D- and F-Block Elements MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: Statement I is correct because native copper contains elemental copper with oxidation state zero. Chromite and pyrolusite, in contrast, are compounds in which the corresponding transition metals already possess positive oxidation states. Statement II is correct because chromium has oxidation state +3 in chromite and +6 in chromate. Each chromium centre must therefore lose three electrons during oxidative processing. Manganese has oxidation state +4 in pyrolusite and +7 in permanganate. Its complete conversion into permanganate similarly requires oxidation, although preparation commonly proceeds through manganese(VI) manganate. Statement III is incorrect because simple dissolution does not automatically cause the necessary increases in oxidation state. Oxidative fusion or another suitable oxidation process is required to form the higher-oxidation-state intermediates. Statement IV is correct because chromate contains chromium(VI), whereas manganate contains manganese(VI). Both metals have the same formal +6 oxidation state despite their different identities. Their d-electron occupancies differ: Cr(VI) is \(d^0\), whereas Mn(VI) is \(d^1\). Therefore, Statements I, II and IV are correct, while Statement III is incorrect.

Explanation: Statement I is correct because Mn2+ has the favourable half-filled \(3d^5\) configuration. Removal of another electron produces Mn3+ with \(3d^4\), disrupting that arrangement. This contributes to manganese's relatively high third ionisation enthalpy. Statement II is correct because Fe2+ has \(3d^6\). Removal of one electron gives Fe3+ with the half-filled \(3d^5\) subshell. Thus, the third ionisation has a different electronic consequence for iron than for manganese. Statement III is correct because Zn2+ has a completely filled \(3d^{10}\) subshell. Further oxidation to formal Zn3+ would remove a d electron and produce \(3d^9\), disrupting the filled arrangement. This helps explain the strong preference of zinc for the +2 oxidation state. Copper(II), meanwhile, has \(3d^9\), and its hypothetical oxidation produces \(3d^8\). Statement IV is incorrect because equal numbers of removed electrons do not imply equal ionisation enthalpies. The starting ions have different nuclear charges, d-electron occupancies and electronic stabilities. Their third ionisation energies therefore need not be identical. Moreover, hypothetical gas-phase ionisation does not establish that every resulting trivalent state is chemically accessible under ordinary aqueous conditions. The comparison demonstrates why successive electron removal must be analysed using the specific initial and final configurations.

Explanation: Oxygen has oxidation state \(-2\) in the four oxides under consideration. In MnO, manganese must therefore possess oxidation state +2. In Mn2O3, the total oxygen contribution is \(-6\), so each manganese has oxidation state +3. In MnO2, manganese is +4, while in Mn2O7 it is +7. Thus, the series represents progressively higher oxidation states of the same metal. Lower transition-metal oxides generally show greater basic character. As the metal oxidation state increases, its oxygen-containing compounds tend to exhibit greater acidic character. Manganese(VII) oxide is an acidic oxide and is associated with the chemistry of permanganic acid. However, the trend is not a rule that every intermediate oxide must be exclusively acidic or basic. Manganese dioxide provides an important qualification because it can display amphoteric behaviour. The formal oxidation state is therefore useful for interpreting the overall pattern but does not replace knowledge of the actual oxide. The sequence also illustrates why transition metals can form several chemically distinct oxides. Their acid–base behaviour reflects differences in oxidation state and metal–oxygen bonding, not simply the number of oxygen atoms.

