A2B has antifluorite structure ( textB forms textFCC lattice and textA occupies tetrahedral voids). If all ions along any one body diagonal are removed, then new formula of compound will be:

Question

A2B has antifluorite structure ( textB forms textFCC lattice and textA occupies tetrahedral voids). If all ions along any one body diagonal are removed, then new formula of compound will be: (A) A 4 B 5  (B) A 8 B 5 (C) A 7 B 6  (D) A 8 B 4

Tardigrade Admin · Accepted Answer

On one body diagonal, 2 A atoms(full contribution) 2 B-atoms (1/8) contributions lie. So, new formula = A8-2 B4-(1/4)=A8 B5

Q. $A_{2} B$ has antifluorite structure ( $B$ forms $FCC$ lattice and $A$ occupies tetrahedral voids). If all ions along any one body diagonal are removed, then new formula of compound will be:

$A_{4} B_{5}$

$A_{8} B_{5}$

$A_{7} B_{6}$

$A_{8} B_{4}$

On one body diagonal, $2 A$ atoms(full contribution) $2 B$ -atoms $\frac{1}{8}$ contributions lie. So, new formula $=$ $A_{8 - 2} B_{4 - \frac{1}{4}} = A_{8} B_{5}$

Q. A2​B has antifluorite structure ( B forms FCC lattice and A occupies tetrahedral voids). If all ions along any one body diagonal are removed, then new formula of compound will be:

A4​B5​

A8​B5​

A7​B6​

A8​B4​