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Chapter 8: Problem 17

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) \(\mathrm{Al}\) and \(\mathrm{F}\), (b) \(\mathrm{K}\) and \(\mathrm{S},(\mathrm{c}) \mathrm{Y}\) and \(\mathrm{O},\) (d) \(\mathrm{Mg}\) and \(\mathrm{N}\).

Short Answer

Expert verified
The chemical formulas for the ionic compounds formed between the given pairs of elements are: (a) \(\mathrm{AlF_3}\), (b) \(\mathrm{K_2S}\), (c) \(\mathrm{Y_2O_3}\), and (d) \(\mathrm{Mg_3N_2}\).

Step by step solution

01

Pair (a): \(\mathrm{Al}\) and \(\mathrm{F}\)

For \(\mathrm{Al}\) (Aluminum), its position in Group 13 of the periodic table means it has three valence electrons. It loses three electrons to obtain a stable octet, forming \(\mathrm{Al^{3+}}\) ion. For \(\mathrm{F}\) (Fluorine), its position in Group 17 means it has seven valence electrons. It gains one electron to obtain a stable octet, forming \(\mathrm{F^{-}}\) ion. Now, to obtain overall electrical neutrality, we need three \(\mathrm{F^{-}}\) ions for each \(\mathrm{Al^{3+}}\) ion. Thus, the chemical formula of the ionic compound formed between \(\mathrm{Al}\) and \(\mathrm{F}\) is \(\mathrm{AlF_3}\).
02

Pair (b): \(\mathrm{K}\) and \(\mathrm{S}\)

For \(\mathrm{K}\) (Potassium), its position in Group 1 means it has one valence electron. It loses this electron to form \(\mathrm{K^{+}}\) ion. For \(\mathrm{S}\) (Sulfur), its position in Group 16 means it has six valence electrons. It gains two electrons to obtain a stable octet, forming \(\mathrm{S^{2-}}\) ion. To obtain overall electrical neutrality, we need two \(\mathrm{K^{+}}\) ions for each \(\mathrm{S^{2-}}\) ion. Thus, the chemical formula of the ionic compound formed between \(\mathrm{K}\) and \(\mathrm{S}\) is \(\mathrm{K_2S}\).
03

Pair (c): \(\mathrm{Y}\) and \(\mathrm{O}\)

For \(\mathrm{Y}\) (Yttrium), its position in Group 3 means it has three valence electrons. It loses three electrons to form \(\mathrm{Y^{3+}}\) ion. For \(\mathrm{O}\) (Oxygen), its position in Group 16 means it has six valence electrons. It gains two electrons to obtain a stable octet, forming \(\mathrm{O^{2-}}\) ion. To obtain overall electrical neutrality, we need three \(\mathrm{O^{2-}}\) ions for every two \(\mathrm{Y^{3+}}\) ions. Thus, the chemical formula of the ionic compound formed between \(\mathrm{Y}\) and \(\mathrm{O}\) is \(\mathrm{Y_2O_3}\).
04

Pair (d): \(\mathrm{Mg}\) and \(\mathrm{N}\)

For \(\mathrm{Mg}\) (Magnesium), its position in Group 2 means it has two valence electrons. It loses two electrons to form \(\mathrm{Mg^{2+}}\) ion. For \(\mathrm{N}\) (Nitrogen), its position in Group 15 means it has five valence electrons. It gains three electrons to obtain a stable octet, forming \(\mathrm{N^{3-}}\) ion. To obtain overall electrical neutrality, we need three \(\mathrm{Mg^{2+}}\) ions for every two \(\mathrm{N^{3-}}\) ions. Thus, the chemical formula of the ionic compound formed between \(\mathrm{Mg}\) and \(\mathrm{N}\) is \(\mathrm{Mg_3N_2}\). In conclusion, the chemical formulas for the ionic compounds formed between the given pairs of elements are: (a) \(\mathrm{AlF_3}\), (b) \(\mathrm{K_2S}\), (c) \(\mathrm{Y_2O_3}\), and (d) \(\mathrm{Mg_3N_2}\).

