Q14.70 P - Chemistry: Molecular Nature Of Matter And Change

Question

Thallium forms the compound $\mathit{T}\mathit{l}{\mathit{I}}_{\mathbf{3}}$. What is the apparent oxidation state of Tl in this compound? Given that the anion is ${{\mathbf{I}}_{\mathbf{3}}}^{\mathbf{-}}$, what is the actual oxidation state of Tl? Draw the shape of the anion, giving its VSEPR class and bond angles. Propose a reason why the compound does not exist as $\left(T{l}^{3+}\right){\left(I^{-}\right)}_{\mathbf{3}}$.

Step-by-Step Solution

Answer

The apparent oxidation state of Tl in $Tl{I}_3$is +3.

The actual oxidation state of Tl in $Tl{l}_3$is +1.

The shape of the anion is given below:

The strength of the bond between$T{l}^{3+}$ and ${\left(I^{-}\right)}_3$ is very low.

Step 1: Apparent oxidation state of Tl in T l I 3

The apparent oxidation state of Tl in $\mathit{T}\mathit{l}{\mathit{I}}_{\mathbf{3}}$is +3 since iodine has -1 oxidation state.

Let the oxidation state of thallium be x. The calculation is given below: 

$$\begin{gathered} x+\left(-1\times 3\right)=0 \\ x-3=0 \\ x=+3 \end{gathered}$$ 

Thus, the apparent oxidation state of Tl in $Tl{I}_3$ is +3.

Step 2: Actual oxidation state of Tl in T l I 3

It is given that the anion is ${I_3}^{-}$ . Let x be the oxidation state of thallium. So the oxidation state of Tl in $Tl{I}_3$ can be calculated as follows:

$$\begin{gathered} x-1=0 \\ x=+1 \end{gathered}$$

So the oxidation state of thallium is +1.

Step 3: Shape of anion

The shape of anion ${I_3}^{-}$ is given below:

There is an excess of one electron at the second iodine, which is the reason for the negative charge. It belongs to $A{X}_2{E}_3$ class and has a bond angle of $180°$.

Step 4: Non-existence of ( T l 3 + ) ( I - ) 3

The bond between thallium and iodine has very low strength. Thus $\left(T{l}^{3+}\right){\left(I^{-}\right)}_3$ does not exist.