Introduction and Unit Cells

ChemistrySolid StateFor JEE aspirants

The solid state is the state of matter in which particles are held at fixed positions and can only vibrate, so a solid keeps its own shape and volume. This concept sorts solids by the force between particles, separates crystalline from amorphous solids, and builds the language of crystals: lattice, unit cell, the seven crystal systems and the 14 Bravais lattices. It ends with counting atoms per unit cell ( $Z$ ) and writing formulas from atom positions. Solid state is now part of the JEE Advanced syllabus only.

On this page1Solid vs liquid vs gas2Types by force3Crystalline vs amorphous4Lattice and unit cell5Primitive and centred cells67 systems, 14 lattices7Counting atoms (Z)8Formula and density

Key Formulas - Quick Reference

★ Must learn Share of an atom in a cubic cell: corner $\frac{1}{8}$ , edge centre $\frac{1}{4}$ , face centre $\frac{1}{2}$ , body centre $1$ .
★ Must learn Atoms per cell: $Z_{sc} = 1$ , $Z_{b cc} = 2$ , $Z_{f cc} = 4$ , end-centred $= 2$ .
★ Must learn Six cell parameters: edges $a, b, c$ and angles $α$ (between $b$ , $c$ ), $β$ (between $a$ , $c$ ), $γ$ (between $a$ , $b$ ).
★ Must learn Density of a unit cell: $d = \frac{Z \times M}{N _{A} \times a ^{3}}$ (use $a$ in cm to get g cm $^{- 3}$ ).
7 crystal systems, 14 Bravais lattices: cubic 3, tetragonal 2, orthorhombic 4, monoclinic 2, rhombohedral 1, hexagonal 1, triclinic 1.
★ Must learn Unit cells in a cube of edge $L$ built from cells of edge $a$ : $(L / a)^{3}$ .
Mass of one unit cell $= \frac{Z \times M}{N _{A}}$ ; 1 u $= 1.66 \times 1 0^{- 24}$ g.

1. What Makes a Solid a Solid

Whether a substance is a solid, liquid or gas depends on two competing things: the intermolecular forces that pull particles together, and the thermal energy that makes them move apart. In a solid the forces win. Each particle is held at a fixed position and can only vibrate about it.

Figure 1: Solid, liquid and gas at the particle level. In a solid the particles are locked at fixed positions and can only vibrate, which is why a solid has a definite shape and volume.

Property	Solid	Liquid	Gas
Motion of particles	no free motion, only vibration	random motion over short distances	completely random
Intermolecular forces	very strong	intermediate	very weak (nearly zero)
Volume	fixed (average separation fixed)	almost fixed	no fixed volume
Shape	definite (positions fixed)	no definite shape	no definite shape
Compressibility	incompressible	almost incompressible	highly compressible
Heat capacity	almost independent of process	almost independent of process	depends on process ( $C_{p} \neq = C_{v}$ )

Because the particles cannot move past one another, solids are rigid, have a definite shape and volume, are incompressible, and diffuse extremely slowly.

Key idea

A solid is a solid because forces beat thermal motion: fixed positions give fixed shape and volume.

2. Classification by Binding Force

Crystalline solids are grouped by the kind of particle at each site and the force that holds the sites together. This single idea predicts hardness, melting point and electrical conduction.

Figure 2: Four families of crystalline solids, sorted by the force that holds the particles together. The stronger the force, the harder the solid and the higher its melting point.

