D- and F-Block Elements MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: The supplied equilibrium involves replacement of water ligands by chloride ligands around cobalt(II). In the pink species, six water molecules coordinate to the cobalt centre. In the blue species, four chloride ions surround cobalt. Therefore, the coordination number changes from six to four when the forward reaction occurs. Increasing chloride-ion activity favours formation of the chloride-containing complex. Consequently, Change P shifts the equilibrium towards the blue tetrachlorocobaltate(II) ion. Change Q decreases chloride-ion activity after the chloride-rich state has been established. Under the stipulated conditions, the equilibrium can then shift back towards the pink hexaaquacobalt(II) ion. However, neither transformation requires oxidation or reduction of cobalt. Water is neutral, whereas each chloride ligand carries charge \(-1\). In the blue complex, four chloride ligands contribute \(-4\), and the complex has charge \(-2\), giving cobalt oxidation state +2. The pink complex also contains cobalt(II) because its water ligands are neutral and its charge is +2. Thus, a pronounced colour change and a change in coordination number occur without a change in metal oxidation state. The equilibrium illustrates how ligand identity and relative concentration can influence the optical properties of transition-metal complexes.

Explanation: \(\textbf{Count the bonds in each ion:}\) A chromate ion has four terminal Cr–O bonds and no bridging bonds. A dichromate ion has six terminal Cr–O bonds. Its one shared bridging oxygen is bonded to two separate chromium atoms. Therefore, one dichromate ion contains two bridging Cr–O bonds, even though it contains only one bridging oxygen atom. \(\textbf{Calculate terminal-bond amount:}\) The chromate contribution is \[ n_{\mathrm{terminal,chromate}}=4(0.080)=0.320\,\mathrm{mol} \] The dichromate contribution is \[ n_{\mathrm{terminal,dichromate}}=6(0.060)=0.360\,\mathrm{mol} \] Adding the two contributions gives \[ n_{\mathrm{terminal,total}}=0.320+0.360=0.680\,\mathrm{mol} \] \(\textbf{Calculate bridging-bond amount:}\) Only dichromate contributes bridging Cr–O bonds. \[ n_{\mathrm{bridging,total}}=2(0.060)=0.120\,\mathrm{mol} \] \(\textbf{Check the structural count:}\) The total Cr–O bond amount is \(0.680+0.120=0.800\,\mathrm{mol}\). Independently, the chromium-centred tetrahedra amount to \(0.080+2(0.060)=0.200\,\mathrm{mol}\), each with four Cr–O bonds. Thus, \(4(0.200)=0.800\,\mathrm{mol}\), confirming that no shared bond has been omitted or counted twice. The important distinction is between counting one bridging oxygen atom and counting its two connections to chromium. \(\textbf{Final answer:}\) \[ \boxed{n(\text{terminal Cr–O bonds})=0.680\,\mathrm{mol},\quad n(\text{bridging Cr–O bonds})=0.120\,\mathrm{mol}} \]

Explanation: \(\textbf{Permanganate reduction in acidic solution:}\) Manganese changes from +7 in \(\mathrm{MnO_4^-}\) to +2 in \(\mathrm{Mn^{2+}}\), accepting five electrons per permanganate ion. Each iron(II) ion loses one electron when converted into iron(III). Hence, the balanced equation requires one mole of permanganate for every five moles of iron(II): \[ n(\mathrm{MnO_4^-})_{\mathrm{Fe^{2+}}}=\frac{0.0250}{5}=0.0050\,\mathrm{mol} \] \(\textbf{Additional chloride oxidation:}\) The separate chloride-to-chlorine oxidation consumes the stated \(0.0010\,\mathrm{mol}\) of permanganate. This additional requirement is not included in the amount used for iron(II). \(\textbf{Total oxidant requirement:}\) \[ n(\mathrm{MnO_4^-})_{\mathrm{total}}=0.0050+0.0010 \] \[ n(\mathrm{MnO_4^-})_{\mathrm{total}}=0.0060\,\mathrm{mol} \] The total consumption includes the permanganate needed to accept electrons from both independently specified reducing species. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{MnO_4^-})_{\mathrm{total}}=0.0060\,\mathrm{mol}} \]

