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Match the following (a) Diamond (p) sp^{2} - h y b r i d i s a t i o n (b) Graphite (q) \(\mathrm{sp}^{3}\)-hybridisation (c) Fullerene (r) Tetrahedral (d) Norbide (s) Carbide

Short Answer

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(a) - q, r; (b) - p; (c) - p; (d) - s

Step by step solution

01

Understand the Elements

First, identify each element mentioned in the exercise: (a) Diamond, (b) Graphite, (c) Fullerene, and (d) Norbide. These are different forms or compounds involving carbon.
02

Identify Hybridization and Structure

Understand the hybridization and structure associated with each element: - Diamonds feature a tetrahedral structure involving \( \mathrm{sp}^{3} \) hybridization.- Graphite is characterized by \( \mathrm{sp}^{2} \)-hybridization and a layered planar structure.- Fullerene is a molecular compound similar to graphite with carbon atoms forming a cage-like structure, involving \( \mathrm{sp}^{2} \) hybridization.- Norbide is related to carbides.
03

Match Each Element to the Corresponding Letter

Using the understanding from Step 2, create the correct matches:- Diamond (a) relates to both tio \( \mathrm{sp}^{3} \)-hybridization (q) and a tetrahedral structure (r).- Graphite (b) is matched with \( \mathrm{sp}^{2} \)-hybridization (p).- Fullerene (c) also involves \( \mathrm{sp}^{2} \)-hybridization due to its molecular similarity to graphite.- Norbide (d) matches with Carbide (s).

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Key Concepts

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

Diamond Structure
Diamonds are one of the hardest substances known to man, primarily due to their unique and durable structure. At the heart of a diamond's strength is its tetrahedral configuration. This means that each carbon atom is bonded to four other carbon atoms. The bonds are equally spaced in three dimensions, forming a rigid lattice.

This structure is a result of the carbon atoms undergoing \( \mathrm{sp}^{3} \)-hybridization. In \( \mathrm{sp}^{3} \)-hybridization, one s orbital and three p orbitals mix to form four equivalent hybrid orbitals. These orbitals arrange in a tetrahedral configuration, which is an essential attribute that gives diamonds their exceptional hardness and optical properties.
  • The diamond's lattice structure allows it to reflect light with great brilliance.
  • The strong covalent bonds in the lattice make diamonds resistant to physical and chemical wear.
Graphite Properties
Graphite is another fascinating form of carbon, quite different from diamond in structure and properties. It is composed of layers of carbon atoms bonded together by \( \mathrm{sp}^{2} \)-hybridization. In this hybridization, one s orbital mixes with two p orbitals to form three \( \mathrm{sp}^{2} \) hybrid orbitals.

These orbitals form a planar structure, resulting in sheets of carbon atoms arranged in a honeycomb-like lattice. The sheets are layered on top of one another, with weak van der Waals forces holding them together.
  • The layered structure allows the sheets to slide over one another, giving graphite its lubricating properties.
  • This structure also makes graphite a good conductor of electricity, as electrons can move freely between the layers.
Graphite's properties make it an excellent material for pencils, batteries, and even as a lubricant.
Fullerene Structure
Fullerenes are a unique form of carbon, known for their distinctive cage-like molecular structure. Most commonly, they resemble a hollow sphere, ellipsoid, or tube. These structures are comprised of carbon atoms bonded together using \( \mathrm{sp}^{2} \)-hybridization.

The fullerenes, especially the spherical molecules known as buckyballs, feature a combination of hexagons and pentagons. This arrangement allows fullerenes to create their characteristic closed, hollow forms.
  • Fullerenes have potential applications in nanotechnology, electronics, and materials science.
  • Due to their unusual shape, they exhibit unique mechanical and chemical properties, such as high resilience and the ability to trap other atoms inside their cages.
Carbide Chemistry
Carbides are chemical compounds that consist of carbon and a less electronegative element. They exhibit a vast range of properties and uses, depending on the nature of the other element involved.

Commonly, carbides are categorized into three types: ionic, covalent, and metallic. For instance, ionic carbides involve a metal with a strong ionic character, while covalent carbides, like silicon carbide, are known for their hardness and thermal conductivity.
  • Covalent carbides are extremely hard and are used in cutting tools and abrasives.
  • Metallic carbides tend to be good conductors of electricity and are utilized in industrial applications due to their durability and toughness.
Through understanding carbide chemistry, scientists can harness their properties for technological advancements in various fields.

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Most popular questions from this chapter

The oxidizing power of \(\mathrm{HXO}_{4}\) type compounds of halogens is in the order (a) \(\mathrm{HIO}_{4}>\mathrm{HBrO}_{4}>\mathrm{HClO}_{4}\) (b) \(\mathrm{HIO}_{4}>\mathrm{HClO}_{4}>\mathrm{HBrO}_{4}\) (c) \(\mathrm{HBrO}_{4}>\mathrm{HIO}_{4}>\mathrm{HClO}_{4}\) (d) \(\mathrm{HClO}_{4}>\mathrm{HBrO}_{4}>\mathrm{HIO}_{4}\)

Match the following: List I List II 1\. most electronegative (i) \(\mathrm{I}_{2}\) 2\. most electron affinity (ii) At 3\. liquid halogen (iii) \(\mathrm{F}_{2}\) 4\. radio active halogen (iv) \(\mathrm{Cl}_{2}\) (v) \(\mathrm{Br}_{2}\) The correct matching is: \(\begin{array}{llll}1 & 2 & 3 & 4\end{array}\) (a) (iii) (iv) (v) (ii) (b) \(\begin{array}{ll}\text { (iii) (iv) } & \text { (v) }\end{array}\) (i) (c) (ii) (i) (iii) (iv) (d) \(\begin{array}{llll}\text { (iii) (i) } & \text { (ii) } & \text { (iv) }\end{array}\)

The decreasing values of bond angles from \(\mathrm{NH}_{3}\) \(\left(106^{\circ}\right)\) to \(\mathrm{SbH}_{3}\left(101^{\circ}\right)\) down group 15 of the periodic table is due to (a) increasing bp-bp repulsion (b) increasing p orbital character in sp \(^{3}\) (c) decreasing lp-bp repulsion (d) decreasing electronegativity

The ease of liquefaction of noble gases increases in the order (a) \(\mathrm{Ar}>\mathrm{Kr}>\mathrm{Xe}>\mathrm{Ne}>\mathrm{He}\) (b) \(\mathrm{Xe}>\mathrm{Kr}>\mathrm{Ne}>\mathrm{Ar}>\mathrm{He}\) (c) \(\mathrm{Kr}>\mathrm{Xe}>\mathrm{He}>\mathrm{Ne}>\mathrm{Ar}\) (d) \(\mathrm{He}>\mathrm{Ne}>\mathrm{Ar}>\mathrm{Kr}>\mathrm{Xe}\)

The elements which occupy the peaks of ionization energy curve are (a) \(\mathrm{Cl}, \mathrm{Br}, \mathrm{I}, \mathrm{F}\) (b) \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}\) (c) \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Cl}, \mathrm{I}\) (d) \(\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}\)

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