Chapter 22: Problem 1

For \(\mathrm{HF}\) and \(\mathrm{HBr}, \delta_{\mathrm{H}}=0.29\) and \(0.09\), respectively. Use electronegativities to explain why the partial charge on \(\mathrm{H}\) in \(\mathrm{HF}\) is more positive than the partial charge on \(\mathrm{H}\) in \(\mathrm{HBr}\).

Short Answer

Expert verified

The partial charge on hydrogen in HF is more positive because fluorine, which is more electronegative than bromine, attracts electrons more strongly. This results in a larger positive charge on the hydrogen in HF than on the hydrogen in HBr.

Step by step solution

Understand the Concept of Electronegativity

Electronegativity refers to the ability of an atom to attract shared electron pairs in a covalent bond. The greater the electronegativity of an atom, the more it tends to attract the shared pair, resulting in polarization and hence the formation of partial positive and negative charges.

Identification of Atoms' Electronegativities

To solve this problem, the electronegativities of Fluorine (\(\mathrm{F}\)) and Bromine (\(\mathrm{Br}\)) need to be identified. Using the tabulated electronegativities, Fluorine is widely known to be the most electronegative atom, with an electronegativity of 3.98, while Bromine has an electronegativity of 2.96.

Analyze the Relationship between Electronegativity and Polarization

The difference in electronegativity between atoms in a molecule generates polarization, leading to the formation of partial positive and negative charges. The atom with the larger electronegativity value attracts more of the charge and gets a partial negative charge, while the other atom gets a partial positive charge.

Explanation via Electronegativity Difference

Since Fluorine (\(\mathrm{F}\)) is more electronegative than Bromine (\(\mathrm{Br}\)), it attracts electrons more strongly than Bromine. Consequently, the polarization is more pronounced for Hydrogen in a \(\mathrm{HF}\) molecule, which implies that it carries a larger positive partial charge in \(\mathrm{HF}\) than in \(\mathrm{HBr}\).

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For \(\mathrm{HF}\) and \(\mathrm{HBr}, \delta_{\mathrm{H}}=0.29\) and \(0.09\), respectively. Use electronegativities to explain why the partial charge on \(\mathrm{H}\) in \(\mathrm{HF}\) is more positive than the partial charge on \(\mathrm{H}\) in \(\mathrm{HBr}\).

Short Answer

Step by step solution

Understand the Concept of Electronegativity

Identification of Atoms' Electronegativities

Analyze the Relationship between Electronegativity and Polarization

Explanation via Electronegativity Difference

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