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Chapter 5: Problem 42

Predict the formula of the ionic compound formed by each of the following pairs of elements: (a) \(\mathrm{Ca}_{,} \mathrm{I}\) (b) \(\mathrm{Ca}, \mathrm{O}\) (c) \(\mathrm{Al}, \mathrm{S}\) (d) \(\mathrm{Ca}, \mathrm{Br}\)

Short Answer

Expert verified
The predicted formulas for the ionic compounds formed by the given pairs of elements are: (a) \(\mathrm{CaI}_{2}\) (b) \(\mathrm{CaO}\) (c) \(\mathrm{Al}_{2}\mathrm{S}_{3}\) (d) \(\mathrm{CaBr}_{2}\)

Step by step solution

01

(Determine the charges of each ion)

To find the charges of each ion, we need to know their group number in the periodic table. The number of valence electrons is the same as the group number for representative elements, and the charge of the ions can be predicted by how many electrons are needed to gain or lose to achieve a stable electronic configuration, i.e., to achieve the nearest noble gas configuration. **Step 2: Find the ratio in which ions combine**
02

(Find the ratio in which ions combine)

The ions combine in such a way that the overall charge of the compound is neutral. To find the ratio, we need to balance charges of the ions involved. Now, let's work on each part of the exercise using the above steps: **(a) \(\mathrm{Ca}_{,} \mathrm{I}\)**
03

(Predict the formula of CaI)

Using step 1, Ca (group 2) is a metal and loses two electrons to form a \(\mathrm{Ca}^{2+}\) ion. I (group 17) is a non-metal that gains one electron to form a \(\mathrm{I}^{-}\) ion. Using step 2, we need two \(\mathrm{I}^{-}\) ions to balance the charge of one \(\mathrm{Ca}^{2+}\) ion. Therefore, the formula for the ionic compound is \(\mathrm{CaI}_{2}\). **(b) \(\mathrm{Ca}, \mathrm{O}\)**
04

(Predict the formula of CaO)

Using step 1, Ca (group 2) loses two electrons to form a \(\mathrm{Ca}^{2+}\) ion and O (group 16) gains two electrons to form a \(\mathrm{O}^{2-}\) ion. Using step 2, the charges of \(\mathrm{Ca}^{2+}\) and \(\mathrm{O}^{2-}\) ions balance each other. Therefore, the formula for the ionic compound is \(\mathrm{CaO}\). **(c) \(\mathrm{Al}, \mathrm{S}\)**
05

(Predict the formula of AlS)

Using step 1, Al (group 13) loses three electrons to form a \(\mathrm{Al}^{3+}\) ion, and S (group 16) gains two electrons to form a \(\mathrm{S}^{2-}\) ion. Using step 2, the lowest common multiple (LCM) of 3 and 2 is 6. To balance charges for the ionic compound, we need two \(\mathrm{Al}^{3+}\) ions and three \(\mathrm{S}^{2-}\) ions to neutralize the charges. Therefore, the formula for the ionic compound is \(\mathrm{Al}_{2}\mathrm{S}_{3}\). **(d) \(\mathrm{Ca}, \mathrm{Br}\)**
06

(Predict the formula of CaBr)

Using step 1, Ca (group 2) loses two electrons to form a \(\mathrm{Ca}^{2+}\) ion, and Br (group 17) gains one electron to form a \(\mathrm{Br}^{-}\) ion. Using step 2, we need two \(\mathrm{Br}^{-}\) ions to balance the charge of one \(\mathrm{Ca}^{2+}\) ion. Therefore, the formula for the ionic compound is \(\mathrm{CaBr}_{2}\).

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