D- and F-Block Elements MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Identify iron(II) oxidation:}\) The reducing process is \[ \mathrm{Fe^{2+}}\longrightarrow\mathrm{Fe^{3+}}+e^- \] Each iron(II) ion releases one electron. \(\textbf{Evaluate oxidation by permanganate:}\) The supplied acidic permanganate reduction potential is \(+1.52\,\mathrm{V}\). The standard reduction potential of the iron(III)/iron(II) couple is \(+0.77\,\mathrm{V}\). Therefore, \[ E^\circ_P=1.52-0.77 \] \[ E^\circ_P=+0.75\,\mathrm{V} \] The positive value favours the oxidation of Fe2+ by permanganate. \(\textbf{Evaluate oxidation by cerium(IV):}\) The Ce(IV)/Ce(III) reduction potential is \(+1.74\,\mathrm{V}\). Thus, \[ E^\circ_Q=1.74-0.77 \] \[ E^\circ_Q=+0.97\,\mathrm{V} \] This reaction is also favourable and has the greater supplied standard driving potential. \(\textbf{Compare electron acceptance by individual oxidant ions:}\) In acidic solution, permanganate is reduced from manganese(VII) to manganese(II). Each permanganate ion consequently accepts five electrons. Cerium(IV) is reduced to cerium(III). Each cerium(IV) ion accepts one electron. \(\textbf{Distinguish driving potential from electron capacity:}\) Cerium(IV) has the more positive reduction potential. However, this does not mean that each cerium(IV) ion accepts more electrons. The reduction potential concerns thermodynamic driving tendency, while electron capacity depends on the specified oxidation-state change. \(\textbf{Chemical interpretation:}\) An oxidising agent can possess a larger positive reduction potential yet accept fewer electrons per formula unit. These properties must not be confused when comparing oxidants. The empty \(4f^0\) configuration of Ce(IV) does not prevent its reduction to the common Ce(III) state.

Explanation: Transformation P converts chromate into dichromate by acidification. Chromium has oxidation state +6 in both oxoanions. Therefore, the transformation changes the form of the chromium-containing species without requiring electron transfer at chromium. Transformation Q converts manganate(VI) into permanganate(VII). Manganese increases in oxidation state from +6 to +7 and loses one electron per manganese centre. This is an oxidation process. Transformation R involves reduction of Ce(IV) to Ce(III). Cerium accepts one electron, and its formal 4f occupancy changes from \(4f^0\) to \(4f^1\). Transformation S involves uranium-containing oxocations. Uranium has oxidation state +5 in \(\mathrm{UO_2^+}\) and +6 in \(\mathrm{UO_2^{2+}}\). Removal of one electron therefore oxidises uranium(V) to uranium(VI). These transformations demonstrate that changes in formula, charge or colour must be interpreted through explicit oxidation-state accounting. The presence of oxygen atoms does not independently establish whether a process is redox. Likewise, the same central element may undergo oxidation or reduction depending on the direction of the transformation. The correct matching requires examining each chemical species and identifying whether the central metal gains electrons, loses electrons or retains its original oxidation state.

Explanation: The equilibrium combines two chromate ions into one dichromate ion with the participation of hydrogen ions. Ordinarily, increasing hydrogen-ion activity favours the dichromate side. However, the question explicitly maintains hydrogen-ion activity at its original value during dilution. The effect of changing hydrogen-ion activity must therefore be separated from the effect of changing the concentrations of the chromium-containing species. When the volume doubles, the concentrations of chromate and dichromate initially decrease. At constant hydrogen-ion activity, the equilibrium relation contains dichromate concentration in the numerator and the square of chromate concentration in the denominator. Therefore, immediately after dilution, the reaction quotient becomes larger than its original equilibrium value. The system responds by shifting towards chromate until equilibrium is restored. This direction is also consistent with dilution favouring the side containing more dissolved chromium-containing ions. The conclusion does not require reduction or oxidation of chromium. Chromium remains in oxidation state +6 in both species. The shift may increase the chromate fraction even though dilution simultaneously reduces the total chromium concentration and can make the solution appear less intensely coloured. Thus, changes in species distribution and overall colour intensity must be distinguished.

