Question

Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Calculate the ionic radius of TI⁺. (The ionic radius of I⁻ is 2.16 Å.)

Answer 1

The ionic radius of Tl+ in TlI, which crystallizes in a CsCl structure and has a unit cell edge length of 4.20 Å, is calculated to be 1.475 Å.

Step-by-step explanation:

To calculate the ionic radius of Tl+, we start with understanding that Thallium(I) iodide has the same structure as CsCl, which is a simple cubic structure where cations sit in the center of the cube and anions are at the corners. The edge length of the unit cell for TlI is 4.20 Å. Since the anion and cation touch each other along the body diagonal in this structure, the body diagonal length is equal to twice the ionic radius of Tl+ plus twice the ionic radius of I−.

The body diagonal can be determined using the Pythagorean theorem in three dimensions (since the unit cell is a cube): body diagonal (d) = √3 * edge length (a). So d = √3 * 4.20 Å = 7.27 Å. With the ionic radius of I− being 2.16 Å, we can find the radius of Tl+ by subtracting two times the radius of I− from the body diagonal length and then divide the remainder by two:

Radius of Tl+ = (d - 2 * radius of I−)/2 = (7.27 Å - 2 * 2.16 Å)/2 = 1.475 Å

Answer 2

The ionic radius of Tl⁺ in thallium(I) iodide (TlI) can be calculated using the edge length of the unit cell and the ionic radius of I⁻.

Step-by-step explanation:

To calculate the ionic radius of Tl⁺ in thallium(I) iodide (TlI), we can use the fact that the edge length of the unit cell of TlI is given as 4.20 Å and the ionic radius of I⁻ is given as 2.16 Å. In CsCl-type structures, each cation is in contact with 8 anions. Since the ratio of Tl⁺ to I⁻ in TlI is 1:1, we can consider that each Tl⁺ ion is in contact with 8 I⁻ ions. Therefore, the radius of Tl⁺ can be calculated using the following formula:

RTl⁺ = RI⁻ * √2

RTl⁺ = 2.16 Å * √2