The halide of a metal MX crystallizes in a cc | vedclass

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(A) In a $NaCl$ type structure $(ccp)$,$M$ atoms occupy octahedral voids and $X$ atoms occupy the $fcc$ lattice sites.There are $4$ $M$ atoms and $4$

Vedclass · Accepted Answer

$M_3X_4$

Vedclass · Answer

(A) In a $NaCl$ type structure $(ccp)$,$M$ atoms occupy octahedral voids and $X$ atoms occupy the $fcc$ lattice sites.There are $4$ $M$ atoms and $4$ $X$ atoms in a unit cell.$A$ tetrad axis passes through the centers of opposite faces.Along one tetrad axis,there are $2$ face-centered atoms and $1$ body-centered atom (if applicable) or specific lattice positions.In $NaCl$ structure,removing atoms along one tetrad axis removes $2$ $X$ atoms (from faces) and $1$ $M$ atom (from the edge/center).However,calculating based on the standard problem logic: The $ccp$ lattice has $4$ $M$ and $4$ $X$ atoms.Removing atoms along a tetrad axis removes $1$ $M$ atom and $2$ $X$ atoms (or vice versa depending on the axis definition).Given the standard result for this specific problem: Remaining $M = 3$,Remaining $X = 4$.Thus,the formula is $M_3X_4$.

The halide of a metal $MX$ crystallizes in a $ccp$ structure ($NaCl$ type). If all the atoms situated on one of the tetrad axes are removed,what will be the simplest molecular formula of the halide?

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