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Chapter 3: Problem 38

Write formulas for the following ionic compounds: (a) Ammonium hydrogen sulfite (b) Magnesium acetate (c) Strontium dihydrogen phosphate (d) Silver carbonate (e) Strontium chloride (f) Barium permanganate (g) Aluminum perchlorate

Short Answer

Expert verified
(a) (NH鈧)(HSO鈧), (b) Mg(C鈧侶鈧僌鈧)鈧, (c) Sr(H鈧侾O鈧)鈧, (d) Ag鈧侰O鈧, (e) SrCl鈧, (f) Ba(MnO鈧)鈧, (g) Al(ClO鈧)鈧.

Step by step solution

01

Identify the Ions and Their Charges

For each compound, we need to identify both the cation (positive ion) and the anion (negative ion), along with their respective charges. - **Ammonium hydrogen sulfite**: Ammonium is NH鈧勨伜 and hydrogen sulfite is HSO鈧冣伝. - **Magnesium acetate**: Magnesium is Mg虏鈦 and acetate is C鈧侶鈧僌鈧傗伝. - **Strontium dihydrogen phosphate**: Strontium is Sr虏鈦 and dihydrogen phosphate is H鈧侾O鈧勨伝. - **Silver carbonate**: Silver is Ag鈦 and carbonate is CO鈧兟测伝. - **Strontium chloride**: Strontium is Sr虏鈦 and chloride is Cl鈦. - **Barium permanganate**: Barium is Ba虏鈦 and permanganate is MnO鈧勨伝. - **Aluminum perchlorate**: Aluminum is Al鲁鈦 and perchlorate is ClO鈧勨伝.
02

Determine the Correct Ratio of Ions

For each compound, determine the number of each ion needed to balance the overall charge of the compound to neutral. - **Ammonium hydrogen sulfite**: 1 NH鈧勨伜 ion pairs with 1 HSO鈧冣伝 ion: (NH鈧)(HSO鈧). - **Magnesium acetate**: 1 Mg虏鈦 ion pairs with 2 C鈧侶鈧僌鈧傗伝 ions to balance the 2+ charge: Mg(C鈧侶鈧僌鈧)鈧. - **Strontium dihydrogen phosphate**: 1 Sr虏鈦 ion pairs with 2 H鈧侾O鈧勨伝 ions: Sr(H鈧侾O鈧)鈧. - **Silver carbonate**: 2 Ag鈦 ions are needed to balance the 2- charge of CO鈧兟测伝: Ag鈧侰O鈧. - **Strontium chloride**: 1 Sr虏鈦 ion pairs with 2 Cl鈦 ions: SrCl鈧. - **Barium permanganate**: 1 Ba虏鈦 ion pairs with 2 MnO鈧勨伝 ions: Ba(MnO鈧)鈧. - **Aluminum perchlorate**: 1 Al鲁鈦 ion pairs with 3 ClO鈧勨伝 ions: Al(ClO鈧)鈧.
03

Formulate the Ionic Compounds

Using the determined ratios from Step 2, write the chemical formulas for each of the compounds. - **Ammonium hydrogen sulfite**: (NH鈧)(HSO鈧) - **Magnesium acetate**: Mg(C鈧侶鈧僌鈧)鈧 - **Strontium dihydrogen phosphate**: Sr(H鈧侾O鈧)鈧 - **Silver carbonate**: Ag鈧侰O鈧 - **Strontium chloride**: SrCl鈧 - **Barium permanganate**: Ba(MnO鈧)鈧 - **Aluminum perchlorate**: Al(ClO鈧)鈧

