A solid is formed by two elements $P$ and $Q$ . The element $Q$ forms cubic close packing and atoms of $P$ occupy one-third of tetrahedral voids. The formula of the compound is:

P_{7} Q_{3}

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P_{3} Q

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P_{2} Q_{3}

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P_{3} Q_{2}

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Solution

The correct option is C $P_{2} Q_{3}$
Effective number of $Q, Z_{Q} = 4$ (FCC Lattice)
Effective number of $P, Z_{P} = \frac{1}{3} \times$ Tetrahedral void $= \frac{1}{3} \times 8 = \frac{8}{3}$
Formula of compound $= P_{\frac{8}{3}} Q_{4} = P_{2} Q_{3}$

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A solid is formed by two elements P and Q. The element Q forms cubic close packing and atoms of P occupy one-third of tetrahedral voids. The formula of the compound is:

The correct option is C P2Q3Effective number of Q,ZQ=4 (FCC Lattice)Effective number of P,ZP=13× Tetrahedral void =13×8=83Formula of compound =P83Q4=P2Q3

A solid is formed by two elements $P$ and $Q$ . The element $Q$ forms cubic close packing and atoms of $P$ occupy one-third of tetrahedral voids. The formula of the compound is:

The correct option is C $P_{2} Q_{3}$
Effective number of $Q, Z_{Q} = 4$ (FCC Lattice)
Effective number of $P, Z_{P} = \frac{1}{3} \times$ Tetrahedral void $= \frac{1}{3} \times 8 = \frac{8}{3}$
Formula of compound $= P_{\frac{8}{3}} Q_{4} = P_{2} Q_{3}$