Test Index

BITSAT 2021 Paper

Section: Chemistry

Question : 52 of 150

Marks: +1, -0

Solution:

No. of lattice points = No. of octahedral voids

=\frac{1}{2} \times

No. of tetrahedral voids in

c\,c\,p

structure

∴ No. of atoms of B = 4

No. of atoms of

A=\frac{1}{2} \times

No. of octahedral voids

=\frac{1}{2} \times 4=2

No. of atoms of

O=

All tetrahedral voids

=2 \times

No. of lattice points

=2 \times 4=8

Hence,

A: B: O = 1: 2: 4

Therefore, the formula of the compound is

\mathrm{AB}_{2} \mathrm{O}_{4}

Go to Question:

Prev Question

Next Question