CsBr (Cesium Bromide) possesses a Body Centered Cubic (BCC) structure. In this structure, the constituent atoms or ions are arranged in a cubic lattice with one atom at each corner of the cube and one atom at the body center.



The given edge length of the unit cell of CsBr is 200 pm. This means that the distance between the opposite corners of the cube is 200 pm.



The interionic distance in CsBr can be calculated using the formula:

Interionic Distance = sqrt(3) * Edge Length / 4

Substituting the given values, we get:

Interionic Distance = sqrt(3) * 200 pm / 4

Interionic Distance = 100 * sqrt(3) pm

Interionic Distance = 173.2 pm

Therefore, the interionic distance in CsBr is 173.2 pm.



The interionic distance is the distance between the centers of the constituent ions in a crystal lattice. In the case of CsBr, the Cs+ and Br- ions are arranged in a BCC structure. The formula for calculating the interionic distance takes into account the edge length of the unit cell and the arrangement of the ions in the lattice. The factor of sqrt(3) arises from the diagonal distance between opposite corners of the cube. The factor of 4 arises from the number of constituent ions in the BCC unit cell. The calculated value of the interionic distance in CsBr is an important parameter that helps to understand the physical and chemical properties of the crystal.

Given edge length =a = 200pm in bcc structure √3a=2(radius of cation + radius of anion)=2*inter-ionic distance hence inter-ionic distance =(√3/2)*a=100√3pm