JEE CHEMISTRY MCQ: SOLID STATE

1 Hard

1 of 1

Analyzing +4 | -1

In a fcc lattice

A, B, C

and

D

atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing, correspondingly then what is the resulted formula of the compound?

View answer and explanation

Correct option: B

{AB}_4 {C}_4 {D}_4

Option B is correct. After removing atoms along the body diagonal, the formula becomes AB₄C₄D₄, calculated by considering the contributions of atoms at corners, faces, octahedral voids, and tetrahedral voids.

More Questions from SOLID STATE

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{CaF}_2

, fluorite type structure,

{F}^{-}

ions occupy Consider a body centred cubic (BCC) arrangement, let

\mathrm{d}_{\mathrm{e}}

\mathrm{d}_{\mathrm{fd}}, \mathrm{d}_{\mathrm{bd}}

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\mathrm{AB}_2

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\mathrm{A}

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