Question

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

table headers: element #1, element #2, forms ionic compound? (yes/no radio buttons), empirical formula of ionic compound (text box), name of ionic compound (text box).

rows:

1. element #1: oxygen, element #2: chlorine
2. element #1: strontium, element #2: bromine
3. element #1: oxygen, element #2: potassium
4. element #1: oxygen, element #2: strontium

Explanation:

To solve this, we analyze each pair of elements:

1. Oxygen (O) and Chlorine (Cl)

• Step 1: Check Ionic Compound Formation

Ionic compounds form between metals and nonmetals. Oxygen (nonmetal) and Chlorine (nonmetal) are both nonmetals, so they form a covalent compound, not ionic.
→ Forms ionic compound? No

2. Strontium (Sr) and Bromine (Br)

• Step 1: Check Ionic Compound Formation

Strontium (metal, Group 2) and Bromine (nonmetal, Group 17) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

Sr loses 2 electrons → \( \text{Sr}^{2+} \); Br gains 1 electron → \( \text{Br}^- \).

• Step 3: Balance Charges

To balance \( 2+ \) and \( 1- \), we need 2 Br⁻ for 1 Sr²⁺.
→ Empirical formula: \( \text{SrBr}_2 \)

• Step 4: Name the Compound

Metal (Sr) + nonmetal (Br, “bromide”) → Strontium bromide

3. Oxygen (O) and Potassium (K)

• Step 1: Check Ionic Compound Formation

Potassium (metal, Group 1) and Oxygen (nonmetal, Group 16) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

K loses 1 electron → \( \text{K}^+ \); O gains 2 electrons → \( \text{O}^{2-} \).

• Step 3: Balance Charges

To balance \( 1+ \) and \( 2- \), we need 2 K⁺ for 1 O²⁻.
→ Empirical formula: \( \text{K}_2\text{O} \)

• Step 4: Name the Compound

Metal (K) + nonmetal (O, “oxide”) → Potassium oxide

4. Oxygen (O) and Strontium (Sr)

• Step 1: Check Ionic Compound Formation

Strontium (metal, Group 2) and Oxygen (nonmetal, Group 16) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

Sr loses 2 electrons → \( \text{Sr}^{2+} \); O gains 2 electrons → \( \text{O}^{2-} \).

• Step 3: Balance Charges

\( 2+ \) and \( 2- \) balance in a 1:1 ratio.
→ Empirical formula: \( \text{SrO} \)

• Step 4: Name the Compound

Metal (Sr) + nonmetal (O, “oxide”) → Strontium oxide

Final Answers (Table Format)

Element #1	Element #2	Forms Ionic Compound?	Empirical Formula	Name of Ionic Compound
Strontium	Bromine	Yes	\( \text{SrBr}_2 \)	Strontium bromide
Oxygen	Potassium	Yes	\( \text{K}_2\text{O} \)	Potassium oxide
Oxygen	Strontium	Yes	\( \text{SrO} \)	Strontium oxide

(For each row, mark “Yes”/“No” and fill the formula/name as above.)

Answer:

To solve this, we analyze each pair of elements:

1. Oxygen (O) and Chlorine (Cl)

• Step 1: Check Ionic Compound Formation

Ionic compounds form between metals and nonmetals. Oxygen (nonmetal) and Chlorine (nonmetal) are both nonmetals, so they form a covalent compound, not ionic.
→ Forms ionic compound? No

2. Strontium (Sr) and Bromine (Br)

• Step 1: Check Ionic Compound Formation

Strontium (metal, Group 2) and Bromine (nonmetal, Group 17) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

Sr loses 2 electrons → \( \text{Sr}^{2+} \); Br gains 1 electron → \( \text{Br}^- \).

• Step 3: Balance Charges

To balance \( 2+ \) and \( 1- \), we need 2 Br⁻ for 1 Sr²⁺.
→ Empirical formula: \( \text{SrBr}_2 \)

• Step 4: Name the Compound

Metal (Sr) + nonmetal (Br, “bromide”) → Strontium bromide

3. Oxygen (O) and Potassium (K)

• Step 1: Check Ionic Compound Formation

Potassium (metal, Group 1) and Oxygen (nonmetal, Group 16) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

K loses 1 electron → \( \text{K}^+ \); O gains 2 electrons → \( \text{O}^{2-} \).

• Step 3: Balance Charges

To balance \( 1+ \) and \( 2- \), we need 2 K⁺ for 1 O²⁻.
→ Empirical formula: \( \text{K}_2\text{O} \)

• Step 4: Name the Compound

Metal (K) + nonmetal (O, “oxide”) → Potassium oxide

4. Oxygen (O) and Strontium (Sr)

• Step 1: Check Ionic Compound Formation

Strontium (metal, Group 2) and Oxygen (nonmetal, Group 16) → metal + nonmetal = ionic compound.
→ Forms ionic compound? Yes

• Step 2: Determine Ions

Sr loses 2 electrons → \( \text{Sr}^{2+} \); O gains 2 electrons → \( \text{O}^{2-} \).

• Step 3: Balance Charges

\( 2+ \) and \( 2- \) balance in a 1:1 ratio.
→ Empirical formula: \( \text{SrO} \)

• Step 4: Name the Compound

Metal (Sr) + nonmetal (O, “oxide”) → Strontium oxide

Final Answers (Table Format)

Element #1	Element #2	Forms Ionic Compound?	Empirical Formula	Name of Ionic Compound
Strontium	Bromine	Yes	\( \text{SrBr}_2 \)	Strontium bromide
Oxygen	Potassium	Yes	\( \text{K}_2\text{O} \)	Potassium oxide
Oxygen	Strontium	Yes	\( \text{SrO} \)	Strontium oxide

(For each row, mark “Yes”/“No” and fill the formula/name as above.)