Question

Crystal lattice is formed by atoms a, b,
c. A' occupies centre of the body, b' occupies face centre, and c' occupies corners. If atoms present at one of the face centre is removed and atoms at two corners are removed, expected formula of compound is a4b5c3a4b5c3 a4b10c3a4b10c3 a4b10c2a4b10c2 a2b5c3a2b5c3

Answer 1

Answer:

If atoms present at one of the face centre is removed and atoms at two corners are removed, expected formula of compound is a4b5c3a4b5c3 ...

Answer 2

Answer:

The empirical formula of the given compound is a₄b₁₀c₃.

Therefore the option (b) is correct.

Explanation:

Given: a, b and c are atoms in crystal lattice.

• Atom a is present in the body center.

        Number of 'a' atoms in crystal lattice $=1atom$

• Atom b is present on the faces of the lattice

       Number of b atoms present $= (6*\frac{1}{2})=3atoms$

• Atom c is present at the corners of the lattice.

       Number of c atoms in lattice $=(8*\frac{1}{8})=1atom$

  So, Empirical formula of the compound is $a_{1}b_{3}c_{1}$

After some atoms are being missing from the lattice:

• If one atom b is missing from face center,

        Number of b atoms in lattice $=(5*\frac{1}{2})=\frac{5}{2}$

• If two atoms from the corners are missing,

       Number of c atoms in lattice $=(6*\frac{1}{8})=\frac{3}{4}$

Therefore, the empirical formula becomes  $a_{1}b_{\frac{5}{2} }c_{\frac{3}{4} }=a_{4}b_{10}c_{3}$