With the help of a labelled diagram,show that there are $4$ octahedral voids per unit cell in a cubic close-packed $(CCP)$ structure.
(N/A) In a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ structure,the number of octahedral voids is equal to the number of atoms present in the unit cell.
$1$. The number of atoms per unit cell in $CCP$ is $4$.
$2$. One octahedral void is present at the body center of the cube. This void is entirely within the unit cell and is formed by $6$ atoms (one at each face center).
$3$. Additionally,there are $12$ octahedral voids located at the edges of the cube. Each edge-centered void is shared by $4$ adjacent unit cells. Therefore,the contribution of each edge-centered void to a single unit cell is $1/4$.
$4$. Total number of octahedral voids = (Number of voids at body center $\times 1$) + (Number of voids at edge centers $\times 1/4$)
$5$. Total number of octahedral voids = $(1 \times 1) + (12 \times 1/4) = 1 + 3 = 4$.
Thus,there are $4$ octahedral voids per unit cell in a $CCP$ structure.
$1$. The number of atoms per unit cell in $CCP$ is $4$.
$2$. One octahedral void is present at the body center of the cube. This void is entirely within the unit cell and is formed by $6$ atoms (one at each face center).
$3$. Additionally,there are $12$ octahedral voids located at the edges of the cube. Each edge-centered void is shared by $4$ adjacent unit cells. Therefore,the contribution of each edge-centered void to a single unit cell is $1/4$.
$4$. Total number of octahedral voids = (Number of voids at body center $\times 1$) + (Number of voids at edge centers $\times 1/4$)
$5$. Total number of octahedral voids = $(1 \times 1) + (12 \times 1/4) = 1 + 3 = 4$.
Thus,there are $4$ octahedral voids per unit cell in a $CCP$ structure.
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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 $= \dots \dots \%$.
$(a)$ In $ccp$ structure,$a = \dots \dots r$.
$(b)$ In $hcp$ structure,packing efficiency $= \dots \dots \%$.
$(c)$ In $bcc$ structure,percentage of packing efficiency $= \dots \dots \%$.
Difficult
View Solution$A$ compound is formed by two elements $X$ and $Y$. The element $Y$ forms a cubic close-packed $(CCP)$ arrangement and the atoms of element $X$ occupy one-third of the tetrahedral voids. What is the formula of the compound?
The number of tetrahedral voids in the unit cell of a face-centered cubic $(fcc)$ lattice of similar atoms is:
Medium
View SolutionThe $CORRECT$ statement$(s)$ for cubic close packed $(ccp)$ three dimensional structure is(are)
$(A)$ The number of the nearest neighbours of an atom present in the topmost layer is $12$
$(B)$ The efficiency of atom packing is $74 \%$
$(C)$ The number of octahedral and tetrahedral voids per atom are $1$ and $2$,respectively
$(D)$ The unit cell edge length is $2\sqrt{2}$ times the radius of the atom
$(A)$ The number of the nearest neighbours of an atom present in the topmost layer is $12$
$(B)$ The efficiency of atom packing is $74 \%$
$(C)$ The number of octahedral and tetrahedral voids per atom are $1$ and $2$,respectively
$(D)$ The unit cell edge length is $2\sqrt{2}$ times the radius of the atom
$A$ face-centred cubic $(FCC)$ unit cell is made up of two types of atoms $A$ and $B$,in which $A$ occupies the corner positions and $B$ occupies the face centres. If atoms along one of the body diagonals are removed,what is the empirical formula of the remaining solid?
Medium
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