Chapter 8, Problem 8.31CE

(a)

Interpretation Introduction

Interpretation:

The hybridization of carbon atoms in carbon nanotubes, and the hybridization of boron and nitrogen atoms in boron nitride nanotubes have to be given.

Answer to Problem 8.31CE

The hybridization of carbon atoms in carbon nanotubes is ${\text{sp}}^{\text{2}}\text{.}$

The hybridization of boron atoms in boron nitride nanotubes is ${\text{sp}}^{\text{2}}\text{.}$

The hybridization of nitrogen atoms in boron nitride nanotubes is ${\text{sp}}^{\text{2}}\text{.}$

Explanation of Solution

Carbon nanotubes are one-dimensional nanomaterial with density ranging from $\text{1}\text{.2-2}\text{.0}\text{g}{\text{cm}}^{\text{-3}}\text{.}$  The hybridization of carbon atoms in carbon nanotubes is ${\text{sp}}^{\text{2}}\text{.}$  Carbon nanotubes shows their electronic properties similar to a metal or semi-conductor.

Boron nitride nanotubes are a polymorph of boron nitride.  They have similar structures to that of carbon nanotubes except that boron and nitride atoms in boron nitride nanotubes replace carbon atoms.  Boron nitride nanotube is an electrical insulator.  The boron and nitrogen atoms in boron nitride nanotubes are ${\text{sp}}^{\text{2}}$ hybridized.

(b)

Interpretation Introduction

Interpretation:

The number of hexagons that must be strung together around the circumference of a nanotube has to be given.

Explanation of Solution

Each unit in the given nanotube contains two hexagons, therefore, sixteen hexagons per circumference.  The number of hexagons that must be strung together around the circumference of a nanotube is eight.

(c)

Interpretation Introduction

Interpretation:

The reason why the hydrogenation of Buckminsterfullerene stops at a certain point has to be explained.

Explanation of Solution

In buckminsterfullerene, the carbon atoms are planar and it is ${\text{sp}}^{\text{2}}$ hybridized.  The curvature of the molecule introduces some strain at the carbon atoms, so that there is a tendency for some of the atoms of carbons undergoes the conversion of ${\text{sp}}^{\text{3}}$ hybridization.  In order to make every atom of carbon ${\text{sp}}^{\text{3}}$ hybridized, more amount of strain has to be introduced into the carbon cage and therefore, after a certain point addition of hydrogen becomes unfavorable.

(d)

Interpretation Introduction

Interpretation:

The structure of buckminsterfullerene has to be explained and the reason why boron nitride cannot form spheres has to be given.

Explanation of Solution

Buckminsterfullerene is a type of fullerene with a formula ${\text{C}}_{\text{60}}\text{.}$  It has a cage-like fused-ring structure (truncated icosahedron), which resembles a soccer ball, that is made of twenty hexagons and twelve pentagons with a carbon atom that has one pi bond and two sigma bonds at each corner of the shape to form a universal vertex.  Each carbon atom in the structure is bonded covalently with three others.

The spherical structures require the formation of five-member rings.  Boron nitride cannot form these rings because they require high-energy boron-boron or nitrogen-nitrogen bonds.

(e)

Interpretation Introduction

Interpretation:

The density of the cubic boron nitride has to be calculated.

Answer to Problem 8.31CE

The density of the cubic boron nitride is $\text{3}\text{.4863}\text{g}{\text{cm}}^{\text{-3}}\text{.}$

Explanation of Solution

Given,

Edge length of the unit cell is $\text{361}\text{.5pm}\text{.}$

The number of atoms in the unit cell of face-centered cubic cell is four.

The mass of four atoms in the unit cell is calculated from the molar mass of boron nitride and Avogadro’s number.

  Mass=$\left(\text{4}\text{atoms}\text{BN}\right)\text{×}\left(\frac{\text{1}\text{mole}}{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}}\right)\text{×}\left(\frac{\text{24}\text{.8}\text{g}}{\text{mole}}\right)\text{=1}{\text{.647×10}}^{\text{-22}}\text{g}$

The volume of the cubic unit cell is the length of the edge raised to its third power.  The volume is calculated as,

  $\text{Volume=}{\left(\text{361}\text{.5pm×}\frac{\text{100}\text{cm}}{{\text{1×10}}^{\text{10}}\text{pm}}\right)}^{\text{3}}\text{=4}{\text{.7241×10}}^{\text{-23}}{\text{cm}}^{\text{3}}$

The density is calculated as,

  $\begin{array}{l}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Density=}\frac{\text{1}{\text{.647×10}}^{\text{-22}}\text{g}}{\text{4}{\text{.7241×10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\\ \text{Density=3}\text{.4863}\text{g}{\text{cm}}^{\text{-3}}\end{array}$

The density of the cubic boron nitride is $\text{3}\text{.4863}\text{g}{\text{cm}}^{\text{-3}}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The form of boron nitride that is favored at high pressures has to be given.

Answer to Problem 8.31CE

At high pressures, the cubic form of boron nitride is more favored.

Explanation of Solution

Cubic boron nitride is formed by high pressure, high temperature treatment of hexagonal boron nitride.

Cubic form of boron nitride is analogue to structure of diamond.

Hexagonal form of boron nitride is analogue to structure of graphite.

At high pressures, the cubic form of boron nitride is more favored since it is analogous to the structure of diamond.