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The most stable form of carbon at high temperature is \(X\). The \(C\) - C bond length in diamond is \(Y\) while C - C bond length in graphite is \(Z\) What are \(X, Y\) and \(Z\) respectively? (a) Graphite, \(1.42 \AA, 1.54 \AA\) (b) Coke, \(1.54 \AA, 1.84 \AA\) (c) Diamond, \(1.54 \AA, 1.42 \AA\) (d) Fullerene, \(1.54 \AA, 1.54 \boldsymbol{A}\)

Short Answer

Expert verified
The most stable form of carbon at high temperature is graphite, so the correct answer is (a) Graphite, with a C-C bond length in diamond of 1.54 Angstroms and a C-C bond length in graphite of 1.42 Angstroms.

Step by step solution

01

Identify the most stable form of carbon at high temperature

Graphite is known to be the most stable form of carbon at high temperature. This eliminates options (b), (c), and (d).
02

Determine the C-C bond length in diamond

In diamond, each carbon atom is tetrahedrally coordinated to four other carbon atoms, and the bond length is known to be relatively short. The C-C bond length in diamond is approximately 1.54 Angstroms.
03

Determine the C-C bond length in graphite

In graphite, each carbon atom is bonded to three other carbon atoms in a plane, forming hexagonal rings. Graphite has a C-C bond length that is slightly different from diamond's, and it's around 1.42 Angstroms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphite Stability
When we talk about graphite's stability, we're looking at carbon's ability to withstand high temperatures without changing its structural integrity or breaking down. At a fundamental level, graphite's superior stability comes down to its unique layered structure, with layers of carbon atoms arranged in a hexagonal lattice.

Each carbon atom forms three strong covalent bonds with three others in the same layer, giving graphite its characteristic flat sheets that easily slide over each other due to the weaker Van der Waals forces between them. This structural feature makes graphite thermodynamically favorable at high temperatures compared to other allotropes of carbon, which contributes greatly to its widespread use in high-temperature applications, such as in refractories and as a lubricant.
Diamond C-C Bond Length
Diamond stands out in the carbon allotrope family due to each carbon atom being tetrahedrally coordinated with four other carbon atoms through strong covalent bonds. This intricate network, often admired for the gemstone's hardness, has a bond length of approximately 1.54 Angstroms.The uniformity and compactness of the bond length across the entire crystal lattice are responsible for diamond's incompressible nature and remarkable physical properties. Although diamond and graphite both consist purely of carbon, it's the variation in the bond lengths that contribute to their differing physical properties, and hence their varied applications in industry and jewelry.
Carbon Allotropes Properties
The properties of carbon allotropes are as fascinating as they are varied. Each allotrope's distinctive bonding arrangements and structures impart unique physical and chemical attributes that are exploited in numerous fields.
  • Graphite: Graphite is known for its electrical conductivity and ability as a lubricant. It is also highly stable at high temperatures, making it useful in applications involving extreme heat.
  • Diamond: Renowned for its hardness and thermal conductivity, diamond is used in cutting tools and for heat sinks in electronics. Its optical transparency also makes it vital in the jewelry industry.
  • Fullerenes and Nanotubes: These structures exhibit special electronic, thermal, and mechanical properties, leading to research into their use in nanotechnology, electronics, and materials science.
Understanding these properties not only drives innovation but also provides the fundamental knowledge necessary for solving problems related to material selection and engineering applications.

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Most popular questions from this chapter

Carbon monoxide acts as a donor and reacts with certain metals to give metal carbonyls. This is due to (a) presence of one sigma and two pi bonds between \(\mathrm{C}\) and \(\mathrm{O}(\cdot \mathrm{C} \equiv \mathrm{O}:)\) (b) presence of a lone pair on carbon atom in CO molecule (c) presence of lone pair on oxygen atom in CO molecule (d) poisonous nature of CO.

Which of the following compounds are formed when \(\mathrm{BCl}_{3}\) is treated with water? (a) \(\mathrm{H}_{3} \mathrm{BO}_{3}\) (b) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{B}_{2} \mathrm{O}_{3}\) (d) \(\mathrm{HBO}_{2}\)

A metal \(X\) reacts with aqueous \(\mathrm{NaOH}\) solution to form \(Y\) and a highly inflammable gas. Solution \(Y\) is heated and \(\mathrm{CO}_{2}\) is poured through it. \(Z\) precipitates out and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is formed. \(Z\) on heating gives \(\mathrm{Al}_{2} \mathrm{O}_{3}\). Identify \(X, Y\) and \(Z\). $$\begin{array}{lll} {\boldsymbol{X}} & {\boldsymbol{Y}} & {\boldsymbol{Z}} \\ (a) \mathrm{Al} &\mathrm{NaAlO}_{2} & \mathrm{Al}(\mathrm{OH})_{3} \\ (b) \mathrm{Al}_{2} \mathrm{O}_{3} & \mathrm{NaAlO}_{2} & \mathrm{Al}_{2} \mathrm{CO}_{3} \\ (c) \mathrm{Al}_{2} \mathrm{O}_{3} & {\left[\mathrm{Na}_{2} \mathrm{AlO}_{2}\right]^{+} \mathrm{OH}^{-}} & \mathrm{Al}(\mathrm{OH})_{3} \\\ (d) \mathrm{Al} & \mathrm{Al}(\mathrm{OH})_{3} & \mathrm{Al}_{2} \mathrm{O}_{3} \end{array}$$

Which of the following is the correct statement about silicones? (a) They are made up of \(\mathrm{SiO}_{4}{ }^{4-}\) units. (b) They are polymers made up of \(R_{2} \mathrm{SiO}\) units. (c) They are water soluble compounds. (d) They are hydrophillic in nature.

In group 13, electronegativity first decreases from B to \(\mathrm{Al}\) and then increases marginally down the group. This is because of (a) non-metallic nature of B (b) discrepancies in atomic size of elements (c) ability of \(\mathrm{B}\) and \(\mathrm{Al}\) to form \(p \pi-p \pi\) multiple bonds (d), irregular trend in electronegativity throughout the periodic table.

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