic charges from the periodic table and electron configurations, you can apply this information to write accurate chemical formulas for compounds.