Which of the following compounds are ionic compounds? Check all of the boxes that apply.
$CaCl _2$
$C _6 H _{12} O _6$
MgO
$Na _2 O$
$SiO _2$
Answer (1)
Calculate the electronegativity difference for C a C l 2 : 1.7"> ∣3.2 − 1.0∣ = 2.2 > 1.7 , so it's ionic.
Calculate the electronegativity differences for C 6 H 12 O 6 : ∣2.6 − 2.1∣ = 0.5 , ∣3.4 − 2.6∣ = 0.8 , ∣3.4 − 2.1∣ = 1.3 . All < 1.7, so it's not ionic.
Calculate the electronegativity difference for MgO: 1.7"> ∣3.4 − 1.3∣ = 2.1 > 1.7 , so it's ionic.
Calculate the electronegativity difference for N a 2 O : 1.7"> ∣3.4 − 0.9∣ = 2.5 > 1.7 , so it's ionic.
Calculate the electronegativity difference for S i O 2 : ∣3.4 − 1.9∣ = 1.5 < 1.7 , so it's not ionic.
The ionic compounds are: C a C l 2 , M g O , N a 2 O .
Explanation
Identifying Ionic Compounds We are asked to identify ionic compounds from a given list, using electronegativity differences to guide our decisions. Ionic compounds typically form between metals and nonmetals, involving a significant transfer of electrons. A large electronegativity difference (usually greater than 1.7) suggests an ionic bond.
Analyzing Calcium Chloride ( C a C l 2 )
First, let's consider C a C l 2 . The electronegativity of Ca is 1.0, and the electronegativity of Cl is 3.2. The difference is: ∣3.2 − 1.0∣ = 2.2 Since 2.2 > 1.7, C a C l 2 is considered an ionic compound.
Analyzing Glucose ( C 6 H 12 O 6 )
Next, let's analyze C 6 H 12 O 6 (glucose). We need to consider the electronegativity differences between C, H, and O. The electronegativity of C is 2.6, H is 2.1, and O is 3.4. ∣2.6 − 2.1∣ = 0.5∣3.4 − 2.6∣ = 0.8∣3.4 − 2.1∣ = 1.3 Since all these differences are less than 1.7, C 6 H 12 O 6 is not considered an ionic compound. It's a covalent compound.
Analyzing Magnesium Oxide (MgO) Now, let's examine MgO. The electronegativity of Mg is 1.3, and the electronegativity of O is 3.4. The difference is: ∣3.4 − 1.3∣ = 2.1 Since 2.1 > 1.7, MgO is considered an ionic compound.
Analyzing Sodium Oxide ( N a 2 O )
Let's analyze N a 2 O . The electronegativity of Na is 0.9, and the electronegativity of O is 3.4. The difference is: ∣3.4 − 0.9∣ = 2.5 Since 2.5 > 1.7, N a 2 O is considered an ionic compound.
Analyzing Silicon Dioxide ( S i O 2 )
Finally, let's consider S i O 2 . The electronegativity of Si is 1.9, and the electronegativity of O is 3.4. The difference is: ∣3.4 − 1.9∣ = 1.5 Since 1.5 < 1.7, S i O 2 is not considered an ionic compound. It's a polar covalent compound.
Conclusion Based on our analysis, the ionic compounds are C a C l 2 , MgO, and N a 2 O .
Examples
Understanding ionic compounds is crucial in various fields. For instance, in agriculture, fertilizers like magnesium oxide (MgO) provide essential nutrients to plants. In medicine, calcium chloride ( C a C l 2 ) is used to treat calcium deficiencies. Recognizing these compounds helps us understand their properties and applications in everyday life.
