JEE Chemistry — Solid State Complete Chapter Guide

Chapter Overview & Weightage

Solid State is one of those chapters where JEE rewards students who understand the geometry, not just memorize formulas. It appears consistently in JEE Main — typically 1 question per paper — and occasionally shows up in JEE Advanced as a part of a multi-concept problem.

Weightage snapshot: Solid State contributes roughly 1 question in JEE Main (4 marks), appearing in almost every shift. JEE Advanced picks it up maybe once every 2-3 years, usually paired with electrochemistry or solutions. NEET students: this chapter has low weightage (~1-2 questions), but they’re almost always scoring questions if you know the unit cell.

Year	JEE Main Questions	Topic Tested
2024	1 (Shift 1)	Packing efficiency, BCC
2023	1	Point defects (Frenkel vs Schottky)
2022	1	Unit cell — number of atoms, density
2021	2	Coordination number + void calculation
2020	1	Density formula application

The pattern is clear: density formula and defects are the two most repeated themes. Know these cold.

Key Concepts You Must Know

Prioritized by how often they appear in actual PYQs:

Tier 1 — Appears almost every year:

Unit cell types: SC, BCC, FCC/CCP — number of atoms per unit cell (1, 2, 4)
Density formula: $\rho = \frac{Z \times M}{N_A \times a^3}$ — every number in this formula has been tested separately
Packing efficiency: SC (52.4%), BCC (68%), FCC (74%)
Point defects: Frenkel (smaller ion displaced to interstitial) vs Schottky (cation-anion pair missing)

Tier 2 — Asked once every 2-3 years:

Tetrahedral and octahedral voids — number and location in FCC
Radius ratio rules for coordination numbers (4, 6, 8)
Ionic crystal structures: NaCl (rock salt), ZnS (zinc blende), CsCl, fluorite, antifluorite
Conductivity in solids: n-type vs p-type semiconductors

Tier 3 — Conceptual, rare but tricky:

Ferromagnetism, ferrimagnetism, antiferromagnetism
Crystal field theory connections (usually in Advanced)
Edge and body diagonal relationships for SC, BCC, FCC

Important Formulas

Simple Cubic (SC): $Z = 8 \times \frac{1}{8} = 1$
BCC: $Z = 8 \times \frac{1}{8} + 1 = 2$
FCC/CCP: $Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$

When to use: Any question asking “how many atoms does one unit cell contain” — or any density/packing question where Z is not given directly.

\rho = \frac{Z \times M}{N_A \times a^3}

Where $Z$ = atoms per unit cell, $M$ = molar mass, $N_A$ = Avogadro’s number, $a$ = edge length.

When to use: Whenever edge length, molar mass, or density is given and one of the others is asked. This is the single most important formula in this chapter.

\text{PE} = \frac{\text{Volume of atoms in unit cell}}{\text{Volume of unit cell}} \times 100

For FCC: $\text{PE} = \frac{4 \times \frac{4}{3}\pi r^3}{a^3}$ , where $a = 2\sqrt{2}\,r$ , giving 74%
For BCC: $a = \frac{4r}{\sqrt{3}}$ , giving 68%
For SC: $a = 2r$ , giving 52.4%

When to use: Direct percentage questions, or to compare structures. FCC is close-packed — highest efficiency among cubic systems.

$r^+/r^-$ Range	Coordination Number	Structure
0.155 – 0.225	3	Triangular
0.225 – 0.414	4	Tetrahedral
0.414 – 0.732	6	Octahedral
0.732 – 1.000	8	Cubic

When to use: Given the ionic radii, predict the crystal structure or coordination number.

Tetrahedral voids: $2Z = 8$ per unit cell, located at $(\frac{1}{4}, \frac{1}{4}, \frac{1}{4})$ type positions
Octahedral voids: $Z = 4$ per unit cell, at edge centers and body center

When to use: Any question about how many voids exist, or what fraction is occupied (like in NaCl all octahedral voids filled by $\text{Na}^+$ ).

Solved Previous Year Questions

PYQ 1 — JEE Main 2024, Shift 1

Q: An element crystallizes in a BCC lattice with edge length $a = 3.5 \text{ Å}$ . The density is $7.92 \text{ g/cm}^3$ . Find the molar mass of the element.

Solution:

We use the density formula rearranged for $M$ :

M = \frac{\rho \times N_A \times a^3}{Z}

For BCC, $Z = 2$ . Convert $a = 3.5 \text{ Å} = 3.5 \times 10^{-8} \text{ cm}$ .

a^3 = (3.5 \times 10^{-8})^3 = 4.288 \times 10^{-23} \text{ cm}^3

M = \frac{7.92 \times 6.022 \times 10^{23} \times 4.288 \times 10^{-23}}{2}

M = \frac{7.92 \times 6.022 \times 0.04288}{2} \approx \frac{2.044}{2} \approx 56 \text{ g/mol}

The element is Iron (Fe). The calculation route: rearrange the formula, plug in, watch the powers of 10 carefully.

Most students forget to cube the edge length properly. $3.5^3 = 42.875$ , not $10.5$ . Write out the cube explicitly — rushing this step is the #1 source of error here.

