Question

A crystalline solid AB has NaCl type structure with radius of B negative ion is 250 pm. Which of the following cation can be made to slip into tetrahedral site of crystals of A positive B negative?

Answer 1

NaCl lattice is a simple cubic unit or FCC unit lattice.

Here B is the negative atom, generally containing the bigger radius and thus positive charged atom A has lower radius.

In case of tetrahedral voids,

a=√2R

or, R=√2a/2....(i)

where, a is one side of the lattice structure and R is the radius of the lattice point.

Here, R=250 pm

Now if 'r' be the radius of the void, for a diagonal of a cube,

2(R+r) = √3a.....(ii)

dividing equation (ii) by (i);

2(R+r)/R = √3a x 2 / √2a

or, 1 + r/R = √3/√2

or, r/R = 0.225

For tetrahedral void, r = 0.225 x 250 pm = 56.25 pm

The value of the radius of the atom to be fitted inside the void must be lesser than(or equal) 56.25 pm.

Answer 2

Answer:option 2 is correct

Explanation:

Q+ and B- ion's radius ration is near 0.225...

we know tetrahedral means Co-ordination number=4 and radius ratio have to be between 0.225- 0.414