Voids in Close Packed Structures — Revision Notes

⚡ 30-Second Revision

Close Packing: — HCP (6 atoms/cell), FCC/CCP (4 atoms/cell).
Voids: — Empty spaces in close-packed structures.
Tetrahedral Voids (TV):

- Surrounded by 4 spheres (tetrahedron). - Coordination number: 4. - Number: $2N$ (where $N$ = number of spheres). - Radius ratio ($r_+/r_-$): $0.225$. - FCC location: 8 voids, along body diagonals (1/4th & 3/4th from corners).

Octahedral Voids (OV):

- Surrounded by 6 spheres (octahedron). - Coordination number: 6. - Number: $N$ (where $N$ = number of spheres). - Radius ratio ($r_+/r_-$): $0.414$. - FCC location: 4 voids, at body center (1) and edge centers (3).

Relative Size: — TV < OV.
Formula Derivation: — Calculate $N_{anions}$, then $N_{voids}$, then $N_{cations}$ (based on occupancy), then simplify ratio.

2-Minute Revision

Voids are the interstitial spaces within close-packed crystal structures like HCP and FCC. These are crucial as smaller atoms or ions often occupy them. There are two main types: tetrahedral voids (TVs) and octahedral voids (OVs).

Tetrahedral voids are smaller, surrounded by four spheres, giving them a coordination number of 4. For every 'N' spheres in close packing, there are '2N' tetrahedral voids. In an FCC unit cell (4 effective atoms), there are 8 TVs, located along the body diagonals.

Octahedral voids are larger, surrounded by six spheres, with a coordination number of 6. For 'N' spheres, there are 'N' octahedral voids. In an FCC unit cell, there are 4 OVs, located at the body center and edge centers.

The stability of an ion occupying a void is determined by the radius ratio ($r_+/r_-$): $0.225$ for tetrahedral and $0.414$ for octahedral voids. To determine the formula of a compound, calculate the effective number of lattice atoms, then the number of voids, and finally the number of occupying ions based on the given occupancy fraction, simplifying the ratio to its lowest whole numbers.

5-Minute Revision

Close-packed structures, such as hexagonal close packing (HCP) and cubic close packing (CCP or FCC), are the most efficient ways for identical spheres to arrange themselves. Despite their efficiency, these arrangements always leave specific empty spaces called voids or interstitial sites. These voids are fundamentally important in solid-state chemistry, particularly for understanding ionic compounds where smaller cations often reside within the voids formed by larger anions.

There are two primary types of voids:

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Tetrahedral Voids (TVs): — These voids are formed by four spheres whose centers define a tetrahedron. An atom or ion occupying a tetrahedral void has a coordination number of 4. These voids are relatively small. For any close-packed structure containing 'N' spheres, there are always '2N' tetrahedral voids. In an FCC unit cell (which effectively contains 4 atoms), there are $2 \times 4 = 8$ tetrahedral voids. They are located along the body diagonals, with two voids on each diagonal, at $1/4$th and $3/4$th of the diagonal length from the corners.
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Octahedral Voids (OVs): — These voids are formed by six spheres whose centers define an octahedron. An atom or ion occupying an octahedral void has a coordination number of 6. These voids are larger than tetrahedral voids. For any close-packed structure containing 'N' spheres, there are always 'N' octahedral voids. In an FCC unit cell, there are $1 \times 4 = 4$ octahedral voids. These are located at the body center (1 void) and at the center of each of the 12 edges (each contributing $1/4$th to the unit cell, totaling 3 voids).

Radius Ratio Rule: The stability of a cation in a void is governed by the radius ratio ($r_+/r_-$). For tetrahedral voids, the ideal radius ratio is $0.225$. For octahedral voids, it is $0.414$. These values help predict the coordination geometry. For example, if $r_+/r_- = 0.3$, the cation would likely occupy a tetrahedral void.

Example: A compound consists of element B forming an FCC lattice, and element A occupying half of the octahedral voids. What is the formula?

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Number of B atoms (N): — In FCC, $N = 4$.
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Total Octahedral Voids: — For N atoms, there are N OVs. So, $4$ OVs.
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Number of A atoms: — A occupies $1/2$ of OVs. So, $1/2 \times 4 = 2$ A atoms.
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Ratio A:B: — $2:4$, which simplifies to $1:2$.
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Formula: — AB$_2$.

Prelims Revision Notes

Voids in Close-Packed Structures (NEET Focus)

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Close Packing Basics:

* HCP (Hexagonal Close Packing): Effective number of atoms per unit cell = 6. * FCC / CCP (Cubic Close Packing): Effective number of atoms per unit cell = 4. * Packing efficiency for both is $approx 74$, leaving $26$ empty space (voids).

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Types of Voids:

* Tetrahedral Voids (TVs): * Geometry: Surrounded by 4 spheres, forming a tetrahedron. * Coordination Number: 4. * Number: For 'N' spheres in close packing, there are 2N tetrahedral voids.

* Relative Size: Smaller than octahedral voids. * **Radius Ratio ($r_+/r_-$):** Ideal value $0.225$. Range: $0.225 le r_+/r_- < 0.414$. * Location in FCC: 8 TVs per unit cell. Located along the 4 body diagonals, two on each diagonal (at $1/4$th and $3/4$th of the diagonal length from the corners).

* Octahedral Voids (OVs): * Geometry: Surrounded by 6 spheres, forming an octahedron. * Coordination Number: 6. * Number: For 'N' spheres in close packing, there are N octahedral voids.

* Relative Size: Larger than tetrahedral voids. * **Radius Ratio ($r_+/r_-$):** Ideal value $0.414$. Range: $0.414 le r_+/r_- < 0.732$. * Location in FCC: 4 OVs per unit cell. One at the body center and three at the center of each edge (each edge center contributes $1/4$th to the unit cell, $12 \times 1/4 = 3$).

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Key Relationships & Formulas:

* If 'N' is the number of atoms forming the close-packed lattice: * Number of OVs = N * Number of TVs = 2N * Stoichiometry Derivation: 1. Identify the number of lattice-forming atoms (e.g., anions) per unit cell (N).

2. Calculate the total number of relevant voids (N for OV, 2N for TV). 3. Determine the number of occupying atoms (e.g., cations) based on the given fraction of occupied voids. 4. Find the simplest whole-number ratio of occupying atoms to lattice-forming atoms to get the chemical formula.

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Common Examples:

* NaCl: Cl$^-$ forms FCC, Na$^+$ occupies all OVs (1:1 ratio). * ZnS (Zinc Blende): S$^{2-}$ forms FCC, Zn$^{2+}$ occupies half of TVs (1:1 ratio). * **CaF$_2$ (Fluorite):** Ca$^{2+}$ forms FCC, F$^-$ occupies all TVs (1:2 ratio).

NEET Strategy: Practice numerical problems involving void occupancy and formula derivation. Be precise with fractions and unit cell atom counts. Visualize void locations for conceptual questions.

Vyyuha Quick Recall

Tiny Tetra Twice, Often One, Radius Rules Really Relevant!

Tiny Tetra: Tetrahedral voids are smaller.
Twice: There are twice as many tetrahedral voids (2N) as spheres (N).
Often One: There is one octahedral void (N) for every sphere (N).
Radius Rules Really Relevant: Remember the radius ratio rules ($0.225$ for Tetra, $0.414$ for Octa) for stability.