Abstract

Manganese (Mn) belongs to the manganese group, Group $VIIB$, which comprises the four transition metals manganese (Mn), technetium (Tc), rhenium (Re), and bohrium (Bh).

Manganese ores occur mainly as oxides:

• Pyrolusite, $\ce{MnO2}$
• Hausmannite, $\ce{Mn3O4}$
• Manganosite, $\ce{MnO}$
• Rhodochrosite, $\ce{MnCO3}$

The ground-state valence-electron configuration of the manganese-group elements is $(n-1)d^5ns^2$, and their highest oxidation state is $+7$. Manganese exhibits the widest range of oxidation states in the group. The $+2$ state is its most common and most stable state; $\ce{Mn^2+}$ occurs in solids, solutions, and coordination compounds, while $\ce{Mn(VII)}$ is strongly oxidizing.

Unlike manganese, Tc and Re most commonly and stably adopt the $+7$ oxidation state, which is only weakly oxidizing. Their $+2$ and lower oxidation states are unstable and strongly reducing.

In summary, down the manganese group:

• The stability of high oxidation states increases
• The stability of low oxidation states decreases

With that context established, this opening article on the manganese group focuses on manganese itself.

The Element

Metallic manganese is silvery white. Manganese exposed to air, as well as powdered manganese, appears gray.

Pure manganese can be prepared by reducing $\ce{MnO2}$ or $\ce{Mn3O4}$ through an aluminothermic reaction.

The electrode potential $E^\ominus(\ce{Mn^{2+}/Mn})=-1.18\ \mathrm{V}$ shows that manganese is an active metal. It dissolves in cold, dilute, non-oxidizing acids, for example:

$$\ce{Mn + 2HCl(aq)->MnCl2 + H2 ^}$$

At room temperature, manganese is not highly reactive toward nonmetals, but it reacts readily on heating:

• Heating in air produces $\ce{Mn3O4}$
• At high temperatures it reacts with halogens, sulfur, carbon, and phosphorus
• It reacts very little with cold water, but reacts with hot water to form $\ce{Mn(OH)2}$ and release $\ce{H2}$, similarly to magnesium

$\ce{Mn(II)}$ Compounds

Common manganese compounds include:

• $\ce{Mn(II)}$ salts
• $\ce{Mn(IV)}$ oxides
• Permanganates

$\ce{Mn(II)}$

Soluble Salts

Salts of $\ce{Mn(II)}$ derived from strong acids are soluble:

• $\ce{MnSO4}$
• $\ce{MnCl2}$
• $\ce{Mn(NO3)2}$

The five d electrons of $\ce{Mn^2+}$ have parallel spins. Because the probability of a d-d transition is low, its compounds are generally only weakly colored.

• Concentrated $\ce{Mn^2+}$ solutions are pink
• Dilute solutions are nearly colorless

Most hydrated $\ce{Mn^2+}$ salts are pink or rose-colored, for example:

• $\ce{MnSO4.7H2O}$
• $\ce{MnCl2.6H2O}$

The following anhydrous salts are white:

• $\ce{MnSO4}$
• $\ce{Mn(NO3)2}$

Sparingly Soluble Compounds

The hydroxide and most weak-acid salts of $\ce{Mn(II)}$ are sparingly soluble.

• $\ce{MnCO3}$: pink
• $\ce{Mn(OH)2}$: white
• $\alpha$-$\ce{MnS}$: green
• $\ce{MnC2O4.2H2O}$: white

These substances dissolve readily in strong acids, a general pattern among transition-metal compounds.

Note that $\ce{MnS}$ is insoluble in water but dissolves in weak acids such as $\ce{HAc}$, so it cannot be precipitated from an acidic solution.

Reducing Properties

In alkaline solution, $\ce{Mn(II)}$ is a fairly strong reducing agent and is readily oxidized to $\ce{Mn(IV)}$.

$$E^\ominus[\ce{MnO2/Mn(OH)2}]=-0.05\ \mathrm{V},\qquad E^\ominus(\ce{O2/OH-})=0.40\ \mathrm{V}$$

$$\mathrm{Mn^{2+}} \xrightarrow{\mathrm{OH^-}} \mathrm{Mn(OH)_2} \text{ (white)} \xrightarrow{\mathrm{O_2}} \mathrm{MnO(OH)} \text{ (brown)} \xrightarrow{\mathrm{O_2}} \mathrm{MnO_2 \cdot nH_2O}$$

Both aqueous ammonia and strong bases convert $\ce{Mn^2+}$ into basic, nearly white$\ce{Mn(OH)2}$.