Explanation: \(\textbf{Identify the starting chromium state:}\) In chromium(III) oxide, oxygen has oxidation state \(-2\). \[ 2x+3(-2)=0 \] \[ x=+3 \] Thus, chromium begins in oxidation state +3. \(\textbf{Examine the product of P:}\) Sodium chromite has the formula \(\mathrm{NaCrO_2}\). Its oxidation-number balance is \[ (+1)+x+2(-2)=0 \] \[ x=+3 \] Chromium therefore retains oxidation state +3. Reaction P is a non-redox transformation involving the amphoteric behaviour of chromium(III) oxide. \(\textbf{Examine the product of Q:}\) In sodium chromate, \(\mathrm{Na_2CrO_4}\), \[ 2(+1)+x+4(-2)=0 \] \[ x=+6 \] Chromium consequently changes from +3 to +6. Each chromium centre loses three electrons. \(\textbf{Identify oxygen's role:}\) Reaction Q contains four chromium centres. Their total electron donation is \[ n(e^-)=4(3)=12 \] Three oxygen molecules accept \[ n(e^-)=3(4)=12 \] electrons when oxygen changes from oxidation state zero to \(-2\). The electron balance therefore agrees with the supplied reaction. \(\textbf{Interpret the contrast:}\) Alkaline conditions alone do not establish whether oxidation occurs. Reaction P changes the chemical form of chromium(III) without oxidising it. Reaction Q additionally contains molecular oxygen, which enables chromium(VI) formation. The distinction depends on the oxidising conditions and the actual product, not merely on the presence of sodium hydroxide. \(\textbf{Final answer:}\) Reaction P is non-redox, whereas Reaction Q oxidises Cr(III) to Cr(VI) through reduction of molecular oxygen.

Explanation: \(\textbf{Identify the oxidation at the anode:}\) Manganese has oxidation state +6 in manganate and +7 in permanganate. Therefore, each manganate ion loses one electron. \[ \mathrm{MnO_4^{2-}}\longrightarrow\mathrm{MnO_4^-}+e^- \] \(\textbf{Relate cathodic hydrogen to electron transfer:}\) The cathodic reaction is \[ 2\mathrm{H_2O}+2e^-\longrightarrow\mathrm{H_2}+2\mathrm{OH^-} \] Two moles of electrons produce one mole of hydrogen. Therefore, \[ n(e^-)_{\mathrm{cathode}}=2n(\mathrm{H_2}) \] \[ n(e^-)_{\mathrm{cathode}}=2(0.050) \] \[ n(e^-)_{\mathrm{cathode}}=0.100\,\mathrm{mol} \] \(\textbf{Use electron conservation between electrodes:}\) Under the stipulated 100% current efficiency, the amount of electrons released at the anode equals the amount consumed at the cathode. \[ n(e^-)_{\mathrm{anode}}=0.100\,\mathrm{mol} \] Each manganate ion releases one electron. Therefore, \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{oxidised}}=\frac{0.100}{1} \] \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{oxidised}}=0.100\,\mathrm{mol} \] \(\textbf{Calculate permanganate formation:}\) The oxidation converts one manganate ion into one permanganate ion. \[ n(\mathrm{MnO_4^-})_{\mathrm{formed}}=0.100\,\mathrm{mol} \] \(\textbf{Calculate manganate remaining:}\) \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{remaining}}=0.180-0.100 \] \[ n(\mathrm{MnO_4^{2-}})_{\mathrm{remaining}}=0.080\,\mathrm{mol} \] \(\textbf{Check the overall chemical process:}\) Combining the electrode reactions gives \[ 2\mathrm{MnO_4^{2-}}+2\mathrm{H_2O} \longrightarrow 2\mathrm{MnO_4^-}+\mathrm{H_2}+2\mathrm{OH^-} \] The corresponding potassium-salt equation is \[ 2\mathrm{K_2MnO_4}+2\mathrm{H_2O} \longrightarrow 2\mathrm{KMnO_4}+\mathrm{H_2}+2\mathrm{KOH} \] This confirms that one mole of hydrogen corresponds to the oxidation of two moles of manganate. \(\textbf{Chemical interpretation:}\) The process differs from manganate disproportionation because the released electrons are transferred through the external circuit. Hydrogen formation provides an independent measure of the electron transfer in the stipulated idealised process. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{MnO_4^-})=0.100\,\mathrm{mol},\quad n(\mathrm{MnO_4^{2-}})_{\mathrm{remaining}}=0.080\,\mathrm{mol}} \]

