Question

Given the electron dot diagram, answer the following question. The electrons in the bond between hydrogen and fluorine are more strongly attracted to the atom of:

A. Hydrogen, which has the higher electronegativity.

B. Fluorine, which has the high electronegativity.

C. Hydrogen, which has the lower electronegativity.

D. Fluorine, which has the lower electronegativity.

Answer 1

The electrons in the hydrogen-fluorine bond are more strongly attracted to fluorine due to its higher electronegativity of 4.0, compared to hydrogen's electronegativity of 2.2, resulting in a polar bond.

Step-by-step explanation:

The electrons in the bond between hydrogen and fluorine are more strongly attracted to the atom of fluorine, which has the higher electronegativity. Fluorine is known in chemistry for having the highest electronegativity of all the elements on the Pauling scale, with a value of 4.0. As opposed to hydrogen, which has an electronegativity of 2.2, the significant difference between the two leads to a polar bond, with the electrons being closer to fluorine. Thus, we can conclude that in the covalent bond of hydrogen fluoride (HF), the electrons are attracted to and spend more time around fluorine, creating a partial negative charge on fluorine (symbolized as δ-) and a partial positive charge on hydrogen (symbolized as δ+).