For \(\mathrm{HF}\) and \(\mathrm{HBr}, \delta_{\mathrm{H}}=0.29\) and \(0.09\), respectively. a) 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 \(\mathrm{HBr}\). b) Which bond is more polar, the bond in HF or the bond in HBr? Explain your reasoning.

Electronegativity is a fundamental chemical concept used to predict the distribution of electrons in a chemical bond. It represents an atom's ability to attract and hold onto the shared electrons in a covalent bond. The periodic table shows a trend in electronegativity: it increases from left to right across a period and decreases down a group.

The value of electronegativity is unitless and is typically measured on the Pauling scale, where the most electronegative element, fluorine, is assigned the value of 4.0. Elements with higher electronegativities tend to attract the bonding electron pair closer to them, leading to a partial negative charge. In contrast, elements with lower electronegativities end up with a partial positive charge because the shared electrons are further away.

To illustrate, considering the exercise in question, fluorine's electronegativity is 4.0 which is significantly higher than that of bromine (2.8), indicating that fluorine has a stronger pull on the shared electrons when bonded with hydrogen.

A partial charge occurs when electrons in a chemical bond are not shared equally between two atoms, leading to a charge separation. This occurs because the atoms involved in the bond have different electronegativities. Specifically, the more electronegative atom pulls the shared electrons closer to itself, resulting in a partial negative charge (denoted as \( \delta^- \) ), while the less electronegative atom develops a partial positive charge (denoted as \( \delta^+ \) ).

In the context of the given exercise, the hydrogen atom in hydrofluoric acid (HF) has a higher partial positive charge than in hydrogen bromide (HBr). This is due to the substantial electronegativity difference between hydrogen (2.1) and fluorine (4.0), compared to the smaller difference between hydrogen and bromine (2.8). This greater electronegativity difference results in a stronger attraction of electrons towards fluorine, hence a higher \( \delta^+ \) on hydrogen in HF.

Polar bonds are a type of covalent bond where the distribution of electrons between two atoms is uneven due to a difference in their electronegativities. This results in one end of the bond being more negative and the other end more positive, establishing a dipole, or a pair of electric charges separated by a distance.

The electronegativity difference determines the degree of polarity: the greater the difference, the more polar the bond is. In the exercise provided, HF has a bond that is more polar than HBr because fluorine is much more electronegative than bromine, leading to a larger difference in electronegativity with hydrogen. This large difference means that the shared electrons are much closer to fluorine, making the HF bond highly polar. In contrast, the smaller electronegativity difference between hydrogen and bromine results in a less polar HBr bond.