Category: Solid State

Q: Fraction of total volume occupied by atoms in a simple cubic cell is

(A) π/2

(B) √3π/2

(C) √2π/2

(D) π/6

Ans: (D)

Sol: In simple cubic arrangement, no. of atoms = 1

a = 2r

Packing fraction = Volume occupied by one atom / Volume of unit cell

$\large = \frac{\frac{4}{3}\pi r^3}{a^3} $

$\large = \frac{\frac{4}{3}\pi r^3}{(2r)^3} = \frac{\pi}{6} $

Q: A solid is made of two elements P & Q . Atoms P are in ccp arrangement and atoms Q occupy all the octahedral voids and half of the tetrahedral voids, then the simplest formula of the compound is

(A) PQ₂

(B) P₂Q

(C) PQ

(D) P₂Q₂

Solution: Four atoms (P) contributes to one unit cell from ccp arrangement and 4-atoms (Q) from the all octahedral voids and 4-atoms (Q) from the half of the tetrahedral void contributes one unit cell.

So, formula of solid is P₄Q₈ so, the simplest formula of the solid is PQ₂

Ans: (A)

Q: Select the correct statement(s)

(A) The C.N. of cation occupying a tetrahedral hole is 4

(B) The C.N. of cation occupying a octahedral hole is 6

(C) In Schotky defects, density of the lattice decreases.

(D) In Frenkel defects, density of the lattice increases.

Solution: Since tetrahedral holes are surrounded by 4 nearest neighbours. So, the C.N. of cation occupying tetrahedral hole is 4. Since octahedral hole is surrounded by six nearest neighbours, so, C.N. of cation occupying octahedral is 6.

In schottky a pair of anion and cation leaves the lattice, so density of lattice decreases.

Ans: (A), (B), (C)

Q: In fluorite structure (CaF₂)

(A) Ca⁺⁺ ions are ccp & ions are present in all the tetrahedral voids

(B) Ca⁺⁺ ions are ccp & ions are present in all the octahedral voids

(C) Ca⁺⁺ ions are ccp & ions are present in all the octahedral voids and half of ions are

(D) None

Solution: Ca²⁺ ions are ccp and F^– ions are present in the tetrahedral voids. So, the no. of Ca²⁺ ions is 4 and no. of F^– ions is 8.

So, the formula of the Calcium fluoride Ca₄F₈ or, the simplest formula of calcium fluoride is CaF₂

Ans: (A)

Q: In a fcc arrangement of A & B atoms, where A atoms are at the Corners of the unit cell, B atoms at the face centers, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is

(A) A₇B₆

(B) A₆B₇

(C) A₇B₂₄

(D) AB₄

Solution: Since there are six atoms (A) in the corner of the unit cell. So, contribution of atoms in 1 unit cell is .

Since 3 face-centered atoms (B) contributes to one unit cell

So, formula is A_6/8 B₃

or A₆ B₂₄, or AB₄

Ans: (D)