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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molybdenum Compounds
Molybdenum, a key component in industrial applications, forms several essential compounds. Two important ones are molybdenum sulfide (\(\text{MoS}_2\)) and molybdenum trioxide (\(\text{MoO}_3\)).
Molybdenum sulfide is primarily extracted as a by-product in copper mining. It is crucial in lubricants due to its excellent performance under high pressure and temperature. On the other hand, molybdenum trioxide is commonly used in steel production, particularly in creating stainless steel because it enhances strength and durability.
The ability to transform \(\text{MoS}_2\) into \(\text{MoO}_3\) signifies molybdenum's adaptable chemical nature. Understanding the different compounds and their applications helps us grasp how a single element can serve various industrial roles.
Naming Chemical Compounds
Naming chemical compounds is a systematic practice that allows chemists to communicate precisely. The names are usually derived from the elements involved and the type of compound.
For binary compounds, which consist of two elements, the naming pattern follows: 'Name of first element' + 'Base name of second element' + 'ide'. For example, the chemical formula \(\text{MoS}_2\) becomes 'Molybdenum sulfide', where 'sulfide' denotes the sulfur element.
Similarly, for \(\text{MoO}_3\), the name is 'Molybdenum trioxide'. The 'tri' prefix indicates the presence of three oxygen atoms, a typical feature in naming polyatomic compounds.
Understanding these rules not only aids in correctly identifying compounds but also in deciphering their chemical properties and uses.
Binary Compounds
Binary compounds are chemical species consisting of two different elements. They are integral to chemistry and include compounds like \(\text{MoS}_2\) and \(\text{MoO}_3\).
  • Simplicity: Binary compounds are simpler than their multi-element counterparts, making them easier to analyze and understand.
  • Versatility: They are found across various chemical classes, including ionic compounds, covalent compounds, and acids.
  • Common Examples: Apart from molybdenum compounds, examples include sodium chloride (\(\text{NaCl}\)), carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)).
The understanding of binary compounds is a stepping stone in learning advanced chemical interactions and synthesis.
Chemical Formula Interpretation
Interpreting chemical formulas involves analyzing the elements present and their respective quantities. This understanding is fundamental in chemistry as it reveals key insights about the compound.
For \(\text{MoS}_2\), the interpretation tells us that each molecule consists of one molybdenum atom and two sulfur atoms. Similarly, \(\text{MoO}_3\) comprises one molybdenum atom and three oxygen atoms.
  • Oxidation State: Crucial for understanding chemical reactivity, the oxidation state is determined through known rules of common ion charges.
  • Balancing Charges: In neutral compounds, the sum of oxidation states equals zero. This rule ensures that chemical formulas represent stable compounds.
  • Example Calculations: From the solution, we derive the oxidation states of molybdenum as +4 in \(\text{MoS}_2\) and +6 in \(\text{MoO}_3\).
Mastering chemical formula interpretation enhances our ability to predict the behavior and utility of different compounds.

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Most popular questions from this chapter

Compounds of copper(II) are generally colored, but compounds of copper(I) are not. Explain. Would you expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) to be colored? Explain.

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When an aqueous solution of KCN is added to a solution containing \(\mathrm{Ni}^{2+}\) ions, a precipitate forms, which redissolves on addition of more \(\mathrm{KCN}\) solution. Write reactions describing what happens in this solution. [Hint: \(\mathrm{CN}^{-}\) is a Bronsted-Lowry base \(\left(K_{\mathrm{b}} \approx 10^{-5}\right)\) and a Lewis base.]

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.
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