Calcium has a charge of +2. The chart lists the charges of different ions.

\begin{tabular}{|l|l|}
\hline
Ions & Charge \\
\hline
Oxygen [tex]$(O)$[/tex] & -2 \\
\hline
Phosphorus [tex]$(P)$[/tex] & -3 \\
\hline
Chlorine [tex]$(Cl)$[/tex] & -1 \\
\hline
Magnesium [tex]$(Mg)$[/tex] & +2 \\
\hline
Sodium [tex]$(Na)$[/tex] & +1 \\
\hline
Fluorine [tex]$(F)$[/tex] & -1 \\
\hline
\end{tabular}

Which are possible equations for an ionic compound with calcium?

A. [tex]$CaO$[/tex], [tex]$CaMg$[/tex], or [tex]$CaF_2$[/tex]
B. [tex]$Ca_2Cl$[/tex], [tex]$CaNa$[/tex], or [tex]$CaP$[/tex]
C. [tex]$CaF$[/tex], [tex]$CaMg$[/tex], or [tex]$CaNa$[/tex]
D. [tex]$CaO$[/tex], [tex]$CaF_2$[/tex], or [tex]$CaCl_2$[/tex]



Answer :

To determine which ionic compounds can be formed with calcium (Ca) which has a charge of +2, we need to balance this charge with the charges of the other ions provided in the chart. Let's examine each ion and see how it can pair with calcium.

1. Oxygen (O[tex]\(-2\)[/tex]):
- Calcium has a charge of +2.
- Oxygen has a charge of -2.
- They balance each other perfectly in a 1:1 ratio, forming [tex]\( \text{CaO} \)[/tex].

2. Phosphorus (P[tex]\( \-3\)[/tex]):
- Calcium has a charge of +2.
- Phosphorus has a charge of -3.
- To balance these charges, the formula would theoretically be [tex]\( \text{Ca}_3(\text{P}_2) \)[/tex], but this is not one of the given options.

3. Chlorine (Cl[tex]\(-1\)[/tex]):
- Calcium has a charge of +2.
- Chlorine has a charge of -1.
- It requires two chlorine ions to balance the charge of one calcium ion, forming [tex]\( \text{CaCl}_2 \)[/tex].

4. Magnesium (Mg[tex]\(+2\)[/tex]):
- Both calcium and magnesium have a +2 charge.
- They cannot form an ionic compound together because they are both metals and have the same charge.

5. Sodium (Na[tex]\(+1\)[/tex]):
- Both calcium and sodium have positive charges, so they cannot form an ionic compound together.

6. Fluorine (F[tex]\(-1\)[/tex]):
- Calcium has a charge of +2.
- Fluorine has a charge of -1.
- To balance the calcium ion, we need two fluorine ions, forming [tex]\( \text{CaF}_2 \)[/tex].

By examining the given options:

- Option 1: [tex]\( \text{CaO}, \text{CaMg}, \text{or} \text{CaF}_2 \)[/tex]:
- [tex]\( \text{CaO} \)[/tex] is correct.
- [tex]\( \text{CaMg} \)[/tex] is incorrect (both are metals with positive charges).
- [tex]\( \text{CaF}_2 \)[/tex] is correct.

- Option 2: [tex]\( \text{Ca}_2\text{Cl}, \text{CaNa}, \text{or} \text{CaP} \)[/tex]:
- [tex]\( \text{Ca}_2\text{Cl} \)[/tex] is not a correct formula (Ca would balance with Cl in [tex]\( \text{CaCl}_2 \)[/tex]).
- [tex]\( \text{CaNa} \)[/tex] is incorrect.
- [tex]\( \text{CaP} \)[/tex] is incorrect as the correct formula is [tex]\( \text{Ca}_3(\text{P}_2) \)[/tex].

- Option 3: [tex]\( \text{CaF}, \text{CaMg}, \text{or} \text{CaNa} \)[/tex]:
- [tex]\( \text{CaF} \)[/tex] is incorrect (CaF[tex]\(_2\)[/tex] is the correct formula).
- [tex]\( \text{CaMg} \)[/tex] is incorrect.
- [tex]\( \text{CaNa} \)[/tex] is incorrect.

- Option 4: [tex]\( \text{CaO}, \text{CaF}_2, \text{or} \text{CaCl}_2 \)[/tex]:
- [tex]\( \text{CaO} \)[/tex] is correct.
- [tex]\( \text{CaF}_2 \)[/tex] is correct.
- [tex]\( \text{CaCl}_2 \)[/tex] is correct.

Thus, the correct answer is:

[tex]\[ \text{CaO, CaF}_2, \text{or CaCl}_2 \][/tex]