## An ionic compound AB has ZnS type of structure, if the radius A+ is  22.5 pm , then the ideal radius of B- is

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)

## Close packing is maximum in the Crystal which is

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)

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

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

(A) mass of 8Ca++ ions & 4 Fions

(B) mass of 4Ca++ ions & 8 Fions

(C) mass of 4Ca++ ions & 4 Fions

(D) mass of 1Ca++ ions & 2 Fions

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)

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

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)

## In α-WCl6 …

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 W1/6Cl, i.e. 1/6  of the octahedral hole is filled by W.

Ans: (A), (B)

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

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

(A) 0.155 r

(B) 0.125 rA

(C) 0.414 rA

(D) 0.732 rA

Solution: For octahedral void

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

or,  rB = 0.414 rA

Ans: (C)

## The arrangement of Cl- ions in CsCl structure is

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)