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 sulfur, \(\mathrm{S}\) b. barium, \(\mathrm{Ba}\), and selenium, Se c. magnesium, \(\mathrm{Mg}\), and bromine, \(\mathrm{Br}\) d. lithium, Li, and nitrogen, \(N\) e. potassium, \(\mathrm{K}\), and hydrogen, \(\mathrm{H}\)

Short answer

The formulas of the binary ionic compounds are as follows: a. \(\mathrm{Na_2S}\) b. \(\mathrm{BaSe}\) c. \(\mathrm{MgBr_2}\) d. \(\mathrm{Li_3N}\) e. \(\mathrm{KH}\)

Step-by-step solution

a. Sodium (Na) and Sulfur (S)

Sodium has an electron configuration of \(1s^2 2s^2 2p^6 3s^1\). It loses one electron to achieve a stable electron configuration and becomes a \(1+\) ion. Sulfur has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^4\). It gains two electrons to achieve a stable electron configuration and becomes a \(2-\) ion. So, the binary ionic compound formed has the formula \(\mathrm{Na_2S}\).

b. Barium (Ba) and Selenium (Se)

Barium has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2\). It loses two electrons to achieve a stable electron configuration and becomes a \(2+\) ion. Selenium has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^4\). It gains two electrons to achieve a stable electron configuration and becomes a \(2-\) ion. So, the binary ionic compound formed has the formula \(\mathrm{BaSe}\).

c. Magnesium (Mg) and Bromine (Br)

Magnesium has an electron configuration of \(1s^2 2s^2 2p^6 3s^2\). It loses two electrons to achieve a stable electron configuration and becomes a \(2+\) ion. Bromine has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\). It gains one electron to achieve a stable electron configuration and becomes a \(1-\) ion. So, the binary ionic compound formed has the formula \(\mathrm{MgBr_2}\).

d. Lithium (Li) and Nitrogen (N)

Lithium has an electron configuration of \(1s^2 2s^1\). It loses one electron to achieve a stable electron configuration and becomes a \(1+\) ion. Nitrogen has an electron configuration of \(1s^2 2s^2 2p^3\). It gains three electrons to achieve a stable electron configuration and becomes a \(3-\) ion. So, the binary ionic compound formed has the formula \(\mathrm{Li_3N}\).

e. Potassium (K) and Hydrogen (H)

Potassium has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1\). It loses one electron to achieve a stable electron configuration and becomes a \(1+\) ion. Hydrogen has an electron configuration of \(1s^1\). It gains one electron to achieve a stable electron configuration (that of Helium) and becomes a \(1-\) ion. So, the binary ionic compound formed has the formula \(\mathrm{KH}\).

Key concepts

Electron Configuration

Understanding electron configuration is like getting a peek into an atom's internal blueprint. It tells how electrons are distributed in an atom's orbitals. Electrons fill the lowest energy levels first, following the order laid out by quantum mechanics.
For example, sodium (\(\text{Na}\)) has the configuration \(1s^2 2s^2 2p^6 3s^1\). The "3s¹" means there is one valence electron in the outermost shell. This is crucial because how electrons are arranged directly affects how elements react with each other.
Knowing the configuration helps predict how elements will interact, forming compounds. For instance, when sulfur (\(\text{S}\)) with \(1s^2 2s^2 2p^6 3s^2 3p^4\) needs two more electrons to achieve a stable configuration, it pairs perfectly with two sodium atoms, leading to the formation of \(\text{Na}_2\text{S}\).
The patterns of filling these orbitals help us understand the chemical properties and reactive potential of different elements.

Chemical Bonding

Chemical bonding refers to the attraction between atoms that allows for the formation of more complex substances. One of the primary types of chemical bonding is ionic bonding, where atoms transfer electrons to achieve stability.
In ionic compounds, like \(\text{Na}_2\text{S}\), electrons are transferred from one atom to another. Sodium gives up its lone electron, and sulfur accepts two. This transfer results in charged ions that attract each other strongly.

• Formation of Ions: Atoms become ions by losing or gaining electrons.
• Ionic Bonds: Occur between metals (like Na) and non-metals (like S).
• Stability: Achieved by adopting noble gas electron configurations.

These forces hold the compounds together with a considerable amount of strength, dictating many physical properties, such as melting points and solubility. Understanding these bonds gives insight into how compounds form and behave.

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. They play a key role in determining how an element will react with others. The behavior of valence electrons is central to understanding bonding and stability in ionic compounds.
For example, barium (\(\text{Ba}\)) has two valence electrons in the 6s orbital \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2\). It tends to lose these electrons, forming a \(2+\) ion. Selenium (\(\text{Se}\)), on the other hand, needs two electrons to fill its valence shell, forming \(\text{BaSe}\).
This outermost electron shell configuration determines how atoms share, lose, or gain electrons, thereby influencing chemical reactions.Understanding valence electrons:

• Reactivity: Elements with unfulfilled valence shells tend to seek stability through electron transfer.
• Noble Gas Mimicry: Atoms strive to resemble the electron configuration of the nearest noble gas.
• Predictive Power: Valence electrons allow us to predict formulas of compounds, like \(\text{MgBr}_2\) for magnesium and bromine.