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In graphite, electrons are: (a) Spread out between the structure (b) Localized on earth C-atom (c) Present in anti-bonding orbital (d) Localized on every third C-atom

Short Answer

Expert verified
Electrons in graphite are spread out between the structure (option a).

Step by step solution

01

Understanding Graphite's Structure

Graphite is made up of layers of carbon atoms arranged in a hexagonal lattice. Within each layer, carbon atoms are bonded with three others, forming a planar structure. This creates a network of overlapping p-orbitals, where electrons can move freely.
02

Delocalization of Electrons

The electrons in graphite are not confined to a single carbon atom. Instead, the overlapping p-orbitals form a 'sea' of electrons, which are delocalized across the plane of carbon atoms. This delocalization allows electrons to move easily, contributing to graphite's electrical conductivity.
03

Considering the Multiple Choice Options

Option (a) 'Spread out between the structure' best describes the state of electrons in graphite as they are delocalized. Options (b), (c), and (d) describe situations that do not match the electronic structure of graphite, as they imply localization or incorrect orbital placement of electrons.

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Key Concepts

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

Carbon Atoms in Graphite
In graphite, carbon atoms play a crucial role in creating its unique structure. These atoms are arranged in a hexagonal lattice, forming a two-dimensional plane. Each carbon atom in this plane forms strong covalent bonds with three other carbon atoms. These bonds result in graphite's flat, sheet-like structure. The arrangement permits each carbon atom to have three bonds, leaving one electron free to participate in electrical conduction. This structure not only makes graphite a good lubricant but also contributes to its excellent electrical conductivity. The layers in graphite can slide past each other easily, giving graphite its characteristic softness and slippiness, which is why it’s often used as a lubricant or in pencil leads.
Delocalized Electrons
A distinctive feature of graphite is the presence of delocalized electrons. In the carbon lattice, the electrons are not confined to individual carbon atoms but are free to move across the plane of atoms. Because of the structure of graphite, the overlapping of p-orbitals allows these electrons to delocalize. Think of these electrons as a 'sea' or 'cloud' that flows over and between the carbon layers. This delocalization is a crucial factor that explains the good electrical conductivity in graphite. Delocalized electrons can move freely across the layers, carrying charge over longer distances compared to situations where electrons are localized on individual atoms.
Electrical Conductivity in Graphite
The ability of graphite to conduct electricity is directly linked to its structure and the presence of delocalized electrons. When an electric field is applied, these electrons can pick up energy and move, creating an electrical current. This property makes graphite useful in a variety of applications, from electrodes in batteries to brushes in electric motors. The planar structure of graphite means these electrons can move with less resistance, enhancing conductivity. As electrons move freely along the plane, they transfer electric charges through the material with high efficiency.
The Role of P-Orbitals
In graphite, p-orbitals are essential for its electrical properties. Each carbon atom uses its p-orbital to overlap with those of adjacent atoms. This overlap creates a continuous, delocalized 'network' of electrons above and below the plane of carbon atoms. This network forms part of the 'sea' of electrons mentioned earlier. The overlapping p-orbitals are responsible for the electron delocalization in graphite, allowing electrons to roam freely over the sheets. Hence, understanding p-orbitals is key to grasping why graphite's structure allows for such efficient electron movement and consequently, electrical conductivity.

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Most popular questions from this chapter

\(\mathrm{X}\) reacts with aqueous \(\mathrm{NaOH}\) solution to form \(\mathrm{Y}\) and \(\mathrm{H}_{2}\). Aqueous solution of \(\mathrm{Y}\) is heated to \(323-333 \mathrm{~K}\) and on passing \(\mathrm{CO}_{2}\) into it, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Z}\) were formed. When \(Z\) is heated to \(1200^{\circ} \mathrm{C}, \mathrm{Al}_{2} \mathrm{O}_{3}\) is formed. \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) respectively are (a) \(\mathrm{Al}, \mathrm{Al}(\mathrm{OH})_{3}, \mathrm{AlCl}_{3}\) (b) \(\mathrm{Al}, \mathrm{NaAlO}_{2}, \mathrm{Al}(\mathrm{OH})_{3}\) (c) \(\mathrm{Zn}, \mathrm{Na}_{2} \mathrm{ZnO}_{2}, \mathrm{Al}(\mathrm{OH})_{3}\) (d) \(\mathrm{Al}, \mathrm{AlCl}_{3}, \mathrm{NaAlO}_{2}\)

The first ionization enthalpies of \(\mathrm{Li}, \mathrm{Be}, \mathrm{B}\) and \(\mathrm{C}\) are given below. Which of these corresponds to Boron? (a) \(520 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(1086 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(800 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(899 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Ionisation of boric acid in aqueous medium gives which one of the following? (a) \(\left[\mathrm{BO}_{3}\right]^{3-}\) (b) \(\left[\mathrm{B}(\mathrm{OH})_{4}\right]\) (c) \(\left[\mathrm{B}(\mathrm{OH})_{2} \mathrm{O}\right]^{-}\) (d) \(\left[\mathrm{B}(\mathrm{OH}) \mathrm{O}_{2}\right]^{2-}\)

\(\mathrm{BCl}_{3}\) is a planar molecule because in this molecule boron is (a) Sp ^{2} - \text { hybridised } (b) Sp-hybridised (c) Sp \(^{3}\) -hybridised (d) Sp \(^{3}\) d-hybridised

\(\mathrm{BCl}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{X}\) The products formed in above reaction are (a) \(\mathrm{B}_{2} \mathrm{H}_{6}+\mathrm{HCl}\) (b) \(\mathrm{B}_{2} \mathrm{O}_{3}+\mathrm{HOCl}\) (c) \(\mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{HCl}\) (d) No reaction

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