Solid State

Q: An ionic compound AB has ZnS type of structure, if the radius A⁺ is  22.5 pm , then the ideal radius of B^– is

(A) 54.35 pm

(B) 100 pm

(C) 145.16 pm

(D) None

Solution:    Since ionic compound AB has ZnS type of structure, therefore it has tetrahedral holes, for which,

$\large \frac{radius \; of\; cation}{radius \; of\; anion} = 0.225 $

$\large \frac{r_+}{r_-} = 0.225 $

$\large \frac{22.5}{r_-} = 0.225 $

Hence, r_– = 100 pm

Ans: (B)   

Q: Close packing is maximum in the Crystal which is

(A) Simple cube

(B) bcc

(C) fcc

(D) none

Solution: The close packing in the crystal is 0.52, 0.68 and 0.74 for simple cubic, body centered cubic, and face-centered cubic respectively.
i.e. the close packing is maximum in fcc.

Ans: (C)

Q: The mass of unit cell of CaF₂ (fluorite structure) corresponds to

(A) mass of 8Ca⁺⁺ ions & 4 F^– ions

(B) mass of 4Ca⁺⁺ ions & 8 F^– ions

(C) mass of 4Ca⁺⁺ ions & 4 F^– ions

(D) mass of 1Ca⁺⁺ ions & 2 F^– ions

Solution:  In CaF₂ (Calcium fluorite) structure 1-unit cell contains 4-Ca²⁺ ions and 8F^– ions.

So, mass of unit cell of CaF₂

= mass of 4 Ca²⁺ ions + mass of 8F ions

Ans:  (B)

Q: The edge length of a cube is 400 pm. Its body diagonal would be

(A) 600 pm

(B) 566 pm

(C) 693 pm

(D) 500 pm

Solution: Since in body center cubic, the body diagonal = √3 a

= √3 × 400 pm = 692.82 pm

= 693 pm

Ans:  (C)

Q: In α-WCl₆

(A) all Cl^– ions are present in cubic close packing

(B) W occupies, 1/6th of the octahedral holes.

(C) W occupies 1/3rd of the tetrahedral holes.

(D) All Cl^– ions are present in all octahedral

Solution: In α-WCl₆, Cl^– ions are arranged in cubic close packing, so, there is only one Cl–ion in 1-unit cell. So, the formula can be written W_1/6Cl, i.e. 1/6  of the octahedral hole is filled by W.

Ans: (A), (B)

Q: In closest packing of A type of atoms (radius, r_A), the radius of atom B that can be fitted into Octahedral void is

(A) 0.155 r_A 

(B) 0.125 r_A

(C) 0.414 r_A

(D) 0.732 r_A

Solution: For octahedral void

$\large \frac{r_B}{r_A} = 0.414 $

or,  r_B = 0.414 r_A

  Ans: (C)

Q: The arrangement of Cl^– ions in CsCl structure is

(A) hcp

(B) fcc

(C) bcc

(D) Simple cubic

Solution:  Arrangement of atoms or, ions in the corner of the unit cell is simple cubic. So in body centered cubic arrangement, Cl^– ions are arranged in the corner of the cube. So, it is a simple cubic.

Ans: (D)