Explanation: \(\textbf{Choose a consistent comparison basis:}\) The host metal is nickel, with radius \[ r_{\mathrm{Ni}}=124\,\mathrm{pm} \] The approximate substitutional-alloy guideline compares the radius difference with the host radius. \[ \%\Delta r=\frac{|r_{\mathrm{solute}}-r_{\mathrm{Ni}}|}{r_{\mathrm{Ni}}}\times100 \] \(\textbf{Evaluate iron and cobalt:}\) \[ \%\Delta r_{\mathrm{Fe}}=\frac{126-124}{124}\times100 \] \[ \%\Delta r_{\mathrm{Fe}}\approx1.61\% \] \[ \%\Delta r_{\mathrm{Co}}=\frac{125-124}{124}\times100 \] \[ \%\Delta r_{\mathrm{Co}}\approx0.81\% \] \(\textbf{Evaluate copper and zinc:}\) \[ \%\Delta r_{\mathrm{Cu}}=\frac{128-124}{124}\times100 \] \[ \%\Delta r_{\mathrm{Cu}}\approx3.23\% \] \[ \%\Delta r_{\mathrm{Zn}}=\frac{134-124}{124}\times100 \] \[ \%\Delta r_{\mathrm{Zn}}\approx8.06\% \] \(\textbf{Apply the approximate size criterion:}\) All four radius differences are below 15%. Therefore, none is excluded by the specified size guideline alone. \(\textbf{Interpret the periodic trend:}\) The supplied metallic radii decrease slightly from iron through nickel. They then increase towards copper and zinc. Thus, increasing nuclear charge does not produce an uninterrupted decrease in metallic radius across these elements. Changes in d-electron occupancy, shielding and metallic bonding contribute to the observed irregularities. \(\textbf{Distinguish compatibility from guaranteed alloy formation:}\) A favourable atomic-size comparison is one factor promoting substitutional alloy formation. Crystal structure, chemical interactions and temperature also influence solubility in the solid state. Therefore, the data do not prove that every composition forms a homogeneous alloy. \(\textbf{Final answer:}\) \[ \boxed{\%\Delta r_{\mathrm{Fe}}=1.61\%,\quad\%\Delta r_{\mathrm{Co}}=0.81\%,\quad\%\Delta r_{\mathrm{Cu}}=3.23\%,\quad\%\Delta r_{\mathrm{Zn}}=8.06\%} \]

Explanation: Statement I is correct because chromium-containing stainless steels derive important corrosion resistance from a protective surface oxide. This film limits continued attack on the underlying metallic material. Consequently, the state of the surface can influence corrosion behaviour independently of the overall alloy composition. Statement II is correct because the two samples are exposed to different environments after scratching. In Experiment P, conditions favour restoration of the protective oxide film. In Experiment Q, oxygen deficiency inhibits that restoration. The latter surface may therefore remain more vulnerable to corrosion despite having the same bulk chromium content. Statement III is incorrect because protective-film formation involves oxidation of chromium at the surface, not complete conversion of chromium compounds into elemental metal. Chromium-containing oxide formation can preserve the underlying bulk metallic alloy. Statement IV is correct because initial surface appearance does not establish how a material will behave after mechanical damage and environmental exposure. The comparison also distinguishes bulk alloying from surface passivation. The presence of chromium is important, but the effectiveness of corrosion protection depends on the formation and persistence of the protective layer. Therefore, material behaviour must be evaluated using both composition and chemical environment. The example illustrates a practical consequence of transition-metal oxide chemistry without requiring an independent detailed corrosion mechanism.

Explanation: Iron(II) has the formal \(d^6\) electronic configuration. In the hexacyanoferrate(II) complex, six cyanide ligands each carry a charge of \(-1\). Their total charge is \(-6\), while the complete complex has charge \(-4\). Iron therefore remains in oxidation state +2. Thus, the formal \(d^6\) designation is correct for both the simple ion and the complex. However, formal d-electron count does not completely determine the arrangement of electrons in every chemical environment. The nature of the surrounding ligands can influence orbital energies and electron pairing. Consequently, two species with identical formal d-electron counts need not have identical numbers of unpaired electrons. Aqueous high-spin Fe2+ is ordinarily paramagnetic. The magnetic behaviour of a particular complex must instead be interpreted using its actual electronic arrangement. The student's mistake is to treat formal oxidation state and d-electron count as sufficient information for every magnetic prediction. This comparison is also distinct from oxidation of Fe2+ to Fe3+, which genuinely removes an electron. Coordination can alter physical behaviour without changing the formal number of metal d electrons.

