In a solid, $P$ atoms form FCC, $Q$ atoms are present in alternate edge centres, $R$ in alternate tetrahedral voids. If $P$ atoms on the two face centres are replaced by $X$ atoms, then the formula of the compound is:

P_{3} Q R_{4} X

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P_{3} Q_{2} R_{4} X

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P_{2} Q_{2} R_{3} X_{2}

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P_{4} Q R_{4} X_{2}

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Solution

The correct option is A $P_{3} Q R_{4} X$
In FCC,
$Z = 4$
$No. of tetrahedral voids = 8$
$No. of edge centres = 12$
$Alternate tetrahedral voids = 4$
$Alternate edge centres = 4$
$Hence, original crystal is = P_{4} Q_{1} R_{4}$
After replacement of $P$ atoms,
$Crystal is = P_{3} Q R_{4} X$

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In a solid, P atoms form FCC, Q atoms are present in alternate edge centres, R in alternate tetrahedral voids. If P atoms on the two face centres are replaced by X atoms, then the formula of the compound is:

The correct option is A P3QR4XIn FCC, Z=4 No. of tetrahedral voids=8 No. of edge centres=12 Alternate tetrahedral voids=4 Alternate edge centres=4 Hence, original crystal is=P4Q1R4 After replacement of P atoms, Crystal is=P3QR4X

In a solid, $P$ atoms form FCC, $Q$ atoms are present in alternate edge centres, $R$ in alternate tetrahedral voids. If $P$ atoms on the two face centres are replaced by $X$ atoms, then the formula of the compound is:

The correct option is A $P_{3} Q R_{4} X$
In FCC,
$Z = 4$
$No. of tetrahedral voids = 8$
$No. of edge centres = 12$
$Alternate tetrahedral voids = 4$
$Alternate edge centres = 4$
$Hence, original crystal is = P_{4} Q_{1} R_{4}$
After replacement of $P$ atoms,
$Crystal is = P_{3} Q R_{4} X$