Explanation: \(\textbf{Determine manganese's electron donation:}\) Manganese changes from oxidation state +4 to +6. Each manganese centre therefore loses two electrons. \[ \mathrm{Mn(IV)}\longrightarrow\mathrm{Mn(VI)}+2e^- \] The manganese dioxide amount is \[ n(\mathrm{MnO_2})=0.240\,\mathrm{mol} \] Hence, \[ n(e^-)_{\mathrm{donated}}=2(0.240) \] \[ n(e^-)_{\mathrm{donated}}=0.480\,\mathrm{mol} \] \(\textbf{Evaluate nitrate reduction in Route P:}\) Nitrogen has oxidation state +5 in nitrate. It has oxidation state +3 in nitrite. Therefore, each nitrate ion accepts two electrons. \[ n(\mathrm{NO_3^-})=\frac{0.480}{2} \] \[ n(\mathrm{NO_3^-})=0.240\,\mathrm{mol} \] \(\textbf{Evaluate chlorate reduction in Route Q:}\) Chlorine has oxidation state +5 in chlorate. It has oxidation state \(-1\) in chloride. The change is a decrease of six oxidation-state units. Therefore, each chlorate ion accepts six electrons. \[ n(\mathrm{ClO_3^-})=\frac{0.480}{6} \] \[ n(\mathrm{ClO_3^-})=0.080\,\mathrm{mol} \] \(\textbf{Verify the different oxidant requirements:}\) In Route P, \[ 2(0.240)=0.480\,\mathrm{mol}\ e^- \] In Route Q, \[ 6(0.080)=0.480\,\mathrm{mol}\ e^- \] Both electron-acceptance amounts equal the manganese electron donation. \(\textbf{Interpret the comparison:}\) The manganese oxidation process is identical in the two routes. However, nitrate and chlorate undergo different reductions and therefore possess different electron-acceptance capacities. Consequently, their required mole amounts are different. The comparison concerns the specified complete reduction products; it does not imply that the reagents have identical reaction rates or practical operating conditions. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{NO_3^-})=0.240\,\mathrm{mol},\quad n(\mathrm{ClO_3^-})=0.080\,\mathrm{mol}} \]

Explanation: \(\textbf{Define the two reducing-agent amounts:}\) Let \[ x=n(\mathrm{C_2O_4^{2-}}) \] and \[ y=n(\mathrm{H_2O_2}) \] The combined amount gives \[ x+y=0.120 \] \(\textbf{Determine gas production from oxalate:}\) The balanced equation shows that five oxalate ions produce ten carbon dioxide molecules. Therefore, one oxalate ion produces two carbon dioxide molecules. \[ n(\mathrm{CO_2})=2x \] \(\textbf{Determine gas production from hydrogen peroxide:}\) The second balanced equation shows that five hydrogen peroxide molecules produce five oxygen molecules. Therefore, \[ n(\mathrm{O_2})=y \] \(\textbf{Use the measured gas amount:}\) The total dry gas consists only of CO2 and O2. \[ 2x+y=0.190 \] \(\textbf{Solve the simultaneous relations:}\) Subtract the total-reductant equation from the gas equation. \[ (2x+y)-(x+y)=0.190-0.120 \] \[ x=0.070\,\mathrm{mol} \] Therefore, \[ y=0.120-0.070 \] \[ y=0.050\,\mathrm{mol} \] \(\textbf{Check the gas amounts:}\) \[ n(\mathrm{CO_2})=2(0.070)=0.140\,\mathrm{mol} \] \[ n(\mathrm{O_2})=0.050\,\mathrm{mol} \] \[ n(\mathrm{gas})_{\mathrm{total}}=0.140+0.050=0.190\,\mathrm{mol} \] The result agrees with the observation. \(\textbf{Calculate electron donation:}\) Each oxalate ion loses two electrons during oxidation to carbon dioxide. Each hydrogen peroxide molecule also loses two electrons during oxidation to oxygen. Therefore, \[ n(e^-)=2x+2y \] \[ n(e^-)=2(0.070)+2(0.050) \] \[ n(e^-)=0.240\,\mathrm{mol} \] \(\textbf{Calculate acidic permanganate consumption:}\) Each permanganate ion accepts five electrons on reduction to Mn2+. \[ n(\mathrm{MnO_4^-})=\frac{0.240}{5} \] \[ n(\mathrm{MnO_4^-})=0.048\,\mathrm{mol} \] \(\textbf{Chemical interpretation:}\) Oxalate and hydrogen peroxide have equal electron-donation capacities per mole in the specified reactions. Consequently, oxidant consumption alone would not determine their individual amounts. The different gas yields provide the additional independent information needed to resolve the mixture composition. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{C_2O_4^{2-}})=0.070\,\mathrm{mol},\quad n(\mathrm{MnO_4^-})=0.048\,\mathrm{mol}} \]

