Question

On the basis of their electron configurations, predict the formula of the simple binary ionic compounds likely to form when the following pairs of elements react with each other. a. aluminum, Al, and sulfur, S b. radium, \(\mathrm{Ra}\), and oxygen, \(\mathrm{O}\) c. calcium, \(\mathrm{Ca},\) and fluorine, \(\mathrm{F}\) d. cesium, \(\mathrm{Cs}\), and nitrogen, \(\mathrm{N}\) e. rubidium, Rb, and phosphorus, \(\mathrm{P}\)

Short answer

The formula of the simple binary ionic compounds are as follows: a. Al₂S₃ b. RaO c. CaF₂ d. Cs₃N e. Rb₃P

Step-by-step solution

Determine the electron configurations

For each given pair of elements, we first need to find their electron configurations. We can look them up in a periodic table or use their atomic numbers to determine their electron configurations. a. Aluminum, Al: [Ne] 3s² 3p¹ Sulfur, S: [Ne] 3s² 3p⁴ b. Radium, Ra: [Rn] 7s² Oxygen, O: [He] 2s² 2p⁴ c. Calcium, Ca: [Ar] 4s² Fluorine, F: [He] 2s² 2p⁵ d. Cesium, Cs: [Xe] 6s¹ Nitrogen, N: [He] 2s² 2p³ e. Rubidium, Rb: [Kr] 5s¹ Phosphorus, P: [Ne] 3s² 3p³

Predict the likely ion charges

Next, we will predict the most likely charges for each ion when forming a binary ionic compound based on the electron configurations. a. Al: 3+ (loses 3 electrons) S: 2- (gains 2 electrons) b. Ra: 2+ (loses 2 electrons) O: 2- (gains 2 electrons) c. Ca: 2+ (loses 2 electrons) F: 1- (gains 1 electron) d. Cs: 1+ (loses 1 electron) N: 3- (gains 3 electrons) e. Rb: 1+ (loses 1 electron) P: 3- (gains 3 electrons)

Determine the compound formulas

Finally, we will use the predicted ion charges to determine the possible formulas for the resulting binary ionic compounds. The compound formula should result in a neutral overall charge. a. Al³⁺ and S²⁻: Al₂S₃ b. Ra²⁺ and O²⁻: RaO c. Ca²⁺ and F⁻: CaF₂ d. Cs⁺ and N³⁻: Cs₃N e. Rb⁺ and P³⁻: Rb₃P

Key concepts

Electron Configurations

Understanding electron configurations is foundational to predicting how atoms will combine to form compounds. Each element has a unique electron configuration that represents how its electrons are distributed among the different orbital shells.

For instance, aluminum (Al) has the electron configuration [Ne] 3s² 3p¹, indicating it has two electrons in the 3s orbital and one in the 3p orbital beyond the core configuration of neon (Ne). Sulfur (S), on the other hand, has [Ne] 3s² 3p⁴, with two electrons in the 3s orbital and four in the 3p orbital. These configurations suggest that aluminum has three electrons it can lose to achieve a more stable configuration, while sulfur needs two more to complete its octet.

It is by examining these configurations we begin to predict how these atoms might react to form ionic bonds, starting with the loss or gain of electrons to achieve noble gas electron configuration.

Predicting Ion Charges

Predicting ion charges from electron configurations is all about balance. Atoms lose or gain electrons to achieve the stable electron configuration of the nearest noble gas. The number of electrons lost or gained becomes the ion charge.

Take calcium (Ca), for example, with an electron configuration of [Ar] 4s², it will lose two electrons to attain the stable configuration of argon (Ar), thus it becomes Ca²⁺. Fluorine (F), with [He] 2s² 2p⁵, gains one electron to complete its outermost p orbital, becoming F⁻. The magnitude of the charges directly influences the ratio of ions in the resulting compound; each compound aims for an overall neutral charge.

A helpful tip for remembering common charges is to look at the group number of elements in the periodic table. Group 1 elements often form 1+ ions, while group 16 elements commonly form 2- ions.

Formation of Ionic Compounds

The formation of ionic compounds is a chemical process where positively charged ions (cations) and negatively charged ions (anions) are held together by electrostatic forces in a lattice structure. By considering the charges we predicted earlier, we can deduce the formula of an ionic compound.

The goal is to combine ions in a way that the total positive charge is balanced by an equal total negative charge. For instance, when radium (Ra²⁺) reacts with oxygen (O²⁻), they form RaO where the charges of Ra and O ions cancel each other out. Meanwhile, aluminum (Al³⁺) requires two sulfur (S²⁻) ions to balance its charge, resulting in Al₂S₃. Ensuring charge neutrality leads us to the correct stoichiometry of these compounds, crucial for predicting the formula of ionic compounds like CaF₂, Cs₃N, and Rb₃P.