Type	Particles	Force	Examples	Nature	Melting point	Conduction
Molecular (non-polar)	molecules	dispersion (London)	$I_{2}$ , Ar, $C H_{4}$ , $CC l_{4}$ , $H_{2}$ , $C_{6} H_{6}$	very soft	very low	insulator
Molecular (polar)	molecules	dipole-dipole	$HCl$ , $S O_{2}$ , $S F_{4}$	soft	low	insulator
Molecular (H-bonded)	molecules	hydrogen bonds	$H_{2} O$ (ice), $H_{3} B O_{3}$	hard	low	insulator
Ionic	ions	coulombic, non-directional, long range	NaCl, ZnS, $Ca F_{2}$ , CsCl	hard, brittle	very high	only molten or in water
Metallic	cations in a sea of electrons	metallic bond	Cu, Al, Zn, Ag, Fe	hard to soft, malleable	low to high	good (solid and molten)
Covalent network	atoms	covalent bonds	diamond, SiC, $Si O_{2}$ , AlN, graphite	very hard (graphite soft)	very high	insulator (graphite conducts)

Graphite is the odd one out among network solids. Each carbon uses three electrons in bonds inside a layer; the fourth electron is delocalised over the layer, so graphite conducts. The layers are held only by weak forces and slide over one another, which is why graphite is soft and used as a lubricant.

Exam Trick Ask two questions: what sits at the lattice point (molecule, ion, metal atom, atom) and what joins them. Molecules → weak forces → low m.p. Everything else → strong bonds → high m.p. Only metals and graphite conduct as solids.

Key idea

The force between particles decides the type of solid, and the type decides hardness, melting point and conduction.

3. Crystalline and Amorphous Solids

Based on how the particles are arranged, solids are crystalline (particles in a definite, repeating pattern throughout) or amorphous (from Greek amorphos, no form; particles arranged at random with only short-range order). Amorphous solids such as glass are called pseudo solids or supercooled liquids, because like liquids they flow, though extremely slowly.

Crystalline (true solid)

Long-range order: pattern repeats through the whole crystal.
Sharp melting point and fixed enthalpy of fusion.
Anisotropic: properties differ with direction.
Cleaves along planes with smooth faces.
Formed by slow cooling. Examples: NaCl, quartz, diamond, sugar, Cu, ice.

Amorphous (pseudo solid)

Only short-range order; no repeating pattern.
Softens over a range of temperature; no fixed enthalpy of fusion.
Isotropic: same properties in every direction.
Irregular cut with curved surfaces.
Formed by rapid cooling. Examples: glass, rubber, plastics, amorphous silica, starch.

3.1 Cooling curves

When a liquid that forms a crystal is cooled, its temperature stays constant at the freezing point until all of it has solidified, because the energy released on forming the ordered lattice balances the heat lost. An amorphous solid has no such step: it stiffens gradually.

Figure 3: Cooling curves computed from Newton cooling. The crystalline liquid holds a constant temperature

T_{f}

while it freezes (sharp melting point, fixed enthalpy of fusion); the amorphous solid has no plateau, only a softening range

T_{1}

T_{2}

3.2 Anisotropy

In a crystal, the arrangement of particles looks different along different directions. Electrical resistance, refractive index or thermal expansion therefore change with direction. This is anisotropy. Amorphous solids, with their random arrangement, are isotropic.

Figure 4: In a crystal the particle spacing depends on direction (dense along A, sparser along B and C), so refractive index and conductivity change with direction: anisotropy. In an amorphous solid every direction is statistically the same: isotropy.

3.3 Polymorphism and devitrification

Polymorphism: one substance can crystallise in more than one structure depending on conditions. Examples: rhombic and monoclinic sulphur; calcite and aragonite ( $CaC O_{3}$ ); diamond and graphite (carbon; strictly, allotropes of an element).
Crystalline and amorphous forms of one substance: quartz is crystalline $Si O_{2}$ ; melting it and cooling rapidly gives quartz glass, which is amorphous.
Devitrification: an amorphous solid can slowly become crystalline on heating or over long times. Very old glass objects turn milky for this reason.

A sharp melting point is the quickest test that separates the two types. Hardness, density and colour do not.

Key idea

Crystalline = long-range order, sharp m.p., anisotropic. Amorphous = short-range order, softening range, isotropic.

Quick Recall: tap to check

Why is glass called a supercooled liquid?

It has only short-range order like a liquid and flows extremely slowly; it was cooled too fast for its particles to order into a lattice.

Which property shows anisotropy in a crystal?