$\ce{Mn(OH)2}$ is oxidized extremely readily. Even the small amount of dissolved oxygen in water can oxidize it to brownish-black$\ce{MnO(OH)2}$, also written as $\ce{MnO2.H2O}$ or $\ce{H2MnO3}$. This reaction can be used to determine dissolved oxygen.

When precipitates of $\ce{MnS}$, $\ce{MnCO3}$, or $\ce{MnC2O4}$ stand in air or are heated, atmospheric oxygen oxidizes them to $\ce{MnO(OH)2}$.

$$\ce{MnS + O2 + H2O -> MnO(OH)2 + S}$$

$$\ce{2MnCO3 + O2 + 2H2O ->[\triangle] 2MnO(OH)2 + 2CO2}$$

In an oxygen-free environment, MnO can be prepared by the thermal decomposition of $\ce{MnC2O4}$ or $\ce{MnCO3}$.

$$\mathrm{MnC_2O_4(s)} \ce{->[\Delta]} \mathrm{MnO(s)} + \mathrm{CO(g)} + \mathrm{CO_2(g)}$$

$$\mathrm{MnCO_3(s)} \ce{->[\Delta]} \mathrm{MnO(s)} + \mathrm{CO_2(g)}$$

In acidic solution, $\ce{Mn(II)}$ is a weaker reducing agent:

$$E^\ominus(\ce{MnO4-/Mn^2+})=1.51\ \mathrm{V}$$

Strong oxidizing agents can oxidize $\ce{Mn^2+}$ to $\ce{MnO4^-}$:

• $\ce{S2O8^2-->[Ag+]SO4^2-}$
• $\ce{BiO3-->Bi^3+}$
• $\ce{PbO2->Pb^2+}$

The first two reactions are commonly used to identify $\ce{Mn^2+}$.

The concentration $c(\ce{Mn^2+})$ must not be too high, especially in the first reaction. Otherwise $\ce{Mn^2+}$ readily comproportionates with $\ce{MnO4^-}$ to form brownish-black$\ce{MnO2}$.

When a $\ce{Mn(II)}$ salt is heated and its anion is oxidizing, $\ce{Mn(II)}$ is oxidized:

$$\ce{Mn(NO3)2->[\triangle]MnO2 + 2NO2 ^}$$

$$\ce{Mn(ClO4)2 ->[\triangle] MnO2 + Cl2 ^ + 3O2 ^}$$

By analogy, one possible decomposition pathway of nitrosyl perchlorate, $\ce{NOClO4}$, is:

$$\ce{2NOClO4 ->[\triangle] N2O4 + Cl2 ^ + 3O2 ^}$$

Coordination Compounds

Weak-Field Ligands: High-Spin Octahedral Complexes

$\ce{Mn^2+}$ has the electron configuration $3d^5$. Its high-spin octahedral complexes with weak-field ligands have the electron arrangement $\ce{(t_{2g})^3(e_g)^2}$ and a crystal-field stabilization energy of zero.

The hydrated ion $\ce{[Mn(H2O)6]^2+}$ is extremely pale pink, almost invisible below a concentration of $1\ \mathrm{M}$. In these high-spin compounds, a d-d transition requires not only promotion from a lower to a higher energy level but also a reversal of electron spin. This is a spin-forbidden transition. Because such transitions are very unlikely, absorption in the visible region is weak.

Weak-Field Ligands: High-Spin Tetrahedral Complexes

In ethanol, $\ce{Mn^2+}$ forms yellow$\ce{[MnX4]^2-}$, where $\ce{X=Cl,Br,I}$. Its electron arrangement is $(e)^2(t_2)^3$, and $CFSE=0$.

