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Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one-eighth of the tetrahedral holes occupied by one type of metal ion and one-half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by \(\mathrm{Zn}^{2+}, \mathrm{Al}^{3+}\) and \(\mathrm{O}^{2-}\). The simplest formula of such spinel is (a) \(\mathrm{ZnAl}_{2} \mathrm{O}_{4}\) (b) \(\mathrm{Zn}_{2} \mathrm{AlO}_{4}\) (c) \(\mathrm{Zn}_{2} \mathrm{Al}_{3} \mathrm{O}_{4}\) (d) \(\mathrm{ZnAlO}_{2}\)

Short Answer

Expert verified
(ZnAl_2O_4)

Step by step solution

01

- Determine the ratio of tetrahedral to octahedral holes in CCP

In a close-packed structure of oxide ions (abla{O}^{2-}), there are 4 oxide ions per unit cell. This results in 8 tetrahedral holes and 4 octahedral holes in one unit cell, because the number of tetrahedral holes is twice the number of anions, and the number of octahedral holes is equal to the number of anions.
02

- Assign the site occupancy of metal ions

According to the problem, one-eighth of the tetrahedral sites are occupied by Zn2+ ions and one-half of the octahedral sites are occupied by Al3+ ions. There are 8 tetrahedral holes so one-eighth occupancy means 1 Zn2+ ion. There are 4 octahedral holes and one-half occupancy means 2 Al3+ ions.
03

- Calculate the simplest formula

Combining the metal ions with oxide ions present in the CCP, we get: 1 Zn2+ ion + 2 Al3+ ions + 4 O2- ions. The ions must balance the overall charge, so the simplest formula unit would be ZnAl2O4. Charge balance: (1 abla{Zn}^{2+}) + (2 abla{Al}^{3+}) + (4 abla{O}^{2-}) = 0, thus maintaining charge neutrality.

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Key Concepts

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

CCP pattern in crystal lattice
The close cubic packed (CCP) pattern, also known as face-centered cubic (FCC) structure, is a highly efficient way of packing spheres in a crystal lattice. In the CCP structure, oxide ions, for example, are arranged in a sequence where each layer is offset and sits in the gaps of the previous one, creating a repeating ABAB... pattern. Each sphere is surrounded by 12 others, touching each of its neighbors at different points.

The CCP pattern in crystals is significant because it allows us to deduce possible locations for other atoms or ions. In spinel compounds, with oxide ions forming a CCP lattice, the metal ions occupy interstitial sites known as 'holes'. Knowing the arrangement of the oxide ions gives insight into the positions where other types of ions can be situated within the structure, which is crucial for understanding spinel structures and stoichiometry.
Tetrahedral and octahedral holes
Tetrahedral and octahedral holes are two types of interstices present in a CCP lattice where additional smaller ions can fit. Tetrahedral holes are formed by four surrounding spheres in the crystal lattice, creating a tetrahedron-shaped space. In contrast, octahedral holes are surrounded by six spheres, forming an octahedron-shaped gap.

In a unit cell with a CCP pattern, the number of tetrahedral holes is twice that of the anions, while the octahedral holes are equal in number to the anions. In spinel structures, different metal ions preferentially occupy these holes. For example, in a normal spinel, specific metal ions occupy only one-eighth of the available tetrahedral holes and one-half of the octahedral holes. This selective occupation directly influences the formula, physical properties, and behavior of the spinel compound.
Stoichiometry and charge balance
Stoichiometry is the relationship between the quantities of reactants and products in chemical reactions. In a spinel structure, it also refers to the ratio of different ions within the crystal lattice to maintain charge neutrality. Charge balance, an essential aspect of stoichiometry, demands that the total positive charge of the cations must equal the total negative charge of the anions within a compound.

Using the principles of stoichiometry and charge balance, we can determine the simplest formula for a compound, as seen in the spinel exercise. Metal ions are allocated to specific tetrahedral and octahedral positions based on their sizes, charges, and the number of available holes. The formula is obtained by ensuring that the overall charge of the unit cell is zero. This combination of stoichiometry and charge balance in crystal structures leads to unique compositions and predictable properties for materials such as spinels.

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Most popular questions from this chapter

The number of closest neighbours of a carbon atom in graphite is (a) 2 (b) 3 (c) 4 (d) 8

A metal having atomic mass \(60.22\) \(\mathrm{g} /\) mole crystallizes in \(\mathrm{ABCABC} \ldots\) type packing. The density of each metal atom if the edge length of unit cell is \(10 \AA\), is \(\left(N_{\mathrm{A}}=6.022 \times 10^{23}\right)\) (a) \(0.4 \mathrm{~g} / \mathrm{cm}^{3}\) (b) \(40 \mathrm{~g} / \mathrm{cm}^{3}\) (c) \(0.54 \mathrm{~g} / \mathrm{cm}^{3}\) (d) \(54 \mathrm{~g} / \mathrm{cm}^{3}\)

For a certain crystal, the unit cell axial lengths are found to be \(a=5.62 \AA, b=7.41\) \(\AA\) and \(c=10.13 \AA\). The three coordinate axes are mutually perpendicular. The crystal system to which the crystal belongs is (a) tetragonal (b) orthorhombic (c) monoclinic (d) cubic

A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because (a) all the domains get oriented in the direction of magnetic field. (b) all the domains get oriented in the direction opposite to the direction of magnetic field. (c) domains get oriented randomly. (d) domains are not affected by magnetic field.

In an ionic solid \(\mathrm{AB}_{2} \mathrm{O}_{4}\), the oxide ions form CCP. 'A' and 'B' are metal ions in which one is bivalent and another is trivalent (not necessarily in given order). If all the bivalent ions occupy octahedral holes and the trivalent ions occupy tetrahedral and octahedral voids in equal numbers, then the fraction of octahedral voids unoccupied is (a) \(\frac{1}{2}\) (b) \(\frac{3}{4}\) (c) \(\frac{1}{4}\) (d) \(\frac{7}{8}\)

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