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Chapter 9: Problem 16

As compared with \(\mathrm{CO}\) and \(\mathrm{O}_{2}, \mathrm{CS}\) and \(\mathrm{S}_{2}\) are very unstable molecules. Give an explanation based on the relative abilities of the sulfur and oxygen atoms to form \(\pi\) bonds.

Short Answer

Expert verified
In short, CO and O2 are more stable than CS and S2 due to the differences in pi bond formation between oxygen and sulfur atoms. Oxygen atoms, with their smaller size, form stronger pi bonds because of better orbital overlap compared to sulfur atoms. As a result, CO and O2 have stronger pi bonds and greater stability than CS and S2.

Step by step solution

01

Electron Configuration of Oxygen and Sulfur

For the explanation, we first need to identify the electron configurations of oxygen and sulfur. Oxygen is in group 16 and period 2, with an electron configuration of \( 1s^{2} 2s^{2} 2p^{4} \). Sulfur is in group 16 and period 3, with an electron configuration of \( 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{4} \).
02

Atomic Size and Orbital Overlap

The size of the atoms plays a crucial role in determining their abilities to form strong pi bonds. Smaller atoms have their electrons more tightly packed around the nucleus, which allows for better overlap of atomic orbitals and, in turn, the formation of strong pi bonds. Oxygen, being in period 2, is smaller than sulfur, which is in period 3. This means that oxygen can form stronger pi bonds than sulfur.
03

Overlap Between Atomic Orbitals

In order to form a pi bond, there should be a strong overlap between the parallel atomic orbitals. Oxygen has one unpaired electron in each of its two 2p orbitals, allowing it to form a pi bond by overlapping these orbitals with another oxygen atom. In the case of sulfur, its unpaired electrons are in the 3p orbital, which is larger and less effective in orbital overlap than the 2p orbital. This results in a weaker pi bond between sulfur atoms.
04

Comparing Stability

The stability of a molecule is directly proportional to the strength of the pi bonds formed between the atoms. As we have seen, oxygen's smaller atomic size and better orbital overlap result in stronger pi bonds than those formed by sulfur. Therefore, CO and O2 (which have oxygen atoms in their composition) are more stable than CS and S2 (which have sulfur atoms). In conclusion, the difference in stability between CO and O2, and CS and S2 can be attributed to the relative abilities of oxygen and sulfur atoms to form strong pi bonds. Oxygen atoms form stronger pi bonds due to their smaller atomic size and the better overlap of their atomic orbitals, which makes CO and O2 more stable compared to CS and S2.

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Most popular questions from this chapter

Biacetyl and acetoin are added to margarine to make it taste more like butter Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. $$ \begin{array}{l}{\text { a. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{SO}_{3}}\end{array} $$ $$ \text {c} \mathrm{s}_{2} \mathrm{O}_{3}^{2-}\left[\begin{array}{c}{\mathrm{o}} \\\ {\mathrm{s}-\mathrm{s}-\mathrm{o}} \\ {\mathrm{o}} \\\ {\mathrm{o}}\end{array}\right]^{2-} $$ e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}+\)

The diatomic molecule OH exists in the gas phase. OH plays an important part in combustion reactions and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 \(\mathrm{pm}\) and 424.7 \(\mathrm{kJ} / \mathrm{mol}\) respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of a \(p_{z}\) orbital from oxygen and the 1\(s\) orbital of hydrogen (the O-H bond lies along the z axis). a. Draw pictures of the sigma bonding and antibonding molecular orbitals in OH. b. Which of the two MOs has the greater hydrogen 1\(s\) character? c. Can the 2\(p_{x}\) orbital of oxygen form MOs with the 1\(s\) orbital of hydrogen? Explain. d. Knowing that only the 2\(p\) orbitals of oxygen interact significantly with the 1\(s\) orbital of hydrogen, complete the MO energy-level diagram for OH. Place the correct number of electrons in the energy levels. e. Estimate the bond order for OH. f. Predict whether the bond order of \(\mathrm{OH}^{+}\) is greater than, less than, or the same as that of OH. Explain.

Do lone pairs about a central atom affect the hybridization of the central atom? If so, how?

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.
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