A face centred cubic lattice is made up of two types of atoms A and B, in which A occupies the comer positions and B occupies the face centres. If atom along an axis joining the diagonally opposite comers on a face are removed, the empirical formula of the remaining solid would be

Question

A face centred cubic lattice is made up of two types of atoms A and B, in which A occupies the comer positions and B occupies the face centres. If atom along an axis joining the diagonally opposite comers on a face are removed, the empirical formula of the remaining solid would be (A)  A2B5  (B)  A3B7  (C)  A3B10  (D)  A7B24

Tardigrade Admin · Accepted Answer

A face of this solid would appear as Missing A=(1/8)× 2=(1/4) A present =1-(1/4)=(3/4) Missing B=(1/2) B present =3-(1/2)=(5/2) formula A(3/4)B(5/2)=A3B10

Q. A face centred cubic lattice is made up of two types of atoms A and B, in which A occupies the comer positions and B occupies the face centres. If atom along an axis joining the diagonally opposite comers on a face are removed, the empirical formula of the remaining solid would be

$A_{2} B_{5}$

$A_{3} B_{7}$

$A_{3} B_{10}$

$A_{7} B_{24}$

A face of this solid would appear as
Missing $A = \frac{1}{8} \times 2 = \frac{1}{4}$ A present $= 1 - \frac{1}{4} = \frac{3}{4}$ Missing $B = \frac{1}{2}$ B present $= 3 - \frac{1}{2} = \frac{5}{2}$ formula $A_{\frac{3}{4}} B_{\frac{5}{2}} = A_{3} B_{10}$

Q. A face centred cubic lattice is made up of two types of atoms A and B, in which A occupies the comer positions and B occupies the face centres. If atom along an axis joining the diagonally opposite comers on a face are removed, the empirical formula of the remaining solid would be

A2​B5​

A3​B7​

A3​B10​

A7​B24​

A face of this solid would appear as Missing A=81​×2=41​ A present =1−41​=43​ Missing B=21​ B present =3−21​=25​ formula A43​​B25​​=A3​B10​

$A_{2} B_{5}$

$A_{3} B_{7}$

$A_{3} B_{10}$

$A_{7} B_{24}$

A face of this solid would appear as
Missing $A = \frac{1}{8} \times 2 = \frac{1}{4}$ A present $= 1 - \frac{1}{4} = \frac{3}{4}$ Missing $B = \frac{1}{2}$ B present $= 3 - \frac{1}{2} = \frac{5}{2}$ formula $A_{\frac{3}{4}} B_{\frac{5}{2}} = A_{3} B_{10}$