Explanation: \(\textbf{Use the common structural assumption:}\) The question assumes geometrically similar structures with equal packing fractions. Under this condition, the volume associated with an atom is proportional to the cube of its metallic radius. \[ V_{\mathrm{atom}}\propto r^3 \] Density is mass divided by volume. Therefore, \[ \rho\propto\frac{M}{r^3} \] where \(M\) is molar mass. \(\textbf{Construct the density ratio:}\) \[ \frac{\rho_{\mathrm{Hf}}}{\rho_{\mathrm{Zr}}} = \frac{M_{\mathrm{Hf}}}{M_{\mathrm{Zr}}} \left(\frac{r_{\mathrm{Zr}}}{r_{\mathrm{Hf}}}\right)^3 \] \(\textbf{Substitute the supplied values:}\) \[ \frac{\rho_{\mathrm{Hf}}}{\rho_{\mathrm{Zr}}} = \frac{178.5}{91.2} \left(\frac{160}{159}\right)^3 \] \(\textbf{Evaluate the mass contribution:}\) \[ \frac{178.5}{91.2}\approx1.957 \] \(\textbf{Evaluate the radius contribution:}\) \[ \left(\frac{160}{159}\right)^3\approx1.019 \] \(\textbf{Combine the contributions:}\) \[ \frac{\rho_{\mathrm{Hf}}}{\rho_{\mathrm{Zr}}} \approx1.957(1.019) \] \[ \frac{\rho_{\mathrm{Hf}}}{\rho_{\mathrm{Zr}}}\approx1.99 \] \(\textbf{Explain the chemical trend:}\) Lanthanoid contraction reduces the size difference normally expected between corresponding 4d and 5d elements. Consequently, zirconium and hafnium have very similar metallic radii. However, hafnium has a much greater atomic mass. Under the stated structural assumptions, the nearly equal atomic volumes therefore accompany substantially different masses per atom. This leads to a much greater density for hafnium. \(\textbf{Qualification:}\) The numerical estimate follows from the deliberately specified equal-packing model. Actual densities also depend on experimentally determined crystal structures and volumes. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_{\mathrm{Hf}}}{\rho_{\mathrm{Zr}}}\approx1.99} \]

Explanation: A complete catalytic cycle requires both reduction of Fe3+ to Fe2+ and subsequent oxidation of Fe2+ back to Fe3+. The two oxidation states act as electron-transfer intermediates between iodide and peroxodisulphate. Mixture P initially contains Fe3+, iodide and peroxodisulphate. Its Fe3+ ions can first oxidise iodide and become Fe2+. Peroxodisulphate can then regenerate Fe3+, allowing repeated turnovers. Mixture Q initially contains Fe2+ rather than Fe3+. However, peroxodisulphate is available to oxidise Fe2+ into Fe3+. The regenerated Fe3+ can then oxidise iodide, and the complete cycle can operate. Thus, starting with the reduced catalytic form does not necessarily prevent catalysis. Mixture R contains Fe2+ and iodide but lacks peroxodisulphate. Neither supplied catalytic stage can establish a sustained electron-transfer cycle from those components alone. Mixture S contains Fe3+ and iodide, so the first step can occur. However, without peroxodisulphate, the resulting Fe2+ cannot be regenerated through the stated second stage. Consequently, S permits a finite stoichiometric reaction involving Fe3+ but not repeated complete turnovers of the specified catalytic cycle. The distinction is between a reaction involving a transition-metal ion and a genuinely regenerating catalytic mechanism. The correct conclusion also shows why the initial oxidation state of the catalyst can differ while the eventual catalytic sequence remains the same.

