Question

decide whether each pair of elements in the table below will form an ionic compound. if they will, write the empirical formula of the compound formed in the space provided.

element #1	bromine	calcium	rubidium	chlorine
element #2	oxygen	oxygen	fluorine	calcium
forms ionic compound?	∘ yes ∘ no	∘ yes ∘ no	∘ yes ∘ no	∘ yes ∘ no
empirical formula of ionic compound	□	□	□	□

Explanation:

To solve this, we analyze each pair of elements:

1. Bromine (Br) and Oxygen (O)

• Bromine is a non - metal, oxygen is a non - metal. Ionic compounds are formed between metals and non - metals (or polyatomic ions). Since both are non - metals, they form covalent compounds, not ionic. So the answer for “Forms ionic compound?” is no.

2. Calcium (Ca) and Oxygen (O)

• Calcium is a metal (group 2, forms \(Ca^{2+}\) ion), oxygen is a non - metal (forms \(O^{2 - }\) ion).
• For ionic compound formation, the charges must balance. The charge of \(Ca\) is \(+2\) and \(O\) is \(-2\). The ratio of \(Ca\) to \(O\) is \(1:1\) (since \(+2+( - 2)=0\)). So the empirical formula is \(CaO\), and the answer for “Forms ionic compound?” is yes.

3. Rubidium (Rb) and Fluorine (F)

• Rubidium is a metal (group 1, forms \(Rb^{+}\) ion), fluorine is a non - metal (forms \(F^{-}\) ion).
• The charge of \(Rb\) is \(+1\) and \(F\) is \(-1\). The ratio of \(Rb\) to \(F\) is \(1:1\) (since \(+1+( - 1)=0\)). So the empirical formula is \(RbF\), and the answer for “Forms ionic compound?” is yes.

4. Chlorine (Cl) and Calcium (Ca)

• Chlorine is a non - metal (forms \(Cl^{-}\) ion), calcium is a metal (forms \(Ca^{2+}\) ion).
• The charge of \(Ca\) is \(+2\) and \(Cl\) is \(-1\). To balance the charges, we need 2 \(Cl^{-}\) ions for 1 \(Ca^{2+}\) ion (since \(+2 + 2\times(-1)=0\)). So the empirical formula is \(CaCl_{2}\), and the answer for “Forms ionic compound?” is yes.

Final Answers (row - by - row for “Forms ionic compound?” and empirical formula):

1. Bromine & Oxygen: No; (no empirical formula for ionic compound as it's covalent)
2. Calcium & Oxygen: Yes; \(CaO\)
3. Rubidium & Fluorine: Yes; \(RbF\)
4. Chlorine & Calcium: Yes; \(CaCl_{2}\)

Answer:

To solve this, we analyze each pair of elements:

1. Bromine (Br) and Oxygen (O)

• Bromine is a non - metal, oxygen is a non - metal. Ionic compounds are formed between metals and non - metals (or polyatomic ions). Since both are non - metals, they form covalent compounds, not ionic. So the answer for “Forms ionic compound?” is no.

2. Calcium (Ca) and Oxygen (O)

• Calcium is a metal (group 2, forms \(Ca^{2+}\) ion), oxygen is a non - metal (forms \(O^{2 - }\) ion).
• For ionic compound formation, the charges must balance. The charge of \(Ca\) is \(+2\) and \(O\) is \(-2\). The ratio of \(Ca\) to \(O\) is \(1:1\) (since \(+2+( - 2)=0\)). So the empirical formula is \(CaO\), and the answer for “Forms ionic compound?” is yes.

3. Rubidium (Rb) and Fluorine (F)

• Rubidium is a metal (group 1, forms \(Rb^{+}\) ion), fluorine is a non - metal (forms \(F^{-}\) ion).
• The charge of \(Rb\) is \(+1\) and \(F\) is \(-1\). The ratio of \(Rb\) to \(F\) is \(1:1\) (since \(+1+( - 1)=0\)). So the empirical formula is \(RbF\), and the answer for “Forms ionic compound?” is yes.

4. Chlorine (Cl) and Calcium (Ca)

• Chlorine is a non - metal (forms \(Cl^{-}\) ion), calcium is a metal (forms \(Ca^{2+}\) ion).
• The charge of \(Ca\) is \(+2\) and \(Cl\) is \(-1\). To balance the charges, we need 2 \(Cl^{-}\) ions for 1 \(Ca^{2+}\) ion (since \(+2 + 2\times(-1)=0\)). So the empirical formula is \(CaCl_{2}\), and the answer for “Forms ionic compound?” is yes.

Final Answers (row - by - row for “Forms ionic compound?” and empirical formula):

1. Bromine & Oxygen: No; (no empirical formula for ionic compound as it's covalent)
2. Calcium & Oxygen: Yes; \(CaO\)
3. Rubidium & Fluorine: Yes; \(RbF\)
4. Chlorine & Calcium: Yes; \(CaCl_{2}\)