Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements. a. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\) b. \(S\) and \(S\) e. \(\mathrm{N}\) and \(\mathrm{P}\) c. \(C\) and \(\bar{F}\) f. \(\mathrm{B}\) and \(\mathrm{H}\)

Consult a periodic table or reference material to find the electronegativity values for the elements involved in each bond. Here are the values we'll need: Rb = 0.82, Cl = 3.16, Ba = 0.89, S = 2.58, N = 3.04, P = 2.19, C = 2.55, F = 3.98, B = 2.04, H = 2.20

Using the electronegativity values, find the DEN for each given pair: a. Rb and Cl: |0.82 - 3.16| = 2.34 b. S and S: |2.58 - 2.58| = 0 c. C and F: |2.55 - 3.98| = 1.43 d. Ba and S: |0.89 - 2.58| = 1.69 e. N and P: |3.04 - 2.19| = 0.85 f. B and H: |2.04 - 2.20| = 0.16

Now that we have the DEN values for each pair, we can determine the bond type: a. Rb and Cl: DEN = 2.34, which is greater than 1.7, so the bond is ionic. b. S and S: DEN = 0, which is less than 0.4, so the bond is covalent (nonpolar). c. C and F: DEN = 1.43, which is between 0.4 and 1.7, so the bond is polar covalent. d. Ba and S: DEN = 1.69, which is slightly less than 1.7, so the bond is polar covalent. e. N and P: DEN = 0.85, which is between 0.4 and 1.7, so the bond is polar covalent. f. B and H: DEN = 0.16, which is less than 0.4, so the bond is covalent (nonpolar). So, the bond types are as follows: a. Rb and Cl: Ionic bond b. S and S: Covalent (nonpolar) bond c. C and F: Polar covalent bond d. Ba and S: Polar covalent bond e. N and P: Polar covalent bond f. B and H: Covalent (nonpolar) bond