Explanation: \(\textbf{Determine manganese's oxidation-state change:}\) Manganese has oxidation state +4 in MnO2. It has oxidation state +6 in potassium manganate. Therefore, each manganese centre loses two electrons during the oxidative fusion. \(\textbf{Identify the required reactant proportions:}\) The balanced equation is \[ 2\mathrm{MnO_2}+4\mathrm{KOH}+\mathrm{O_2} \longrightarrow 2\mathrm{K_2MnO_4}+2\mathrm{H_2O} \] Two moles of MnO2 require four moles of KOH and one mole of oxygen. \(\textbf{Compare available potassium hydroxide:}\) Complete processing of \(0.300\,\mathrm{mol}\) MnO2 would require \[ n(\mathrm{KOH})_{\mathrm{required}}=2(0.300) \] \[ n(\mathrm{KOH})_{\mathrm{required}}=0.600\,\mathrm{mol} \] Only \(0.440\,\mathrm{mol}\) KOH is present. \(\textbf{Check the oxygen requirement:}\) Complete processing of MnO2 would require \[ n(\mathrm{O_2})_{\mathrm{required}}=\frac12(0.300) \] \[ n(\mathrm{O_2})_{\mathrm{required}}=0.150\,\mathrm{mol} \] The available oxygen amount is \(0.200\,\mathrm{mol}\). Thus, oxygen is sufficient while KOH is limiting. \(\textbf{Calculate potassium manganate formation:}\) Four moles of KOH produce two moles of potassium manganate. \[ n(\mathrm{K_2MnO_4})=\frac24n(\mathrm{KOH}) \] \[ n(\mathrm{K_2MnO_4})=\frac12(0.440) \] \[ n(\mathrm{K_2MnO_4})=0.220\,\mathrm{mol} \] \(\textbf{Calculate residual manganese dioxide:}\) The reaction converts one mole of MnO2 into one mole of manganate. \[ n(\mathrm{MnO_2})_{\mathrm{remaining}}=0.300-0.220 \] \[ n(\mathrm{MnO_2})_{\mathrm{remaining}}=0.080\,\mathrm{mol} \] \(\textbf{Check oxygen consumption:}\) \[ n(\mathrm{O_2})_{\mathrm{consumed}}=\frac12(0.220) \] \[ n(\mathrm{O_2})_{\mathrm{consumed}}=0.110\,\mathrm{mol} \] This is less than the supplied \(0.200\,\mathrm{mol}\), confirming the limiting-reactant decision. \(\textbf{Chemical interpretation:}\) The reaction generates green manganese(VI) manganate from manganese(IV) dioxide. The presence of excess oxygen does not overcome an insufficient alkali supply. All three reactant amounts must therefore be compared before calculating the maximum product yield. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{K_2MnO_4})=0.220\,\mathrm{mol},\quad n(\mathrm{MnO_2})_{\mathrm{remaining}}=0.080\,\mathrm{mol}} \]

Explanation: Manganate has the formula \(\mathrm{MnO_4^{2-}}\). Four oxygen atoms contribute a total oxidation number of \(-8\), and the ion carries charge \(-2\). Manganese therefore has oxidation state +6. Permanganate has the formula \(\mathrm{MnO_4^-}\). Its manganese oxidation state is +7. The transformation consequently involves removal of one electron per manganese centre. This is a genuine oxidation process rather than a simple protonation equilibrium. By comparison, chromate and dichromate both contain chromium(VI). Their interconversion can occur through changes in hydrogen-ion concentration without changing chromium's oxidation state. The visual similarity of two colour-changing systems therefore does not establish a common reaction mechanism. The absence of manganese dioxide is also compatible with the explicitly proposed oxidation pathway rather than manganate disproportionation. In disproportionation, manganese(VI) produces both Mn(VII) and Mn(IV) under suitable conditions. The present question specifies formation of permanganate without that competing product. The correct explanation must therefore track the actual ionic charges and oxidation states. Colour observations are useful evidence, but they cannot replace electron-balance reasoning.