PYQ 2 — JEE Main 2023

Q: Which of the following statements about Frenkel defect is correct? (A) It increases the density of the solid (B) It is shown by NaCl (C) The smaller ion is displaced to an interstitial position (D) It involves equal numbers of cation and anion vacancies

Solution:

Work through elimination:

(A) Wrong. In Frenkel defect, no ion leaves the crystal — density stays the same.
(B) Wrong. NaCl shows Schottky defect (both ions are similar in size). Frenkel is shown by AgCl, AgBr, ZnS where there’s a big size difference.
(C) Correct. The smaller ion (usually cation) moves from its lattice site to an interstitial position, creating a “hole” at the original site.
(D) Wrong — that’s Schottky defect.

Answer: (C)

The key distinguisher: Frenkel = one ion hops to interstitial (no density change). Schottky = both ions leave (density decreases).

PYQ 3 — JEE Main 2021

Q: In a face-centred cubic lattice, atom A is at corner positions and atom B is at face-centre positions. What is the formula of the compound?

Solution:

Count atoms per unit cell for each type:

Atom A (corners): $8 \times \frac{1}{8} = 1$
Atom B (face centres): $6 \times \frac{1}{2} = 3$

Ratio A:B = 1:3, so the formula is AB₃.

This is the NiF₃ / ReO₃ structure type. The logic is purely geometric — just apply the corner/face/edge/body contributions systematically.

For ionic solids like these, JEE sometimes twists the question by removing some atoms (e.g., “atoms at alternate corners are removed”). Just recalculate Z for each species and find the ratio. Same method, slight modification.

Difficulty Distribution

For JEE Main, the chapter difficulty breaks down roughly like this:

Difficulty	% of Questions	What Gets Asked
Easy	40%	Z calculation, identifying SC/BCC/FCC, naming defect types
Medium	45%	Density formula with unit conversions, packing efficiency derivation, void counting
Hard	15%	Combined problems (given density + defect, find new density), radius ratio + structure prediction

Most students lose marks on Medium questions — not because the concept is hard, but because of arithmetic errors in the density formula. Unit conversion ( $\text{Å} \to \text{cm}$ ) is where marks slip.

Expert Strategy

Week 1 — Build the geometric picture: Draw all three cubic unit cells (SC, BCC, FCC) by hand. Mark where atoms sit, which fractions they contribute. Do this once properly and you’ll never confuse Z values again.

Week 2 — Formula drilling: The density formula has 5 variables. Practice solving for each one — given 4, find the 5th. JEE has asked for $a$ , $\rho$ , $M$ , and $Z$ individually. You should be able to rearrange the formula in under 10 seconds.

The fast shortcut for packing efficiency: Don’t re-derive it every time. Memorize the three values (52.4%, 68%, 74%) and the relationship $a = f(r)$ for each structure. The derivation is useful to understand once — after that, just apply.

For ionic structures (NaCl, ZnS, CsCl, fluorite), focus on: which ion goes where, what is each ion’s coordination number, and what is Z for each ion. A table comparing all four structures side-by-side is the best revision tool — make one yourself before the exam.

Defects — learn the logic, not just the names: Schottky: both ions leave to maintain electrical neutrality → density decreases → NaCl, KCl type (similar sizes). Frenkel: small ion jumps to interstitial → density unchanged → AgCl, ZnS type (big size difference). If you remember why each defect happens, you won’t confuse them.

Semiconductors in this chapter: n-type = extra electrons (Group 15 impurity in Si), p-type = holes (Group 13 impurity). JEE Main occasionally asks this as a one-liner. 2 minutes of reading, guaranteed 4 marks if it appears.

Common Traps

Trap 1 — Forgetting to cube the edge length properly. $a = 4 \text{ Å} = 4 \times 10^{-8} \text{ cm}$ . Then $a^3 = 64 \times 10^{-24} \text{ cm}^3$ . Students often write $4 \times 10^{-24}$ . Always expand the cube explicitly.

Trap 2 — Misidentifying which defect NaCl shows. NaCl → Schottky (both Na⁺ and Cl⁻ leave). AgCl → Frenkel (Ag⁺ jumps to interstitial). The trap: both are ionic crystals, but the size difference determines the defect type. This exact distinction appeared in JEE Main 2023.

Trap 3 — Packing efficiency of HCP vs FCC. Both are close-packed, both have 74% packing efficiency. JEE sometimes asks students to compare, expecting them to say FCC is “more packed” — it’s not. ABCABC (FCC) and ABABAB (HCP) have identical packing efficiency. The difference is only in the stacking sequence.

Trap 4 — Void calculation in FCC. There are 8 tetrahedral voids and 4 octahedral voids per FCC unit cell (where Z = 4). The ratio is always 2:1 (tetrahedral:octahedral). Students flip this. In NaCl, Na⁺ fills all octahedral voids — 4 Na⁺ and 4 Cl⁻ per unit cell, giving 1:1 ratio. Check your void assignment against the known structure.

Trap 5 — Treating Schottky defect as affecting only one ion. Schottky requires equal numbers of cations AND anions to leave (to maintain electrical neutrality). If a question says “10¹⁸ Schottky defects per mole,” that means $10^{18}$ cation vacancies AND $10^{18}$ anion vacancies — 2 × $10^{18}$ total missing sites. This distinction appears in numerical problems.