To determine the type of bond between bromine (Br) and fluorine (F), we must consider their electronegativities. Electronegativity is a measure of an atom’s tendency to attract electrons in a bond.
Bromine has an electronegativity of about 2.96, while fluorine is more electronegative, with a value of around 3.98. The difference in electronegativity between these two elements is:
- Electronegativity difference = 3.98 (F) – 2.96 (Br) = 1.02
In general, the type of bond is classified based on the difference in electronegativity:
- Nonpolar covalent bonds
- Polar covalent bonds
- Ionic bonds
With a difference of 1.02, the bond between Br and F is classified as polar covalent. This means that the electrons are shared between the two atoms, but they are not shared equally. The more electronegative fluorine atom will attract the shared electrons more strongly than the bromine atom, creating a dipole moment where the fluorine has a slight negative charge and the bromine has a slight positive charge.
In summary, the bond between bromine and fluorine is polar covalent due to the difference in their electronegativities, which leads to an unequal sharing of electrons.