Strong-Field Ligands: Low-Spin Octahedral Complexes

Only $\ce{Mn(II)}$ with certain strong-field ligands forms colored, low-spin complexes. An example is blue-violet$\ce{[Mn(CN)6]^4-}$, with electron arrangement $(e)^5(t_2)^0$ and $CFSE=(2\Delta-2P)$. In air, this compound is readily oxidized to brownish-red$\ce{[Mn(CN)6]^3-}$.

$\ce{Mn(III)}$ Compounds

$$\mathrm{MnO_2} \xrightarrow{+0.95\ \mathrm{V}} \mathrm{Mn^{3+}} \xrightarrow{+1.51\ \mathrm{V}} \mathrm{Mn^{2+}}$$

The Latimer diagram shows that $\ce{Mn(III)}$ is strongly oxidizing. It is unstable in solution and readily disproportionates.

Important $\ce{Mn(III)}$ compounds include:

• $\ce{Mn(CH3COO)3.3H2O}$: reddish purple
• $\ce{MnF3}$: reddish purple, prepared by reacting $\ce{MnF2}$ with $\ce{F2}$
• $\ce{Mn2O3}$: black, prepared by heating $\ce{MnO2}$ below $800\ ^\circ\mathrm{C}$

Important $\ce{Mn(III)}$ coordination compounds include:

• $\ce{[Mn(CN)6]^3-}$: brownish red
• $\ce{[Mn(PO4)2]^3-}$: purple

$\ce{Mn(IV)}$ Compounds

$\ce{MnO2}$ powder is black. The precipitate formed in solution is the brownish-black hydrate $\ce{MnO2.H2O}$.

Under ordinary conditions, $\ce{MnO2}$ is very stable. It is insoluble in $\ce{H2O}$, dilute acids, and dilute bases, and does not disproportionate in acid or base. However, $\ce{MnO2}$ is amphoteric and reacts slowly with concentrated acids and bases.

Fusion of $\ce{MnO2}$ with $\ce{NaOH}$ in the absence of air produces the manganite $\ce{Na2MnO3}$:

$$\ce{MnO2 + 2NaOH ->[\Delta]Na2MnO3 + H2O}$$

The fact that $\ce{MnO2}$ reacts with $\ce{NaOH}$ in this way demonstrates its acidic character.

Because $\ce{Mn(IV)}$ is an intermediate oxidation state, it can act as either an oxidizing or a reducing agent.

In strong acid, $\ce{MnO2}$ is a powerful oxidizing agent. It oxidizes $\ce{I^-}$ to $\ce{I2}$ and $\ce{Fe^2+}$ to $\ce{Fe^3+}$; heating it with concentrated hydrochloric acid produces chlorine.

Heat a test tube containing a mixture of $\ce{MnO2}$ powder and concentrated $\ce{H2SO4}$ in a water bath. After cooling and standing, the upper part of the test tube becomes reddish purple, indicating the formation of $\ce{Mn^3+}$:

$$4\mathrm{MnO_2} + 6\mathrm{H_2SO_4}(\text{conc.}) \xlongequal{\Delta} 2\mathrm{Mn_2(SO_4)_3} + 6\mathrm{H_2O} + \mathrm{O_2} \uparrow$$

$\ce{Mn^3+}$ is unstable. At higher temperatures it converts into the more stable $\ce{Mn^2+}$:

$$2\mathrm{Mn_2(SO_4)_3} + 2\mathrm{H_2O} \xlongequal{\Delta} 4\mathrm{MnSO_4} + \mathrm{O_2} \uparrow + 2\mathrm{H_2SO_4}$$

Under alkaline conditions, $\ce{MnO2}$ can be oxidized to $\ce{Mn(VI)}$. Mix $\ce{MnO2}$ with a base and an oxidizing agent such as potassium chlorate (2026 Beijing Gaokao chemistry) or potassium nitrate, or use oxygen from air, and heat the mixture to fusion:

$$3\mathrm{MnO_2} + 6\mathrm{KOH} + \mathrm{KClO_3} \xlongequal{\text{fusion}} 3\mathrm{K_2MnO_4} + \mathrm{KCl} + 3\mathrm{H_2O}$$

$$2\mathrm{MnO_2} + 4\mathrm{KOH} + \mathrm{O_2} \xlongequal{\text{fusion}} 2\mathrm{K_2MnO_4} + 2\mathrm{H_2O}$$

$\ce{Mn(VI)}$ Compounds

Among $\ce{Mn(VI)}$ compounds, potassium manganate, $\ce{K2MnO4}$, is relatively stable. It forms deep-green crystals and decomposes at $190\ ^\circ\mathrm{C}$ into potassium manganite, $\ce{K2MnO3}$, and oxygen. $\ce{Mn(VI)}$ is relatively stable in strongly alkaline solution.

