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Chapter 20: Problem 98

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 \(1.45 \mathrm{M}\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and 1.50 atm?

Short Answer

Expert verified
The oxidation state of iron is +6, and the electron configuration of iron in the ferrate ion is \([Ar] 3d^0 4s^2\). The volume of nitrogen gas formed at 25°C and 1.50 atm is V = nRT/P, where n is the moles of nitrogen gas formed, R is the ideal gas constant (0.0821 \(L\cdot atm /(mol\cdot K)\)), T is the temperature in Kelvin (298 K), and P is the pressure (1.50 atm).

Step by step solution

01

Identify the charges of elements in \(\mathrm{FeO}_{4}^{2-}\)

First, find the charges of each element in \(\mathrm{FeO}_{4}^{2-}\). Oxygen has a charge of -2. Since there are 4 oxygen atoms, the total charge represented by oxygen in the compound is -8. As the overall charge of the ion \(\mathrm{FeO}_{4}^{2-}\) is -2, we can find the charge of iron (Fe).
02

Calculate the charge of iron

To find the charge of iron, balance the charges of oxygen and iron. It can be formulated as: \[x + (-8) = -2\] \[x = -2 + 8\] \[x= 6\] The charge of iron is thus +6.
03

Determine the electron configuration of iron

To obtain the electron configuration of iron in \(\mathrm{FeO}_{4}^{2-}\), remove 6 electrons from the ground state electron configuration of iron, since it has an oxidation state of +6. The ground state electron configuration of iron is: \([Ar] 3d^6 4s^2\). Removing 6 electrons gives the electron configuration of iron in \(\mathrm{FeO}_{4}^{2-}\): \([Ar] 3d^0 4s^2\) #a. Answer:# The oxidation state of iron is +6, and the electron configuration of iron in the ferrate ion is \([Ar] 3d^0 4s^2\). #b. Calculating the volume of nitrogen gas:#
04

Find the limiting reactant

First, find the limiting reactant between \(\mathrm{FeO}_{4}^{2-}\) and \(\mathrm{NH}_{3}\). The balanced chemical equation of this reaction is: \[2\mathrm{NH}_3 (aq) + \mathrm{FeO}_4^{2-} (aq) \rightarrow N_2 (g) + \mathrm{Fe}^{3+} + 4 \mathrm{OH}^{-}\] Using the molarity and volume, calculate the moles of \(\mathrm{FeO}_{4}^{2-}\) and \(\mathrm{NH}_{3}\) as follows: Moles of \(\mathrm{FeO}_{4}^{2-} = 0.243 \mathrm{M} \times 0.025 L\) Moles of \(\mathrm{NH}_{3} = 1.45 \mathrm{M} \times 0.055 L\)
05

Compare the mole ratio

Compare the mole ratio with the balanced equation to find the limiting reactant. Mole ratio of \(\mathrm{FeO}_{4}^{2-}\) and \(\mathrm{NH}_{3}\): \(1 : 2\)
06

Determine the moles of nitrogen gas formed

The moles of nitrogen gas formed will be limited by the moles of the limiting reactant (which was found in the other note). Calculate the moles of nitrogen gas formed through the mole ratio of reactants and products. \(N_{2}\) moles \(=\frac{1}{2}\) * moles of the limiting reactant
07

Apply the ideal gas law

Use the ideal gas law (PV = nRT) to determine the volume of nitrogen gas formed at the given conditions (1.50 atm, 25°C). R = 0.0821 \(L\cdot atm /(mol\cdot K)\) T = 298 K Calculate the volume of nitrogen gas: V = nRT/P #b. Answer# The volume of nitrogen gas formed at 25°C and 1.50 atm is V.

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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: $$ \begin{aligned} \operatorname{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q) &+\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\ & \longrightarrow \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned} $$ 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 inter- mediate? Explain.

Which of the following molecules exhibit(s) optical isomerism? a. \(c i s-\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) b. \(t r a n s \) - Ni(en)_Br_(en is ethylenediamine) c. \(cis-Ni\)(\mathrm{en})_{2} \mathrm{Br}_{2}$ (en is ethylenediamine)

The complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is paramagnetic with one unpaired electron. The complex ion \(\mathrm{Fe}(\mathrm{SCN})_{6}^{3-}\) has five unpaired electrons. Where does SCN \(^{-}\) lie in the spectrochemical series relative to \(\mathrm{CN}^{-} ?\)

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

Tetrahedral complexes of \(\mathrm{Co}^{2+}\) are quite common. Use a \(d\) -orbital splitting diagram to rationalize the stability of \(\mathrm{Co}^{2+}\) tetrahedral complex ions.
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