Question

If the radius of a metal atom (A) is 5 pm and the radius of an electronegative atom (B) is 20 pm, then in the unit cell

(1) A in octahedral voids, B in FCC unit

(2) A in FCC unit, B in the tetrahedral void

(3) A in BCC unit, B in the cubic void

(4) A in the tetrahedral void, B in FCC unit

Answer 1

The explanation for the correct option:

Option(4), A in the tetrahedral void, and B in the FCC unit

• The metal atom with less electronegativity is A and hence, it will be a cation

• The metal atom with high electronegativity is B and hence, it will be an anion.
• If the ratio of the radius of cation to anion is between $0.225<\frac{r_{+}}{r_{-}}>0.414$ then the cation has a tetrahedral void and the anion will have an FCC arrangement
• Thus, the ratio of the radius of cation to anion can be calculated as:

$\frac{r_{+}}{r_{-}}=\frac{5}{20}=\frac{1}{4}=0.25$

• Hence the ratio of 0.25 falls in between ​$0.225<\frac{r_{+}}{r_{-}}>0.414$.
• Therefore, cation A of the compound occupies the tetrahedral void and anion B will have FCC arrangement.

Tetrahedral void

• when a sphere of the second layer is above the void of the first layer a tetrahedral void is formed, where the cation of the compound lies.
• Tetrahedral void is basically the empty space that is found in substances having tetrahedral crystal systems, So in the case of tetrahedral voids, these are present among four spheres that have a tetrahedral arrangement

FCC arrangement

• A face-centered cubic unit cell structure consists of atoms arranged in a cube where each corner of the cube has a fraction of an atom with six additional full atoms positioned at the center of each cube face.
• The atoms at the corner of the cube are shared with eight other unit cells
  

Hence, option(4), A in the tetrahedral void, and B in the FCC unit is the correct one.