Why do tetrahedral voids ($2N$) double octahedral voids ($N$) in close-packed lattices?

In a close-packed arrangement of $N$ spheres, each sphere forms $2$ tetrahedral voids along body diagonals (one above and one below the sphere) and $1$ octahedral void located at edge centers and body center. Thus, for $N$ close-packed lattice points, total tetrahedral voids equal $2N$ and octahedral voids equal $N$.

How do I instantly distinguish between Frenkel and Schottky defects in exam questions?

Look at relative ion sizes and crystal density changes. Schottky defects occur in compounds with similar cation-anion sizes (e.g., $NaCl$, $KCl$), creating equal cation-anion vacancies and decreasing density. Frenkel defects occur when cations are significantly smaller than anions (e.g., $AgCl$, $ZnS$), dislocating cations to interstitial sites with zero change in density.

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Question #1NEET 2024

An element crystallizes in a face-centered cubic ( $FCC$ ) lattice with an edge length of $400 \text{ pm}$ . If the atomic mass of the element is $60 \text{ g mol}^{-1}$ , what is the density of the crystal? (Take $N_A = 6.0 \times 10^{23} \text{ mol}^{-1}$ )

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Step-by-Step Solution & Concept

Step 1

Identify given parameters for $FCC$ : effective number of atoms $z = 4$ , edge length $a = 400 \text{ pm} = 4 \times 10^{-8} \text{ cm}$ , molar mass $M = 60 \text{ g mol}^{-1}$ , and $N_A = 6.0 \times 10^{23} \text{ mol}^{-1}$ .

Step 2

Calculate unit cell volume: $V_{\text{cell}} = a^3 = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3$ .

Step 3

Substitute parameters into the unit cell density formula: $\rho = \frac{z \cdot M}{N_A \cdot a^3} = \frac{4 \times 60}{(6.0 \times 10^{23}) \times (64 \times 10^{-24})} = \frac{240}{38.4} = 6.25 \text{ g cm}^{-3}$ .

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Question #2JEE Main 2023

A compound formed by elements $A$ and $B$ crystallizes in a cubic structure where atoms of $A$ occupy the corners of the cube and atoms of $B$ occupy the face centers. If one atom of $A$ is missing from one corner position, what is the empirical formula of the compound?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate the contribution of $A$ atoms. Since $1$ corner atom is missing out of $8$ , there are $7$ corners occupied. Each corner contributes $\frac{1}{8}$ , giving $A = 7 \times \frac{1}{8} = \frac{7}{8}$ .

Step 2

Calculate the contribution of $B$ atoms. All $6$ face centers are fully occupied. Each face center contributes $\frac{1}{2}$ , giving $B = 6 \times \frac{1}{2} = 3$ .

Step 3

Determine the simplest integral stoichiometry ratio: $A : B = \frac{7}{8} : 3 = 7 : 24$ . Therefore, the empirical formula is $A_7 B_{24}$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Density of a Cubic Unit Cell

Atomic Radius and Edge Length Relations

Radius Ratio Rules for Coordination Geometry

Bragg's Law of X-Ray Diffraction

Packing Efficiency (Atomic Packing Factor)

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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