Explain, in terms of electronegativity, why an H-F bond is expected to be more polar than an H-I bond.
The electronegativity of an element measures the ability of that element to attract electrons. The EN for F is 4.0, that of I is about 2.7; thus the electrons of the HF bond are attracted more to the F than to the H. That's also true for HI; however, the I has less EN and it doesn't attract as much so the HI bond is less polar than HF.
The concept of electronegativity refers to the ability of an atom to attract electron density towards itself in a covalent bond. In general, electronegativity increases from left to right across a period and decreases from top to bottom within a group on the periodic table.
In the case of comparing the H-F and H-I bonds, fluorine (F) has a higher electronegativity value compared to iodine (I). Fluorine is located on the right side of the periodic table in period 2, while iodine is in period 5.
The difference in electronegativity between the two atoms of a bond determines the polarity of the bond. With a larger electronegativity difference, the bond skew towards the more electronegative atom, creating a partial positive charge on the less electronegative atom and a partial negative charge on the more electronegative atom.
Since the electronegativity difference between fluorine and hydrogen (H-F) is larger than the electronegativity difference between iodine and hydrogen (H-I), the H-F bond is expected to be more polar. This means that the fluorine atom in H-F will have a partial negative charge, and the hydrogen atom will have a partial positive charge. In the H-I bond, the electronegativity difference is smaller, resulting in a less polar bond.
To understand why an H-F bond is expected to be more polar than an H-I bond, we need to consider the concept of electronegativity. Electronegativity is the measure of an atom's ability to attract electrons in a chemical bond. The higher the electronegativity of an atom, the more strongly it attracts the shared electrons towards itself.
In the case of the H-F bond, fluorine (F) has a much higher electronegativity compared to hydrogen (H). Fluorine is the most electronegative element on the periodic table. It has a greater pull on the electron pair in the bond, resulting in a significant charge imbalance. This means that the electron pair is unevenly shared between hydrogen and fluorine.
On the other hand, iodine (I) has a lower electronegativity value compared to hydrogen. This means that the iodine atom has a weaker tendency to attract the electron pair in the H-I bond. As a result, the electron pair is more evenly shared between hydrogen and iodine.
The greater electronegativity difference between hydrogen and fluorine creates a larger dipole moment, which results in a more polar bond. In other words, the charge separation between hydrogen and fluorine is more significant in an H-F bond due to the high electronegativity difference. This difference in electronegativity is what makes the H-F bond more polar compared to the H-I bond.