Explanation: Statement I is correct because Fe2+ loses one electron when it becomes Fe3+. Therefore, \(0.200\,\mathrm{mol}\) Fe2+ releases \(0.200\,\mathrm{mol}\) electrons. Oxalate undergoes oxidation from carbon oxidation state +3 to +4 in carbon dioxide. Each oxalate ion contains two carbon atoms and releases two electrons during complete oxidation. Consequently, \(0.100\,\mathrm{mol}\) oxalate also releases \(0.200\,\mathrm{mol}\) electrons. Statement II is correct because acidified permanganate accepts five electrons per ion when reduced to Mn2+. Both experiments therefore require \(0.040\,\mathrm{mol}\) permanganate under the specified complete-reaction conditions. Statement III is incorrect because warming does not change the electron count associated with the stated oxalate-to-carbon-dioxide transformation. The carbon oxidation-state increase remains the same. The temperature condition concerns the reaction's kinetic behaviour, not a different formal oxidation state of carbon in the specified products. Statement IV is correct because reaction rate and reaction stoichiometry describe different aspects of a chemical process. Heating can help a reaction proceed at a suitable rate without changing the overall electron-balanced equation. The comparison illustrates why equal electron-equivalent capacities do not imply identical kinetic behaviour. It also shows that the required amount of oxidant must be determined from the stated oxidation products and reduction medium rather than from temperature alone.

Explanation: Equilibrium P contains hydrogen ions on the reactant side. Increasing hydrogen-ion activity therefore favours formation of dichromate under the specified conditions. This corresponds to a shift to the right. Chromium retains oxidation state +6 in both chromate and dichromate. Thus, the chromate–dichromate equilibrium does not involve electron transfer at chromium. Equilibrium Q has a different arrangement because hydrogen ions appear on the product side. Increasing their activity therefore opposes the forward disproportionation reaction. The equilibrium shifts to the left under the stated assumptions. In Q, manganese(III) is both oxidised and reduced when the forward reaction occurs. Some Mn(III) becomes Mn(II), while other Mn(III) becomes Mn(IV) in manganese dioxide. This is a redox process involving disproportionation. Therefore, the two equilibria respond in opposite directions to the same change in hydrogen-ion activity. Their difference cannot be predicted by a universal rule that acid always stabilises the higher oxidation state. Instead, the effect depends on the actual position of hydrogen ions in the balanced chemical equation. The comparison requires interpreting both equilibrium direction and oxidation-state changes. It also demonstrates that proton-dependent chemistry includes both non-redox and redox transformations.

Explanation: Hydrogen peroxide contains oxygen in oxidation state \(-1\), which lies between the common oxidation states of oxygen in water and molecular oxygen. In Reaction P, chromium(II) is oxidised to chromium(III). Each chromium centre loses one electron. The two chromium(II) ions therefore release two electrons in total. Hydrogen peroxide accepts these electrons while forming water, in which oxygen has oxidation state \(-2\). Thus, peroxide undergoes reduction and acts as the oxidising agent in P. In Reaction Q, manganese(VII) in permanganate is reduced to manganese(II). Each manganese centre accepts five electrons, giving ten electrons accepted by two permanganate ions. The five hydrogen peroxide molecules supply these electrons when their oxygen atoms change from \(-1\) to zero in molecular oxygen. Therefore, peroxide undergoes oxidation and acts as the reducing agent in Q. The identity of an oxidising or reducing agent must be determined from its actual electron-transfer role in the specified reaction. A compound cannot be classified as permanently oxidising or permanently reducing without considering its reaction partner and products. Hydrogen peroxide provides a characteristic example of a substance capable of playing either role. The two reactions also demonstrate why oxidation-state accounting is more reliable than identifying a reagent's role from its formula alone.

