Question

CsCl crystallizes in a unit cell that contains the Cs ion at the center of a cube that has a Cl- at each corner. Each unit cell contains ________ Cs ion(s) and ________ Cl- ion(s), respectively.

Answer 1

The given question is incomplete, the complete question is:

CsCl crystallizes in a unit cell that contains the Cs+ ion at the center of a cube that has a Cl- at each corner. Each unit cell contains __________ Cs+ ions and __________ Cl-, ions, respectively.

a. 1 and 8

b. 2 and 2

c. 2 and 1

d. 2 and 4

e. 1 and 1

Answer:

The correct answer is option e, that is, 1 and 1.

Explanation:

A regular and a small geometric unit of a crystal that illustrates the repeating pattern of the crystal lattice is termed as a unit cell. Based on the given question, crystallization of CsCl takes place in a unit cell that comprises the ion of Cs+ at the center of a cube, which possesses Cl- at every corner.  

In the given case, in the center of the lattice one atom of Cs is present, therefore, there is one ion of Cs for one unit cell. It is known that there are 8 corners of a cube, and Cl- ion is situated at each corner. So,  

8 corners * (1 Cl ion/8 corners) = 1 Cl ion

Hence, the ratio of Cs+ : Cl- is 1:1.  

Answer 2

Each unit cell contains 1 $Cs^{+}$ ion(s) and 1 $Cl^{-}$ ion(s), respectively.

CsCl has a cubic structure that consists of an infinite chain of ions. The Cs (or Cl) ions sit at the eight corners of the cube and the Cl (or Cs) sit at the center of the cube.

It is given that:

The Cs ion($Cs^{+}$) is present at the center of a cube and  $Cl^{-}$ is present at each corner.

It is known that there are total 8 corners  in a cube and $Cl^{-}$ is present at each corner.

So, 8 corners * (1  $Cl^{-}$  ion/ 8 corners)=1  $Cl^{-}$  ion

And also, Cs ion at present at the center of a cube thus 1 $Cs^{+}$ ion for one unit cell.

Thus the ration of $Cs^{+}$: $Cl^{-}$ is 1:1 and each unit cell contains 1 $Cs^{+}$ ion(s) and 1 $Cl^{-}$ ion(s), respectively.

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