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Chapter 4: Problem 116

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
The instability of CS and S2 as compared to CO and O2 can be attributed to the differences in the ability of oxygen and sulfur atoms to form π bonds. Oxygen atoms have a greater ability to form strong π bonds due to their unpaired electrons in two half-filled p orbitals and smaller atomic size, which allows for better orbital overlaps and stronger bonding. In contrast, sulfur has fewer unpaired electrons in its p orbitals and a larger atomic size, leading to weaker bonding and instability in molecules like CS and S2.

Step by step solution

01

Electron Configurations of Oxygen and Sulfur

First, we need to recall the electron configurations of oxygen and sulfur. Oxygen, atomic number 8, has an electron configuration of: 1s² 2s² 2p⁴. Sulfur, atomic number 16, has an electron configuration of: 1s² 2s² 2p⁶ 3s² 3p⁴. In both cases, the p orbitals play a crucial role in the formation of π bonds. Oxygen has two half-filled p orbitals remaining, while sulfur has two half-filled and one completely filled p orbitals.
02

Atomic Orbitals Involved in π Bonds

π bonds are formed through the sideways overlap of p orbitals. Due to the presence of unpaired electrons in the p orbitals, oxygen can form two π bonds (in O2) while sulfur can form only one such π bond (in S2). Therefore, sulfur atoms are less capable of forming strong π bonds compared to oxygen.
03

Overlap and Bonding Strength

The strength of a π bond is governed by the extent of overlap between atomic orbitals. Since Oxygen is in the second period and Sulfur is in the third period, the size of the oxygen atom is smaller than the sulfur atom, leading to better orbital overlaps and stronger π bonds in oxygen-containing molecules (CO and O2) as compared to sulfur containing molecules (CS and S2).
04

Conclusion

In summary, the instability of CS and S2 as compared to CO and O2 can be explained by the differences in the formation of π bonds between sulfur and oxygen atoms. Oxygen has a greater ability to form π bonds due to its unpaired electrons and smaller atomic size, which allows for better orbital overlaps and stronger bonding. On the other hand, sulfur has a lower ability to form π bonds due to fewer unpaired electrons and larger atomic size, resulting in weaker bonding and instability of the molecules CS and S2.

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Most popular questions from this chapter

The three NO bonds in \(\mathrm{NO}_{3}^{-}\) are all equivalent in length and strength. How is this explained even though any valid Lewis structure for \(\mathrm{NO}_{3}^{-}\) has one double bond and two single bonds to nitrogen?

Write Lewis structures and predict whether each of the following is polar or nonpolar. a. HOCN (exists as HO-CN) b. cos c. \(\mathrm{XeF}_{2}\) d. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) e. \(\operatorname{SeF}_{6}\) f. \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{C}\) is the central atom.)

Which of the following molecules have net dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule. a. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}, \mathrm{CCl}_{4}\) b. \(\mathrm{CO}_{2}, \mathrm{N}_{2} \mathrm{O}\) c. \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\)

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{aligned} & \mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta E &=375 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) & \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & & \Delta E=158 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Rationalize the difference in the values of \(\Delta E\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? a. \(\mathrm{Li}_{2}\) b. \(C_{2}\) c. \(S_{2}\)
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