In a crystal of an ionic compound, the ions of element B form the close-packed lattice, and the ions of element A occupy all the tetrahedral voids. The formula of the compound is:

A B_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A_{2} B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $A_{2} B$
The formula is $A_{2} B$ . There are two tetrahedral holes for every B ion. These tetrahedral holes are occupied by A ions. Thus the ratio of A to B is 2:1.

Suggest Corrections

Similar questions

Q. Atoms of an element 'A' occupy

\frac{2}{3}

tetrahedral voids in the hexagonal close-packed (hcp) unit cell lattice formed by the element 'B'. The formula of the compound formed by 'A ' and 'B' is :

In a compound, oxide ions constitute cubic close packing. Cations A occupy 50% of tetrahedral holes while cations B occupy all the octahedral voids. The empirical formula of the compound is ?

Q. In a hypothetical ionic crystal,

B

is arranged in cubic close packing and

A

occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound?

In which of the following structures coordination number for cations and anions in the packed structure will be same ?

(a) $C l^{-}$ ions form fcc lattice and $N a^{+}$ ions occupy all octahedral voids of the unit cell

(b) $C a^{2 +}$ ions form fcc lattice and $F^{-}$ ions occupy all the eight tetrahedral voids of the unit cell

(d) $S^{2 -}$ ions form fcc lattice and $Z n^{2 +}$ ions go into alternate tetrahedral voids of the unit cell

Q. Atoms of element B form hcp lattice and those of element A occupy two-third of tetrahedral voids. The formula of compound is:

Join BYJU'S Learning Program

In a crystal of an ionic compound, the ions of element B form the close-packed lattice, and the ions of element A occupy all the tetrahedral voids. The formula of the compound is:

The correct option is B A2BThe formula is A2B. There are two tetrahedral holes for every B ion. These tetrahedral holes are occupied by A ions. Thus the ratio of A to B is 2:1.

The correct option is B $A_{2} B$
The formula is $A_{2} B$ . There are two tetrahedral holes for every B ion. These tetrahedral holes are occupied by A ions. Thus the ratio of A to B is 2:1.