$\ce{MnO4^2-}$ is stable only in concentrated strong base; acidification or a decrease in alkalinity causes disproportionation. The same reaction can be written in acidic-medium and aqueous forms as follows:

$$\ce{3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 v + 2H2O}$$

$$\ce{3MnO4^2- + 2H2O -> 2MnO4^- + MnO2 v + 4OH^-}$$

Because the latter form produces $\ce{OH^-}$, increasing alkalinity shifts the equilibrium to the left. Potassium manganate is stable only in a concentrated strong base, at $\mathrm{pH}>14$.

Heat a mixture of $\ce{MnO2}$, $\ce{KClO3}$, and $\ce{KOH}$ to fusion in a dry test tube. A green product forms. After cooling to room temperature, adding a small amount of water gives a green solution, showing that potassium manganate has formed. Adding a large amount of water immediately turns the solution reddish purple and produces a brown precipitate. Dilution lowers the base concentration, allowing manganate to disproportionate into $\ce{MnO4^-}$ and $\ce{MnO2}$.

Note that producing potassium permanganate by this route wastes one-third of the manganese (2026 Beijing Gaokao chemistry).

$\ce{Mn(VII)}$ Compounds

The most important $\ce{Mn(VII)}$ compound is potassium permanganate, $\ce{KMnO4}$, which forms purple-black crystals. The color of its aqueous solution depends on concentration. From low to high concentration, the solution appears pink, red, reddish purple, purple, and finally purple-black. Sodium permanganate, $\ce{NaMnO4}$, is deliquescent and difficult to purify.

Strong Oxidizing Properties

$\ce{KMnO4}$ is one of the most important and widely used oxidizing agents. Its oxidizing power and reduction product depend on the acidity of the medium. It is reduced to $\ce{Mn^2+}$ in strongly acidic solution, to $\ce{MnO4^2-}$ in strongly alkaline solution, and to $\ce{MnO2}$ in nearly neutral solution because these products are stable in their respective media. Its reactions with $\ce{S(IV)}$ illustrate this behavior:

Acidic:

$$2\mathrm{MnO_4^-} + 5\mathrm{H_2SO_3} \xlongequal{} 2\mathrm{Mn^{2+}} + 5\mathrm{SO_4^{2-}} + 4\mathrm{H^+} + 3\mathrm{H_2O}$$

Neutral:

$$2\mathrm{MnO_4^-} + \mathrm{H_2O} + 3\mathrm{SO_3^{2-}} \xlongequal{} 2\mathrm{MnO_2} \downarrow + 3\mathrm{SO_4^{2-}} + 2\mathrm{OH^-}$$

Alkaline:

$$2\mathrm{MnO_4^-} + 2\mathrm{OH^-} + \mathrm{SO_3^{2-}} \xlongequal{} 2\mathrm{MnO_4^{2-}} + \mathrm{SO_4^{2-}} + \mathrm{H_2O}$$

In acidic solution, $\ce{KMnO4}$ is a very strong oxidizing agent:

$$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{e^-} \xlongequal{} \mathrm{Mn^{2+}} + 4\mathrm{H_2O} \quad E^\ominus = 1.51 \text{ V}$$

It can oxidize $\ce{Cl^-}$, $\ce{Cr^3+}$, $\ce{I2}$, and many other species:

$$2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 10\mathrm{Cl^-} \xlongequal{} 2\mathrm{Mn^{2+}} + 5\mathrm{Cl_2} \uparrow + 8\mathrm{H_2O}$$

$$6\mathrm{MnO_4^-} + 10\mathrm{Cr^{3+}} + 11\mathrm{H_2O} \xlongequal{} 6\mathrm{Mn^{2+}} + 5\mathrm{Cr_2O_7^{2-}} + 22\mathrm{H^+}$$

$$2\mathrm{MnO_4^-} + \mathrm{I_2} + 4\mathrm{H^+} \xlongequal{} 2\mathrm{Mn^{2+}} + 2\mathrm{IO_3^-} + 2\mathrm{H_2O}$$

