Question

In BeO( Zinc Blende structure), Mg2+ is introduced in available tetrahedral voids. Then ions are removed from a single body diagonal of the unit cell. What will be the molecular formula of the unit cell?

Answer 1

There are eight tetrahedral voids in BeO. If Mg²⁺ ions occupy all tetrahedral voids then,

Number of Be²⁺ ions in tetrahedral voids=4

Number of O²⁻ ions in tetrahedral voids=4

Number of Mg²⁺ ions in tetrahedral voids=4

If ions are removed from a single body diagonal of the unit cell, then

Number of Be²⁺ ions in tetrahedral voids=4

Number of O²⁻ ions in tetrahedral voids=3/4

Number of Mg²⁺ ions in tetrahedral voids=4

Then the molecular formula of the unit cell will be $Be_{4}Mg_{4}O_{\frac{3}{4}}$.

Answer 2

• In BeO , Oxide ions are at ccp. Thus o2- ions are 4 in number.
• Be2+ ions occupy 4 out of 8 tetrahedral voids.
• Thus we are left with 4 vacant tetrahedral voids which are occupied by Mg2+ ions.
• Now on removing one body diagonal ions we remove one mg2+ ion and one Be2+ ion. Thus their number reduce to 3 each.
• Also we remove two o2- ions present at corners. As each atom at corner contributes 1/8th to lattice so no of o2- ions become 4-2/8 = 3.75.
• Thus resulting formula is Be3Mg3O3.75.