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The following rules will help us to determine the oxidation states of all atoms in a compound: 1. The oxidation state of an element in its elemental form is 0. 2. The oxidation state of a monatomic ion is equal to its charge. 3. Oxygen usually has an oxidation state of -2 (with some exceptions, like peroxides featuring -1 each). 4. Hydrogen in most compounds has an oxidation state of +1, while in metal hydrides its oxidation state is -1. 5. The sum of the oxidation states of all atoms in a molecule or ion must equal its overall charge. Now let's analyze each compound and identify the oxidation states for each of its atoms.

a. \(\mathrm{BiO}^{+}\) Since Bi is bonded to O, the oxidation state of O is assumed to be -2, according to rule 3. Therefore, for the overall molecule to have a charge of +1, the oxidation state of Bi must be: \[+1 = (+x) + (-2)\] Solving for x, we get the oxidation state of Bi: \[x = +3\] b. \(\mathrm{PO}_{4}^{3-}\) The oxidation state of O is -2, as stated earlier. Since we have 4 O atoms, their combined oxidation state is -8. The overall charge of the ion is -3, so the oxidation state of P must be: \[(-3) = (+x) + (-8)\] Solving for x, we get the oxidation state of P: \[x = +5\] c. \(\mathrm{NO}_{2}^{-}\) Again, O has an oxidation state of -2 (x2 for two O atoms) which sums to -4. The overall charge of the ion is -1. The oxidation state of N must be: \[((-1) - (-4)) = (+x)\] Solving for x, we get the oxidation state of N: \[x = +3\] d. \(\mathrm{Hg}_{2}^{2+}\) Since the overall charge of the ion is +2, the oxidation state of the metals must be equal but opposite to account for the charge. The total oxidation state of the two Hg is +2, so the oxidation state of each Hg is: \[+2/2 = +1\]