Any direction-dependent property: refractive index, electrical or thermal conductivity, thermal expansion, mechanical strength.

Classify: SiC,

C O_{2}

(s), Cu, KCl.

SiC covalent network;

C O_{2}

molecular (non-polar); Cu metallic; KCl ionic.

4. Lattice, Lattice Points and the Unit Cell

To describe a crystal, replace every repeating particle (an atom, an ion, or a whole molecule) by a single point. Each point is a lattice point, and the regular 3D array of points is the crystal lattice or space lattice. Every lattice point has exactly the same surroundings.

Unit cell: the smallest portion of a crystal lattice which, when repeated in different directions by translation (never rotation), generates the whole lattice. The most symmetric, smallest-volume cell is chosen.

Figure 5: A space lattice is a regular 3D array of points; the shaded box is the unit cell. Translating it by its own edge lengths in all three directions rebuilds the whole crystal.

4.1 Lattices in one and two dimensions

A 1D lattice is a row of equally spaced points, fixed by one parameter (the spacing). A 2D lattice needs three parameters: two edge lengths $a$ , $b$ and the angle $γ$ between them. Only five 2D lattices exist.

Figure 6: The five 2D lattices. Each needs three parameters (edges

a

b

and angle

γ

). The square cell is the most symmetric; the oblique cell is the least.

2D lattice	Edges	Angle
Square (most symmetric)	$a = b$	$γ = 9 0^{\circ}$
Rectangular	$a \neq = b$	$γ = 9 0^{\circ}$
Hexagonal	$a = b$	$γ = 12 0^{\circ}$ (the $6 0^{\circ}$ rhombus)
Rhombic (centred rectangular)	$a = b$	$γ \neq = 9 0^{\circ}, 6 0^{\circ}, 12 0^{\circ}$
Oblique (parallelogram)	$a \neq = b$	$γ \neq = 9 0^{\circ}$

4.2 Parameters of a 3D unit cell

In three dimensions a unit cell is a parallelepiped fixed by six parameters: the three edge lengths $a$ , $b$ , $c$ and the three angles between them. The angle $α$ lies between $b$ and $c$ , $β$ between $a$ and $c$ , and $γ$ between $a$ and $b$ . Each angle is named after the edge it does not touch.

Key idea

A unit cell is the repeat unit: six numbers (

a, b, c, α, β, γ

) describe it completely.

5. Primitive and Centred Unit Cells

Primitive (simple, P): lattice points only at the corners. Found in all seven crystal systems.
Body-centred (I): corners plus one point at the body centre. Found in cubic, tetragonal and orthorhombic.
Face-centred (F): corners plus one point at the centre of every face. Found in cubic and orthorhombic.
End-centred (base-centred, C): corners plus points at the centres of one pair of opposite faces. Found in orthorhombic and monoclinic.

Figure 7: The four kinds of unit cell. Only the primitive cell has lattice points at the corners alone; the centred cells carry extra points, which is why their

Z

values are 2, 4 and 2.

Primitive cell

Lattice points at corners only. One lattice point per cell ( $Z = 1$ ). The smallest possible repeat unit.

Centred (non-primitive) cell

Extra points at the body, faces or ends. $Z = 2$ (I, C) or $4$ (F). Chosen because it shows the full symmetry of the lattice.

6. Seven Crystal Systems and 14 Bravais Lattices

Restrictions on the six parameters give exactly seven crystal systems. Combining each system with the centrings its symmetry allows gives 14 distinct 3D lattices, the Bravais lattices.

Figure 8: The seven crystal systems, each drawn from its actual cell parameters. Moving from cubic to triclinic, the restrictions on

a, b, c

and

α, β, γ

drop away one by one; triclinic has none.

