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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

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In summary, the C-C bond strength and ability to form stable π bonds allow carbon to form diverse and stable compounds with various molecular structures. However, Si-Si bonds are weaker and less diverse due to the larger atomic size and limited π bonding, resulting in extended network structures. Therefore, the bond strengths and π bonding significantly impact the properties and chemical behavior of carbon and silicon.

Step by step solution

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1. The Carbon-Carbon (C-C) bond strength and properties of Carbon

Carbon has a unique ability to form strong, stable covalent C-C bonds due to its small atomic size and half-filled valence shell. As a result, carbon forms a diverse range of compounds with single, double, and triple bonds between carbon atoms. These strong C-C bonds are responsible for the stability of organic compounds and the formation of complex molecular structures.
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2. The Silicon-Silicon (Si-Si) bond strength and properties of Silicon

Silicon, also a group 14 element like carbon, has a larger atomic radius and, therefore, forms longer and weaker Si-Si bonds compared to C-C bonds. Silicon compounds tend to be relatively less stable and show a preference for single Si-Si bonds, resulting in extended network structures in many solid silicon compounds. This difference in bond strength is one reason why silicon does not possess the same level of chemical diversity and complexity as carbon.
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3. Role of π bonding in Carbon

π bonding is the overlap of p orbitals along the inter-nuclear axis, leading to the formation of double and triple bonds between carbon atoms. The presence of π bonds strongly influences the geometry and stability of carbon compounds, allowing carbon to form planar and linear structures in cases of double and triple bonds, respectively. Moreover, π bonds in conjugated systems add stability and unique properties to organic compounds, such as color and reactivity. Therefore, π bonding significantly affects the chemical behavior and characteristics of carbon compounds.
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4. Role of π bonding in Silicon

Silicon also has p orbitals capable of forming π bonds, but due to the larger atomic size and a low degree of p orbital overlap, π bonding in silicon is weaker than in carbon. Consequently, multiple bonds like double and triple bonds between Si atoms are relatively less common and less stable compared to those in Carbon. This limited π bonding in silicon restricts the diversity of compounds formed by silicon compared to carbon.
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5. Conclusion

In summary, the distinct bond strengths of C-C and Si-Si bonds and the role of π bonding play essential roles in determining the properties and chemical behavior of carbon and silicon. Carbon's strong C-C bonds and ability to form stable π bonds contribute to the vast diversity and stability of organic compounds, while silicon's weaker Si-Si bonds and limited π bonding result in fewer complex structures, with an emphasis on extended frameworks and network arrangements in solid-state compounds.

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Most popular questions from this chapter

The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide \(\left(\mathrm{CO}_{2}\right)\), and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right)\). The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between \(7.1\) and \(7.2\). What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

Compare the Lewis structures with the molecular orbital view of the bonding in \(\mathrm{NO}, \mathrm{NO}^{+}\), and \(\mathrm{NO}^{-}\). Account for any discrepancies between the two models.

The heaviest member of the alkaline earth metals is radium (Ra), a naturally radioactive element discovered by Pierre and Marie Curie in \(1898 .\) Radium was initially isolated from the uranium ore pitchblende, in which it is present as approximately \(1.0 \mathrm{~g}\) per \(7.0\) metric tons of pitchblende. How many atoms of radium can be isolated from \(1.75 \times 10^{8} \mathrm{~g}\) pitchblende \((1\) metric ton \(=\) \(1000 \mathrm{~kg}) ?\) One of the early uses of radium was as an additive to paint so that watch dials coated with this paint would glow in the dark. The longest-lived isotope of radium has a half-life of \(1.60 \times 10^{3}\) years. If an antique watch, manufactured in 1925, contains \(15.0 \mathrm{mg}\) radium, how many atoms of radium will remain in \(2025 ?\)

Many structures of phosphorus-containing compounds are drawn with some \(\mathrm{P}=\mathrm{O}\) bonds. These bonds are not the typical \(\pi\) bonds we've considered, which involve the overlap of two \(p\) orbitals. Instead, they result from the overlap of a \(d\) orbital on the phosphorus atom with a \(p\) orbital on oxygen. This type of \(\pi\) bonding is sometimes used as an explanation for why \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has the first structure below rather than the second: Draw a picture showing how a \(d\) orbital and a \(p\) orbital overlap to form a \(\pi\) bond.

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