Which of the following atom pairs would most likely be connected by an ionic bond?
K and Br
S and Br
O and Br
F and Br
Ionic bonds are more prevalent between metals and non-metals, such as KBr. Covalent bonds are more prevalent between non-metals and non-metals such as S and Br or O and Br. Said another way, compounds between elements on different sides of the periodic table (K and Br) usually are ionic; compounds between elements on the same side of the table usually are covalent (F and Br, O and Br, S and Br). To add to that, compounds between elements in the center and either side usually are covalent (CCl4, CH4, SiO2).
The atoms that are most likely to form an ionic bond are those that have a large difference in electronegativity. Electronegativity is a measure of an atom's attraction for electrons in a chemical bond. When two atoms with significantly different electronegativities come together, one atom will attract the electrons more strongly and as a result, the bond will be polarized. This polarization leads to the formation of an ionic bond.
Out of the given atom pairs, the one with the largest difference in electronegativity is F and Br. Fluorine (F) is the most electronegative element, and Bromine (Br) has a moderately high electronegativity. The significant difference in electronegativities indicates that F and Br are likely to form an ionic bond.
To determine which atom pairs would most likely be connected by an ionic bond, we need to consider their electronegativity values. Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond.
In general, ionic bonds form between atoms with a large difference in electronegativity. When the difference is significant, one atom will attract the electrons more strongly, leading to the formation of ions.
Let's look at the electronegativity values:
K (Potassium) = 0.82
Br (Bromine) = 2.96
S (Sulfur) = 2.58
O (Oxygen) = 3.44
F (Fluorine) = 3.98
Now, let's calculate the electronegativity difference for each atom pair:
K and Br: |0.82 - 2.96| = 2.14
S and Br: |2.58 - 2.96| = 0.38
O and Br: |3.44 - 2.96| = 0.48
F and Br: |3.98 - 2.96| = 1.02
The larger the electronegativity difference, the more likely an ionic bond is. Among the given options, the atom pair with the largest electronegativity difference is:
F (Fluorine) and Br (Bromine), with a difference of 1.02.
Therefore, the atom pair F and Br is most likely to be connected by an ionic bond.