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)   

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)

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)

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)