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Chapter 3: 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 \(\mathrm{Cl}\) b. \(S\) and \(S\) c. \(C\) and \(F\) d. Ba and S e. \(\mathrm{N}\) and \(\mathrm{P}\) f. \(B\) and \(H\)

Short Answer

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

Step by step solution

01

Finding Electronegativity Values

Find the electronegativity values of Rb and Cl using a periodic table or other resources. For Rb, the value is 0.82, and for Cl, the value is 3.16.
02

Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.16 - 0.82 = 2.34.
03

Determine the Bond Type

As the electronegativity difference is greater than 1.7, the bond between Rb and Cl is ionic. b. S and S
04

Finding Electronegativity Values

Find the electronegativity values of S, which is 2.58.
05

Calculate the Electronegativity Difference

As both elements are the same, the difference is zero (2.58 - 2.58 = 0).
06

Determine the Bond Type

Since the electronegativity difference is less than 0.4, the bond between S and S is covalent. c. C and F
07

Finding Electronegativity Values

Find the electronegativity values of C and F. For C, the value is 2.55, and for F, the value is 3.98.
08

Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.98 - 2.55 = 1.43.
09

Determine the Bond Type

As the electronegativity difference is between 0.4 and 1.7, the bond between C and F is polar covalent. d. Ba and S
10

Finding Electronegativity Values

Find the electronegativity values of Ba and S. For Ba, the value is 0.89, and for S, the value is 2.58.
11

Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 2.58 - 0.89 = 1.69.
12

Determine the Bond Type

Since the electronegativity difference is greater than 1.7, the bond between Ba and S is ionic. e. N and P
13

Finding Electronegativity Values

Find the electronegativity values of N and P. For N, the value is 3.04, and for P, the value is 2.19.
14

Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.04 - 2.19 = 0.85.
15

Determine the Bond Type

As the electronegativity difference is between 0.4 and 1.7, the bond between N and P is polar covalent. f. B and H
16

Finding Electronegativity Values

Find the electronegativity values of B and H. For B, the value is 2.04, and for H, the value is 2.20.
17

Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 2.20 - 2.04 = 0.16.
18

Determine the Bond Type

Since the electronegativity difference is less than 0.4, the bond between B and H is covalent.

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Most popular questions from this chapter

Write Lewis structures for the following. Show all resonancestructures where applicable.a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{N}_{2} \mathrm{O}_{4} \text { exists as } \mathrm{O}_{2} \mathrm{N}-\mathrm{NO}_{2} .\right)\) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{SCN}^{-} .\) )

Write electron configurations for a. the cations \(\mathrm{Sr}^{2+}, \mathrm{Cs}^{+}, \mathrm{In}^{+},\) and \(\mathrm{Pb}^{2+}\) b. the anions \(P^{3-}, S^{2-},\) and \(B r^{-}\)

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