Question

Write down the important formula of chapter solid state class 12th .​

Answer 1

• ★ Solid state ★
• Cubic crystal system

a = 2r

effective no. of per unit cell "Z" = 1

paking fraction = Z × 4/3 π r³ / a³

In simple cubic Packing fraction :- 1 × 4/3 ×πr³

/ (2r)³

π/6 ≈ 52%

∴ Packing fraction = 52%

In simple cubic

d= 1×M/a³×Na = 1×M/(2r)³×Na

Coordination no. in simple cubic lattice will be 6

• Body centre cubic lattice (BCC)
• √3a = 4r
• a = 4r/√3
• r = √3a/4r

Atoms are in contact with body diagonal

∴ Z = 2

PE = 2× 4/3×πr³ / (a)³ = √3π/8 ≈ 68%

d = zm /a³ N ₐ = 2×M/(4r /√3)³×Nₐ

★ Face centred cubic lattice fcc or cubic closest packing ccp

→ √2a = 4r

a = 2√2r

Effective no. of atoms present unit shell(Z) =

1/8×8 + 1/2×6 = 4

→ PF = 4×4/3 × πr³/(a)³

co-ordination no. in fcc system is 12

• ★ Hexagonal closest packing ★

Volume of unit cell = 24√2r³

Effective no. = 6

Pf≈ 74%

Co-ordination no. = 12

• ★ Octahydral voids ★

Effective no. Z = 4

ra + rc = a/2 {if cation is present in Octahydral void and anion is present at fcc latice point )

Cordination number of cation present in Octahydral void is 6

coordination number of anion present in fcc lattice point is 6

★ Tetrahydral voids in fcc

8 TTV in fcc unit

ra + rc = √3a/4

→ Cordination no. of cation present in Tetrahydral void is 4

→ coordination no. of anion present in fcc is 8

→2 × Octahydral void = no. of Tetrahydral voids

→ In hexagonal closest packing (HCP)

effective no. = 6

Octahydral void = 6

Tetrahydral void = 12

• ★Nacl rock salt structure ★

(Z ) eff no. cl⁻ = 4

eff no. of Na⁺ = 4

eff. no. Of per unit cell of Nacl is 4

→ ra+ rc = a/2

→ rc /ra = 0.414

→ PF = 4× 4/3 × πra³ + 4×4/3 × πrc³ /a³

→ d = 4× Mnacl / a³ ×Na

→ Cn of Na⁺ = 6

→ Cn of cl⁻= 6

Nearest distace between cation and aniom = a/2

Nearest next distance between cation and anions = √3a/2

nearest distance between two anions = a/ √2

nearest distance between two cations = a/ √2

nexr nearst distance between two cation or anions a

• ★ Cscl structure★

eff no. of A⁺ / eff no. of B⁻ = cn of B⁻ / Cn of A⁺

rc /ra = 0.732 or ra + rc = √3a/2

Pf = 1× 4/3 × πr³ + 1 × 4/3 × πrc³ / a³

d = 1 × Mcscl / a³ × Na

★ CN of cs⁺ = 8

★ CN of cl⁻ = 8

nearest distance between cation and anion = √3a/2

nearest distance between anion = a

nearest distance between anions a

Floride structure (Caf2)

Z(eff of ca⁺ )= 4

Z (eff of f⁻ ) = 8

Eff no. Of caf₂ per unit cell = 4

ra + rc = √3a/4= 0.225 (ideal case)

Answer 2

Answer:1. Number of atoms in different unit cells:

a. Primitive unit cell: 1 atom

b. Face centered unit cell: 4 atoms

c. Body centered unit cell: 2 atoms

2. Let the number of close packed spheres = N

Number of octahedral voids generated = N

Number of tetrahedral voids generated = 2N

3. Packing efficiency is the percentage of total space occupied by

constituent particles (atoms, molecules or ions).

Packing efficiency= Volume occupied by spheres in the unit cell x100% upon  Total volume of unit cell

a. Packing efficiency for face centered cubic unit cell =74%

b. Packing efficiency for body centered cubic unit cell = 68%

c. Packing efficiency for simple cubic unit cell =52.4%

Explanation: