Atoms C and D form fcc crystalline structure. Atom C is present at the

1.	Atoms C and D form fcc crystalline structure. Atom C is present at the corners of the cube and D is at the faces of the cube. What is the formula of the compound?
Answer» Given: Atoms C and D form fcc lattice. C is present at corners of the cube. D is at the faces of the cube. To find: Formula of the compound Calculation: As C is present at the 8 corners of the cube, number of atoms of C in the unit cell\(=\frac{1}{8}\times8=1\) As D atoms are present at the face centres of the 6 faces of the cube, the number of atoms of D in the unit cell =\(=\frac{1}{2}\times6=3\) \(\therefore\) Ratio of atoms C : D = 1 : 3. Ans: The formula of the compound is CD₃.

Atoms C and D form fcc crystalline structure. Atom C is present at the corners of the cube and D is at the faces of the cube. What is the formula of the compound?

Answer»

Given: Atoms C and D form fcc lattice.

C is present at corners of the cube.

D is at the faces of the cube.

To find: Formula of the compound

Calculation: As C is present at the 8 corners of the cube, number of atoms of C in the unit cell\(=\frac{1}{8}\times8=1\)

As D atoms are present at the face centres of the 6 faces of the cube, the number of atoms of D in the unit cell =\(=\frac{1}{2}\times6=3\)

\(\therefore\) Ratio of atoms C : D = 1 : 3.

Ans: The formula of the compound is CD₃.

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