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Chapter 9: Problem 84

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 conclusion, the instability of CS and S2 molecules is attributed to the weak π bonds formed between sulfur atoms, due to their larger size and less effective orbital overlap compared to oxygen atoms. On the other hand, CO and O2 molecules are more stable because of the strong π bonds formed between oxygen atoms, which have smaller atomic radii and better p orbital overlap.

Step by step solution

01

Understanding the concept of π bonds

π bonds are formed by the sideways overlap of two parallel p orbitals on adjacent atoms. Multiple bonds such as double and triple bonds have one or more π bonds in addition to sigma (σ) bonds. In this exercise, we need to understand how π bond formation is influenced by sulfur and oxygen atoms.
02

Comparing electron configuration and atomic size of sulfur and oxygen atoms

The electronic configurations of sulfur (S) and oxygen (O) atoms are given as: Sulfur: \( [He]3s^2 3p^4 \) Oxygen: \( [He]2s^2 2p^4 \) From the electron configurations, we can see that both elements have their valence electrons in p orbitals. However, sulfur is in the third period while oxygen is in the second period of the periodic table. Atoms in the third period have larger atomic radii and more diffused orbitals as compared to the second period elements, leading to weaker π bond formation in sulfur.
03

Examining the orbital interactions of sulfur and oxygen atoms

The formation of π bonds is influenced by the interaction of the p orbitals in sulfur and oxygen atoms. Due to the larger size of sulfur, the overlap between the p orbitals of sulfur is less effective, resulting in weaker π bonds. Conversely, oxygen atoms have smaller atomic radii, leading to better p orbital overlap and stronger π bond formation.
04

Relating π bond strength to molecular stability

The strength and stability of a molecule are highly dependent on the strength of π bonds. In CO and O2 molecules where oxygen atoms form strong π bonds, the molecules are more stable. However, in CS and S2 molecules, sulfur atoms form weak π bonds due to their larger size and less effective overlap of p orbitals, resulting in unstable molecules.
05

Conclusion

In conclusion, the instability of CS and S2 molecules is attributed to the weak π bonds formed between sulfur atoms, due to their larger size and less effective orbital overlap compared to oxygen atoms. On the other hand, CO and O2 molecules are more stable because of the strong π bonds formed between oxygen atoms, which have smaller atomic radii and better p orbital overlap.

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Most popular questions from this chapter

Describe the bonding in the first excited state of \(\mathrm{N}_{2}\) (the one closest in energy to the ground state) using the molecular orbital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.)

Determine the molecular structure and hybridization of the central atom \(\mathrm{X}\) in the polyatomic ion \(\mathrm{XY}_{3}{ }^{+}\) given the following information: A neutral atom of \(X\) contains 36 electrons, and the element Y makes an anion with a 1 - charge, which has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6}\).

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d\), or \(6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

Consider three molecules: \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). Molecule A has a hybridization of \(s p^{3}\). Molecule B has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule C consists of two \(\sigma\) bonds and two \(\pi\) bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.
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