Inorganic Chemistry
Inorganic Chemistry
5th Edition
ISBN: 9781292134147
Author: Housecroft, Catherine E.
Publisher: Pearson,
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Chapter 6, Problem 1P

(a)

Interpretation Introduction

Interpretation:The similarities and differences between the coordination number of cubic and hexagonal close packed arrangement should be interpreted.

Concept introduction:Solids have strong intermolecular forces between constituent atoms therefore they have certain pattern in which the repeating unit repeat itself to form complete crystal lattice. This repeating unit is called as unit cells.

(a)

Expert Solution
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Answer to Problem 1P

Both cubic and hexagonal closed packed structures have same coordination number that is 12.

Explanation of Solution

The cubic closest packed structure has ABC….pattern in which third layer is placed in the depression of second layer. Hence, each atom is surrounded by six atoms of same layer with three atoms of first and third layers. Therefore, the coordination number of nickel atom is 12.

The hexagonal packed structure follows the ABAB…. pattern in which the first and third layer have same arrangement of metal atoms. The metal atom of second layer is surrounded by six metal atoms of same layer and three metal atoms of first and third layers respectively. Hence the coordination number of each atom is 12.

Thus, both cubic and hexagonal closed packed structures have same coordination number with different arrangement of atoms.

(b)

Interpretation Introduction

Interpretation: The similarities and differences between the interstitial holes of cubic and hexagonal close packed arrangement should be interpreted.

Concept introduction: Solids have strong intermolecular forces between constituent atoms therefore, they have certain pattern in which the repeating unit repeat itself to form complete crystal lattice. This repeating unit is called as unit cells.

(b)

Expert Solution
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Answer to Problem 1P

Both cubic and hexagonal closed packed structures have octahedral and tetrahedral holes but in case of cubic close packing, octahedral holes are covered by a layer of sphere.

Explanation of Solution

In the layered structure of solids, the atom of one-layer placed in the depression of previous layer. The space between these voids is called interstitial holes. There are two types of holes; tetrahedral holes (surrounded by four atoms) and octahedral holes (surrounded by six atoms).

In both the hexagonal and cubic close packing, there are two tetrahedral holes and one octahedral hole per packing atom but in case of cubic close packing, octahedral holes are covered by a layer of sphere. Thus, both show similarities in number of interstitial holes.

(c)

Interpretation Introduction

Interpretation: The similarities and differences between the unit cells of cubic and hexagonal close packed arrangement should be interpreted.

Concept introduction: Solids have strong intermolecular forces between constituent atoms therefore they have certain pattern in which the repeating unit repeat itself to form complete crystal lattice. This repeating unit is called as unit cells.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

The cubic closed packed structures has three layers of atoms with four atoms in the unit cell whereas the hexagonal close packing has three layers with six atoms in each unit cell.

Explanation of Solution

The unit cell is the basic unit of the solid materials. It is the smallest unit that is composed of certain number of the atoms.

In the hexagonal and cubic close packing arrangement, the number of unit cells are different. The hexagonal packing has six atoms in the unit cell as the repeating unit has two layers of spheres. On the contrary, the cubic close packing consists of four atoms in the unit cell as the repeating unit of has three layers of spheres.

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Students have asked these similar questions
(a) A hypothetical compound MB2 crystallises in the rutile lattice structure. (i) Draw the lattice structure of the unit cell MB2. (ii) Show by calculation that the stoichiometry of the formula is consistent with the structure. (iii) Determine the coordination number of the MB₂ unit cell.
(a) Draw an orthorhombic unit cell, and within that cell a (210) plane. (b) Draw a monoclinic unit cell, and within that cell a (002) plane.
(a) Mark the sites in the unit cell. b (b) List the Miller indices of all the octahedral interstitial sites in the FCC unit cell.

Chapter 6 Solutions

Inorganic Chemistry

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