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Chapter 18: Problem 90

The crystal structure of olivine- \(\mathrm{M}_{2} \mathrm{SiO}_{4}(\mathrm{M}=\mathrm{Mg}\) Fe) - can be viewed as a ccp arrangement of oxide ions with silicon(IV) ions in tetrahedral holes and metal ions in octahedral holes. a. What fraction of each type of hole is occupied? b. The unit cells of \(\mathrm{Mg}_{2} \mathrm{SiO}_{4}\) and \(\mathrm{Fe}_{2} \mathrm{SiO}_{4}\) have volumes of \(2.91 \times 10^{-26} \mathrm{cm}^{3}\) and \(3.08 \times 10^{-26} \mathrm{cm}^{3} .\) Why is the volume of \(\mathrm{Fe}_{2} \mathrm{SiO}_{4}\) larger?

Short Answer

Expert verified
Answer: In the olivine structure, the fraction of tetrahedral holes occupied is 1/2, and the fraction of octahedral holes occupied is 2. The volume of Fe鈧係iO鈧� is larger than Mg鈧係iO鈧� because the larger ionic radius of Fe虏鈦� ions causes the crystal lattice to expand to accommodate the larger ions.

Step by step solution

01

Understanding the crystal structure of olivine

Olivine structure can be viewed as a ccp (cubic close-packed) arrangement of oxide ions (O虏鈦�) with silicon(IV) ions (Si鈦粹伜) in tetrahedral holes and metal ions (M虏鈦� = Mg虏鈦� or Fe虏鈦�) in octahedral holes. In a ccp arrangement, there are two tetrahedral holes and one octahedral hole per oxide ion.
02

Calculate the fraction of each type of hole occupied

Since there is one Si鈦粹伜 ion and two M虏鈦� ions per formula unit of M鈧係iO鈧�, the number of tetrahedral holes occupied = number of Si鈦粹伜 ions = 1, and the number of octahedral holes occupied = number of M虏鈦� ions = 2. As there are two tetrahedral and one octahedral hole per oxide ion, the fraction of tetrahedral holes occupied = 1/2, and the fraction of octahedral holes occupied = 2/1 = 2. a. Fraction of tetrahedral holes occupied = 1/2. Fraction of octahedral holes occupied = 2.
03

Explain the volume difference

Fe鈧� has a larger ionic radius (0.92 脜) compared to Mg虏鈦� (0.89 脜). When the M ion is replaced with Fe (which is larger) in an M鈧係iO鈧� crystal, it causes the crystal lattice to expand, accommodating the larger ionic radius of Fe虏鈦�. This results in an increased volume of the unit cell for Fe鈧係iO鈧�. b. The volume of Fe鈧係iO鈧� is larger than Mg鈧係iO鈧� because the larger ionic radius of Fe虏鈦� ions causes the crystal lattice to expand to accommodate the larger ions.

