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Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl}\) c. \(\mathrm{Sr}\) and \(\mathrm{E}\) b. \(\mathrm{Na}\) and \(\mathrm{O}\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Short Answer

Expert verified
The empirical formulas and names of the ionic compounds formed from the given pairs of elements are: a. AlCl3 - Aluminum Chloride b. Na2O - Sodium Oxide c. SrI2 - Strontium Iodide (assuming E is an error and should be I) d. CaSe - Calcium Selenide

Step by step solution

01

Determine the Charges of Ions

Determine the charges of the ions from their group number in the periodic table. Notice that elements in group 1 (alkali metals) form ions with a +1 charge, group 2 (alkaline earth metals) form ions with a +2 charge, and elements in group 17 (halogens) form ions with a -1 charge. Elements in group 16 form ions with a -2 charge.
02

Balance the Overall Charge

Balance the overall charge by finding the simplest ratio of ions in each empirical formula. This is done by crossing the absolute values of the charges.
03

Name the Compound

Name each compound using the rules for naming binary ionic compounds. This includes using the full name of the metal followed by the root of the second element's name, ending in "ide." Now, let's apply these steps to the given pairs of elements. a. Al and Cl - Step 1: Al forms a 3+ ion, while Cl forms a 1- ion. - Step 2: Empirical formula is AlCl3. - Step 3: Name the compound Aluminum Chloride. b. Na and O - Step 1: Na forms a 1+ ion, while O forms a 2- ion. - Step 2: Empirical formula is Na2O. - Step 3: Name the compound Sodium Oxide. c. Sr and E is likely an error. It should be Sr and I (Strontium and Iodine). - Step 1: Sr forms a 2+ ion, while I forms a 1- ion. - Step 2: Empirical formula is SrI2. - Step 3: Name the compound Strontium Iodide. d. Ca and Se - Step 1: Ca forms a 2+ ion, while Se forms a 2- ion. - Step 2: Empirical formula is CaSe. - Step 3: Name the compound Calcium Selenide.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Compounds
Ionic compounds are formed when metals transfer electrons to nonmetals, resulting in a compound held together by ionic bonds. These bonds are quite strong due to the electrostatic attraction between the positively charged ions and negatively charged ions.

Here are a few key characteristics of ionic compounds:
  • They are usually crystalline solids at room temperature.
  • They have high melting and boiling points due to strong interionic forces.
  • They often dissolve in water and other polar solvents, conducting electricity when dissolved or molten.
Understanding how ionic compounds are formed helps grasp their stability and other properties. This begins with knowing how metals and nonmetals interact based on their electron configurations.
Periodic Table
The periodic table is a vital tool in chemistry, providing a framework for organizing elements according to their atomic number and properties. Elements are arranged in rows called periods and columns called groups. Each group shares common properties, including the number of electrons in their outermost shell.

An essential concept for understanding ionic compounds is recognizing periodic table groups and their typical charges. Elements in:
  • Group 1: Exhibit a +1 charge (e.g., Na)
  • Group 2: Exhibit a +2 charge (e.g., Ca)
  • Group 16: Exhibit a -2 charge (e.g., O)
  • Group 17: Exhibit a -1 charge (e.g., Cl)
Comprehending these groups can aid in predicting the empirical formulas of ionic compounds.
Ion Charges
Ion charges result from the gain or loss of electrons. Metals, tending to have fewer electrons in their outer shell, lose electrons to become positively charged cations. Nonmetals gain electrons to form negatively charged anions.

Here is a simple way to remember about ion formation:
  • Cations: Positively charged, typically metals like Al (3+)
  • Anions: Negatively charged, typically nonmetals like Cl (1-)
To form neutral ionic compounds, the total positive and negative charges must balance. For instance, in aluminum chloride (AlCl3), one Al3+ pairs with three Cl ions, balancing the charges.
Binary Ionic Compounds
Binary ionic compounds consist of two different elements, typically a metal and a nonmetal. Their empirical formulas represent the simplest whole-number ratio of ions that results in a neutral charge for the compound.

The naming process for binary ionic compounds is straightforward:
  • Start with the full name of the metal.
  • Follow with the root of the nonmetal's name.
  • Add the ending "ide" to the nonmetal.
For example, Na2O is named sodium oxide. Knowing how to construct and name binary ionic compounds is crucial for clear scientific communication.

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Most popular questions from this chapter

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{~F}_{2}\left(\mathrm{O}_{2} \mathrm{~F}_{2}\right.\) exists as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F}\) ). Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{~F}_{2}\). This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

In general, the higher the charge on the ions in an ionic compound, the more favorable the lattice energy. Why do some stable ionic compounds have \(+1\) charged ions even though \(+4,+5\), and \(+6\) charged ions would have a more favorable lattice energy?

Predict the molecular structure for each of the following. (See Exercises 111 and \(112 .\) ) a. \(\mathrm{BrFI}_{2}\) b. \(\mathrm{XeO}_{2} \mathrm{~F}_{2}\) c. \(\mathrm{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Use Coulomb's law, $$ V=\frac{Q_{1} Q_{2}}{4 \pi \epsilon_{0} r}=2.31 \times 10^{-19} \mathrm{~J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right) $$ to calculate the energy of interaction for the following two arrangements of charges, each having a magnitude equal to the electron charge. a. \(\stackrel{1 \times 10^{-10} \mathrm{~m}}{\longrightarrow(-1) \longleftrightarrow \infty \longrightarrow(+1) \longleftrightarrow 10^{-10} \mathrm{~m}}{\longleftrightarrow} \stackrel{\leftarrow}{\longleftrightarrow}\) b.

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\), \(\mathrm{O}-\mathrm{H}, \mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\).

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