Explanation: \(\textbf{Identify the first oxidation-state change:}\) Manganese begins in oxidation state +2. Its first product, MnO2, contains manganese(IV). Each manganese centre therefore loses two electrons during this stage. \[ \mathrm{Mn(II)}\longrightarrow\mathrm{Mn(IV)}+2e^- \] \(\textbf{Use the supplied first-stage electron transfer:}\) \[ n(e^-)_1=0.240\,\mathrm{mol} \] The amount of manganese processed is \[ n(\mathrm{Mn})=\frac{0.240}{2} \] \[ n(\mathrm{Mn})=0.120\,\mathrm{mol} \] \(\textbf{Track manganese conservation:}\) The question states that no manganese is lost during separation or subsequent treatment. Therefore, the second stage also processes \[ n(\mathrm{MnO_2})=0.120\,\mathrm{mol} \] \(\textbf{Identify the second oxidation-state change:}\) Manganese changes from +4 in MnO2 to +7 in permanganate. Each manganese centre must lose three electrons. \[ \mathrm{Mn(IV)}\longrightarrow\mathrm{Mn(VII)}+3e^- \] \(\textbf{Calculate the second-stage electron transfer:}\) \[ n(e^-)_2=3n(\mathrm{MnO_2}) \] \[ n(e^-)_2=3(0.120) \] \[ n(e^-)_2=0.360\,\mathrm{mol} \] \(\textbf{Verify the complete oxidation requirement:}\) The total electron loss through both stages is \[ n(e^-)_{\mathrm{total}}=0.240+0.360 \] \[ n(e^-)_{\mathrm{total}}=0.600\,\mathrm{mol} \] Direct oxidation of manganese from +2 to +7 requires five electrons per manganese centre. \[ 5(0.120)=0.600\,\mathrm{mol} \] Thus, the two routes have the same overall electron requirement. \(\textbf{Chemical interpretation:}\) Introducing an intermediate oxidation state changes how electron transfer is distributed between stages. It does not change the net electron loss associated with the same initial and final manganese oxidation states. The first-stage electron measurement therefore provides a reverse route to determining the amount of manganese processed. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Mn})=0.120\,\mathrm{mol},\quad n(e^-)_2=0.360\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the selective electron-transfer process:}\) Only manganate(VI) is oxidised. Its manganese oxidation state increases from +6 to +7. Therefore, each manganate ion loses one electron. \[ \mathrm{MnO_4^{2-}}\longrightarrow\mathrm{MnO_4^-}+e^- \] \(\textbf{Determine the amount of manganate converted:}\) The total electron removal is \[ n(e^-)=0.050\,\mathrm{mol} \] Since the process transfers one electron per manganate ion, \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{converted}}=0.050\,\mathrm{mol} \] \(\textbf{Calculate the manganate amount remaining:}\) \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{final}}=0.090-0.050 \] \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{final}}=0.040\,\mathrm{mol} \] \(\textbf{Calculate the additional permanganate:}\) Each converted manganate ion forms one permanganate ion. \[ n(\mathrm{MnO_4^-})_{\mathrm{formed}}=0.050\,\mathrm{mol} \] \(\textbf{Include the initial permanganate:}\) \[ n(\mathrm{MnO_4^-})_{\mathrm{final}}=0.030+0.050 \] \[ n(\mathrm{MnO_4^-})_{\mathrm{final}}=0.080\,\mathrm{mol} \] \(\textbf{Account for manganese dioxide:}\) Manganese dioxide remains unchanged because the stipulated oxidant does not oxidise Mn(IV). Therefore, \[ n(\mathrm{MnO_2})_{\mathrm{final}}=0.060\,\mathrm{mol} \] \(\textbf{Verify manganese conservation:}\) Initially, \[ n(\mathrm{Mn})_{\mathrm{total}}=0.030+0.090+0.060 \] \[ n(\mathrm{Mn})_{\mathrm{total}}=0.180\,\mathrm{mol} \] Finally, \[ 0.080+0.040+0.060=0.180\,\mathrm{mol} \] The manganese-atom balance is satisfied. \(\textbf{Chemical interpretation:}\) The final permanganate amount includes both newly formed permanganate and material present initially. Manganese dioxide contributes to total manganese conservation but not to the specified electron transfer. This distinction prevents treating every manganese-containing species as an equivalent participant in the oxidation. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{MnO_4^{2-}})_{\mathrm{final}}=0.040\,\mathrm{mol},\quad n(\mathrm{MnO_4^-})_{\mathrm{final}}=0.080\,\mathrm{mol}} \]

Explanation: The supplied radii show a progressive decrease from La3+ to Lu3+. This is consistent with lanthanoid contraction. Nuclear charge increases across the lanthanoid series, while the added 4f electrons shield that increase relatively poorly. Consequently, the outer electron cloud experiences a progressively greater effective nuclear attraction. The student correctly recognises the contraction but incorrectly predicts its consequence for hydroxide basicity. Lanthanoid hydroxides generally decrease in basic character across the series. Larger, more ionic trivalent hydroxides towards the beginning of the series are more strongly basic. As ionic radius decreases, the greater polarising influence of the metal ion changes the character of metal–oxygen interactions. This contributes to the reduction in basic character. Therefore, Lu(OH)3 is less basic than La(OH)3 under the ordinary comparative trend. The common formula Ln(OH)3 does not imply identical base strength for every lanthanoid. Equal numbers of hydroxide groups determine neutralisation stoichiometry for complete reaction, not the relative chemical tendency associated with hydroxide basicity. The question illustrates how a correct periodic observation can lead to an incorrect property prediction if the underlying chemical relationship is applied in the wrong direction.