The reaction between $\ce{KMnO4}$ and hydrochloric acid can be used to prepare chlorine in the laboratory, but gas generation cannot be stopped on demand. Because $\ce{KMnO4}$ is also more expensive than $\ce{MnO2}$, laboratories more commonly prepare chlorine by reacting $\ce{MnO2}$ with concentrated hydrochloric acid.

Under acidic conditions, $\ce{KMnO4}$ reacts quantitatively with $\ce{H2C2O4}$ and can therefore be standardized using oxalic acid:

$$2\mathrm{MnO_4^-} + 6\mathrm{H^+} + 5\mathrm{H_2C_2O_4} \xlongequal{} 2\mathrm{Mn^{2+}} + 10\mathrm{CO_2} \uparrow + 8\mathrm{H_2O}$$

$\ce{KMnO4}$ also reacts quantitatively with $\ce{Fe^2+}$ and can be used to determine its concentration.

In volumetric analysis, $\ce{KMnO4}$ is commonly used as a redox titrant. In acidic solution, $\ce{MnO4^-}$ is reduced to $\ce{Mn^2+}$. A slight excess of $\ce{MnO4^-}$ immediately turns the solution red, whereas a dilute $\ce{Mn^2+}$ solution is essentially colorless and does not obscure the endpoint. The titrant therefore acts as its own indicator.

As an oxidizing agent, $\ce{KMnO4}$ is used in many organic syntheses, including the preparation of saccharin, ascorbic acid (vitamin C), and niacin. It is also used for disinfection and for treating drinking and industrial water.

Instability

Permanganates are strongly oxidizing and unstable. They decompose appreciably in acidic solution and slowly in neutral or mildly alkaline solution:

$$4\mathrm{MnO_4^-} + 4\mathrm{H^+} \xlongequal{} 4\mathrm{MnO_2} \downarrow + 3\mathrm{O_2} \uparrow + 2\mathrm{H_2O}$$

$$4\mathrm{MnO_4^-} + 4\mathrm{OH^-} \xlongequal{} 4\mathrm{MnO_4^{2-}} + \mathrm{O_2} \uparrow + 2\mathrm{H_2O}$$

Light catalyzes the decomposition of potassium permanganate, so its solutions should be stored in brown bottles. Because decomposition causes the concentration to change over time, a standard potassium permanganate solution must be re-standardized before use.

Permanganates are more stable as solids than in solution, but they still decompose on heating. At about $200\ ^\circ\mathrm{C}$, $\ce{KMnO4}$ forms $\ce{K2MnO4}$, $\ce{MnO2}$, and $\ce{O2}$:

$$2\mathrm{KMnO_4(s)} \xlongequal{200\ ^\circ\mathrm{C}} \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2} \uparrow$$

Heat $\ce{KMnO4}$ gently in a dry test tube. Popping sounds are heard, and after they stop the solid has lost its original crystalline luster. Add a small amount of water and shake: the test-tube wall becomes green, showing that $\ce{K2MnO4}$ is among the decomposition products. Adding a large amount of water immediately turns the solution purple because $\ce{K2MnO4}$ disproportionates to form $\ce{KMnO4}$.

Cold concentrated sulfuric acid reacts with $\ce{KMnO4}$ to form oily, dark-green manganese heptoxide, $\ce{Mn2O7}$:

$$2\mathrm{KMnO_4} + \mathrm{H_2SO_4(conc.)} \xlongequal{\text{low temperature}} \mathrm{Mn_2O_7} + \mathrm{K_2SO_4} + \mathrm{H_2O}$$

$\ce{Mn2O7}$ ignites on contact with organic matter, decomposes explosively when heated, and slowly releases $\ce{O2}$ at room temperature while converting to $\ce{MnO2}$.

Preparation of Potassium Permanganate

Potassium permanganate is commonly prepared from pyrolusite, $\ce{MnO2}$.

