Question

identify the charge of the ions in each compound, then type the formula below:
ca s
calcium sulfide
( type final formula )
al n
aluminum nitride
( type final formula )
answer key

Explanation:

Step1: Determine the charge of calcium ion

Calcium (Ca) is in Group 2 of the periodic table. Elements in Group 2 tend to lose 2 electrons to form cations. So, the charge of calcium ion (\(Ca^{2+}\)) is \(2+\).

Step2: Determine the charge of sulfide ion

Sulfur (S) is in Group 16 of the periodic table. Elements in Group 16 tend to gain 2 electrons to form anions. So, the charge of sulfide ion (\(S^{2 -}\)) is \(2-\).

Step3: Write the formula of calcium sulfide

Using the criss - cross method (the magnitude of the charge of one ion becomes the subscript of the other ion). For \(Ca^{2+}\) and \(S^{2 -}\), the formula is \(CaS\) (since \(\frac{2}{2}=1\) for both sub - scripts).

Step4: Determine the charge of aluminum ion

Aluminum (Al) is in Group 13 of the periodic table. Elements in Group 13 tend to lose 3 electrons to form cations. So, the charge of aluminum ion (\(Al^{3+}\)) is \(3+\).

Step5: Determine the charge of nitride ion

Nitrogen (N) is in Group 15 of the periodic table. Elements in Group 15 tend to gain 3 electrons to form anions. So, the charge of nitride ion (\(N^{3 -}\)) is \(3-\).

Step6: Write the formula of aluminum nitride

Using the criss - cross method. For \(Al^{3+}\) and \(N^{3 -}\), the formula is \(AlN\) (since \(\frac{3}{3}=1\) for both sub - scripts).

Answer:

For calcium sulfide: \(Ca^{2+}\), \(S^{2 -}\), \(CaS\); For aluminum nitride: \(Al^{3+}\), \(N^{3 -}\), \(AlN\)