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Chapter 2: Problem 51

Which of the following compounds are likely to be ionic? Which are likely to be molecular? \(\mathrm{SiCl}_{4}\), LiF, \(\mathrm{BaCl}_{2}, \mathrm{~B}_{2} \mathrm{H}_{6}, \mathrm{KCl}, \mathrm{C}_{2} \mathrm{H}_{4}\)

Short Answer

Expert verified
Molecular compounds: \(\mathrm{SiCl}_{4}, \mathrm{B}_{2}\mathrm{H}_{6}, \mathrm{C}_{2}\mathrm{H}_{4}\). Ionic Compounds: LiF, \(\mathrm{BaCl}_{2}\), \(\mathrm{KCl}\).

Step by step solution

01

Categorizing ionic and molecular compounds

Begin by identifying the elements in each compound and determining if they are metals or nonmetals. A periodic table can be used to find this information.
02

Determining the type of compound

If a compound is composed of a metal and a nonmetal, categorize it as an ionic compound. If a compound is composed of only nonmetals, categorize it as a molecular compound.
03

Applying to each compound

For \(\mathrm{SiCl}_{4}\), both Si and Cl are nonmetals, so it is a molecular compound. For LiF, Li is a metal and F is a nonmetal, so it is an ionic compound. For \(\mathrm{BaCl}_{2}\), Ba is a metal and Cl is a nonmetal, so it is an ionic compound. For \(\mathrm{B}_{2} \mathrm{H}_{6}\), both B and H are nonmetals, so it is a molecular compound. For \(\mathrm{KCl}\), K is a metal and Cl is a nonmetal, so it is an ionic compound. For \(\mathrm{C}_{2} \mathrm{H}_{4}\), both C and H are nonmetals, so it is a molecular compound.

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Most popular questions from this chapter

Methane, ethane, and propane are shown in Table \(2.8 .\) Show that the following data are consistent with the law of multiple proportions. $$ \begin{array}{lll} & \begin{array}{l} \text { Mass of Carbon } \\ \text { in } 1 \mathrm{~g} \text { Sample } \end{array} & \begin{array}{l} \text { Mass of Hydrogen } \\ \text { in } 1 \mathrm{~g} \text { Sample } \end{array} \\ \hline \text { Methane } & 0.749 \mathrm{~g} & 0.251 \mathrm{~g} \\ \text { Ethane } & 0.799 \mathrm{~g} & 0.201 \mathrm{~g} \\ \text { Propane } & 0.817 \mathrm{~g} & 0.183 \mathrm{~g} \\ \hline \end{array} $$

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