Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3} \quad\) g. \(\mathrm{PbSO}_{4}\) b. \(\mathrm{CuCl}_{2} \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\) j. \(\mathrm{CO}_{2}\) e. $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \quad \mathrm{k} .\left(\mathrm{NH}_{0}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}$ f. \(\mathrm{Ag} \quad\) l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

The oxidation number of an atom is assigned according to the following rules: 1. The oxidation number of an element in its free (uncombined) state is zero. 2. The oxidation number of a monatomic ion is equal to its charge. 3. The sum of all oxidation numbers in a neutral compound is zero. 4. The sum of all oxidation numbers in a polyatomic ion is equal to the charge of the ion. 5. The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where it is -1. 6. The oxidation number of oxygen in a compound is -2, except in peroxides, where it is -1. 7. The oxidation number of an atom in a compound can also be determined by considering the electronegativity of the elements. Now let's assign the oxidation numbers to the atoms in each compound: a. \(\mathrm{HNO}_{3}\)

Using the rules explained above, we can determine the oxidation numbers of H, N, and O in \(\mathrm{HNO}_{3}\): 1. Hydrogen (H): +1 2. Oxygen (O): -2 3. Nitrogen (N): Unknown

Now we can use the rule that the sum of oxidation numbers in a neutral compound is zero to find the oxidation number of nitrogen: (+1) + x + (-2)(3) = 0 Solving for x, x = +5 Therefore, the oxidation numbers for \(\mathrm{HNO}_{3}\) are: H(+1), N(+5), and O(-2). We will now follow the same procedure for the remaining compounds: b. \(\mathrm{CuCl}_{2}\) Cu: +2 (from the charge of the ion) Cl: -1 (from the charge of the ion) c. \(\mathrm{O}_{2}\) (element in its free state) O: 0 d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) (remember that this is a peroxide) H: +1 O: -1 (since it's a peroxide) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) H: +1 O: -2 C: +4 (calculated based on the rules) f. \(\mathrm{Ag}\) (element in its free state) Ag: 0 g. \(\mathrm{PbSO}_{4}\) Pb: +2 (from the charge of the ion) S: +6 (calculated based on the rules) O: -2 h. \(\mathrm{PbO}_{2}\) Pb: +4 (from the charge of the ion) O: -2 i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) Na: +1 C: +3 (calculated based on the rules) O: -2 j. \(\mathrm{CO}_{2}\) C: +4 (calculated based on the rules) O: -2 k. \(\left(\mathrm{NH}_{0}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) N: -3 (calculated based on the rules) H: +1 O: -2 Ce: +3 (from the charge of the ion) S: +6 (calculated based on the rules) l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) Cr: +3 (calculated based on the rules) O: -2