Let's complete the first row of the table by filling in the formula for the ionic compound formed between each cation and the oxide ion [tex]\(O^{2-}\)[/tex].
To form a neutral compound, the total positive charge must balance the total negative charge. This is done by combining the appropriate number of cations and anions so that their charges cancel out.
1. For [tex]\(Na^+\)[/tex]:
- Sodium ion has a charge of [tex]\(+1\)[/tex].
- Oxide ion has a charge of [tex]\(-2\)[/tex].
- To balance these charges, we need 2 sodium ions for every oxide ion.
- The formula is [tex]\(Na_2O\)[/tex].
2. For [tex]\(Ca^{2+}\)[/tex]:
- Calcium ion has a charge of [tex]\(+2\)[/tex].
- Oxide ion has a charge of [tex]\(-2\)[/tex].
- One calcium ion and one oxide ion is sufficient to balance the charges.
- The formula is [tex]\(CaO\)[/tex].
3. For [tex]\(Fe^{2+}\)[/tex]:
- Iron(II) ion has a charge of [tex]\(+2\)[/tex].
- Oxide ion has a charge of [tex]\(-2\)[/tex].
- One iron(II) ion and one oxide ion is sufficient to balance the charges.
- The formula is [tex]\(FeO\)[/tex].
4. For [tex]\(Al^{3+}\)[/tex]:
- Aluminum ion has a charge of [tex]\(+3\)[/tex].
- Oxide ion has a charge of [tex]\(-2\)[/tex].
- To balance these charges, we need 2 aluminum ions for every 3 oxide ions.
- The formula is [tex]\(Al_2O_3\)[/tex].
So, the completed first row of the table is:
[tex]\[
\begin{array}{|l|l|l|l|l|}
\hline \text{Ion} & Na^+ & Ca^{2+} & Fe^{2+} & Al^{3+} \\
\hline O^{2-} & Na_2O & CaO & FeO & Al_2O_3 \\
\hline
\end{array}
\][/tex]
The answer is:
[tex]\(Na_2O, CaO, FeO, Al_2O_3\)[/tex]