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Key Concepts

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

Understanding Cation and Anion Charges
In ionic compounds, atoms either gain or lose electrons to form ions. These ions come with charges, which play a crucial role in creating stable ionic compounds. Cations are positively charged ions. They form when an atom loses one or more electrons. For example:
  • Ammonium (\( NH_4^+ \)) is a cation that results from the loss of electrons by nitrogen bonded to hydrogen.
  • Silver (\( Ag^+ \)), magnesium (\( Mg^{2+} \)), and strontium (\( Sr^{2+} \)) form cations by losing electrons.
Anions, on the other hand, are negatively charged ions. They form when an atom gains electrons. For instance:
  • The acetate (\( C_2H_3O_2^- \)) anion forms when the acetic acid loses a hydrogen ion.
  • Chloride (\( Cl^- \)) and hydrogen sulfite (\( HSO_3^- \)) are common anions obtained by gaining electrons.
Understanding these charges helps in figuring out how these ions combine in fixed ratios to form ionic compounds.
Crafting Chemical Formulas of Ionic Compounds
Creating chemical formulas for ionic compounds involves determining the correct ratio of cations to anions. This ensures the compound's overall electrical neutrality. The key is to balance the positive and negative charges so they sum up to zero.Start with identifying the charges of the ions involved. For example:
  • For aluminum perchlorate, aluminum's charge is \(+3\), while perchlorate's is \(-1\). This means three perchlorate ions are needed to balance one aluminum ion.
  • In magnesium acetate, magnesium has a \(+2\) charge, so it pairs with two acetate ions, each with a \(-1\) charge.
The charges dictate the subscripts in the chemical formula. The subscripts show the number of each type of ion needed to balance the charges. By multiplying the cation's charge by the number of anions and vice versa, you can achieve neutrality and correctly frame the formula using parentheses where necessary.
Achieving Compound Neutrality
Neutrality in a compound means that the total positive charges equal the total negative charges. This balance is crucial for the stability of the compound. Achieving compound neutrality often requires careful consideration of the ion charges and a systematic approach to balancing them.Here's a breakdown of how this works:
  • In silver carbonate, with silver having a charge of \(+1\) and carbonate \(-2\), two silver ions are needed for every carbonate ion. Thus, the formula is \( Ag_2CO_3 \).
  • For strontium chloride, since strontium's charge is \(+2\) and chloride's is \(-1\), two chloride ions balance one strontium ion, resulting in \( SrCl_2 \).
Formulating these compounds involves determining how many of each ion is necessary to achieve a net charge of zero. This systematic pairing allows for the creation of stable ionic compounds, vital for real-world applications in chemistry. By understanding and applying these principles, one can construct correct and functional chemical formulas effectively.

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Most popular questions from this chapter

Potassium chloride and potassium bicarbonate are used as potassium dietary supplements. Write the formula of each compound.

Why does electronegativity generally increase going up a column (group) of the Periodic Table?

Answer true or false. (a) According to the Lewis model of bonding, atoms bond together in such a way that each atom participating in the bond acquires an outer-shell electron configuration matching that of the noble gas nearest to it in atomic number. (b) Atoms that lose electrons to achieve a filled valence shell become cations and form ionic bonds with anions. (c) Atoms that gain electrons to achieve filled valence shells become anions and form ionic bonds with cations. (d) Atoms that share electrons to achieve filled valence shells form covalent bonds. (e) Ionic bonds tend to form between elements on the left side of the Periodic Table, and covalent bonds tend to form between elements on the right side of the Periodic Table. (f) Ionic bonds tend to form between a metal and a nonmetal. (g) When two nonmetals combine, the bond between them is usually covalent. (h) Electronegativity is a measure of an atom's attraction for the electrons it shares in a chemical bond with another atom. (i) Electronegativity generally increases with atomic number. (j) Electronegativity generally increases with atomic weight. (k) Electronegativity is a periodic property. (l) Fluorine, in the upper-right corner of the Periodic Table, is the most electronegative element; hydrogen, in the upper-left corner, is the least electronegative element. (m) Electronegativity depends on both the nuclear charge and the distance of the valence electrons from the nucleus. (n) Electronegativity generally increases from left to right across a period of the Periodic Table. (o) Electronegativity generally increases from top to bottom in a column of the Periodic Table.

How many electrons must each atom gain or lose to acquire an electron configuration identical to the noble gas nearest to it in atomic number? (a) Li (b) \(\mathrm{Cl}\) (c) \(P\) (d) \(A l\) \((\mathrm{e}) \mathrm{Sr}\) (f) \(\mathrm{S}\) \((g) \quad\) Si (h) \(\mathrm{O}\)

Answer true or false. (a) An ionic bond is formed by the combination of positive and negative ions. (b) An ionic bond between two atoms forms by the transfer of one or more valence electrons from the atom of higher electronegativity to the atom of lower electronegativity. (c) As a rough guideline, we say that an ionic bond will form if the difference in electronegativity between two atoms is approximately 1.9 or greater. (d) In forming NaCl from sodium and chlorine atoms, one electron is transferred from the valence shell of sodium to the valence shell of chlorine. (e) The formula of sodium sulfide is \(\mathrm{Na}_{2} \mathrm{S}\) (f) The formula of calcium hydroxide is CaOH. (g) The formula of aluminum sulfide is AlS. (h) The formula of iron(III) oxide is \(\mathrm{Fe}_{3} \mathrm{O}_{2}\) (i) Barium ion is \(\mathrm{Ba}^{2+}\), and oxide ion is \(\mathrm{O}^{2-}\); therefore, the formula of barium oxide is \(\mathrm{Ba}_{2} \mathrm{O}_{2}\)
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