The ferrate ion, \(\mathrm{FeO}_{4}{ }^{2-}\), is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental nitrogen along with the formation of the iron(III) ion. a. What is the oxidation state of iron in \(\mathrm{FeO}_{4}{ }^{2-}\), and what is the electron configuration of iron in this polyatomic ion? b. If \(25.0 \mathrm{~mL}\) of a \(0.243 \mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution is allowed to react with \(55.0 \mathrm{~mL}\) of \(1.45 M\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and \(1.50 \mathrm{~atm}\) ?

Short Answer

Expert verified
The oxidation state of iron in the ferrate ion, \(\mathrm{FeO}_{4}{ }^{2-}\), is +6, and its electron configuration is \([Ar]3d^2\). Upon reacting \(25.0 \mathrm{~mL}\) of \(0.243 \mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution with \(55.0 \mathrm{~mL}\) of \(1.45 \mathrm{M} \mathrm{NH}_{3}\) solution, the volume of nitrogen gas formed at \(25^{\circ} \mathrm{C}\) and \(1.50 \mathrm{~atm}\) is \(0.201\ \mathrm{L}\).

Step by step solution

01

Determine the oxidation state

To find the oxidation state of iron in \(\mathrm{FeO}_{4}{ }^{2-}\), we'll assign oxidation states to oxygen first and then solve for the oxidation state of iron. In general, the oxidation state of oxygen is -2. Let x be the oxidation state of iron (Fe). Then, we have the equation: x + 4(-2) = -2 Solving for x: x = -2 + 8 = +6 Therefore, the oxidation state of iron in the ferrate ion is +6.
02

Determine the electron configuration

Since the oxidation state of iron is +6 in the ferrate ion, it means that iron has lost 6 electrons as compared to its neutral state. The electron configuration of neutral iron (Fe, atomic number 26) is: \([Ar]3d^6 4s^2\) When iron loses 6 electrons, 2 are removed from the 4s orbital and 4 are removed from the 3d orbital. Therefore, the electron configuration of iron in the ferrate ion will be: \([Ar]3d^2\)
03

Write the balanced chemical equation

To find the volume of nitrogen gas at given temperature and pressure, we need to determine the balanced chemical equation representing the reaction between aqueous ammonia (\(\mathrm{NH}_{3}\)) and ferrate ion (\(\mathrm{FeO}_{4}{ }^{2-}\)). The balanced chemical equation is: 3 \(\mathrm{FeO}_{4}{ }^{2-}\) + 12 \(\mathrm{NH}_{3}\) → 3 \(\mathrm{Fe}^{3+}\) + 6 \(\mathrm{N}_{2}\) + 24 \(\mathrm{OH}^-\)
04

Calculate the moles of the reactants

Given that we have \(25.0 \mathrm{~mL}\) of \(0.243 \mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution and \(55.0 \mathrm{~mL}\) of \(1.45 \mathrm{M} \mathrm{NH}_{3}\) solution, we can find the moles of the reactants. For \(\mathrm{FeO}_{4}{ }^{2-}\): moles = molarity × volume (in liters) moles of \(\mathrm{FeO}_{4}{ }^{2-} = 0.243 \times \frac{25.0}{1000} = 0.006075\ \mathrm{mol}\) For \(\mathrm{NH}_{3}\): moles of \(\mathrm{NH}_{3} = 1.45 \times \frac{55.0}{1000} = 0.07975\ \mathrm{mol}\)
05

Determine the limiting reactant

To find the limiting reactant, we must compare the mole ratios of the reactants to their coefficients in the balanced chemical equation: \(\frac{\text{moles of }\mathrm{FeO}_{4}{ }^{2-}}{3} = \frac{0.006075}{3} = 0.002025\) \(\frac{\text{moles of }\mathrm{NH}_{3}}{12} = \frac{0.07975}{12} = 0.006646\) The smallest value represents the limiting reactant. In this case, the limiting reactant is \(\mathrm{FeO}_{4}{ }^{2-}\).
06

Calculate the moles of nitrogen gas

From the balanced chemical equation, we know that 3 moles of \(\mathrm{FeO}_{4}{ }^{2-}\) produce 6 moles of \(\mathrm{N}_{2}\). Since \(\mathrm{FeO}_{4}{ }^{2-}\) is the limiting reactant, we can use its moles to calculate the moles of nitrogen gas formed: moles of \(\mathrm{N}_{2} = \frac{6}{3} \times \text{moles of }\mathrm{FeO}_{4}{ }^{2-}\) moles of \(\mathrm{N}_{2} = 2 \times 0.006075= 0.01215\ \mathrm{mol}\)
07

