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Zincblende Structure (ZnS)

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Zincblende (ZnS)
The cubic form of zinc sulfide with Zn2+ and S2- in tetrahedral coordination, similar to diamond structure.
Zincblende (also called sphalerite) is the cubic polymorph of ZnS and is structurally related to diamond. The structure consists of two interpenetrating face-centered cubic (fcc) sublattices - one of Zn2+ and one of S2-, displaced by one-quarter of the body diagonal. Each ion is tetrahedrally coordinated by four ions of the opposite type. The structure belongs to space group F4̄3m with 4 formula units per unit cell. The ABCABC cubic close-packing sequence of sulfur differs from wurtzite's hexagonal ABAB sequence. This structure is characteristic of many important III-V and II-VI semiconductors such as GaAs, InP, CdTe, and HgS.

Problem 1

Question: In the zinc blende (ZnS) structure, how many formula units of ZnS are present per unit cell?

Solution

Calculation:

S²⁻ ions in FCC: 8 × (1/8) + 6 × (1/2) = 4

Zn²⁺ ions in tetrahedral holes: 4 (completely inside)

Formula units = 4 ZnS per unit cell

Problem 2

Question: If the edge length of zinc blende unit cell is 5.41 Å, and Zn²⁺ and S²⁻ touch along the body diagonal of a smaller cube, calculate the sum of ionic radii (r₊ + r₋).

[Hint: Body diagonal of small cube = a√3/2]

Solution

Calculation:

Body diagonal of 1/8th cube = (a/2)√3

This diagonal = 2(r₊ + r₋)

(a/2)√3 = 2(r₊ + r₋)

(5.41/2)√3 = 2(r₊ + r₋)

4.68 = 2(r₊ + r₋)

r₊ + r₋ = 2.34 Å

Problem 3

Question: How is the zinc blende structure related to the diamond structure?

Solution

Answer:

Zinc blende structure is similar to diamond, but with two different atoms.

If all Zn²⁺ and S²⁻ were replaced by carbon atoms, we would get the diamond structure.

Both have tetrahedral coordination and FCC-based structure.

This structural similarity gives zinc blende significant covalent character.

Numerical Problem

Question: In the zincblende (ZnS) structure, sulfur atoms form a face-centered cubic (fcc) arrangement, and zinc atoms occupy alternate tetrahedral holes.

(a) In an fcc unit cell, how many atoms are effectively present per unit cell? (Corner atoms contribute 1/8 each, face-centered atoms contribute 1/2 each)

(b) If there are 4 sulfur atoms per unit cell in zincblende, how many tetrahedral holes are present in total?

(c) Since zinc occupies alternate (half) tetrahedral holes, how many zinc atoms are present per unit cell? Verify that this gives the correct ZnS formula.

Solution

(a) Number of atoms in fcc unit cell:

In an fcc unit cell:

• Corner atoms: 8 corners × (1/8) = 1 atom

• Face-centered atoms: 6 faces × (1/2) = 3 atoms

Total atoms per fcc unit cell = 1 + 3 = 4 atoms

(b) Number of tetrahedral holes:

In any close-packed structure (including fcc):

Number of tetrahedral holes = 2 × number of atoms

Given: 4 sulfur atoms per unit cell

Number of tetrahedral holes = 2 × 4 = 8 tetrahedral holes

(c) Number of zinc atoms per unit cell:

In zincblende, zinc occupies alternate (half) tetrahedral holes

Number of Zn atoms = (1/2) × total tetrahedral holes

Number of Zn atoms = (1/2) × 8 = 4 Zn atoms

Verification:

• Zn atoms per unit cell = 4

• S atoms per unit cell = 4

• Ratio Zn : S = 4 : 4 = 1 : 1 ✓

This confirms the formula ZnS

Note: The unit cell contains 4 formula units of ZnS, which is typical for fcc-based structures.

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