Question

1. simplify if possible. ionic compound formulas are always written in the empirical formula, meaning they are written with the minimum number of atoms required. if you could balance out the charges with fewer atoms, rewrite the formula. this is the same process as reducing fractions. example name cation anion chemical formula barium sulfide ba²⁺ s²⁻ bas a. barium sulfide is made of ba²⁺ and s²⁻ions. using the crisscross shortcut method, the barium ions charge (2) is equal to the number of sulfur ions, and the sulfur ions charge (2) is equal to the number of barium ions. this gives us the formula ba₂s₂. however, you do not need this many atoms to balance the charges. write the 2 numbers as a fraction and simplify: 2 barium atoms / 2 sulfur atoms = 1/1, so the correct formula is bas. practice exercises write the formula for each of the following compounds. 1. aluminum telluride 2. cadmium carbide 3. zinc chloride 4. silver bromide 5. barium hydride 6. cesium fluoride 7. sodium oxide 8. magnesium arsenide 9. silver phosphide 10. cadmium iodide

Explanation:

Step1: Determine ion charges

Aluminum forms $Al^{3 +}$ and tellurium forms $Te^{2 -}$.

Step2: Apply criss - cross method

The charge of aluminum (3) is the number of tellurium ions and the charge of tellurium (2) is the number of aluminum ions, giving $Al_{2}Te_{3}$.

Step3: For cadmium carbide

Cadmium forms $Cd^{2+}$ and carbide is $C^{4 -}$. Using criss - cross, we get $Cd_{2}C$. Simplifying (dividing by 2), the formula is $Cd_{2}C$.

Step4: For zinc chloride

Zinc forms $Zn^{2+}$ and chlorine forms $Cl^{-}$. Criss - cross gives $ZnCl_{2}$.

Step5: For silver bromide

Silver forms $Ag^{+}$ and bromine forms $Br^{-}$, so the formula is $AgBr$.

Step6: For barium hydride

Barium forms $Ba^{2+}$ and hydrogen forms $H^{-}$, giving $BaH_{2}$.

Step7: For cesium fluoride

Cesium forms $Cs^{+}$ and fluorine forms $F^{-}$, so the formula is $CsF$.

Step8: For sodium oxide

Sodium forms $Na^{+}$ and oxygen forms $O^{2 -}$, giving $Na_{2}O$.

Step9: For magnesium arsenide

Magnesium forms $Mg^{2+}$ and arsenic forms $As^{3 -}$. Criss - cross gives $Mg_{3}As_{2}$.

Step10: For silver phosphide

Silver forms $Ag^{+}$ and phosphorus forms $P^{3 -}$, giving $Ag_{3}P$.

Step11: For cadmium iodide

Cadmium forms $Cd^{2+}$ and iodine forms $I^{-}$, giving $CdI_{2}$.

Answer:

1. $Al_{2}Te_{3}$
2. $Cd_{2}C$
3. $ZnCl_{2}$
4. $AgBr$
5. $BaH_{2}$
6. $CsF$
7. $Na_{2}O$
8. $Mg_{3}As_{2}$
9. $Ag_{3}P$
10. $CdI_{2}$