Problem 119
Question
Predict which bond is the most polar $$\begin{array}{ll}{\text { a. } C-O} & {\text { c. } C-C l} \\ {\text { b. } S i-O} & {\text { d. } C-B r}\end{array}$$
Step-by-Step Solution
Verified Answer
Si-O bond is the most polar with an electronegativity difference of 1.54.
1Step 1: Understanding Polarity
Polarity is determined by the difference in electronegativity between two atoms. The greater the difference, the more polar the bond.
2Step 2: Look Up Electronegativity Values
Consult the periodic table for electronegativity values: Carbon (C) is 2.55, Oxygen (O) is 3.44, Silicon (Si) is 1.90, Chlorine (Cl) is 3.16, and Bromine (Br) is 2.96.
3Step 3: Calculate Electronegativity Differences
For bond a (C-O): \[3.44 - 2.55 = 0.89\]For bond b (Si-O): \[3.44 - 1.90 = 1.54\]For bond c (C-Cl): \[3.16 - 2.55 = 0.61\]For bond d (C-Br): \[2.96 - 2.55 = 0.41\]
4Step 4: Compare the Differences
Among the calculated electronegativity differences, bond b (Si-O) has the largest difference at 1.54, indicating the highest polarity among the given options.
Key Concepts
ElectronegativityPeriodic TablePolar Covalent Bonds
Electronegativity
Electronegativity is a key concept in chemistry essential for understanding how atoms interact with one another. It refers to the tendency of an atom to attract shared electrons within a covalent bond. The more electronegative an atom, the stronger its pull on electrons.
Electronegativity values are typically assigned based on a scale developed by Linus Pauling, with fluorine being the most electronegative element at 3.98. Other elements are compared to fluorine to determine their ability to attract electrons.
Electronegativity values are typically assigned based on a scale developed by Linus Pauling, with fluorine being the most electronegative element at 3.98. Other elements are compared to fluorine to determine their ability to attract electrons.
- Atoms with higher electronegativity, like oxygen and fluorine, tend to pull electrons towards themselves more strongly.
- Atoms with lower electronegativity, like sodium and potassium, have less pull on electrons.
Periodic Table
The periodic table is a powerful tool that organizes the chemical elements based on their atomic number, electron configurations, and recurring chemical properties. It is also instrumental in predicting the behaviors of elements, including electronegativity values.
- As you move from left to right across a period, electronegativity values generally increase. This means elements on the right side of the table are usually more electronegative than those on the left.
- Conversely, as you move down a group, the electronegativity values decrease. This is why elements lower in a group, such as bromine, are less electronegative compared to elements higher up, like chlorine.
Polar Covalent Bonds
Polar covalent bonds arise when two atoms with different electronegativities share electrons. When this happens, the shared electrons are pulled closer toward the more electronegative atom, creating a dipole moment characterized by a partial charge distribution across the molecule.
- If the electronegativity difference between two atoms is significant but not enough to transfer electrons completely, a polar covalent bond forms. This is the case in water (\( \text{H}_2\text{O} \)), where oxygen’s higher electronegativity pulls electrons away from hydrogen.
- In contrast, if the electronegativity difference is very small, the bond is non-polar covalent, as seen in molecules like nitrogen gas (\( \text{N}_2 \)), where there is little to no charge separation.
Other exercises in this chapter
Problem 117
For each pair, indicate the more polar bond by circling the negative end of its dipole. $$ \begin{array}{l}{\text { a. } C-S, C-O} \\ {\text { b. } C-F, C-N} \\
View solution Problem 118
For each of the bonds listed, tell which atom is more negatively charged. $$\begin{array}{ll}{\text { a. } C-H} & {\text { c. } C-S} \\ {\text { b. } C-N} & {\t
View solution Problem 120
Rank the bonds according to increasing polarity. $$\begin{array}{ll}{\text { a. } \mathrm{C}-\mathrm{H}} & {\text { d. } \mathrm{O}-\mathrm{H}} \\ {\text { b. }
View solution Problem 121
Refrigerant The refrigerant known as freon-14 is an ozone-damaging compound with the formula CF\(_{4}\) . Why is the CF \(_{4}\) molecule nonpolar even though i
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