Crystal system	Edges	Angles	Bravais lattices	Examples
Cubic	$a = b = c$	$α = β = γ = 9 0^{\circ}$	P, I, F (3)	NaCl, ZnS, Cu, Fe, KCl, diamond, alum
Tetragonal	$a = b \neq = c$	$α = β = γ = 9 0^{\circ}$	P, I (2)	white tin, $Sn O_{2}$ , $Ti O_{2}$ , $CaS O_{4}$
Orthorhombic	$a \neq = b \neq = c$	$α = β = γ = 9 0^{\circ}$	P, I, F, C (4)	rhombic sulphur, $KN O_{3}$ , $BaS O_{4}$ , aragonite
Rhombohedral (trigonal)	$a = b = c$	$α = β = γ \neq = 9 0^{\circ}$	P (1)	calcite ( $CaC O_{3}$ ), cinnabar (HgS), $NaN O_{3}$
Hexagonal	$a = b \neq = c$	$α = β = 9 0^{\circ}, γ = 12 0^{\circ}$	P (1)	graphite, ZnO, CdS, Mg, $Pb I_{2}$
Monoclinic	$a \neq = b \neq = c$	$α = γ = 9 0^{\circ}, β \neq = 9 0^{\circ}$	P, C (2)	monoclinic sulphur, $PbCr O_{4}$ , $N a_{2} S O_{4} \cdot 10 H_{2} O$
Triclinic	$a \neq = b \neq = c$	$α \neq = β \neq = γ \neq = 9 0^{\circ}$	P (1)	$K_{2} C r_{2} O_{7}$ , $CuS O_{4} \cdot 5 H_{2} O$ , $H_{3} B O_{3}$

Figure 9: Which centring types each system allows. Orthorhombic is the only system with all four; cubic lacks end-centring. The ticks add up to the 14 Bravais lattices.

Exam Trick Remember the Bravais count as 3-2-4-2-1-1-1 (cubic, tetragonal, orthorhombic, monoclinic, then the three single ones). Orthorhombic is the only system that allows all four centrings, and all three single-lattice systems are primitive only.

JEE Advanced Why is there no end-centred cubic or face-centred tetragonal lattice? Centring two faces of a cube destroys the three-fold symmetry along the body diagonals, so the result is no longer cubic (it becomes tetragonal). A face-centred tetragonal lattice is not new at all: a smaller body-centred tetragonal cell, rotated by

4 5^{\circ}

with edge

a / 2

, describes the same points. Bravais showed that only 14 lattices are truly different.

Key idea

Seven systems come from the parameter restrictions; adding allowed centrings gives 14 Bravais lattices.

Quick Recall: tap to check

What are the parameters of a hexagonal cell?

a = b \neq = c

;

α = β = 9 0^{\circ}

γ = 12 0^{\circ}

Which systems have an end-centred lattice?

Orthorhombic and monoclinic.

Which crystal system has the least symmetry?

Triclinic:

a \neq = b \neq = c

and

α \neq = β \neq = γ \neq = 9 0^{\circ}

; only a primitive lattice.

7. Counting Atoms in a Cubic Unit Cell

An atom on a corner, edge or face does not belong to one cell alone. It is shared with the neighbouring cells, and each cell counts only its share. The number of atoms that truly belong to one cell is written $Z$ (the effective number of atoms).

Figure 10: Why the fractions are 1/8, 1/4, 1/2 and 1. Count how many cubes meet at the atom's position; the shaded cube owns that fraction of it.

Position	Cells sharing it	Share per cell	Number of such sites
Corner	8	$\frac{1}{8}$	8
Edge centre	4	$\frac{1}{4}$	12
Face centre	2	$\frac{1}{2}$	6
Body centre	1	1	1

Applying these shares:

Simple cubic: $Z = 8 \times \frac{1}{8} = 1$ .
Body-centred cubic: $Z = 8 \times \frac{1}{8} + 1 = 2$ .
Face-centred cubic: $Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ .
End-centred cell: $Z = 8 \times \frac{1}{8} + 2 \times \frac{1}{2} = 2$ .

7.1 Formula of a compound from positions

Count $Z$ for each kind of atom separately; their ratio is the formula. If an atom is missing from a site, subtract its share. The flowchart below is the route for every such question.