Calculate the volume of nitrogen gas

We can use the ideal gas law to find the volume of nitrogen gas formed at the given temperature and pressure. Ideal gas law equation: PV = nRT Where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. Given P = \(1.50\ \mathrm{atm}\), T = \(25^{\circ} \mathrm{C} = 298.15 \ \mathrm{K}\), n = \(0.01215\ \mathrm{mol}\), and R = \(0.0821\ \frac {\mathrm{L\ atm}}{\mathrm{mol\ K}}\) We need to find the volume (V). Rearranging the equation to solve for V: V = \( \frac {nRT} {P} \) Plugging in the values: V = \(\frac{(0.01215)(0.0821)(298.15)}{1.5} = 0.201\ \mathrm{L}\) So, the volume of nitrogen gas formed at \(25^{\circ} \mathrm{C}\) and \(1.50\ \mathrm{atm}\) is \(0.201\ \mathrm{L}\).

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Key Concepts

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

Oxidation State Calculation
Understanding the oxidation state of an element within a compound is crucial for predicting its chemical behavior. The oxidation state signifies the hypothetical charge an atom would have if all bonds were purely ionic. In the instance of the ferrate ion, \(\mathrm{FeO}_{4}{ }^{2-}\), we calculate the oxidation state of iron by acknowledging that oxygen typically has an oxidation state of -2. Because there are four oxygen atoms, their combined charge is -8. Since the overall charge of the ion is -2, we infer that iron must carry a +6 charge to balance the oxygens. Simplified, the equation is:
\[ x + 4(\textnormal{-2}) = -2 \]. By solving for \(x\), we find that the oxidation state of iron is +6.
It's beneficial for students to grasp this concept to understand redox reactions, predicting compound stability, and the systematic naming of compounds. Remember, the sum of oxidation states always equals the net charge of the molecule or ion.
Electron Configuration
The electron configuration is an essential tool for understanding an atom's behavior in chemical bonding. For iron in the ferrate ion \(\mathrm{FeO}_{4}{ }^{2-}\), we've determined its oxidation state is +6, which implies iron loses six electrons relative to its neutral state. The electron configuration for a neutral iron atom (atomic number 26) is \( [Ar]3d^6 4s^2 \). Following the loss of six electrons, two from the 4s and four from the 3d orbitals, we're left with the configuration of \( [Ar]3d^2 \). This configuration informs us about the likely types of chemical bonds iron can form and its magnetic properties. A firm understanding of electron configurations empowers students to predict and explain trends in the periodic table, such as elements' reactivity or why certain configurations lead to an element's color.
Stoichiometry and Gas Laws
Stoichiometry bridges the gap between the atomic scale and the practical world by using the balanced chemical equation to relate substances' qualitative identities with their quantitative information. In our problem, the reaction between ferrate ion and ammonia was scaled using a balanced chemical equation to find the volume of nitrogen gas produced. Calculating moles, we determine the limiting reagent, which then tells us how much product can be formed.
Utilizing gas laws, specifically the ideal gas law expressed as \( PV = nRT \), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant (\(0.0821\ \frac{\text{L atm}}{\text{mol K}}\)), and \(T\) is temperature in Kelvin, we can calculate the volume of the nitrogen gas produced at a given temperature and pressure. This example shows how stoichiometry and gas laws enable us to predict outcomes in chemical reactions, crucial for fields such as chemical engineering and environmental science.

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Most popular questions from this chapter

Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: \(\begin{aligned} \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+& \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\\ \longrightarrow & \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned}\) Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an intermediate? Explain.

Qualitatively draw the crystal field splitting for a trigonal bipyramidal complex ion. (Let the \(z\) axis be perpendicular to the trigonal plane.)

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

BAL is a chelating agent used in treating heavy metal poisoning. It acts as a bidentate ligand. What type of linkage isomers are possible when BAL coordinates to a metal ion?

Use the data in Appendix 4 for the following. a. Calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction $$3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)$$ that occurs in a blast furnace. b. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature. Calculate \(\Delta G^{\circ}\) at \(800 .{ }^{\circ} \mathrm{C}\) for this reaction.

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