3.6.1: Bond Strength (Problems)

PROBLEM $$\PageIndex{1}$$

Which bond in each of the following pairs of bonds is the strongest?

a. C–C or $$\mathrm{C=C}$$
b. C–N or $$\mathrm{C≡N}$$
c. $$\mathrm{C≡O}$$
or $$\mathrm{C=O}$$
d. H–F or H–Cl
e. C–H or O–H
f. C–N or C–O

$$\mathrm{C=C}$$

$$\mathrm{C≡N}$$

$$\mathrm{C≡O}$$

H–F

O–H

C–O

PROBLEM $$\PageIndex{2}$$

When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:

The greater bond energy is in the figure on the left. It is the more stable form.

PROBLEM $$\PageIndex{3}$$

Use principles of atomic structure to answer each of the following:1

a. The radius of the Ca atom is 197 pm; the radius of the Ca2+ ion is 99 pm. Account for the difference.
b. Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.

Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol)
K 419 3050
Ca 590 1140

c. The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.

When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius.

Removal of the 4s electron in Ca requires more energy than removal of the 4s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level.