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Quartz Structure (SiO2)

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Quartz (SiO2)
A framework silicate with Si4+ in tetrahedral coordination with O2- ions, forming a continuous 3D network.
Quartz is the most common polymorph of silica (SiO₂) and features a three-dimensional network of SiO₄ tetrahedra connected by shared oxygen atoms at each corner. Each silicon atom is tetrahedrally coordinated by 4 oxygen atoms, while each oxygen bridges two silicon atoms. The structure exists in two enantiomorphic forms (right-handed and left-handed) and belongs to space group P3₂21 or P3₁21. The framework contains helical chains of tetrahedra running parallel to the c-axis, creating open channels. Alpha-quartz is stable below 573°C and transforms to beta-quartz at higher temperatures.

Problem 1

Question: In the quartz (SiO₂) structure, what is the coordination number of silicon and oxygen atoms?

Solution

Answer:

Coordination number of Si = 4 (each Si is surrounded by 4 O atoms in tetrahedral arrangement)

Coordination number of O = 2 (each O bridges between 2 Si atoms)

The coordination ratio is 4:2

Problem 2

Question: How many SiO₄ tetrahedra share each oxygen atom in the quartz structure?

Solution

Answer:

Each oxygen atom is shared by 2 SiO₄ tetrahedra.

This is called corner-sharing, where tetrahedra connect at their corners (oxygen atoms).

This gives the formula SiO₂ (each Si has 4 O, but each O is shared by 2 Si atoms).

Problem 3

Question: What type of 3D network does quartz form?

Solution

Answer:

Quartz forms a framework silicate structure.

The SiO₄ tetrahedra share all corners to create a continuous 3D network.

This framework structure gives quartz its hardness and high melting point.

Numerical Problem

Question: In the quartz structure (SiO₂), each silicon atom is tetrahedrally coordinated by 4 oxygen atoms, and each oxygen atom bridges two silicon atoms.

(a) If a quartz crystal contains 3.01 × 10²³ silicon atoms, calculate the total number of oxygen atoms in the crystal.

(b) How many SiO₄ tetrahedral units are present in this crystal?

(c) Each Si-O bond length in quartz is approximately 1.61 Å. If we consider a single SiO₄ tetrahedron as a regular tetrahedron, calculate the O-O distance (edge length of the tetrahedron).

[Given: For a regular tetrahedron with Si at center and O at vertices, if Si-O = r, then O-O = r√(8/3)]

Solution

(a) Number of oxygen atoms:

Formula of quartz: SiO₂

Ratio of Si : O = 1 : 2

Number of Si atoms = 3.01 × 10²³

Number of O atoms = 2 × (number of Si atoms)

Number of O atoms = 2 × 3.01 × 10²³

Number of O atoms = 6.02 × 10²³

(b) Number of SiO₄ tetrahedral units:

Each SiO₄ tetrahedron has one Si atom at the center

Therefore, number of SiO₄ units = number of Si atoms

Number of SiO₄ units = 3.01 × 10²³

Note: Although each tetrahedron has 4 oxygen atoms, each oxygen is shared between 2 tetrahedra. So the ratio remains SiO₂ in the overall structure (4 oxygens per Si, but each oxygen is ½ owned = 2 oxygens per Si).

(c) O-O distance in tetrahedral unit:

Given: Si-O bond length (r) = 1.61 Å

For a regular tetrahedron: O-O distance = r√(8/3)

O-O distance = 1.61 × √(8/3)

O-O distance = 1.61 × √(2.667)

O-O distance = 1.61 × 1.633

O-O distance = 2.63 Å

Verification: In a regular tetrahedron, the edge length is typically about 1.63 times the distance from center to vertex, which matches our calculation.

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