Question

Thallium(I) iodide crystallizes with the same structure a \({\rm{CsCl}}\). The edge length of the Unit cell of Tll is\(4.20\mathop A\limits^o \). Calculate the ionic radius of \(T{I^ + }\). (The ionic radius of \({{\rm{I}}^ - }\)is \(2.16\mathop A\limits^o \)

Short answer

The ionic radius of Tl+ is \(1.48\mathop A\limits^o \)

Step-by-step solution

Define the ionic radius

Ionic radius is defined as the internuclear distance between the closest anion and cation.

The ionic radius is given by:

\(2{r^ - } + 2{r^ + }\)

where \(r\) - is radius of anion and \(r + \) is radius of cation.

 Find the ionic radius of NaH

Since, the structure of \({\rm{CsCl}}\) is body-centered cubic so, the structure of thallium(I) iodide is body-centered. The length of the face diagonal of the cube is calculated using the Pythagorean Theorem:

The diagonal of the cube is 4r.

\(\begin{aligned}{}{(4r)^2} &= {d^2} + {e^2}{\rm{ }}\\{\rm{But }}{d^2} &= {e^2} + {e^2}\\{d^2} &= {e^2} + {e^2}\\ &= 2{e^2}\\ &= 2{(4.20{A°})^2}\\ &= 35.28{{A°}^2}\end{aligned}\)

The diagonal of the cube is:

\(\begin{aligned}{}{d^2} + {e^2} &= 35.28 + {(4.20)^2}\\35.28 + 17.64 &= 52.92\\d + e &= 7.27{A°}\end{aligned}\)

The ionic radius is:

\(2{r^ - } + 2{r^ + }\)

Value of \(r\) - is \(2.16{A°}\) (given).

\(\begin{aligned}{}{r^ + } &= \frac{{(7.27 - 2(2.16))}}{2}\\ &= 1.48{A°}\end{aligned}\)