Name the empirical formula and name of each ionic compound that could be formed from the ions in this table:
| cation | anion | empirical formula | name of compound |
|---|---|---|---|
| [tex]$Ba ^{2+}$[/tex] | [tex]$P ^{3-}$[/tex] | | |
| [tex]$Ba ^{2+}$[/tex] | [tex]$N ^{3-}$[/tex] | | |
| [tex]$Ba ^{2+}$[/tex] | [tex]$S ^{2-}$[/tex] | | |
| [tex]$Ba ^{2+}$[/tex] | [tex]$O ^{2-}$[/tex] | | |
Answer (2)
Determine the empirical formula for each cation-anion pair by balancing the charges.
Write the empirical formula with the cation first and the anion second.
Name the compound by stating the name of the cation followed by the name of the anion, modified to end in '-ide'.
The empirical formulas and names are: B a 3 P 2 (Barium phosphide), B a 3 N 2 (Barium nitride), B a S (Barium sulfide), and B a O (Barium oxide).
Explanation
Understanding the Problem We are given a table with B a 2 + as the cation and P 3 − , N 3 − , S 2 − , O 2 − as anions. We need to determine the empirical formula and name of the ionic compound formed by each cation-anion pair.
Solution Plan For each cation-anion pair, we need to determine the smallest whole number ratio of ions that results in a neutral compound. Then, we write the empirical formula based on the ion ratio, with the cation listed first. Finally, we name the compound by stating the name of the cation followed by the name of the anion, with the anion name modified to end in '-ide'.
Barium and Phosphorus
Barium and Phosphorus:
Barium ion: B a 2 +
Phosphide ion: P 3 − To form a neutral compound, we need to balance the charges. The least common multiple of 2 and 3 is 6. Therefore, we need 3 barium ions ( 3 × + 2 = + 6 ) and 2 phosphide ions ( 2 × − 3 = − 6 ).
Empirical formula: B a 3 P 2
Name of compound: Barium phosphide
Barium and Nitrogen
Barium and Nitrogen:
Barium ion: B a 2 +
Nitride ion: N 3 − Similar to the previous case, we need to balance the charges. The least common multiple of 2 and 3 is 6. Therefore, we need 3 barium ions ( 3 × + 2 = + 6 ) and 2 nitride ions ( 2 × − 3 = − 6 ).
Empirical formula: B a 3 N 2
Name of compound: Barium nitride
Barium and Sulfur
Barium and Sulfur:
Barium ion: B a 2 +
Sulfide ion: S 2 − In this case, the charges are equal in magnitude but opposite in sign. Therefore, we need 1 barium ion ( 1 × + 2 = + 2 ) and 1 sulfide ion ( 1 × − 2 = − 2 ).
Empirical formula: B a S
Name of compound: Barium sulfide
Barium and Oxygen
Barium and Oxygen:
Barium ion: B a 2 +
Oxide ion: O 2 − Similar to the previous case, the charges are equal in magnitude but opposite in sign. Therefore, we need 1 barium ion ( 1 × + 2 = + 2 ) and 1 oxide ion ( 1 × − 2 = − 2 ).
Empirical formula: B a O
Name of compound: Barium oxide
Final Answer Here's the completed table:
cation
anion
empirical formula
name of compound
B a 2 +
P 3 −
B a 3 P 2
Barium phosphide
B a 2 +
N 3 −
B a 3 N 2
Barium nitride
B a 2 +
S 2 −
B a S
Barium sulfide
B a 2 +
O 2 −
B a O
Barium oxide
Examples
Ionic compounds are everywhere! For example, sodium chloride (NaCl), or table salt, is an ionic compound that we use every day to season our food. Another example is calcium carbonate (CaCO3), which is the main component of limestone and is used in building materials and as an antacid. Understanding how ionic compounds are formed and named helps us to identify and utilize these materials effectively in various applications.
The empirical formulas and names for the given ionic compounds formed from B a 2 + and various anions are as follows: B a 3 P 2 (Barium phosphide), B a 3 N 2 (Barium nitride), B a S (Barium sulfide), and B a O (Barium oxide). This result is found by balancing the ionic charges appropriately to form neutral compounds. Each compound's name reflects the cation followed by the modified name of the anion.
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