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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

Describe the bonding in the \(\mathrm{O}_{3}\) molecule and the \(\mathrm{NO}_{2}^{-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in these two species?

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triodide \(\left(\mathrm{N} \mathrm{I}_{3}\right)\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction \((-307 \mathrm{kJ})\) and the enthalpies of formation for $\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \operatorname{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol})\( and \)\mathrm{BF}_{3}(g)(-1136 \mathrm{kJ} / \mathrm{mol}) ?$ b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using 4 moles of IF for every 1 \(\mathrm{mole}\) of \(\mathrm{BN}\) , one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .\) What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CN}^{+} \quad \text { b. CN } \quad \text { c. } \mathrm{CN}^{-} $$

Use the MO model to determine which of the following has the smallest ionization energy: $\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .$ Explain your answer.

The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.
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