Problem 42 - Chemistry An Atoms First Approach

Question

Rank the following bonds in order of increasing ionic character: \(\mathrm{N}-\mathrm{O}, \mathrm{Ca}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Br}-\mathrm{Br}, \mathrm{K}-\mathrm{F}\)

Step-by-Step Solution

Answer

The given bonds ranked in order of increasing ionic character are: \(\mathrm{Br}-\mathrm{Br}, \mathrm{N}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Ca}-\mathrm{O}, \mathrm{K}-\mathrm{F}\).

Step 1: Know Electronegativity Values

We must find the electronegativity values for the elements involved in the given bonds: - Nitrogen (N): 3.04 - Oxygen (O): 3.44 - Calcium (Ca): 1.00 - Carbon (C): 2.55 - Fluorine (F): 3.98 - Bromine (Br): 2.96 - Potassium (K): 0.82

Step 2: Calculate Electronegativity Difference

Now, we need to find the difference in electronegativity values for each bond: - N-O: \(|3.44-3.04|\) = 0.40 - Ca-O: \(|3.44-1.00|\) = 2.44 - C-F: \(|3.98-2.55|\) = 1.43 - Br-Br: \(|2.96-2.96|\) = 0 - K-F: \(|3.98-0.82|\) = 3.16

Step 3: Rank Bonds

Rank the bonds in order of increasing ionic character based on the electronegativity differences calculated in Step 2: 1. Br-Br (0.00) 2. N-O (0.40) 3. C-F (1.43) 4. Ca-O (2.44) 5. K-F (3.16) So, the given bonds ranked in order of increasing ionic character are: \(\mathrm{Br}-\mathrm{Br}, \mathrm{N}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Ca}-\mathrm{O}, \mathrm{K}-\mathrm{F}\).

Key Concepts

Electronegativity

Electronegativity is a measure of how strongly an atom attracts electrons in a chemical bond. The higher the electronegativity value, the stronger an atom pulls on electrons.
These values help predict bond types and the ionic character of compounds.

• Fluorine has the highest electronegativity at 3.98, making it a powerful electron attractor.
• Elements like potassium (0.82) and calcium (1.00) have low electronegativity.

Understanding electronegativity is crucial for determining how atoms will interact in molecules.

Chemical Bonds

Chemical bonds are connections between atoms, created by the sharing or transferring of electrons. There are different types of bonds based on how the electrons are distributed:

• Covalent Bonds: Electrons are shared equally or nearly equally between atoms, as seen in the Br-Br bond.
• Ionic Bonds: Electrons are transferred from one atom to another, resulting in opposite charges that attract, seen in the K-F bond.

Whether a bond is ionic, covalent, or somewhere in between depends largely on the electronegativity of the atoms involved.

Electronegativity Difference

The electronegativity difference between two atoms in a bond indicates how electrons are distributed:

• When the difference is 0, like in Br-Br, the bond is purely covalent.
• A small difference, like 0.40 in N-O, suggests a polar covalent bond, where electrons are shared unequally.
• Large differences, like 3.16 in K-F, lead to ionic bonds, with electrons transferred to the more electronegative atom.

The greater the electronegativity difference, the higher the ionic character of the bond.