In a fcc arrangement of A and B atoms where A atoms are at the corners of the unit cell and B atoms at the face centres, one of the A atom is missing from one corner in each unit cell. What is the simplest formula of compound?

A_{7} B_{24}

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A_{8} B_{27}

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A_{2} B_{5}

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A_{5} B_{24}

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Solution

The correct option is C $A_{7} B_{24}$
Number of atoms of A with one atom removed from one corner of unit cell $= \frac{7}{8}$
Number of atom B from the face of unit cell $= 3$
Thus, $A : B :: \frac{7}{8} : 3$ or $A : B :: 7 : 24$ .

Thus, the formula is $A_{7} B_{24}$ .

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In a fcc arrangement of A and B atoms where A atoms are at the corners of the unit cell and B atoms at the face centres, one of the A atom is missing from one corner in each unit cell. What is the simplest formula of compound?

The correct option is C A7B24Number of atoms of A with one atom removed from one corner of unit cell =78Number of atom B from the face of unit cell =3Thus, A:B::78:3 or A:B::7:24.Thus, the formula is A7B24.

The correct option is C $A_{7} B_{24}$
Number of atoms of A with one atom removed from one corner of unit cell $= \frac{7}{8}$
Number of atom B from the face of unit cell $= 3$
Thus, $A : B :: \frac{7}{8} : 3$ or $A : B :: 7 : 24$ .

Thus, the formula is $A_{7} B_{24}$ .