Figure 11: Problem-solving route for every 'find the formula' question. The one step students skip is the missing-atom check.

7.2 Density of a unit cell

Mass of one cell = (number of atoms) × (mass of one atom) $= \frac{Z M}{N _{A}}$ . Volume of a cubic cell $= a^{3}$ . So

d = \frac{Z \times M}{N _{A} \times a ^{3}}

Keep units consistent: with $M$ in g mol $^{- 1}$ and $a$ in cm, $d$ comes out in g cm $^{- 3}$ (1 Å $= 1 0^{- 8}$ cm, 1 pm $= 1 0^{- 10}$ cm). The same density applies to the whole crystal, because the crystal is only the unit cell repeated.

7.3 How many unit cells?

If $x$ unit cells fit along one edge of a cubic crystal, the crystal holds $x^{3}$ cells (2 along an edge gives $2^{3} = 8$ ). Equally, the number of cells in a given mass is (number of atoms) $/ Z$ .

Exam Trick 1-2-4: sc, bcc, fcc hold 1, 2, 4 atoms; each step doubles. And for any cubic lattice,

d a^{3} N_{A} = Z M

: know any three of

d

a

Z

M

and you can find the fourth.

Key idea

Z

= sum of (number at a site) × (share of that site); the formula is the ratio of the

Z

values.

Quick Recall: tap to check

Z for a cell with atoms at corners and edge centres only?

8 \times \frac{1}{8} + 12 \times \frac{1}{4} = 1 + 3 = 4

One corner atom of an fcc cell is removed. New Z?

4 - \frac{1}{8} = \frac{31}{8}

Units of

d = Z M / (N_{A} a^{3})

a

is in pm?

Convert: 1 pm

= 1 0^{- 10}

cm, then

d

is in g cm

^{- 3}

7.4 The whole concept at a glance

Figure 12: The whole concept on one page. Revise from the centre outwards.

8. Solved Examples

Solved Example 1

Identify the molecular, covalent and ionic solids:

P_{4}

(s),

S_{8}

(s), SiC(s),

A l_{2} O_{3}

(s), He(s),

A l_{2} C l_{6}

(s).

Solution:

Molecular solids (discrete molecules or atoms held by weak forces): $P_{4}$ , $S_{8}$ , He, $A l_{2} C l_{6}$ .

Covalent network solid: SiC.

Ionic solid: $A l_{2} O_{3}$ .

Solved Example 2

The lattice parameters of a crystal are

a = 5.62

Å,

b = 7.41

Å and

c = 9.48

Å, and the three axes are mutually perpendicular. The crystal is
(A) tetragonal
(B) orthorhombic
(C) monoclinic
(D) trigonal

Solution:

Answer: (B). All three edges differ and $α = β = γ = 9 0^{\circ}$ , which is the orthorhombic system.

Solved Example 3

The tetragonal crystal system has the unit-cell dimensions:
(A)

a = b = c

α = β = γ = 9 0^{\circ}

(B)

a = b \neq = c

α = β = γ = 9 0^{\circ}

(C)

a \neq = b \neq = c

α = β = γ = 9 0^{\circ}

(D)

a = b \neq = c

α = β = 9 0^{\circ}

γ = 12 0^{\circ}

Solution:

Answer: (B). Two equal edges and all angles $9 0^{\circ}$ . Option (A) is cubic, (C) orthorhombic, (D) hexagonal.

Solved Example 4

In a face-centred cubic arrangement of A and B atoms, A atoms are at the corners and B atoms at the face centres. One A atom is missing from one corner. The simplest formula is
(A)

A_{7} B_{3}

(B)

A B_{3}

(C)

A_{7} B_{24}

(D)

A_{2} B_{3}

Solution:

Answer: (C).

A = 7 \times \frac{1}{8} = \frac{7}{8}, B = 6 \times \frac{1}{2} = 3

Ratio $\frac{7}{8} : 3 = 7 : 24$ , so the formula is $A_{7} B_{24}$ .

