Question

At temperatures above ~175 °C, thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Assuming contact between the body-centered Tl+ ion and the cube-corner I− ions of a unit cell, calculate the ionic radius of TI+. (The ionic radius of I− is 2.16 Å.)

Answer 1

The ionic radius of Tl+ in Thallium(I) iodide (TlI) can be calculated by considering the contact between the body-centered Tl+ ion and the cube-corner I- ions.

Step-by-step explanation:

Thallium(I) iodide (TlI) crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å.

To calculate the ionic radius of Tl+, we can consider the contact between the body-centered Tl+ ion and the cube-corner I- ions.

Since the body-centered Tl+ ion touches eight cube-corner I- ions, the distance between the Tl+ and I- ions is equal to twice the ionic radius of I-. Therefore, the ionic radius of Tl+ can be calculated as:

R(Tl+) = (edge length of cube - 2 * ionic radius of I-) / 2

Substituting the given values:

R(Tl+) = (4.20 Å - 2 * 2.16 Å) / 2 = 4.20 Å - 4.32 Å / 2 = -0.12 Å / 2 = -0.06 Å

The ionic radius of Tl+ is -0.06 Å.