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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 \(\mathrm{Cl} \quad\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O} \quad\) 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. AlCl鈧: Aluminium Chloride b. Na鈧侽: Sodium Oxide c. SrF鈧: Strontium Fluoride d. CaSe: Calcium Selenide

Step by step solution

01

Identify the charges of the ions

Since all the elements provided are metals and non-metals, we can know their usual ionic charges by observing their groups in the periodic table. Metals tend to lose electrons and form positive ions (cations), while the non-metals tend to gain electrons and form negative ions (anions). a. Al: Aluminium is in group 13 and forms +3 cation, Al鈦郝. Cl: Chlorine is in group 17 and forms -1 anion, Cl鈦宦. b. Na: Sodium is in group 1 and forms +1 cation, Na鈦郝. O: Oxygen is in group 16 and forms -2 anion, O鈦宦. c. Sr: Strontium is in group 2 and forms +2 cation, Sr鈦郝. F: Fluorine is in group 17 and forms -1 anion, F鈦宦. d. Ca: Calcium is in group 2 and forms +2 cation, Ca鈦郝. Se: Selenium is in group 16 and forms -2 anion, Se鈦宦.
02

Combine ions to create the empirical formula

We need to balance the positive and negative charges in each compound by finding the least common multiple (LCM) of the charges, and then adjusting the number of ions so that their charges are equal. a. Al鈦郝 and Cl鈦宦: The LCM of +3 and -1 is 3, so 1 Al鈦郝 ion should combine with 3 Cl鈦宦 ions to form AlCl鈧. b. Na鈦郝 and O鈦宦: The LCM of +1 and -2 is 2, so 2 Na鈦郝 ions should combine with 1 O鈦宦 ion to form Na鈧侽. c. Sr鈦郝 and F鈦宦: The LCM of +2 and -1 is 2, so 1 Sr鈦郝 ion should combine with 2 F鈦宦 ions to form SrF鈧. d. Ca鈦郝 and Se鈦宦: The charges of Ca鈦郝 and Se鈦宦 are equal, so 1 Ca鈦郝 ion should combine with 1 Se鈦宦 ion to form CaSe.
03

Name each compound

We can name the compounds using the names of the positively charged ion followed by the negatively charged ion with an -ide suffix. a. AlCl鈧: Aluminium Chloride b. Na鈧侽: Sodium Oxide c. SrF鈧: Strontium Fluoride d. CaSe: Calcium Selenide

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Key Concepts

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

Empirical Formula
The empirical formula of a compound shows the simplest whole-number ratio of atoms in the compound. It's like the recipe that tells us how many of each type of atom is needed to form the substance. For ionic compounds, the empirical formula is found by balancing the charges of the ions involved.

- **Determining charges**: Identify the charge of each ion. For example, in forming AlCl鈧: - Aluminium ion ( Al^3+ ) has a charge of +3. - Chloride ion ( Cl^{-1} ) has a charge of -1.

- **Balanced charges**: Use the principle of charge neutrality, which states that the total positive charge must balance the total negative charge. With AlCl鈧, the 3 chlorine ions each contribute -1 charge, perfectly cancelling out the +3 from one aluminium ion, showing the compound's simplest form is AlCl鈧.

- **Empirical formula representation**: Once you have balanced the overall charges by using the lowest ratio of ion numbers, you have the empirical formula, such as AlCl鈧, Na鈧侽, SrF鈧, and CaSe in our examples.
Cation and Anion Charges
Cation and Anion charges are key to predicting the formation and formula of ionic compounds. Cations are positively charged ions, while anions carry a negative charge. They are formed when atoms lose or gain electrons, respectively.

- **Formation of cations**: Metals like sodium (Na), strontium (Sr), calcium (Ca), and aluminium (Al) tend to lose electrons. This results in positively charged ions or cations. Each element's position in the periodic table helps predict its charge. For instance, aluminium forms Al^3+ because it's in group 13.

- **Formation of anions**: Non-metals such as chlorine (Cl), oxygen (O), fluorine (F), and selenium (Se) tend to gain electrons, forming negatively charged ions or anions. For example, chlorine becomes Cl^-, as it is in group 17.

- **Combining charges**: In compounds, like attracts like, and opposite charges attract. The formula for the compound results from finding the least common multiple between the charges of the cations and anions involved. For instance, combining the ions Na^+1 and O^2- results in Na鈧侽 after balancing the two charges.
Naming Ionic Compounds
Naming ionic compounds follows straightforward rules. The compound's name derives from the cation first, then the anion, often with modifications like suffixes.

- **Name the cation first**: The name of the metal or positively charged ion retains its elemental name in the compound. For example, Na in Na鈧侽 remains sodium and Al in AlCl鈧 remains aluminium.

- **Modify the anion's name**: For the non-metal or negatively charged ion, take the element's root name and add the suffix '-ide.' Hence, oxygen becomes oxide, as in sodium oxide (Na鈧侽), and chlorine becomes chloride, as in aluminium chloride (AlCl鈧).

- **Complete name**: The full name of the ionic compound combines these two elements. So, the combination of calcium (Ca) and selenium (Se) would yield calcium selenide (CaSe). This helps in easily identifying the type of ions present and their proportion within the compound.

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Most popular questions from this chapter

Predict the molecular structure for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{BrFI}_{2} \quad\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2} \quad\) 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 bond energies (Table \(8.5 ),\) values of electron affinities (Table 7.7\()\) , and the ionization energy of hydrogen \((1312 \mathrm{kJ} / \mathrm{mol})\) to estimate \(\Delta H\) for each of the following reactions. a. \(\mathrm{HF}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{F}^{-}(g)\) b. \(\mathrm{HCl}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{Cl}^{-}(g)\) c. \(\mathrm{HI}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{I}^{-}(g)\) d. \(\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{OH}^{-}(g)\)

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being 鈥減ulled鈥 from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?

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