Explanation: Process P concerns catalytic poisoning of a nickel surface. A heterogeneous hydrogenation reaction depends on suitable interactions between reactant molecules and accessible catalytic sites. Strong adsorption of impurities can block some of those sites. Consequently, the initial hydrogenation rate may decrease even though the bulk nickel has not been chemically consumed. Process Q concerns passivation of a chromium-containing alloy. An adherent protective oxide film can restrict access of reactants to the underlying metal and slow subsequent corrosion. Thus, both processes involve kinetic consequences of altered surface accessibility. However, their chemical roles are different. In P, the desired catalytic surface becomes less effective. In Q, an initially reactive metallic surface gains protection against further unwanted oxidation. Neither observation independently establishes a change in the equilibrium constant of the underlying chemical reaction at fixed temperature. The equilibrium constant depends on the thermodynamic properties of the specified reaction. Surface poisoning and passivation principally influence the available reaction pathway or the rate at which reactants can reach the reactive region. Nor does the similarity in surface involvement require nickel and chromium to undergo identical oxidation-state changes. The correct conclusion therefore identifies a common kinetic principle while distinguishing catalytic-site blockage from protective-film formation.

Explanation: \(\textbf{Identify the condition for iodide oxidation:}\) The oxidised mediator must accept electrons from iodide. Therefore, its reduction potential must exceed that of the iodine/iodide couple. \[ E^\circ(\mathrm{M_{ox}/M_{red}})\gt +0.54\,\mathrm{V} \] \(\textbf{Identify the condition for mediator regeneration:}\) Peroxodisulphate must oxidise the reduced mediator. Therefore, the peroxodisulphate/sulphate reduction potential must exceed that of the mediator couple. \[ E^\circ(\mathrm{M_{ox}/M_{red}})\lt +2.01\,\mathrm{V} \] \(\textbf{Combine the two requirements:}\) A suitable mediator must satisfy \[ +0.54\lt E^\circ(\mathrm{M_{ox}/M_{red}})\lt +2.01 \] with potentials expressed in volts. \(\textbf{Test the lower-potential mediators:}\) Mediator P has \(E^\circ=+0.40\,\mathrm{V}\). Mediator Q has \(E^\circ=+0.48\,\mathrm{V}\). Both values are below the iodine/iodide potential. Consequently, their oxidised forms cannot favour the required iodide-oxidation step under the stated standard conditions. \(\textbf{Test mediator R:}\) Mediator R has \(E^\circ=+0.95\,\mathrm{V}\). For iodide oxidation, \[ E^\circ_{\mathrm{step\ P}}=0.95-0.54 \] \[ E^\circ_{\mathrm{step\ P}}=+0.41\,\mathrm{V} \] For regeneration by peroxodisulphate, \[ E^\circ_{\mathrm{step\ Q}}=2.01-0.95 \] \[ E^\circ_{\mathrm{step\ Q}}=+1.06\,\mathrm{V} \] Both values are positive. \(\textbf{Test mediator S:}\) Mediator S has \(E^\circ=+2.17\,\mathrm{V}\). It can favour iodide oxidation, but regeneration would have \[ E^\circ_{\mathrm{regeneration}}=2.01-2.17 \] \[ E^\circ_{\mathrm{regeneration}}=-0.16\,\mathrm{V} \] Thus, the required regeneration step is not favourable under standard conditions. \(\textbf{Chemical interpretation:}\) An appropriate redox mediator must be strong enough in its oxidised state to oxidise the substrate but capable of being regenerated by the supplied terminal oxidant. A more positive mediator potential does not always make a better catalyst because it can obstruct regeneration. The two-step requirement is essential for a complete catalytic cycle. \(\textbf{Final answer:}\) Mediator R alone satisfies both standard-potential conditions.

Explanation: Lanthanoid contraction describes the general decrease in ionic radii across the lanthanoid series, especially when ions of the same oxidation state are compared. In the trivalent sequence, La3+ is larger than Eu3+, and Eu3+ is larger than Lu3+. These changes are associated with increasing effective nuclear attraction and imperfect shielding by 4f electrons. The corresponding increase in cation polarising tendency contributes to the general decrease in basicity of lanthanoid(III) hydroxides across the series. However, the student's comparison includes Eu2+ rather than Eu3+. The supplied data show that Eu2+ is larger than Eu3+. This difference arises partly because formation of the divalent ion removes fewer electrons and leaves a lower positive charge. Therefore, the trivalent ionic-radius trend cannot simply be transferred to a comparison involving a divalent ion. The basicity of Eu(OH)2 must be assessed using its actual charge, bonding and chemical behaviour. Its relative basicity cannot be established solely from the statement that europium lies after lanthanum in the periodic table. Equally, it would be unjustified to claim that every divalent hydroxide is necessarily more basic than every trivalent hydroxide under all conditions. The important distinction is between a well-defined trend within a comparable oxidation-state series and a comparison across different oxidation states. The supplied evidence supports the former but does not independently determine the latter.