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Key Concepts

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

Chemical Formulas
Chemical formulas are crucial in identifying the composition of ionic compounds. An ionic compound consists of positive and negative ions that attract each other due to opposite charges.

For instance, in understanding the formula for aluminum fluoride (\( \text{AlF}_3 \)), aluminum (\( \text{Al}^{3+} \)) ions combine with fluoride (\( \text{F}^- \)) ions. The balancing of charges occurs by using three fluoride ions for each aluminum ion.

This results in a balanced formula: \( \text{AlF}_3 \).

Some important pointers to keep in mind regarding chemical formulas include:
  • The positive ion (cation) is always written first.
  • The subscripts in the formula indicate the number of each ion needed for electronic neutrality.
Understanding chemical formulas helps predict how different elements interact to form compounds.
Valence Electrons
Valence electrons are the outermost electrons of an atom, playing a critical role in chemical bonding. They determine how atoms interact and the types of bonds they form. Each group in the periodic table reflects the number of valence electrons.

For example:
  • Elements in Group 1, like potassium (\( \text{K} \)), have one valence electron.
  • Elements in Group 17, such as fluorine (\( \text{F} \)), have seven valence electrons.
  • Elements in Group 2, like magnesium (\( \text{Mg} \)), possess two valence electrons.
When these electrons are lost, gained, or shared, atoms form ions or bonds to reach a stable electron configuration similar to that of noble gases.

For magnesium and nitrogen yielding \( \text{Mg}_3\text{N}_2 \), magnesium's two electrons are shared with nitrogen needing three electrons, balancing to achieve stability.
Electrical Neutrality
Electrical neutrality is a fundamental principle in forming ionic compounds. It ensures that the sum of the charges in a compound is zero, maintaining a stable and neutral compound.

To achieve neutrality, the total positive charge must equal the total negative charge. This ensures the compound is balanced. For example, when \( \text{K}^{+} \) ions, each with a positive charge, bond with \( \text{S}^{2-} \) ions, the potassium ions must be added in pairs to balance the sulfur ion's charge, forming \( \text{K}_2\text{S} \).

Here鈥檚 how it typically works:
  • Determine the charges of the ions involved.
  • Adjust the ratio of cations to anions to achieve a net charge of zero.
By balancing the charges, compounds remain stable and neutral, forming reliable chemical structures.

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Most popular questions from this chapter

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) \(\mathrm{Ru}\) and \(24.0 \% \mathrm{O}\) (by mass), Compound 2: \(61.2 \% \mathrm{Ru}\) and \(38.8 \%\) O (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound \(2 ?\) (c) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C},\) while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending on whether the compound is better described as a molecular or ionic compound.

Draw the Lewis structures for each of the following molecules or ions. Which do not obey the octet rule? (a) \(\mathrm{NO},\) (b) \(\mathrm{BF}_{3}\), (c) \(\mathrm{ICl}_{2}^{-}\) (d) \(\mathrm{OPBr}_{3}\) (the \(\mathrm{P}\) is the central atom), (e) XeF 4 .

(a) Use the concept of resonance to explain why all six \(\mathrm{C}-\mathrm{C}\) bonds in benzene are equal in length. (b) Are the \(\mathrm{C}-\mathrm{C}\) bond lengths in benzene shorter than \(\mathrm{C}-\mathrm{C}\) single bonds? Are they shorter than \(\mathrm{C}=\mathrm{C}\) double bonds?

(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) Explain why it is necessary to form a double bond in the Lewis structure. (c) The bond in \(\mathrm{O}_{2}\) is shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond. Explain this observation.

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+}\) and is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-}\). Why, then, is \(\mathrm{CaO}\) stable relative to the free elements?
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