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Key Concepts

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

Olivine
Olivine is a mineral that naturally occurs in the earth's mantle and is known for its distinct green color. It is usually composed of two end-members: forsterite (Mg-rich) and fayalite (Fe-rich). The chemical formula of olivine is generally represented as \( \text{M}_2\text{SiO}_4 \) where \( \text{M} \) can be magnesium (Mg) or iron (Fe). This mineral is important not only in geology but also in various industrial applications due to its unique properties.
  • It's a common mineral in the upper mantle and is also found in some meteorites.
  • Olivine is important for understanding mantle convection, plate tectonics, and volcanic eruptions.
  • Its ability to withstand high temperatures makes it suitable for refractory applications.
Grasping the crystal structure of olivine helps in understanding its properties related to strength, density, and stability under earth's geological processes.
Close-packed Arrangement
The close-packed arrangement, often seen in crystal structures, refers to how atoms, ions, or molecules pack together in a highly efficient manner to minimize unused space. In the case of olivine, the oxide ions \((\text{O}^{2-})\) are packed in a cubic close-packed (ccp) structure. This means that the ions are organized in a way that each oxide ion is surrounded by others in a repeating pattern that maximizes spatial efficiency.
  • The ccp arrangement is sometimes called face-centered cubic (FCC) because of its repeating three-dimensional pattern.
  • Every layer of ions is offset from the one below it, increasing packing efficiency.
  • This specific packing helps in stabilizing the structure of minerals like olivine under high pressures and temperatures.
Understanding this concept is essential in explaining the strength and formation of different mineral structures, as well as their capabilities in various geological conditions.
Tetrahedral Holes
In a crystal lattice, the spaces between the packed ions where other ions can fit are referred to as 'holes.' Tetrahedral holes are those created by four surrounding ions. In olivine's ccp arrangement, the smaller silicon ions (\( \text{Si}^{4+} \)) occupy these tetrahedral holes. This positioning within the structure ensures that silicon ions are securely held in place, contributing to the stability of the mineral.
  • There are two tetrahedral holes for every oxide ion in a typical ccp arrangement.
  • Silicon ions filling these holes account for crucial bonding interactions that define the mineral's properties.
  • The occupation of tetrahedral holes in olivine is half, meaning one out of the two available holes is filled per formula unit.
Understanding how tetrahedral holes are utilized helps clarify the mineral鈥檚 density and strength as well.
Octahedral Holes
Octahedral holes are another type of space found in close-packed arrangements, created by six surrounding ions. These spaces are typically larger than tetrahedral holes. In the structure of olivine, larger metal ions such as magnesium (\( \text{Mg}^{2+} \)) or iron (\( \text{Fe}^{2+} \)) occupy these positions. The formula unit of olivine indicates that two metal ions fit into these holes, effectively utilizing the space.
  • Each oxide ion in a ccp arrangement has one octahedral hole.
  • The fraction of octahedral holes filled in olivine is 2, indicating that this type of hole sees full and additional occupation for the metal ions per formula unit.
  • Occupation of octahedral holes contributes to the mineral鈥檚 volume and affects its density.
Understanding the occupation of these holes helps explain differences in unit cell volumes between different olivine compositions, where ionic radii can induce volumetric changes in the crystal lattice as seen between magnesium and iron-based olivine.

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Most popular questions from this chapter

Some scientists believe that the solid hydrogen that forms at very low temperatures and high pressures may conduct electricity. Is this hypothesis supported by band theory?

Nickel has an fcc unit cell with an edge length of \(350.7 \mathrm{pm}\). Use this information to calculate the radius of a nickel atom.

The bec unit cell of an intermetallic compound has atoms of element \(A\) at the corners of the unit cell and an atom of element \(\mathrm{B}\) at the center of the unit cell. a. What is the formula of the compound? b. What would be the formula if the positions of the two elements were reversed in the unit cell?

An interstitial alloy is prepared from metals \(\mathrm{A}\) and \(\mathrm{B}\), where \(\mathrm{B}\) has the smaller atomic radius. The unit cell of metal \(\mathrm{A}\) is fec. What is the formula of the alloy if \(\mathrm{B}\) occupies (a) all of the octahedral holes; (b) half of the octahedral holes; (c) half of the tetrahedral holes?

In the fullerene known as buckminsterfullerene, molecules of \(\mathrm{C}_{60}\) form a cubic closest-packed array of spheres with a unit cell edge length of \(1410 \mathrm{pm}\). a. What is the density of crystalline \(C_{60} ?\) b. If we treat each \(\mathrm{C}_{60}\) molecule as a sphere of 60 carbon atoms, what is the radius of the \(C_{60}\) molecule? c. \(C_{60}\) reacts with alkali metals to form \(\mathrm{M}_{3} \mathrm{C}_{60}\) (where \(\mathrm{M}=\mathrm{Na} \text { or } \mathrm{K}) .\) The crystal structure of \(\mathrm{M}_{3} \mathrm{C}_{60}\) contains cubic closest-packed spheres of \(C_{60}\) with metal ions in holes. If the radius of a \(\mathrm{K}^{+}\) ion is \(138 \mathrm{pm},\) which type of hole is a \(\mathrm{K}^{+}\) ion likely to occupy? What fraction of the holes will be occupied? d. Under certain conditions, a different substance, \(\mathrm{K}_{6} \mathrm{C}_{60}\) can be formed in which the \(\mathrm{C}_{60}\) molecules have a bcc unit cell. Calculate the density of a crystal of \(\mathrm{K}_{6} \mathrm{C}_{60}\)
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