Explanation: Statement I is correct because oxygen has oxidation state \(-2\) in each ion. In \(\mathrm{MO_2^+}\), the two oxygen atoms contribute \(-4\), so the metal must have oxidation state +5. In \(\mathrm{MO_2^{2+}}\), the overall charge is +2, giving metal oxidation state +6. Statement II is correct because oxidation removes one electron from the formal \(5f^1\) population to produce \(5f^0\). Statement III is correct because the two ions contain the same number of oxygen atoms. Their charges differ because the metal oxidation state changes, not because an oxygen atom is added or removed. Statement IV is incorrect because its reasoning improperly equates electronic configuration, magnetism and colour. A formal \(5f^0\) metal centre has no unpaired f electron in the elementary description, but this does not establish that every compound containing that oxidation state is colourless. Electronic absorption can involve charge-transfer processes and other transitions. The existence of an empty f subshell therefore does not automatically determine optical appearance. The comparison demonstrates why oxidation-number calculation requires both oxygen contribution and overall ionic charge. It also shows how an actinoid oxocation can undergo a one-electron redox transformation without changing its oxygen stoichiometry.

Explanation: \(\textbf{Establish the stipulated reaction priority:}\) Chromium(III) undergoes oxidation before any iron(II) reacts. Therefore, the oxidant's available electron-acceptance capacity must first be assigned to chromium. \(\textbf{Determine chromium's electron loss:}\) Chromium changes from oxidation state +3 to +6. Each chromium centre releases three electrons. \[ \mathrm{Cr(III)}\longrightarrow\mathrm{Cr(VI)}+3e^- \] \(\textbf{Calculate complete chromium oxidation's demand:}\) \[ n(e^-)_{\mathrm{Cr}}=3n(\mathrm{Cr^{3+}}) \] \[ n(e^-)_{\mathrm{Cr}}=3(0.060) \] \[ n(e^-)_{\mathrm{Cr}}=0.180\,\mathrm{mol} \] \(\textbf{Compare with the oxidant capacity:}\) The available oxidising capacity is \[ n(e^-)_{\mathrm{available}}=0.180\,\mathrm{mol} \] This exactly equals the requirement for complete chromium oxidation. \(\textbf{Determine the chromium(VI) amount:}\) Every chromium(III) centre converted into chromium(VI) remains one chromium atom. Therefore, \[ n(\mathrm{Cr(VI)})=0.060\,\mathrm{mol} \] \(\textbf{Calculate oxidising capacity remaining for iron:}\) \[ n(e^-)_{\mathrm{remaining}}=0.180-0.180 \] \[ n(e^-)_{\mathrm{remaining}}=0 \] No oxidising capacity remains for iron(II). \(\textbf{Determine iron(II) remaining:}\) \[ n(\mathrm{Fe^{2+}})_{\mathrm{final}}=0.090\,\mathrm{mol} \] \(\textbf{Interpret the selectivity condition:}\) The result depends on the question's explicit oxidation order. Without that condition, the total electron balance would not uniquely specify the distribution of oxidation between chromium and iron. The important calculation is therefore not merely the amount of oxidant supplied but how its electron-acceptance capacity is allocated. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Cr(VI)})=0.060\,\mathrm{mol},\quad n(\mathrm{Fe^{2+}})_{\mathrm{remaining}}=0.090\,\mathrm{mol}} \]

Explanation: Chromium(III) has the formal \(d^3\) configuration with three unpaired electrons in its ordinary free-ion description. Manganese(II) has the half-filled \(d^5\) configuration and contains five unpaired electrons under the high-spin approximation. Zinc(II) has the filled \(d^{10}\) configuration, with no unpaired d electrons. Therefore, the three ions begin with different electronic populations. Coordination by the same ligand does not automatically make those populations identical. The metal's nuclear charge, oxidation state and d-electron occupancy remain relevant to the resulting electronic structure. Chromium(III) and manganese(II) can display paramagnetic behaviour, whereas zinc(II) is ordinarily diamagnetic. Their optical behaviour can also differ because the energies and intensities of available electronic transitions depend on both metal identity and ligand environment. A shared ligand may influence all three complexes, but it does not guarantee identical colours. Moreover, zinc(II) can form coordination compounds without changing its oxidation state or becoming a transition element under the usual electronic definition. The student's proposal therefore confuses one common chemical feature with complete electronic equivalence. The appropriate conclusion requires comparing both ligand identity and the electronic configuration of each metal centre.

