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A solid has a structure in which ‘W’ atoms are located at the corners of a cubic lattice ‘O’ atom at the corner of edge and Na atoms at the center of the cube. The formula for the compound is:
 a. ${ Na }_{ 2 }{ WO }_{ 3 }$
 b. ${ Na }_{ 2 }{ WO }_{ 2 }$
 c. ${ Na }{ WO }_{ 2 }$
 d. ${ Na }{ WO }_{ 3 }$

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Last updated date: 28th Mar 2024
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MVSAT 2024
Answer
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Hint: As we know that in different unit cells, constituent particles are present at the corners, at the centre of faces or at the centre of body. It is also clear that each unit cell of crystalline solid is present near the other unit cell. So, the constituent particles of a cell are also shared by neighbouring cells.

Complete answer:
We know that a crystal lattice is formed from many numbers of unit cells and a constituent particle is present at each lattice point. Each unit cell resembles the geometry of a cube.

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There are three types of cubic centered unit cells.

i. Simple cubic unit cell: constituent particles are present at each corner of a simple cubic unit cell. In this unit cell, each constituent particle is distributed in eight simple unit cells from which four are situated toward the lower and four are situated toward the upper end. So, number of particles per unit cell is $8\times \dfrac{1}{8}=1$

ii. Body centred cubic unit cell- in this cell, one constituent particle is present at each corner and one additional particle is also present at the centre of the cube. So, the number of particles in a body centred cubic unit cell is-
 1) At corners $8\times \dfrac{1}{8}=1$ (as there are 8 edges in a cube).
 2) At centre of body = 1 particle.
Therefore, total no. of particles per unit cell = 2

iii. Face centred unit cell- in this cell, constituent articles are present at all corners and one additional particle is also present at the centre of all faces. So, the number of particles will be-
1) At corners $8\times \dfrac{1}{8}=1$ (as there are 8 edges in a cube).
 2) At faces-$6\times \dfrac{1}{2}=3$ ( as there are 6 faces in a cube).
Therefore, total no. of particles per unit cell = 4. So, according to question As W atoms are situated at the corners and each corner atom contributes $\dfrac{1}{8}$ and there are 8 corner atoms in a cube, so the total number of W atoms = $\dfrac{1}{8}\times 8=1$
As Na is located at center and each center atom contributes fully to the unit cell, the total number of Na atoms =1
As O atom is located at edge centers and each such atom contributes $\dfrac{1}{4}$ and there are 12 edge centers in a cube the total number of W atoms = $\dfrac{1}{4}\times 12=3$ So, the formula of compound is ${ Na }{ WO }_{ 3 }$,
So, the correct answer is “Option D”.

Note: Don’t confuse with the term unit cell and crystal lattice as both are different like crystal lattice is a three dimensional pattern of points that describes the arrangement of constituent particles (atoms, ions or molecules) in a crystal while unit cell is the smallest repeating unit in a crystal lattice which when repeated again and again in different direction.
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