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Chapter 8: Problem 16

Which ionic compound is expected to form from combining the following pairs of elements: (a) barium and fluorine, (b) cesium and chlorine, (c) lithium and nitrogen, (d) aluminum and oxygen?

Short Answer

Expert verified
The ionic compounds formed by combining the given pairs of elements are: (a) Barium Fluoride: \(BaF_2\) (b) Cesium Chloride: \(CsCl\) (c) Lithium Nitride: \(Li_3N\) (d) Aluminum Oxide: \(Al_2O_3\)

Step by step solution

01

(Step 1: Determine the charge of each ion)

Using the periodic table, we can determine the charge of each ion in the given pairs of elements. For example, metals in Group 1 have a charge of +1, metals in Group 2 have a charge of +2, and nonmetals in Group 17 have a charge of -1.
02

(a) Barium and Fluorine)

Barium (Ba) is a Group 2 element, so its ion has a charge of +2 (Ba^2+). Fluorine (F) is a Group 17 element, so its ion has a charge of -1 (F^-).
03

(b) Cesium and Chlorine)

Cesium (Cs) is a Group 1 element, so its ion has a charge of +1 (Cs^+). Chlorine (Cl) is a Group 17 element, so its ion has a charge of -1 (Cl^-).
04

(c) Lithium and Nitrogen)

Lithium (Li) is a Group 1 element, so its ion has a charge of +1 (Li^+). Nitrogen (N) is a Group 15 element, so its ion has a charge of -3 (N^3-).
05

(d) Aluminum and Oxygen)

Aluminum (Al) is a Group 13 element, so its ion has a charge of +3 (Al^3+). Oxygen (O) is a Group 16 element, so its ion has a charge of -2 (O^2-).
06

(Step 2: Combine the ions to form ionic compounds)

Now that we have the charges, we can combine each pair of elements in a way that their overall charge is neutral, forming the ionic compound.
07

(a) Barium and Fluorine)

To achieve a neutral charge, we need two fluorine ions for each barium ion: \[Ba^{2+} + 2F^{-} \rightarrow BaF_2\]
08

(b) Cesium and Chlorine)

To achieve a neutral charge, we need one chlorine ion for each cesium ion: \[Cs^{+} + Cl^{-} \rightarrow CsCl\]
09

(c) Lithium and Nitrogen)

To achieve a neutral charge, we need three lithium ions for each nitrogen ion: \[3Li^{+} + N^{3-} \rightarrow Li_3N\]
10

(d) Aluminum and Oxygen)

To achieve a neutral charge, we need two aluminum ions for each three oxygen ions: \[2Al^{3+} + 3O^{2-} \rightarrow Al_2O_3\] So, the ionic compounds formed by combining the given pairs of elements are: (a) Barium Fluoride: \(BaF_2\) (b) Cesium Chloride: \(CsCl\) (c) Lithium Nitride: \(Li_3N\) (d) Aluminum Oxide: \(Al_2O_3\)

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Most popular questions from this chapter

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \%\) N. Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} \mathrm{~mol}^{-1} .\) The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(S_{8}\) ring is \(2.05 \AA\).) (d) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} \mathrm{~mol}^{-1}\). \(\Delta H_{f}^{\circ}\) of \(S(g)\) is \(222.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Estimate the average bond enthalpy in the compound.

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ?

(a) Explain the following trend in lattice energy: \(\mathrm{BeH}_{2}\), \(3205 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{CaH}_{2}, 2410 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{Sr} \mathrm{H}_{2}, 2250 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{~kJ} / \mathrm{mol}\). (b) The lattice energy of \(\mathrm{ZnH}_{2}\) is \(2870 \mathrm{~kJ} / \mathrm{mol}\). Based on the data given in part (a), the radius of the \(\mathrm{Zn}^{2+}\) ion is expected to be closest to that of which group \(2 \mathrm{~A}\) element?

Which of the following bonds are polar: (a) \(\mathrm{B}-\mathrm{F}\), (b) \(\mathrm{Cl}-\mathrm{Cl}\), (c) \(\mathrm{Se}-\mathrm{O}\), (d) \(\mathrm{H}-\mathrm{I}\) ? Which is the more electronegative atom in each polar bond?

To address energy and environmental issues, there is great interest in powering vehicles with hydrogen rather than gasoline. One of the most attractive aspects of the "hydrogen economy" is the fact that in principle the only emission would be water. However, two daunting obstacles must be overcome before this vision can become a reality. First, an economical method of producing hydrogen must be found. Second, a safe, lightweight, and compact way of storing hydrogen must be found. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. One of the most attractive hydrides is \(\mathrm{NaAlH}_{4}\), which can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(\mathrm{s})\), Al(s), and \(\mathrm{H}_{2}(\mathrm{~g})\). NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\). (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.
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