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Question

# A crystal is made up of metal iron $\text{'}{\mathrm{M}}_{1}\text{'}$ and $\text{'}{\mathrm{M}}_{2}\text{'}$ and oxide ions. Oxide ions form a $\mathrm{ccp}$ lattice structure. The cation ‘$\text{'}{\mathrm{M}}_{1}\text{'}$ occupies 50% of octahedral voids and the cation $\text{'}{\mathrm{M}}_{2}\text{'}$occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation number of $\text{'}{\mathrm{M}}_{1}\text{'}$ and $\text{'}{\mathrm{M}}_{2}\text{'}$ are, respectively:

A

$+2,+4$

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B

$+3,+2$

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C

$+4,+2$

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D

$+1,+3$

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Solution

## The correct option is A $+2,+4$Answer:(A)Explanation for correct option: In $\mathrm{ccp}$ lattice oxide ion $=$corners $+$ face centered $=\frac{1}{8}×8+\frac{1}{2}×6\phantom{\rule{0ex}{0ex}}=4$ Number of ($\mathrm{O}$) $=4\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{O}\right)}_{4}$In $\mathrm{ccp}$ lattice ,Number of octahedral void$=4$Number of tetrahedral void$=8$ $\left({\mathrm{M}}_{1}\right)\to 50%$ of octahedral void $\mathrm{Number}\mathrm{of}\left({\mathrm{M}}_{1}\right)=\frac{50}{100}×4\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}⇒{\left({\mathrm{M}}_{1}\right)}_{2}$$\left({\mathrm{M}}_{2}\right)\to 12.5%$of tetrahedral void$\mathrm{Number}\mathrm{of}\left({\mathrm{M}}_{2}\right)=\frac{12.5}{100}×8\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}={\left({\mathrm{M}}_{2}\right)}_{1}$Hence formula must be ${\left({\mathrm{M}}_{1}\right)}_{2}\left({\mathrm{M}}_{2}\right){\mathrm{O}}_{4}$For whole atom to be neutral(as oxidation state of $\mathrm{O}=-2$), $\mathrm{O}.\mathrm{N}.\mathrm{of}{\mathrm{M}}_{1}=+2\phantom{\rule{0ex}{0ex}}\mathrm{O}.\mathrm{N}.\mathrm{of}{\mathrm{M}}_{2}=+4$As, ${\left({\mathrm{M}}_{1}\right)}_{2}\left({\mathrm{M}}_{2}\right){\mathrm{O}}_{4}\phantom{\rule{0ex}{0ex}}=2×\left(+2\right)+1×\left(+4\right)+4×\left(-2\right)\phantom{\rule{0ex}{0ex}}=4+4+\left(-8\right)\phantom{\rule{0ex}{0ex}}=8-8\phantom{\rule{0ex}{0ex}}=0$Therefore, correct option is A

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