Question

The structure of a crystalline lattice AB is rock salt type where B atom is occupying all the CCP positions but A atoms are occupying all the octahedral voids.

• A. If atoms lying on one tetrad axis of symmetric are removed, then the molecular formula of the compound becomes $AB$
• B. If atoms lying on one tetrad axis of symmetric are removed, then the molecular formula of the compound becomes $A_4{B}_6$
• C. If atoms lying on one tetrad axis of symmetric are removed, then the molecular formula of the compound becomes $A_5{B}_9$
• D. If atoms lying on one tetrad axis of symmetric are removed, then the molecular formula of the compound becomes $A$

Answer 1

The correct option is A If atoms lying on one tetrad axis of symmetric are removed, then the molecular formula of the compound becomes $AB$

A cube contains to total 13 axis of symmetry out of which

3 are tetrad-axis.

$\begin{array}{c}\text{Given,}{A}^{\oplus }\text{occupy octahedrat voids and}B\text{-atoms occupy}ccp\text{.}\\ \text{we know, octahedral void is located at centre of}ccP\\ \text{and at each edge centre.}\end{array}$

$\begin{array}{c}\text{When atoms lying on one tetrad axis is removed, it implies that}\\ \text{one atom of}A\text{is completely removed from lathice ab}H\text{is centre}\\ \text{of cube, possing all three tetrad ares.}\end{array}$

$\text{New formula :}{A}_{12\times \frac{1}{4}}{B}_4\equiv A_3{B}_4\text{. (Arter removing}A)$

New, we alco remove all B-atems one a two opposise Faces.

$\text{Final Fownula}\to A_3{B}_{4-2\times 1}=A_3{B}_3\equiv AB$

So, cerrect option is (A)