
In a compound atom of element ‘Y’ form CCP lattice and those of element ‘X’ occupy 2/3rd of tetrahedral voids. The formula of the compound will be-
(A)- ${{X}_{2}}{{Y}_{3}}$
(B)- ${{X}_{2}}Y$
(C)- ${{X}_{3}}{{Y}_{4}}$
(D)- ${{X}_{4}}{{Y}_{3}}$
Answer
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Hint: In solids, the constituent particles are packed such that there is minimum vacant space and constitutes a regular arrangement known as ‘Closed packing’. Voids are gaps or the vacant space between the constituent particles. Voids in respect of solid states mean the vacant space between the constituent particles in a closely packed structure.
Complete Step by step answer:
-Closed packing in two dimensional and three dimensional gives rise to two types of voids, viz. tetrahedral voids and octahedral voids.
-In three-dimensional structure, cubic close packing as well as hexagonal close packing both have 26% of the total space as vacant space and contribute to the formation of tetrahedral or octahedral voids. These empty spaces are also known as interstitial voids, interstices or holes.
-In cubic close packing structure, the sphere of the second layer lies above the triangular voids created by the first layer. Each sphere touches the three spheres of the first layer and by joining the centre of these four spheres forms a tetrahedron. The space left by joining the centre of these spheres forms a tetrahedral void.
-In a cubic close packing, let the number of spheres be n. Then the number of tetrahedral voids will be 2n.
-According to question,
Y atom occupies the CCP lattice, hence the effective number of Y atoms = 4
(As the effective number of atoms in CCP is 4)
The number of tetrahedral voids generated in any unit cell = 2n (n is the number of atoms)
$=2\times 4=8$
X occupies 2/3rd of tetrahedral voids.
Therefore, the effective number of X $=\dfrac{2}{3}\times 8=\dfrac{16}{3}$
-Therefore the ratio of X: Y from the calculation is $\dfrac{16}{3}:4$
Converting the ratio into the simplest integer, we get
$X:Y\Rightarrow 4:3$
Hence the formula of the compound will be ${{X}_{4}}{{Y}_{3}}$.
Therefore, the correct answer is option D.
Note: Another type of void present in the closed packing is octahedral void. When the triangular voids of the first layer coincide with the triangular voids of the layer present above or below it, we get a void which is enclosed by six spheres. This vacant space formed by combining the triangular voids of the first and second layers is known as Octahedral voids. If the number of the sphere in a closed packed structure is ‘n’, then the number of octahedral voids is also equal to ‘n’.
Complete Step by step answer:
-Closed packing in two dimensional and three dimensional gives rise to two types of voids, viz. tetrahedral voids and octahedral voids.
-In three-dimensional structure, cubic close packing as well as hexagonal close packing both have 26% of the total space as vacant space and contribute to the formation of tetrahedral or octahedral voids. These empty spaces are also known as interstitial voids, interstices or holes.
-In cubic close packing structure, the sphere of the second layer lies above the triangular voids created by the first layer. Each sphere touches the three spheres of the first layer and by joining the centre of these four spheres forms a tetrahedron. The space left by joining the centre of these spheres forms a tetrahedral void.
-In a cubic close packing, let the number of spheres be n. Then the number of tetrahedral voids will be 2n.
-According to question,
Y atom occupies the CCP lattice, hence the effective number of Y atoms = 4
(As the effective number of atoms in CCP is 4)
The number of tetrahedral voids generated in any unit cell = 2n (n is the number of atoms)
$=2\times 4=8$
X occupies 2/3rd of tetrahedral voids.
Therefore, the effective number of X $=\dfrac{2}{3}\times 8=\dfrac{16}{3}$
-Therefore the ratio of X: Y from the calculation is $\dfrac{16}{3}:4$
Converting the ratio into the simplest integer, we get
$X:Y\Rightarrow 4:3$
Hence the formula of the compound will be ${{X}_{4}}{{Y}_{3}}$.
Therefore, the correct answer is option D.
Note: Another type of void present in the closed packing is octahedral void. When the triangular voids of the first layer coincide with the triangular voids of the layer present above or below it, we get a void which is enclosed by six spheres. This vacant space formed by combining the triangular voids of the first and second layers is known as Octahedral voids. If the number of the sphere in a closed packed structure is ‘n’, then the number of octahedral voids is also equal to ‘n’.
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