Explanation: Statement I is incorrect because the oxidation state of a metal cannot be determined from its metal-to-oxygen ratio without considering the overall charge. Dichromate has the formula \(\mathrm{Cr_2O_7^{2-}}\), giving chromium oxidation state +6. Manganese(VII) oxide has the neutral formula \(\mathrm{Mn_2O_7}\), giving manganese oxidation state +7. Statement II is correct because each structure contains a single oxygen atom connecting its two metal-centred tetrahedral units. That atom must be counted only once in the complete formula. Consequently, both structures contain seven oxygen atoms rather than eight. Statement III is correct because the oxygen contribution is \(-14\) in both species. The two chromium atoms together contribute +12 to give the dichromate ion its charge of \(-2\). The two manganese atoms together contribute +14 to make Mn2O7 electrically neutral. Statement IV is correct because dichromate is associated with the acidic chemistry of chromium(VI), while manganese(VII) oxide is a strongly acidic oxide. However, their different metal identities, formal charges and oxidation states prevent their chemistry from being identical. A shared structural arrangement therefore provides useful information about atom connectivity without establishing equal redox behaviour. The correct comparison integrates structural counting, ionic charge and oxidation-state determination.

Explanation: \(\textbf{Identify the competing energetic contributions:}\) Conversion of a solid metal into its aqueous divalent ion involves breaking metallic interactions, removing two electrons and hydrating the resulting ion. Atomisation and ionisation contribute positive enthalpies. Hydration contributes a negative enthalpy. The model requires addition of all four supplied terms. \(\textbf{Evaluate manganese:}\) \[ \Delta H_{\mathrm{Mn}}=279+717+1509-1862 \] \[ \Delta H_{\mathrm{Mn}}=643\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Evaluate nickel:}\) \[ \Delta H_{\mathrm{Ni}}=431+736+1752-2121 \] \[ \Delta H_{\mathrm{Ni}}=798\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Evaluate copper:}\) \[ \Delta H_{\mathrm{Cu}}=339+745+1958-2121 \] \[ \Delta H_{\mathrm{Cu}}=921\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Establish the numerical ordering:}\) \[ 643\lt 798\lt 921 \] Thus, the approximate transformation enthalpy increases in the order \[ \mathrm{Mn\lt Ni\lt Cu} \] \(\textbf{Explain the copper result:}\) Copper(II) receives a strongly favourable hydration contribution. However, copper also has a particularly large second ionisation enthalpy. The combined positive contributions more than offset the hydration advantage in the stated model. \(\textbf{Connect the calculation to electrochemical behaviour:}\) A relatively costly transformation of copper metal into aqueous Cu2+ is consistent with copper's comparatively noble behaviour. However, standard reduction potentials are determined by Gibbs-energy changes, not enthalpy alone. Therefore, the calculated enthalpies cannot be substituted directly into an electrode-potential equation without additional thermodynamic information. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{\mathrm{Mn}}=643,\quad\Delta H_{\mathrm{Ni}}=798,\quad\Delta H_{\mathrm{Cu}}=921\ \mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct because vanadium has oxidation state +5 in \(\mathrm{VO_2^+}\), chromium has +6 in dichromate, and manganese has +7 in permanganate. Statement II is correct because the tendency of an oxidising species to accept electrons depends on the thermodynamic stability of both its initial and reduced forms. The stability of the lower oxidation states contributes to the trend described in NCERT. For example, reduction to manganese(II) involves formation of the favourable \(3d^5\) configuration. Statement III is incorrect because oxidising strength and electron-acceptance capacity per mole are different quantities. Reduction of one V(V) centre to V(IV) involves one electron. One dichromate ion accepts six electrons when its two chromium(VI) centres become chromium(III). One permanganate ion accepts five electrons on reduction from Mn(VII) to Mn(II). Therefore, the electron capacities do not increase monotonically in the stated order. Statement IV is correct because the three oxidants contain different metals and undergo different oxidation-state changes. Their balanced reductions also require different numbers of hydrogen ions and electrons. Consequently, the ranking of oxidising strength cannot replace individual half-reaction analysis. This comparison distinguishes thermodynamic tendency, oxidation state and stoichiometric electron capacity.