Explanation: Each ion has the formula \(\mathrm{MO_4^{2-}}\). Four oxygen atoms contribute a total oxidation number of \(-8\), while the ionic charge is \(-2\). Therefore, the metal must possess oxidation state +6 in every ion. Chromium, manganese and iron nevertheless have different atomic numbers. Chromium has atomic number 24 and six electrons outside the argon core in its neutral ground-state configuration. Formal chromium(VI) consequently has a \(d^0\) description. Manganese has atomic number 25 and seven electrons outside the argon core in its neutral configuration. Formal manganese(VI) retains one d electron and has a \(d^1\) description. Iron has atomic number 26 and eight electrons outside the argon core in its neutral configuration. Formal iron(VI) has a \(d^2\) description. Thus, equal oxidation states do not require equal d-electron counts when different elements are compared. In the simple specified electron-counting model, the numbers of unpaired d electrons would be zero, one and two, respectively. However, this formal count must not be converted automatically into an exact prediction of experimental magnetic behaviour for every compound. Nor does it independently establish the colours or redox potentials of the oxoanions. The comparison illustrates how nuclear charge, formal oxidation state and remaining d-electron population must be considered together. The chemical formula fixes the oxidation state, while the identity of the element determines the electronic occupancy.

Explanation: Copper has oxidation state +1 in copper(I) oxide because oxygen contributes \(-2\) and there are two copper centres per formula unit. In the first reaction, one copper(I) centre becomes copper(II), while the other becomes copper metal. Thus, oxidation and reduction occur simultaneously among copper centres initially having the same oxidation state. This is a disproportionation reaction. The acid participates in the overall transformation, but hydrogen ions are not themselves converted into hydrogen gas in the stated equation. In the second reaction, copper(II) is reduced to copper(I). Iodide is oxidised to molecular iodine. The copper(I) formed is incorporated into solid CuI, which has low aqueous solubility. Precipitation removes copper(I) from the dissolved ionic population and contributes to the favourability of the overall process. Therefore, the instability of free aqueous Cu+ towards disproportionation does not imply that every solid copper(I) compound must be unstable. Copper(I) has the formal \(3d^{10}\) configuration, whereas copper(II) has \(3d^9\). Their different electron occupancies help explain characteristic differences in magnetic and optical behaviour but do not independently determine every redox equilibrium. Both reactions must be evaluated using the actual product species and chemical environment. The comparison demonstrates how oxidation-state stability can depend strongly on precipitation and the physical form of a compound.

Explanation: \(\textbf{Identify the chromium-conservation requirement:}\) Each chromate ion contains one chromium atom. Each dichromate ion contains two chromium atoms. Consequently, the amount of dichromate formed cannot be equated directly with the amount of chromate consumed. \(\textbf{Apply the oxoanion conversion:}\) \[ 2\mathrm{CrO_4^{2-}}+2\mathrm{H^+} \longrightarrow \mathrm{Cr_2O_7^{2-}}+\mathrm{H_2O} \] Two moles of chromate produce one mole of dichromate. \(\textbf{Determine chromate consumption:}\) \[ n(\mathrm{CrO_4^{2-}})_{\mathrm{consumed}} =2n(\mathrm{Cr_2O_7^{2-}})_{\mathrm{formed}} \] \[ =2(0.060) \] \[ =0.120\,\mathrm{mol} \] \(\textbf{Calculate chromate remaining:}\) \[ n(\mathrm{CrO_4^{2-}})_{\mathrm{remaining}} =0.180-0.120 \] \[ =0.060\,\mathrm{mol} \] \(\textbf{Calculate hydrogen-ion consumption:}\) Two hydrogen ions are consumed for every two chromate ions converted. \[ n(\mathrm{H^+})_{\mathrm{consumed}} =2n(\mathrm{Cr_2O_7^{2-}}) \] \[ =2(0.060) \] \[ =0.120\,\mathrm{mol} \] \(\textbf{Check chromium conservation:}\) Initially, the solution contains \[ n(\mathrm{Cr})_{\mathrm{initial}}=0.180\,\mathrm{mol} \] Finally, \[ n(\mathrm{Cr})_{\mathrm{final}} =0.060+2(0.060) \] \[ =0.180\,\mathrm{mol} \] The chromium balance is satisfied. \(\textbf{Interpret the chemical change:}\) Chromium remains in oxidation state +6 throughout acidification. The conversion changes oxoanion composition and the relative amounts of yellow chromate and orange dichromate. It does not require electron transfer. The quantitative inference therefore depends on oxoanion stoichiometry rather than oxidation-state changes. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{CrO_4^{2-}})_{\mathrm{remaining}}=0.060\,\mathrm{mol},\quad n(\mathrm{H^+})_{\mathrm{consumed}}=0.120\,\mathrm{mol}} \]

