Question

A solid AB has ZnS-type structure. If the edge
length of the unit cell is 400 pm and the radius of
Bion is 0.130 nm then the radius of A+ ion is
(1) 35.8 pm
(2) 43.2 pm .
3) 60.5 pm
(4) 532 pm
​

Answer 1

Solution :

a=2(r++r−)a=2(r++r−)

500=2(80+r−)500=2(80+r−)

500=160+r−500=160+r−

500−160=2r−500−160=2r−

340=2r−340=2r−

34023402=r−=r−

r−=170r−=170

∴∴ the radius of anion =170pm

Answer 2

answer : option (2)

explanation : A solid AB has ZnS- type structure . where edge length of unit cell is 400pm. i.e., a = 400 pm

and radius of $B^-$ ion = 0.130nm or 130 pm.i.e., $r^-$ = 130pm.

we have to find radius of $A^+$ ion.

we know, $4r^++4r^-=\sqrt{3}a$

so, $4r^++4\times130=400\sqrt{3}$

or, $r^++130=100\sqrt{3}$

or, $r^+=100\sqrt{3}-130$

or, $r^+=173.2-130=43.2pm$

hence, radius of $A^+$ ion is 43.2pm.

hence, option (2) is correct.