Problem 38 - Chemistry The Molecular Science

Question

For each pair of bonds, identify the more polar one and use \(\delta+\) or \(\delta-\) to indicate the partial charge on each atom. (a) \(\mathrm{B}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{O}\) (b) \(\mathrm{O}-\mathrm{F}\) and \(\mathrm{O}-\mathrm{Se}\) (c) \(\mathrm{S}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{F}\) (d) \(\mathrm{N}-\mathrm{H}\) and \(\mathrm{N}-\mathrm{F}\)

Step-by-Step Solution

Answer

More polar bonds: (a) B-O, (b) O-Se, (c) B-F, (d) N-F.

Step 1: Determine Electronegativity Values

To decide on the polarity, first identify the electronegativity values of the involved elements: - Electronegativity increases across a period and decreases down a group. - Approximate values are: B (2.0), Cl (3.0), O (3.5), F (4.0), Se (2.4), S (2.5), N (3.0), and H (2.1).

Step 2: Compare Bond Polarities

Determine the difference in electronegativity for each bond to assess polarity. The bond with the larger difference is more polar. Consider each pair:

Step 3: Part (a): B-Cl vs. B-O

- B-Cl: |2.0 - 3.0| = 1.0- B-O: |2.0 - 3.5| = 1.5B-O is more polar than B-Cl. Assign partial charges: for B-O, B: \( \delta+ \), O: \( \delta- \).

Step 4: Part (b): O-F vs. O-Se

- O-F: |3.5 - 4.0| = 0.5- O-Se: |3.5 - 2.4| = 1.1O-Se is more polar than O-F. Assign partial charges: for O-Se, O: \( \delta- \), Se: \( \delta+ \).

Step 5: Part (c): S-Cl vs. B-F

- S-Cl: |2.5 - 3.0| = 0.5- B-F: |2.0 - 4.0| = 2.0B-F is more polar than S-Cl. Assign partial charges: for B-F, B: \( \delta+ \), F: \( \delta- \).

Step 6: Part (d): N-H vs. N-F

- N-H: |3.0 - 2.1| = 0.9- N-F: |3.0 - 4.0| = 1.0N-F is slightly more polar than N-H. Assign partial charges: for N-F, N: \( \delta+ \), F: \( \delta- \).

Key Concepts

Electronegativity

Electronegativity is a fundamental aspect to consider when understanding chemical bond polarity. It refers to an atom's ability to attract and hold onto electrons in a bond.

- Within the periodic table, electronegativity increases from left to right across a period. - It decreases as you move down a group.
The more electronegative an atom is, the more strongly it will attract the bond's electron pair. For instance, in the bonds in our exercise, fluorine ( F ) is a highly electronegative element with a value of 4.0, making it very effective in attracting electrons. Understanding these values helps us predict which atom in a bond will exhibit a partial negative charge.

Partial Charges

Partial charges (\(\delta+\) or \(\delta-\)) arise due to differences in electronegativity between bonded atoms. These charges are not full charges like those found in ions but rather small offsets due to uneven electron distribution.

- An atom with higher electronegativity in a bond gains a partial negative charge (\(\delta-\)).- Conversely, the atom with lower electronegativity takes on a partial positive charge (\(\delta+\)).
For instance, in the B-F bond, boron (B) with a lower electronegativity compared to fluorine (F) will have a partial positive charge. Recognizing these charges is crucial in understanding molecule behavior, such as interactions and reactions.

Polar Bonds

A chemical bond is considered polar when there is a significant difference in electronegativity between the two atoms involved, leading to an uneven distribution of charge. A larger difference results in a more polar bond.

- Polarity in a bond creates dipoles, indicated by an arrow pointing from positive to negative partial charge. - The magnitude of a bond's polarity can be assessed by comparing the differences in electronegativity values.
In the examples from the exercise, B-O bond is more polar than B-Cl because of a greater difference in electronegativity (1.5 vs 1.0). As a result, B-O has a more pronounced separation of charge, affecting the bond's behavior and properties. Polar bonds are essential in understanding molecular polarity, which impacts solubility, boiling and melting points, and intermolecular interactions.