MHT CET · Chemistry · Solid State
In a solid, \(\mathrm{B}^{-}\)ions occupy corners of a cube forming ccp structure. If \(A^{+}\)ion occupy half the tetrahedral voids, formula of the solid is
- A \(\mathrm{A}_2 \mathrm{~B}\)
- B \(\mathrm{AB}_2\)
- C \(A_2 B_3\)
- D AB
Answer & Solution
Correct Answer
(D) AB
Step-by-step Solution
Detailed explanation
\(\mathrm{B}^{-}\)ions occupy corners of cube and forms ccp structure.
The number of tetrahedral voids generated is twice the number of \(\mathrm{B}^{-}\)ions.
Thus, number of tetrahedral voids \(=2 \mathrm{~B}\)
\(\mathrm{A}^{+}\)ions occupy ( \(1 / 2\) ) of these tetrahedral voids.
Hence, number of \(\mathrm{A}^{+}\)ions \(=2 \mathrm{~B} \times 1 / 2=1 \mathrm{~B}\)
Ratio of \(\mathrm{A}^{+}\)and \(\mathrm{B}^{-}\)ions \(=1 \mathrm{~B}: 1 \mathrm{~B}=1: 1\)
\(\therefore \quad\) Formula of compound \(=\mathrm{AB}\)
The number of tetrahedral voids generated is twice the number of \(\mathrm{B}^{-}\)ions.
Thus, number of tetrahedral voids \(=2 \mathrm{~B}\)
\(\mathrm{A}^{+}\)ions occupy ( \(1 / 2\) ) of these tetrahedral voids.
Hence, number of \(\mathrm{A}^{+}\)ions \(=2 \mathrm{~B} \times 1 / 2=1 \mathrm{~B}\)
Ratio of \(\mathrm{A}^{+}\)and \(\mathrm{B}^{-}\)ions \(=1 \mathrm{~B}: 1 \mathrm{~B}=1: 1\)
\(\therefore \quad\) Formula of compound \(=\mathrm{AB}\)
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