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Formation of NaCl Lattice

HCP crystal i...

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HCP crystal in made of A ions and B ions are arranged in 2/3 of octahedral voids. The simplest formula of unit cell 1)A4B6 2)A6B4 3) A2B3 4) A3B

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Q. In a crystal oxide ions are arranged in fcc and

A^{2 +}

1 / 8^{t h}

of the tetrahedral voids, and ions

B^{3 +}

1 / 2

of the octahedral voids. Calculate the formula of the oxide.

Q. A compound with formula

A B_{2} O_{4}

adopts a crystal structure in which

O^{2 -}

ions forms ccp arrangement.

A^{2 +}

ions occupy octahedral voids while

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ions are distributed equally between tetrahedral and octahedral voids. The fraction of octahedral and tetrahedral voids occupied respectively are:

Q. If oxide ions are arranged in hcp and the aluminium ions occupy two thirds of octahedral voids. What will be the formula of the compound?

Q. A compound with formula

A B_{2} O_{4}

adopts a crystal structure in which

O^{2 -}

ions forms ccp arrangement.

A^{2 +}

ions occupy octahedral voids while

B^{3 +}

ions are distributed equally between tetrahedral and octahedral voids. The fraction of octahedral and tetrahedral voids occupied respectively are:

Q. A certain oxide of metal M crystallizes in such a way that

O^{2 -}

ions adopt an HCP arrangement that follows the ABABAB pattern. The metal ions, occupy

\frac{2}{3} r d

of the octahedral voids. The formula of the compound is:

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