Questions: Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements.
a. Cs and Cl
b. N and N
c. C and F
d. Rb and Br
e. H and H
f. Si and Cl
Transcript text: Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements.
a. Cs and Cl
b. N and N
c. C and F
d. Rb and Br
e. H and H
f. Si and Cl
Solution
Solution Steps
Step 1: Understanding Bond Types
To predict the type of bond between two elements, we need to consider the difference in electronegativity between them. The general guidelines are:
Ionic Bonds: Formed when the electronegativity difference is greater than 2.0. Typically occurs between metals and non-metals.
Covalent Bonds: Formed when the electronegativity difference is 0.0, indicating equal sharing of electrons. Typically occurs between two identical non-metals.
Polar Covalent Bonds: Formed when the electronegativity difference is between 0.5 and 2.0, indicating unequal sharing of electrons.
Step 2: Analyze Each Pair
Let's analyze each pair of elements based on their electronegativity values:
a. Cs and Cl:
Cesium (Cs) is a metal with low electronegativity.
Chlorine (Cl) is a non-metal with high electronegativity.
The electronegativity difference is large, suggesting an ionic bond.
b. N and N:
Both are nitrogen atoms with identical electronegativity.
The electronegativity difference is 0, indicating a covalent bond.
c. C and F:
Carbon (C) and fluorine (F) are both non-metals.
The electronegativity difference is significant but not enough for an ionic bond, suggesting a polar covalent bond.
d. Rb and Br:
Rubidium (Rb) is a metal with low electronegativity.
Bromine (Br) is a non-metal with higher electronegativity.
The electronegativity difference is large, suggesting an ionic bond.
e. H and H:
Both are hydrogen atoms with identical electronegativity.
The electronegativity difference is 0, indicating a covalent bond.
f. Si and Cl:
Silicon (Si) and chlorine (Cl) are both non-metals.
The electronegativity difference is moderate, suggesting a polar covalent bond.