If a compound (oxide) has an hcp arrangement | vedclass

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Question

(C) In an $hcp$ arrangement,the number of atoms per unit cell is $6$.The number of octahedral voids is equal to the number of atoms in the lattice,whi

Vedclass · Accepted Answer

$M_2O_3$

Vedclass · Answer

(C) In an $hcp$ arrangement,the number of atoms per unit cell is $6$.The number of octahedral voids is equal to the number of atoms in the lattice,which is $6$.The cations $M$ occupy two-thirds of the octahedral voids,so the number of $M$ cations per unit cell is $\frac{2}{3} 	imes 6 = 4$.The number of oxide ions $(O)$ per unit cell is $6$.Therefore,the ratio of $M:O$ is $4:6$,which simplifies to $2:3$.Thus,the formula of the compound is $M_2O_3$.

If a compound (oxide) has an $hcp$ arrangement and the cations $M$ occupy two-thirds of the octahedral voids,the formula of the compound is:

Similar Questions

Fill in the blanks:
$(a)$ In $ccp$ structure,$a = \dots \dots r$.
$(b)$ In $hcp$ structure,packing efficiency $= \dots \dots \%$.
$(c)$ In $bcc$ structure,percentage of packing efficiency $= \dots \dots \%$.

Atoms of element $B$ form $hcp$ lattice and those of the element $A$ occupy $2/3$ of tetrahedral voids. What is the formula of the compound formed by the elements $A$ and $B$?

An alloy of metals $A$,$B$,and $C$ is found to have $A$ constituting a $ccp$ lattice. If $B$ atoms occupy the edge centers and $C$ is present at the body center,the formula of the alloy is:

In a crystalline solid,anion $B$ are arranged in $ccp$ lattice and cation $A$ occupy $50\%$ of the octahedral voids and $50\%$ of tetrahedral voids. What will be the formula of the crystalline solid?

Which of the following statements is correct for an $fcc$ lattice?

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