Solved Example 5

A cubic unit cell has X atoms at 6 corners, Y atoms at the remaining 2 corners and at those 3 face centres that are not opposite to each other, and Z atoms at the remaining face centres and the body centre. Find (i) the formula and (ii) the density if the edge length is 2 Å. Atomic masses: X = 40 u, Y = 60 u, Z = 80 u.

Solution:

Figure 13: Solved Example 5 drawn out. Y sits on three faces that meet at one corner, so Z takes the three opposite faces plus the body centre.

(i) Shares per cell:

X = 6 \times \frac{1}{8} = \frac{3}{4}, Y = 2 \times \frac{1}{8} + 3 \times \frac{1}{2} = \frac{7}{4}, Z = 3 \times \frac{1}{2} + 1 = \frac{10}{4}

Ratio $3 : 7 : 10$ , so the formula is $X_{3} Y_{7} Z_{10}$ .

(ii) Mass in one cell $= \frac{3}{4} (40) + \frac{7}{4} (60) + \frac{10}{4} (80) = 30 + 105 + 200 = 335$ u.

d = \frac{335 \times 1.66 \times 1 0 ^{- 24} g}{( 2 \times 1 0 ^{- 8} cm ) ^{3}} = \frac{5.56 \times 1 0 ^{- 22}}{8 \times 1 0 ^{- 24}} = 69.5 g cm^{- 3}

This is a made-up compound: a real density near 70 g cm $^{- 3}$ is impossible, but the method is exactly what is tested.

Solved Example 6

A crystal has

a = b = 4.0

Å,

c = 6.5

Å,

α = β = 9 0^{\circ}

and

γ = 12 0^{\circ}

. Its crystal system is
(A) tetragonal
(B) hexagonal
(C) orthorhombic
(D) rhombohedral

Solution:

Answer: (B). $a = b \neq = c$ with $γ = 12 0^{\circ}$ is the signature of the hexagonal system. Tetragonal would need $γ = 9 0^{\circ}$ .

Solved Example 7

A solid has A atoms at the corners and face centres of a cube, and B atoms at the edge centres and body centre. (a) Write its formula. (b) What is the formula if all atoms on one pair of opposite faces (the two face centres cut by one axis) are removed?

Solution:

(a) $A = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ ; $B = 12 \times \frac{1}{4} + 1 = 4$ . Formula AB.

(b) Two face-centre A atoms are lost: $A = 4 - 2 \times \frac{1}{2} = 3$ ; $B = 4$ . Formula $A_{3} B_{4}$ .

Solved Example 8

Which property best tells a crystalline solid from an amorphous one?
(A) hardness
(B) a sharp melting point
(C) high density
(D) colour

Solution:

Answer: (B). Crystalline solids melt at one temperature with a fixed enthalpy of fusion; amorphous solids soften over a range. Hardness, density and colour vary within both groups.

Solved Example 9

A cubic crystal of edge 1.0 mm is built from unit cells of edge 0.50 nm. How many unit cells does it contain?

Solution:

Cells along one edge: $x = \frac{1.0 \times 1 0 ^{- 3} m}{0.50 \times 1 0 ^{- 9} m} = 2.0 \times 1 0^{6}$ .

Total cells $= x^{3} = (2.0 \times 1 0^{6})^{3} = 8.0 \times 1 0^{18}$ .

