1. **Stating the problem:** We need to find the chemical formula for rhodium (IV) sulfite, which involves rhodium in the +4 oxidation state and the sulfite ion. 2. **Understanding the ions involved:** - Rhodium (IV) means rhodium has a charge of +4, written as $\mathrm{Rh^{4+}}$. - Sulfite ion is $\mathrm{SO_3^{2-}}$ with a charge of -2. 3. **Formula determination rule:** The total positive charge must balance the total negative charge in the compound. 4. **Calculating the ratio:** - Let the number of sulfite ions be $x$. - Total positive charge from rhodium: $+4$. - Total negative charge from sulfite ions: $-2x$. 5. **Setting up the charge balance equation:** $$+4 + (-2x) = 0$$ 6. **Solving for $x$:** $$4 = 2x$$ $$x = \frac{4}{2} = 2$$ 7. **Writing the formula:** - Rhodium (IV) ion: $\mathrm{Rh^{4+}}$ - Sulfite ion: $\mathrm{SO_3^{2-}}$ - Number of sulfite ions needed: 2 Therefore, the formula is: $$\mathrm{Rh(SO_3)_2}$$ This means one rhodium ion combines with two sulfite ions to neutralize the charge. **Final answer:** $\mathrm{Rh(SO_3)_2}$