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Chapter 11: Problem 21

Which member in each pair has the larger dispersion forces: (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S},(\mathbf{b}) \mathrm{CO}_{2}\) or \(\mathrm{CO}, \mathrm{C}\) (c) \(\mathrm{SiH}_{4}\) or \(\mathrm{GeH}_{4} ?\)

Short Answer

Expert verified
In each pair, the members with the larger dispersion forces are (a) H₂S, due to its larger size and greater polarizability, (b) CO₂, because of its higher electron count and increased polarizability, and (c) GeH₄, as it has more electrons leading to stronger charge fluctuations and dispersion forces.

Step by step solution

01

Pair (a): H₂O and H₂S

Let us first analyze the properties of H₂O and H₂S. H₂S is a larger molecule with more electrons, making it more polarizable compared to H₂O. As a result, H₂S will be more prone to charge fluctuations that induce dispersion forces. Therefore, H₂S has larger dispersion forces than H₂O.
02

Pair (b): CO₂ and CO

Now, let's compare CO₂ and CO. CO has 10 electrons, while CO₂ has 22 electrons. Though CO₂ is a linear and non-polar molecule, it has a higher electron count than CO, making it more polarizable. The increased electron count results in stronger charge fluctuations and subsequently larger dispersion forces. Therefore, CO₂ has larger dispersion forces than CO.
03

Pair (c): SiH₄ and GeH₄

Finally, let us analyze and compare SiH₄ and GeH₄. SiH₄ has 20 electrons, while GeH₄ has 32 electrons. Both SiH₄ and GeH₄ have similar structures, but GeH₄ contains more electrons due to germanium's higher atomic number. The increased electron count leads to larger charge fluctuations and, as a result, stronger dispersion forces. Therefore, GeH₄ has larger dispersion forces than SiH₄. In summary, the members with the larger dispersion forces in each pair are (a) H₂S, (b) CO₂, and (c) GeH₄.

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