First prepare potassium manganate by heating a fused mixture of $\ce{MnO2}$, $\ce{KClO3}$, and $\ce{KOH}$ (2026 Beijing Gaokao chemistry). The product is green potassium manganate:

$$3\mathrm{MnO_2} + 6\mathrm{KOH} + \mathrm{KClO_3} \xlongequal{\text{fusion}} 3\mathrm{K_2MnO_4} + \mathrm{KCl} + 3\mathrm{H_2O}$$

Oxidation of $\ce{K2MnO4}$ with a strong oxidizing agent gives $\ce{KMnO4}$. Chlorine, for example, can be used:

$$2\mathrm{MnO_4^{2-}} + \mathrm{Cl_2} \xlongequal{} 2\mathrm{MnO_4^-} + 2\mathrm{Cl^-}$$

Industrial production commonly uses electrolysis of a $\ce{K2MnO4}$ solution:

Anode reaction $\quad \mathrm{MnO_4^{2-}} \xlongequal{} \mathrm{MnO_4^-}+\mathrm{e^-}$

Cathode reaction $\quad 2\mathrm{H_2O}+2\mathrm{e^-} \xlongequal{} \mathrm{H_2}+2\mathrm{OH^-}$

Overall reaction $\quad 2\mathrm{K_2MnO_4}+2\mathrm{H_2O} \xlongequal{\text{electrolysis}} 2\mathrm{KMnO_4}+2\mathrm{KOH}+\mathrm{H_2} \uparrow$

Exam Practice

Effect of Alkalinity on the Reducing Behavior of $\ce{Mn(II)}$

(2022 Beijing Gaokao Chemistry, Question 19; adapted from the original paper)

An experimental group investigated the reactions of chlorine with manganese(II) compounds under different conditions.

Information:

i. Under suitable conditions, $\ce{Mn^2+}$ can be oxidized by $\ce{Cl2}$ or $\ce{ClO^-}$ to $\ce{MnO2}$ (brown-black), $\ce{MnO4^2-}$ (green), or $\ce{MnO4^-}$ (purple).

ii. In strongly alkaline solution, $\ce{MnO4^-}$ can be reduced by $\ce{OH^-}$ to $\ce{MnO4^2-}$.

iii. The oxidizing power of $\ce{Cl2}$ is independent of the acidity of the solution, whereas the oxidizing power of $\ce{NaClO}$ decreases as alkalinity increases.

The apparatus is shown below (supports omitted):

Vessel C was charged with $10\ \mathrm{mL}$ of substance a and five drops of $0.1\ \mathrm{mol\cdot L^{-1}}$ $\ce{MnSO4}$. Chlorine generated in A and washed in B was then passed into C.

(1) The reagent in vessel B is saturated $\ce{NaCl}$ solution.

(2) Before chlorine was introduced, the equation for the conversion of the white precipitate into a brown-black precipitate in experiments II and III is $\ce{2Mn(OH)2 + O2 -> 2MnO2 + 2H2O}$.

(3) Comparing experiments I and II after chlorine was introduced shows that manganese(II) compounds can be oxidized only to $\ce{MnO2}$ under neutral or weakly acidic conditions, but to higher oxidation states under alkaline conditions.

(4) According to information ii, experiment III should have produced a green solution, but a purple solution was obtained. Two explanations were proposed:

• introducing $\ce{Cl2}$ may have reduced the alkalinity;
• excess oxidant may have oxidized $\ce{MnO4^2-}$ further to $\ce{MnO4^-}$.

① The equation for the reaction that could reduce the alkalinity is $\ce{2NaOH + Cl2 -> NaCl + NaClO + H2O}$. Measurement showed that the alkalinity changed very little.

② To $1\ \mathrm{mL}$ of the suspension from experiment III after standing, $4\ \mathrm{mL}$ of $40\%$ $\ce{NaOH}$ was added. The solution rapidly changed from purple to green, after which the green color slowly deepened. The ionic equation for the rapid color change is $\ce{4MnO4^- + 4OH^- -> 4MnO4^2- + O2 ^ + 2H2O}$; the green color slowly deepened because $\ce{MnO2}$ was oxidized by $\ce{NaClO}$, proving that the oxidant was in excess.

