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Chapter 3: Problem 15

Why is calcium dichloride not the correct systematic name for \(\mathrm{CaCl}_{2} ?\)

Short Answer

Expert verified
The correct systematic name for CaCl2 isn't "calcium dichloride" because numeric prefixes like "di-" are not used for ionic compounds containing a metal and a non-metal. The cation, calcium (Ca), keeps its element name and the anion, chlorine (Cl), has its ending replaced with "-ide," resulting in "chloride." Thus, the correct name is "calcium chloride."

Step by step solution

01

Identify the cation and anion

In the given compound, CaCl2, the cation is calcium (Ca) and the anion is chlorine (Cl).
02

Determine the charge of the cation and anion

For ionic compounds, we need to know the charges of the cation and anion. Calcium (Ca) is an alkaline earth metal and is in group 2 of the periodic table. It forms a cation with a charge of +2. The anion is Chlorine (Cl) which belongs to the halogens and forms an anion with a charge of -1.
03

Name the cation and anion

The cation, calcium (Ca), keeps its element name. The anion, chlorine (Cl), has its ending replaced with "-ide." Thus, the anion is named "chloride."
04

Combine the names without a numerical prefix

For ionic compounds containing a metal and a non-metal, we do not use numerical prefixes like "di-" or "tri-." Instead, the name of the compound is simply a combination of the cation and anion names. In this case, the correct name for CaCl2 should be "calcium chloride." Since numeric prefixes are not used for this type of compound, "calcium dichloride" isn't the correct systematic name. The correct name for CaCl2 is "calcium chloride."

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Most popular questions from this chapter

Write Lewis structures for the following. Show all resonancestructures where applicable.a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{N}_{2} \mathrm{O}_{4} \text { exists as } \mathrm{O}_{2} \mathrm{N}-\mathrm{NO}_{2} .\right)\) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{SCN}^{-} .\) )

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