Question

a binary solid A+B- has a structure with B- ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. formula of the solid is?
a. A2B b. AB c. AB2 d. AB4

Answer 1

Let the number of $B^{-}$ ions be 'N'.
We know that if the number of particles is 'N', the number of tetrahedral voids will be 2N.

However, $A^{+}$ ions are occupying only 25% of the tetrahedral holes. This means that the number of $A^{+}$ ions will be 25% of 2N, i.e., (1/4)*2N which is equal to N/2.

Therefore, the formula of the solid will be:
A - N/2
B - N

Multiplying both by 2, the formula of the solid becomes
A - N
B - 2N

Now, we can cancel out N present in both A and B. Thus, the formula of the solid will be AB$_{2}$

Hence, (c) will be the correct option.

Answer 2

Answer:

A - N/2

B - N

Multiplying both by 2, the formula of the solid becomes

A - N

B - 2N

Now, we can cancel out N present in both A and B ,givesAB

Hence, (c) will be the correct option.