Chapter 18, Problem 46R

To determine

To identify: the formula for the compound if thallium has an oxidation number of 1+.

Explanation of Solution

Introduction:

Thallium carbonate contains thallium and carbonate ions. Thallium belongs to group 13 and has 3 electrons in its valence shell. The atomic number of thallium is 81.

Its electronic configuration is ${\left[\text{Xe}\right]}^{\text{54}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{1}}$ .

Thallium possesses 2 electrons in 6s and 1 electron in 6p subshells.

As we move down the group, the tendency of inner electrons to take part in the bond formation decreases. It is called the inert pair effect.

Therefore, only an electron present in 6p subshell can take part in the bond formation and form the compound.

Thus, thallium shows the oxidation number of +1. Carbonate ion is having overall -2 charges on it. Therefore, the formula of the compound is

Ti	${\text{CO}}_{\text{3}}$
+1	-2	charges

On cross multiplying the charges, the formula is found to be ${\text{Ti}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Conclusion:

Hence, the formula of thallium carbonate is ${\text{Ti}}_{\text{2}}{\text{CO}}_{\text{3}}$ and it is used in making artificial diamonds.