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Chapter 21: Problem 29

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 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 com- pounds mentioned above?

Short Answer

Expert verified
(a) The names of the compounds $\mathrm{MoS}_{2}$ and $\mathrm{MoO}_{3}$ are Molybdenum sulfide and Molybdenum trioxide, respectively. (b) The oxidation states of Mo in the mentioned compounds are as follows: - In $\mathrm{MoS}_{2}$, the oxidation state of Mo is +4. - In $\mathrm{MoO}_{3}$, the oxidation state of Mo is +6. - In $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}$, the oxidation state of Mo is +10. - In $\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}$ , the average oxidation state of Mo is approximately +6.86.

Step by step solution

01

Part (a) - Naming compounds

The naming of binary compounds follows the pattern of "element name" + "element name" + "ide." Since Mo stands for Molybdenum and S stands for sulfur, the name for the compound MoS2 is Molybdenum sulfide. Similarly, since Mo stands for Molybdenum and O stands for oxygen, the name for the compound MoO3 is Molybdenum trioxide.
02

Part (b) - Assign oxidation states

In order to assign the oxidation state to Mo in the given compounds, we first need to determine the oxidation states of other elements. Sulfur's oxidation state in MoS2 can be determined from the fact that it forms an ion with a charge of -2, following the ium rule: the charge of the cation is equal to the positive oxidation states combined. Oxygen's oxidation state is usually -2 in compound. MoS2: Let x be the oxidation state of Mo in MoS2. The sum of oxidation states in a neutral compound is equal to 0. Therefore, x + 2(-2) = 0 x - 4 = 0 x = 4 So, the oxidation state of Mo in MoS2 is +4. MoO3: Let x be the oxidation state of Mo in MoO3. The sum of oxidation states in a neutral compound is equal to 0. Therefore, x + 3(-2) = 0 x - 6 = 0 x = 6 So, the oxidation state of Mo in MoO3 is +6. For the other compound mentioned, let's first find the oxidation state of Mo in (NH4)2Mo2O7: Let x be the oxidation state of Mo in (NH4)2Mo2O7. The sum of oxidation states in a neutral compound is equal to 0. In this case, the oxidation state of N in NH4 is -3, and the oxidation state of O is -2. Therefore, 2(+1) + 2(-3) + 2x + 7(-2) = 0 x - 10 = 0 x = 10 So, the oxidation state of Mo in (NH4)2Mo2O7 is +10. Finally, let's find the oxidation state of Mo in (NH4)6Mo7O24·4H2O: Let x be the oxidation state of Mo in (NH4)6Mo7O24·4H2O. The sum of oxidation states in a neutral compound is equal to 0. In this case, the oxidation state of N in NH4 is -3, and the oxidation state of O is -2. Since H2O does not contribute to the overall charge, we do not have to account for it. Therefore, 6(+1) + 6(-3) + 7x + 24(-2) = 0 x - 48 = 0 x = 48 So, the oxidation state of Mo in (NH4)6Mo7O24·4H2O is +48. However, since there are seven Mo atoms in the compound, their individual average oxidation state is +48/7, which is approximately +6.86.

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