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Question

Answers

A. $1$

B. $2$

C. $3$

D. $4$

Answer
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In $NaCl$ the $C{l^ - }$ ions are present at the corners of the unit cell and at the face-centers of the unit cell and the Na+ ions are present at the octahedral voids of the unit cell. Imagining a unit cell as a cube there are $8$ corners and $6$ face-centers. The octahedral voids present in a unit cell are “$n$”, where ”$n$” is the number of atoms present in a unit cell.

Before calculating the number of carbon atoms in the unit cell one must know about the contribution of the atoms towards a unit cell in the lattice. An atom present at the corner contributes $\dfrac{1}{8}$to a unit cell as it is shared to $8$ different unit cells attached together, whereas the one present at the face-center contributes $\dfrac{1}{2}$to a unit cell, and the contribution of atoms present at octahedral voids is $1$.

Now, let’s calculate the number of $Na$ atoms present in $NaCl$:

As there are $4$ atoms present at octahedral voids therefore,

$1 \times 4 = 4$,where $n = 4$ for fcc lattice.

So, the total number of $Na$ atoms is $4$.

A student can be confused about the number of octahedral (“ $n$ ”) and tetrahedral voids (“ $2n$ ”) present in a lattice so a student should keep the values in mind. And when it comes to the contribution of the octahedral and the tetrahedral voids, they have a contribution of $1$, as they are totally a part of the lattice.

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