③ Another $1\ \mathrm{mL}$ portion of the suspension was diluted with $4\ \mathrm{mL}$ of water. The purple color slowly deepened. The reaction involved is $\ce{2MnO2 + 3ClO^- + 2OH^- -> 2MnO4^- + 3Cl^- + H2O}$.

④ In terms of reaction rates, experiment III did not yield a green solution because under strongly alkaline conditions, $\ce{2MnO4^2- + ClO^- + H2O -> 2MnO4^- + Cl^- + 2OH^-}$ is faster than $\ce{4MnO4^- + 4OH^- -> 4MnO4^2- + O2 ^ + 2H2O}$.

Focus on the competition between green and purple implied by the second explanation: $\ce{MnO4^-}$ is formed rapidly and consumed slowly, whereas $\ce{MnO4^2-}$ is formed slowly and consumed rapidly. Part (4)② shows that the oxidation of $\ce{MnO2}$ by $\ce{ClO^-}$ is slow, so it is not the principal factor.

Preparation, Purification, and Yield of Potassium Permanganate

(2026 Beijing Gaokao Chemistry, Question 19; adapted from the original paper)

An experimental group prepared potassium permanganate.

Information:

i. $\ce{K2MnO4}$ is a dark-green solid. It is soluble in water, stable under strongly alkaline conditions, and disproportionates under suitable conditions:

$$\ce{3MnO4^2- + 2H2O -> 2MnO4^- + MnO2 v + 4OH^-}$$

ii. In strongly alkaline solution, $\ce{MnO4^-}$ can be reduced by $\ce{OH^-}$ to $\ce{MnO4^2-}$.

iii. The solubility of $\ce{KMnO4}$ increases with temperature.

(1) Preparation of $\ce{KMnO4}$

① Complete the equation for the reaction in step I:

$\ce{KClO3 + 3MnO2 + 6KOH} \xlongequal{\text{molten}} \ce{KCl + 3K2MnO4 + 3H2O}$

② In step II, $\ce{K2MnO4}$ disproportionates as pH decreases. If the reducing power of $\ce{MnO4^2-}$ remains unchanged, its oxidizing power increases.

③ When a small sample of Z was dissolved, a purple-red$\ce{KMnO4}$ solution and a small amount of brown-black$\ce{MnO2}$ were obtained.

(2) Investigating the source of $\ce{MnO2}$ in step III

① Student A proposed that $\ce{CH3COO^-}$ reduced $\ce{MnO4^-}$. In experiment i, Y was replaced with a solution containing $0.2\ \mathrm{mol\cdot L^{-1}}$ $\ce{KMnO4}$ and $0.6\ \mathrm{mol\cdot L^{-1}}$ $\ce{CH3COOK}$ at $\mathrm{pH}=10$. Repeating step III produced a solid containing $\ce{MnO2}$. Student B argued that this result alone did not prove that acetate reduced permanganate because the experiment did not rule out interference from reduction of $\ce{MnO4^-}$ by $\ce{OH^-}$.

② In blank experiment ii, Y was replaced with a solution containing $0.2\ \mathrm{mol\cdot L^{-1}}$ $\ce{KMnO4}$ and adjusted to $\mathrm{pH}=10$ with $\ce{KOH}$. Repeating step III produced no $\ce{MnO2}$; experiments i and ii therefore supported Student A's proposal.

③ Student C replaced $\ce{CH3COOK}$ with $0.15\ \mathrm{mol\cdot L^{-1}}$ $\ce{KCl}$ in experiment i. This confirmed that $\ce{Cl^-}$ in step III could not reduce $\ce{MnO4^-}$.

(3) Determination of $\ce{KMnO4}$ purity and yield

Product was prepared from the quantities used in step I using the optimized procedure. One quarter of the product was made up to $250\ \mathrm{mL}$ as the test solution. Under acidic conditions, the test solution was used to titrate a standard $\ce{H2C2O4}$ solution; $\ce{MnO4^-}$ was reduced to $\ce{Mn^2+}$. The measured concentration was $c(\ce{KMnO4})=a\ \mathrm{mol\cdot L^{-1}}$.