Practice Questions

Classify as molecular, ionic, metallic or covalent: $Si O_{2}$ (quartz), Ar, $CC l_{4}$ , $N H_{4} N O_{3}$ , brass, graphite, Rb.Answer: covalent, molecular, molecular, ionic, metallic, covalent, metallic.
A cell has $a \neq = b \neq = c$ , $α = γ = 9 0^{\circ}$ , $β = 10 4^{\circ}$ . Name the crystal system.Answer: Monoclinic.
M atoms occupy the corners and N atoms the face centres of a cube. Atoms at two corners are missing. Find the formula.Answer: $M = 6 \times \frac{1}{8} = \frac{3}{4}$ , $N = 3$ , so $M N_{4}$ .
How many lattice points belong to one end-centred unit cell?Answer: 2.
How many Bravais lattices does the tetragonal system have? Name them.Answer: Two: primitive and body-centred.
An element ( $M = 60$ g mol $^{- 1}$ ) forms an fcc lattice with $a = 400$ pm. Find its density.Answer: $d = \frac{4 \times 60}{6.022 \times 1 0 ^{23} \times ( 4 \times 1 0 ^{- 8} ) ^{3}} = 6.23$ g cm $^{- 3}$ .
Why does very old glass sometimes turn milky?Answer: It slowly crystallises (devitrifies); the tiny crystals scatter light.

Common Mistakes to Avoid

Watch out

Counting every atom drawn on a cell as a whole atom. Corner, edge and face atoms are shared: use $\frac{1}{8}$ , $\frac{1}{4}$ , $\frac{1}{2}$ .
Forgetting to subtract a missing corner or face atom before writing the formula.
Mixing up the angles: $α$ is between $b$ and $c$ , not between $a$ and $b$ .
Calling hexagonal $a = b = c$ . Hexagonal is $a = b \neq = c$ with $γ = 12 0^{\circ}$ ; $a = b = c$ with equal non- $9 0^{\circ}$ angles is rhombohedral.
Saying cubic has an end-centred lattice. Cubic has only P, I and F.
Using $a$ in Å or pm directly in $d = Z M / (N_{A} a^{3})$ . Convert to cm first.
Treating amorphous solids as anisotropic, or glass as having a sharp melting point.
Calling graphite an insulator like diamond. Graphite conducts along its layers.

Frequently Asked Questions

What is a unit cell in solid state chemistry?

A unit cell is the smallest repeating portion of a crystal lattice. Repeating it by translation along its three edges builds the whole crystal. It is described by six parameters: edge lengths a, b, c and the angles alpha, beta and gamma between them.

What is the difference between crystalline and amorphous solids?

Crystalline solids have long-range order, a sharp melting point, a fixed enthalpy of fusion and anisotropic properties. Amorphous solids such as glass and rubber have only short-range order, soften over a temperature range and are isotropic. Amorphous solids are also called pseudo solids or supercooled liquids.

How many atoms are there in sc, bcc and fcc unit cells?

A simple cubic cell has 8 corners × 1/8 = 1 atom. A body-centred cubic cell has 1 + 1 = 2 atoms. A face-centred cubic cell has 1 from the corners plus 6 × 1/2 = 3 from the faces, so 4 atoms in all.

Why are there only 14 Bravais lattices?

Each of the seven crystal systems allows only the centrings that keep its symmetry. Other combinations either break the symmetry or repeat a lattice already counted, for example face-centred tetragonal is the same as body-centred tetragonal. This leaves 3 + 2 + 4 + 2 + 1 + 1 + 1 = 14 lattices.

What is anisotropy in crystals?

Anisotropy means that a physical property such as refractive index, electrical conductivity or thermal expansion has different values in different directions. It arises because the spacing of particles in a crystal depends on direction. Amorphous solids are isotropic.

Why is graphite soft and a good conductor although it is a covalent solid?

In graphite each carbon bonds to three others in a flat layer, and the fourth electron is delocalised over the layer, so it conducts electricity. The layers are held by weak forces and slide over each other, which makes graphite soft and slippery.

Is Solid State in the NEET and JEE Main syllabus for 2026?

No. NCERT removed the Solid State chapter from Class 12 in 2023, and it was dropped from the NEET and JEE Main syllabi from 2024 onward. It remains in the JEE Advanced syllabus, which lists classification of solids, crystal systems, close packing, voids, radius ratio and point defects.

What is asked from unit cells in JEE Advanced?

JEE Advanced questions test the seven crystal systems and their parameters, counting atoms per cell, writing formulas when atoms are removed or replaced, and density calculations with d = ZM divided by N_A times a cubed. Integer-type questions on Z and density are common.

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