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Boron nitride can be represented by the given structure. The structure of \(\mathrm{BN}\) is similar to (a) graphite (b) diamond (c) benzene (d) Pyridine.

Short Answer

Expert verified
The structure of \text{BN} is similar to (a) graphite.

Step by step solution

01

Understand the structure of boron nitride

Boron nitride (BN) has a structure where each boron atom is covalently bonded to three nitrogen atoms and vice versa. This forms a planar hexagonal lattice similar to the carbon atoms in graphite.
02

Compare BN structure to the options provided

Both graphite and diamond are allotropes of carbon. Graphite has a hexagonal planar structure whereas diamond has a tetrahedral structure. Benzene is a hydrocarbon with a hexagonal ring and delocalized electrons, while pyridine is a benzene molecule with a nitrogen atom replacing one of the carbon atoms in the ring.
03

Analyze the similarities

Boron nitride's planar and hexagonal lattice structure is most closely related to that of graphite. Unlike diamond, it does not have a three-dimensional tetrahedral structure, and unlike benzene or pyridine, it is not a hydrocarbon with delocalized electrons or a heteroatom in the ring structure.

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Key Concepts

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

Graphite Structure
Graphite, a naturally occurring form of carbon, exhibits a layered structure where carbon atoms are arranged in a hexagonal lattice. In this configuration, each carbon atom forms three covalent bonds with adjacent carbon atoms, creating flat, two-dimensional layers. These layers can slide over each other due to weak van der Waals forces between them, which accounts for graphite's lubricating properties. Graphite's electrical conductivity is facilitated through the delocalized electrons within its layers.

Graphite vs Boron Nitride

The similarity between graphite and boron nitride lies in their two-dimensional hexagonal frameworks. However, in boron nitride, the lattice is made up of alternating boron and nitrogen atoms, rather than just carbon. This subtle difference changes the properties significantly, making boron nitride more chemically and thermally stable than graphite.
Diamond Structure
In stark contrast to graphite, diamond represents an alternate structure of carbon known as a tetrahedral lattice. Each carbon atom in a diamond is covalently bonded to four other carbon atoms, forming a rigid three-dimensional framework. This spatial arrangement of atoms gives diamond its characteristic hardness and high thermal conductivity. In essence, the diamond structure is renowned for being one of the strongest and most durable configurations on Earth.

Diamond vs Boron Nitride

When comparing diamond to boron nitride, it's clear that boron nitride's similarity to graphite's planar structure holds greater than any likeness to diamond's three-dimensional tetrahedral framework. This fundamental difference in geometry between the structures of diamond and boron nitride accounts for their contrasting physical properties.
Allotropes of Carbon
Carbon is an exceptional element capable of forming various structures known as allotropes. Allotropes of carbon include graphite and diamond, as discussed, but also encompass other forms such as amorphous carbon, fullerenes (like C60 'buckyballs'), and nanotubes. The allotropes' varying properties arise due to the different ways carbon atoms can bond to each other and their arrangement in space, from the flat hexagonal lattices of graphite to the robust tetrahedra of diamonds.

Significance in Material Science

The study of carbon allotropes has greatly impacted material science, yielding substances with unique electrical, mechanical, and chemical characteristics, suitable for diverse applications from electronics to aerospace.
Hexagonal Lattice
A hexagonal lattice is a geometrical arrangement of points in a two-dimensional plane where each point has six neighbors at equal distances around it. This pattern resembles a honeycomb or hexagonal tiling and is prevalent in many crystalline materials. The consistency and balance provided by this arrangement often lead to materials that are structurally stable and may exhibit anisotropic properties, meaning their characteristics can vary depending on the direction of measurement.

Application in Advanced Materials

Materials with hexagonal lattice structures, like graphite and boron nitride, are widely used in various high-tech applications, including semiconductors, thermal management systems, and reinforcing materials in composites, highlighting their importance in both fundamental science and technology advancement.

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Most popular questions from this chapter

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline {\text { Column I is }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { Galena } & \text { (i) } & \text { Abrasive } \\\ \hline \text { (B) } & \text { Diamond } & \text { (ii) } & \text { Metal carbonyls } \\ \hline \text { (C) } & \text { Carbon monoxide } & \text { (iii) } & \text { Hydrides of Si } \\ \hline \text { (D) } & \text { Silanes } & \text { (iv) } & \text { An ore of lead } \\ \hline \end{array} $$ (a) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow(\mathrm{ii}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow\) (iii) (b) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow(\mathrm{ii}),(\mathrm{D}) \rightarrow(\mathrm{iii})\) (c) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow(\mathrm{iii}),(\mathrm{D}) \rightarrow\) (iv) (d) (A) \(\rightarrow(\mathrm{i}),(\mathrm{B}) \rightarrow(\mathrm{ii}),(\mathrm{C}) \rightarrow(\mathrm{iii}),(\mathrm{D}) \rightarrow(\mathrm{iv})\)

Silicon is an important constituent of (a) sand (b) atmosphere (c) plants (d) water bodies.

Boric acid is the trival name for (a) orthoboric acid (b) metaboric acid (c) pyroboric acid (d) None of these.

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 \(Z\). $$\begin{array}{lll} {\boldsymbol{X}} & {\boldsymbol{Y}} & {\boldsymbol{Z}} \\ (a) \mathrm{Al} &\mathrm{NaAlO}_{2} & \mathrm{Al}(\mathrm{OH})_{3} \\ (b) \mathrm{Al}_{2} \mathrm{O}_{3} & \mathrm{NaAlO}_{2} & \mathrm{Al}_{2} \mathrm{CO}_{3} \\ (c) \mathrm{Al}_{2} \mathrm{O}_{3} & {\left[\mathrm{Na}_{2} \mathrm{AlO}_{2}\right]^{+} \mathrm{OH}^{-}} & \mathrm{Al}(\mathrm{OH})_{3} \\\ (d) \mathrm{Al} & \mathrm{Al}(\mathrm{OH})_{3} & \mathrm{Al}_{2} \mathrm{O}_{3} \end{array}$$

Maximum ability of catenation is shown by (a) silicon (b) lead (c) germanium (d) carbon.

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