Question

A solid is formed by two elements p and q. The element q forms cubic close packing and atoms of p occupy one-third of tetrahedral voids. The formula of the compound is

Answer 1

Answer: The formula of the compound will be $p_2q_3$.

Solution:

Let the number of atoms in cubic close packing = N

Then the total number of tetrahedral void in cubic close packing = 2N

Given that element 'q' form cubic packing

So number of atoms of element q= N

Element p occupies 1/3 of tetrahedral voids

So the number of atoms of element p= $\frac{1}{3}\times 2N=\frac{2}{3}N$

Number of atoms of element p:number of atoms of element q

                       $\frac{2}{3}N:1N=\frac{2}{3}:1=2:3$

Hence formula will be $p_2q_3$

Answer 2

p=1/3of tetrahedral void

q=CCP

no of atom in CCP =4

So no of TV=8

P=1/3×8=8/3

Q=4

P 8/3 Q4. (we must multiply by 3)

P8Q12

P2Q3