Explanation: \(\textbf{Identify the first-stage electron requirement:}\) Manganese changes from +6 in manganate to +7 in permanganate. Each manganate ion therefore loses one electron. \[ \mathrm{MnO_4^{2-}}\longrightarrow\mathrm{MnO_4^-}+e^- \] \(\textbf{Calculate manganate oxidised directly:}\) \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{oxidised}}=n(e^-) \] \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{oxidised}}=0.120\,\mathrm{mol} \] The first stage consequently produces \[ n(\mathrm{MnO_4^-})_1=0.120\,\mathrm{mol} \] \(\textbf{Determine the manganate entering the second stage:}\) \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{remaining}}=0.300-0.120 \] \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{remaining}}=0.180\,\mathrm{mol} \] \(\textbf{Apply the disproportionation ratio:}\) Three manganate ions produce two permanganate ions and one manganese dioxide formula unit. \[ n(\mathrm{MnO_4^-})_2=\frac23(0.180) \] \[ n(\mathrm{MnO_4^-})_2=0.120\,\mathrm{mol} \] \(\textbf{Calculate manganese dioxide formation:}\) \[ n(\mathrm{MnO_2})=\frac13(0.180) \] \[ n(\mathrm{MnO_2})=0.060\,\mathrm{mol} \] \(\textbf{Combine the permanganate populations:}\) \[ n(\mathrm{MnO_4^-})_{\mathrm{total}}=0.120+0.120 \] \[ n(\mathrm{MnO_4^-})_{\mathrm{total}}=0.240\,\mathrm{mol} \] \(\textbf{Verify manganese conservation:}\) \[ 0.240+0.060=0.300\,\mathrm{mol} \] The total manganese amount is unchanged. \(\textbf{Interpret the two pathways:}\) Direct oxidation converts every participating Mn(VI) centre into Mn(VII). Disproportionation instead produces both Mn(VII) and Mn(IV). Consequently, the two stages have different permanganate yields per mole of manganate consumed. Their contributions must be calculated separately before determining the overall recovery. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{MnO_4^-})=0.240\,\mathrm{mol},\quad n(\mathrm{MnO_2})=0.060\,\mathrm{mol}} \]

Explanation: Statement I is incorrect because chromium has oxidation state +6 in chromate, while manganese also has oxidation state +6 in manganate. Both ions have charge \(-2\), and both contain four oxygen atoms. However, their metal d-electron configurations differ because chromium and manganese have different atomic numbers. Statement II is correct because manganate and permanganate both have four oxygen atoms around manganese. Manganate contains Mn(VI), with the formal \(d^1\) configuration, while permanganate contains Mn(VII), with \(d^0\). Therefore, manganate has one unpaired d electron in the elementary description, whereas permanganate has none. Statement III is correct because chromate and permanganate are characteristically coloured despite having formally empty metal d subshells. Their colours cannot be explained through ordinary transitions between occupied metal d orbitals. Charge-transfer electronic transitions account for their characteristic intense absorption. Statement IV is incorrect because ionic charge and oxygen count determine the formal oxidation state, but not the d-electron count independently of metal identity. Chromium(VI) and manganese(VI) have different numbers of d electrons. The comparison therefore separates structural similarity, formal oxidation state, electronic configuration, magnetism and colour. No one of these properties can replace the others in describing the oxoanions.

Explanation: Statement I is correct because vanadium has oxidation state +5 in V2O5. Two vanadium atoms must balance the contribution of five oxide ions, giving +5 per vanadium centre. In \(\mathrm{VO_2^+}\), the two oxygen atoms contribute \(-4\), while the ion has charge +1. Vanadium must therefore again have oxidation state +5. Statement II is correct because \(\mathrm{VO^{2+}}\) contains vanadium(IV). The conversion from the vanadium(V) oxocation to the vanadium(IV) oxocation involves acceptance of one electron by the vanadium-containing species. Statement III is correct because vanadium(V) has the formal \(d^0\) configuration, whereas vanadium(IV) has \(d^1\). Thus, the one-electron reduction creates a singly occupied metal d subshell. Statement IV is incorrect because vanadium(V) oxide is amphoteric, although its acidic character predominates. Its reactions with suitable acids and alkalies must therefore be considered separately. Predominantly acidic behaviour does not mean the complete absence of reactions expressing amphoterism. The comparison illustrates why oxides and oxocations must be interpreted using their actual formulas and charges. It also distinguishes an acid–base property of an oxide from a redox transformation involving its metal centre.