Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{O}^{2-}\) (b) \(\mathrm{Li}_{3} \mathrm{~N}\) (c) \(\mathrm{MgSO}_{4}\) (d) \(\mathrm{MnO}_{2}\)

Since none of the given compounds are free elements, we can skip this rule for this exercise.

a) In \(\mathrm{O}^{2-}\), the oxygen atom has a charge of -2. Therefore, the oxidation state of oxygen in this ion is -2.

b) For \(\mathrm{Li}_{3} \mathrm{~N}\), lithium is an alkali metal, and its oxidation state in compounds is always +1. So, in this compound, the oxidation state of each Li atom is +1. c) In \(\mathrm{MgSO}_{4}\), magnesium is an alkaline earth metal, meaning its oxidation state in compounds is +2. Now we need to find the oxidation state of N in \(\mathrm{Li}_{3} \mathrm{~N}\) and other atoms in \(\mathrm{MgSO}_{4}\).

There are no hydrogen atoms in any of the compounds, so we don't need to apply this rule. For oxygen, the oxidation state is usually -2 in compounds. So, c) In \(\mathrm{MgSO}_{4}\), each oxygen atom has an oxidation state of -2. d) In \(\mathrm{MnO}_{2}\), each oxygen atom has an oxidation state of -2. Now we need to find the oxidation states of N in \(\mathrm{Li}_{3} \mathrm{~N}\), S in \(\mathrm{MgSO}_{4}\), and Mn in \(\mathrm{MnO}_{2}\).

This rule helps us find the oxidation states of the remaining atoms. b) In \(\mathrm{Li}_{3} \mathrm{~ N}\), the algebraic sum of oxidation states should be 0. So, \((+1) \times 3 + \text{oxidation state of N} = 0\). Solving for the oxidation state of N, we get -3. c) In \(\mathrm{MgSO}_{4}\), we have +2 (for Mg) + oxidation state of S + 4 times the oxidation state of O (-2), which is equal to 0. Therefore, the oxidation state of S is +6. d) In \(\mathrm{MnO}_{2}\), the algebraic sum of oxidation states should also be 0. So, we have \(\text{oxidation state of Mn} + 2\times (-2) = 0\). Solving for the oxidation state of Mn, we get +4. In summary, the oxidation states for each atom in the given compounds are: (a) \(\mathrm{O}^{2-}\): O = -2 (b) \(\mathrm{Li}_{3} \mathrm{~N}\): Li = +1, N = -3 (c) \(\mathrm{MgSO}_{4}\): Mg = +2, S = +6, O = -2 (d) \(\mathrm{MnO}_{2}\): Mn = +4, O = -2