Based on the differences in electronegativities, with which atom is francium (Fr) most likely to form an ionic bond?(1 point)

Question

Responses

Fr–At = 1.50
Fr–At = 1.50

Fr–Cl = 2.46
Fr–Cl = 2.46

Fr–I = 1.96
Fr–I = 1.96

Fr–Se = 1.76

Answer 1

To determine with which atom francium (Fr) is most likely to form an ionic bond, we compare the differences in electronegativities provided:

• Fr–At: 1.50
• Fr–Cl: 2.46
• Fr–I: 1.96
• Fr–Se: 1.76

Ionic bonds typically form between atoms with a large difference in electronegativity (generally greater than 1.7). The largest electronegativity difference listed here is between Fr and Cl (2.46), which suggests that this combination is most likely to form an ionic bond.

Thus, the atom with which francium is most likely to form an ionic bond is chlorine (Cl) due to the highest electronegativity difference.