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Chapter 8: Problem 39

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 \(\mathrm{Cl} \quad\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\) b. \(\mathrm{S}\) and \(\mathrm{S} \quad\) e. \(\mathrm{N}\) and \(\mathrm{P}\) c. \(\mathrm{C}\) and \(\mathrm{F} \quad\) f. \(\mathrm{B}\) and \(\mathrm{H}\)

Short Answer

Expert verified
a. Rb and Cl: ionic bond b. S and S: nonpolar covalent bond c. C and F: polar covalent bond d. Ba and S: ionic bond e. N and P: polar covalent bond f. B and H: polar covalent bond

Step by step solution

01

List Electronegativity Values

First, we need to find the electronegativity values of all elements involved in the exercise. These values can be found in a periodic table. Here are the electronegativity values for the elements: - 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 Step 2: Calculate the Electronegativity Differences
02

Calculate the Electronegativity Differences

Subtract the electronegativity values of the two elements in each pair to find the difference: 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: \(|2.04 - 2.20| = 0.16\) Step 3: Determine the Type of Bond
03

Determine the Type of Bond

Based on the electronegativity differences, we can determine the type of bond for each pair: a. Rb and Cl: ionic (\(\Delta EN = 2.34 > 1.7\)) b. S and S: nonpolar covalent (\(\Delta EN = 0 < 0.5\)) c. C and F: polar covalent (\(0.5 \leq \Delta EN = 1.43 \leq 1.7\)) d. Ba and S: ionic (\(\Delta EN = 1.69 > 1.7\)) e. N and P: polar covalent (\(0.5 \leq \Delta EN = 0.85 \leq 1.7\)) f. B and H: polar covalent (\(0.5 \leq \Delta EN = 0.16 \leq 1.7\))

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