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Chapter 19: Problem 33

Discuss the importance of the \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{Si}-\mathrm{Si}\) bond strengths and of \(\pi\) bonding to the properties of carbon and silicon.

Short Answer

Expert verified
In summary, the stronger C-C bond is a result of the smaller size of the C atom compared to Si, allowing it to form tighter and robust covalent bonds. π bonding plays a crucial role in determining the properties of carbon-containing compounds, such as increased reactivity and planar structures. In contrast, silicon does not form stable π bonds due to the weak overlap of p orbitals, resulting in silicon-based compounds exhibiting different properties, such as extended networks of covalent bonds in silica and silicates.

Step by step solution

01

Basic Understanding of Ionic and Covalent Bonds

Ionic and covalent bonds are two of the most common types of chemical bonds. Ionic bonds form when one atom transfers its electron(s) to another atom, which results in the formation of positive and negative ions that are held together by electrostatic forces. Covalent bonds, on the other hand, form when atoms share one or more pairs of electrons with each other to achieve stable electronic configurations.
02

Comparing C-C and Si-Si Bond Strengths

Carbon (C) and silicon (Si) are in the same group in the periodic table, and both can form four covalent bonds with other atoms due to the presence of four valence electrons in their outermost electron shells. However, the C-C bond is stronger than the Si-Si bond due to the smaller size of the C atom compared to the Si atom. The smaller size of C enables its valence electrons to be more tightly held by the nucleus, leading to stronger covalent bonds.
03

Discussing the Role of π Bonding in Carbon-Containing Compounds

In carbon-containing compounds, π bonding plays a crucial role in determining their properties. π bonds are formed by the sideways overlap of p orbitals, which allows the electrons to be shared between the bonding atoms. Compounds that contain π bonds, like alkenes and alkynes, have unique properties due to this type of bonding. For example, they are more reactive compared to their saturated counterparts, have higher bond energies, and often exhibit planar structures that result in delocalization of the π electrons, which gives rise to aromatic properties in some cases (e.g., benzene).
04

Explaining the Lack of π Bonds in Silicon-based Compounds

Silicon, being a larger atom in comparison with carbon, does not form stable or strong π bonds. The reason for this is the larger distance between the p orbitals in Si atoms, which leads to weak overlap, thus resulting in weak π bonds if they are formed. As a result, silicon-based compounds do not have many of the unique properties observed in carbon-containing compounds with π bonding. Instead, most silicon-based compounds form extended networks of covalent bonds, such as silica and silicates, which result in different physical and chemical properties from their carbon-based counterparts.

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Most popular questions from this chapter

Consider element \(113 .\) What is the expected electron configuration for element \(113 ?\) What oxidation states would be exhibited by element 113 in its compounds?

All the Group \(1 \mathrm{A}\) (1) and \(2 \mathrm{A}\) ( 2 ) metals are produced by electrolysis of molten salts. Why?

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{CO}(g)\). b. Silicon tetrachloride is reacted with very pure magnesium, producing silicon and magnesium chloride. c. \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{NaF}(s)\).

Diagonal relationships in the periodic table exist as well as the vertical relationships. For example, Be and Al are similar in some of their properties as are \(\mathrm{B}\) and \(\mathrm{Si}\). Rationalize why these diagonal relationships hold for properties such as size, ionization energy, and electron affinity.

For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom. a. \(\mathrm{KrF}_{2}\) c. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) b. \(\mathrm{KrF}_{4}\) d. \(\mathrm{XeO}_{2} \mathrm{F}_{4}\)
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