Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2},\) which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\) b. What is the oxidation state of Mo in each of the compounds mentioned above?

To name the compounds MoS2 and MoO3, we follow the common rules for naming binary compounds. Binary compounds consist of two elements. The element with the lower group number is named first, and the one with the higher group number is named second with an '-ide' suffix. Mo is Molybdenum and S is Sulfur while O is Oxygen.

Using the aforementioned naming rules: 1. MoS2: Molybdenum Sulfide 2. MoO3: Molybdenum Trioxide

The oxidation state (or oxidation number) refers to the charge assigned to an atom in a compound due to a loss or gain of electrons. The oxidation state of an element in a compound is determined by the sharing of electrons to satisfy the electrostatic charges in the compound. The total charge of a compound should equal zero.

We will determine the oxidation state of Mo in the following compounds: MoS2, MoO3, (NH4)2Mo2O7, and (NH4)6Mo7O24·4H2O. 1. MoS2: For sulfide(S), the oxidation state is -2. Since there are two sulfide ions, the total charge contributed from sulfide ions would be -4. Therefore, the oxidation state of Mo (Molybdenum) in this compound should counteract the Sulfide ions charge and be +4. 2. MoO3: For oxide (O), the oxidation state is -2. Since there are three oxide ions, the total charge contributed from oxide ions would be -6. Therefore, the oxidation state of Mo (Molybdenum) in this compound should counteract the oxide ions charge and be +6. 3. (NH4)2Mo2O7: In the ammonium ion (NH4)+, the oxidation state of nitrogen (N) is -3, and that of hydrogen (H) is +1. Therefore, the total charge for a single ammonium ion is +1, and since there are two ammonium ions, the total charge contributed by both is +2. Oxygen has an oxidation state of -2, thus total charge due to the seven oxide ions will be -14. So, the oxidation states of the two Mo atoms should counteract both the positive charge from ammonium ions (+2) and the negative charge from the oxide ions (-14) hence 2x(Mo oxidation state) = 12 or Mo oxidation state = +6. 4. (NH4)6Mo7O24·4H2O: The oxidation states for ammonium (NH4+) and oxide(O) are the same as described previously. The total charge contributed by six ammonium ions is +6, and total charge due to twenty-four oxide ions will be -48. Therefore, the oxidation states of the seven Mo atoms should counteract both the positive charge from ammonium ions (+6) and the negative charge from the oxide ions (-48) hence 7x(Mo oxidation state) = 42 or Mo oxidation state = +6. In summary, the oxidation states of Mo in the mentioned compounds are: MoS2 (+4), MoO3 (+6), (NH4)2Mo2O7 (+6), and (NH4)6Mo7O24·4H2O (+6).