Explanation: Statement I is correct because oxygen is assigned oxidation state \(-2\) in each oxide. Chromium consequently has oxidation state +2 in CrO, +3 in Cr2O3 and +6 in CrO3. Statement II is also correct because the acid–base character of transition-metal oxides is related to the oxidation state of the metal. Chromium(II) oxide is predominantly basic, whereas chromium(III) oxide is amphoteric. Chromium(VI) oxide exhibits predominantly acidic character. This progression reflects changes in metal–oxygen bonding as the metal oxidation state increases. Statement III is incorrect because reaction of an amphoteric oxide with an alkali need not involve oxidation of the metal. Chromium(III) oxide can form a chromium(III)-containing product under suitable alkaline conditions without changing chromium's oxidation state. An oxidising reagent would be required for a transformation that actually converts chromium(III) into chromium(VI). Statement IV is correct because the formal chromium d-electron populations are \(d^4\), \(d^3\) and \(d^0\) for the +2, +3 and +6 oxidation states, respectively. Their decrease is consistent with progressive electron removal from the metal. However, the d-electron count alone does not give a complete explanation of oxide acidity. The chemical distinction involves both oxidation state and the nature of the metal–oxygen interaction. The correct interpretation therefore connects charge balance, electronic configuration and oxide behaviour while rejecting the assumption that every acid–base reaction of a transition-metal oxide is redox.

Explanation: \(\textbf{Identify chromium's electron-transfer direction:}\) Chromium(II) acts as the reducing species in both proposed reactions. \[ \mathrm{Cr^{2+}}\longrightarrow\mathrm{Cr^{3+}}+e^- \] The formal configuration changes from \(3d^4\) to \(3d^3\). \(\textbf{Evaluate Reaction P:}\) Iron(III) accepts one electron. \[ \mathrm{Fe^{3+}}+e^-\longrightarrow\mathrm{Fe^{2+}} \] The standard driving potential is \[ E^\circ_P=0.77-(-0.41) \] \[ E^\circ_P=+1.18\,\mathrm{V} \] Therefore, Reaction P is thermodynamically favourable under standard conditions. \(\textbf{Evaluate Reaction Q:}\) Copper(II) accepts one electron to form dissolved copper(I). \[ \mathrm{Cu^{2+}}+e^-\longrightarrow\mathrm{Cu^+} \] The standard driving potential is \[ E^\circ_Q=0.15-(-0.41) \] \[ E^\circ_Q=+0.56\,\mathrm{V} \] Reaction Q is also thermodynamically favourable. \(\textbf{Compare the two redox driving tendencies:}\) \[ +1.18\gt +0.56 \] Thus, P has the larger standard driving potential, but both have positive values. \(\textbf{Interpret the configurations:}\) Chromium(III) has the favourable \(d^3\) arrangement. Iron(III) is reduced from \(d^5\) to iron(II), \(d^6\). Copper(II) is reduced from \(d^9\) to copper(I), \(d^{10}\). The favourable electronic contribution associated with copper(I)'s filled d subshell is not by itself sufficient to determine every aqueous redox reaction. The supplied reduction potentials are needed to establish the actual thermodynamic directions. \(\textbf{Final answer:}\) Both reactions are thermodynamically favourable under the stated standard conditions.