Explanation: Statement I is correct because titanium has oxidation state +4 in TiO2. The two oxide ions contribute a total oxidation number of \(-4\), so titanium must contribute +4 in the neutral compound. In \(\mathrm{TiO^{2+}}\), one oxygen atom contributes \(-2\), while the overall ionic charge is +2. Titanium must again possess oxidation state +4. Statement II is correct because the same oxidation-number calculation applies to \(\mathrm{VO^{2+}}\). Vanadium also has oxidation state +4 in this oxocation. Statement III is incorrect because titanium does not change oxidation state between the two specified species. Their formulas differ in oxygen content and ionic charge, but these differences do not require electron transfer at titanium. Under suitable conditions, formation of an oxocation may involve acid–base or dissolution-related processes. Such transformations cannot automatically be classified as reductions merely because the product contains fewer oxygen atoms. Statement IV is correct because the oxidation state of a metal in an oxygen-containing species depends on both the oxidation numbers of oxygen and the total electrical charge. TiO2 and TiO2+ must not be confused by reading the superscript ionic charge as an oxygen subscript. The correct conclusion illustrates how accurate chemical notation prevents false oxidation-state assignments. It also connects the behaviour of high-oxidation-state transition-metal oxides with the formation of characteristic oxocations.

Explanation: The first permanganate couple describes reduction of manganese(VII) to manganese(IV) in manganese dioxide. This involves acceptance of three electrons per manganese centre. The second couple describes reduction of manganese(VII) to manganese(II), which involves five electrons. However, the supplied standard reduction potential for the three-electron process is \(+1.69\,\mathrm{V}\), greater than the \(+1.52\,\mathrm{V}\) value for the five-electron process. Therefore, the student's proposed proportionality between electron count and reduction potential is incorrect. Standard electrode potential is related to the free-energy change per unit of transferred electrical charge, not merely to the total number of electrons involved in a balanced half-reaction. Both supplied potentials exceed that of the Fe3+/Fe2+ couple. Consequently, oxidation of Fe2+ can be thermodynamically favoured in an appropriate comparison with either specified reduction pathway under the corresponding standard conditions. Nevertheless, the actual manganese product depends on the chemical environment, especially hydrogen-ion activity. Reaction kinetics may also influence observed behaviour. The larger potential of one formal half-reaction does not establish that it must always be the exclusive pathway in every solution. Correct prediction requires distinguishing thermodynamic data, electron equivalence, speciation and medium-dependent reaction products. The example demonstrates why numerical reduction potentials must be interpreted using the actual half-reaction rather than ranked by electron count.

Explanation: Reduction involves gain of an electron, so the formal f-electron count increases by one in the specified transformations. Thulium(III) has the \(4f^{12}\) configuration. Its reduction to Tm2+ produces \(4f^{13}\), which is one electron short of a completely filled f subshell. Ytterbium(III) has the \(4f^{13}\) configuration. Its reduction to Yb2+ produces the filled \(4f^{14}\) subshell. This supplies an important electronic stabilising contribution to the divalent ytterbium state. Consequently, the two reductions have different subshell-stability outcomes despite both involving gain of one electron. The comparison does not mean that the formation of Yb2+ must be spontaneous in every solution. Lanthanoid oxidation-state stability also depends on ionisation energies, solvation and the chemical environment. Ytterbium(II) can act as a reducing agent and return to the predominant trivalent state under suitable conditions. Thus, a favourable filled subshell does not prohibit subsequent oxidation. Tm2+ and Yb3+ share the formal \(4f^{13}\) occupancy but possess different nuclear charges and oxidation states. Their similar f-electron counts therefore do not establish identical chemical properties. The correct interpretation connects electron transfer to subshell occupancy while preserving the distinction between electronic stabilisation and overall redox thermodynamics.

Explanation: Curium(III) has the half-filled \(5f^7\) configuration. Its oxidation to curium(IV) removes one f electron and produces \(5f^6\). Consequently, the half-filled subshell is disrupted during this transformation. Berkelium(III), by contrast, has the \(5f^8\) configuration. Oxidation to berkelium(IV) removes one f electron and produces \(5f^7\). The berkelium transformation therefore creates a half-filled f subshell. These electronic consequences are opposite despite both processes involving oxidation from +3 to +4. The half-filled configuration can provide a stabilising contribution to the corresponding ionic state. However, that contribution is not a complete measure of the thermodynamic favourability of oxidation. Actinoid redox behaviour is influenced by the relative energies of 5f, 6d and 7s electrons and by their participation in bonding. Solvation and compound formation also affect the stability of particular oxidation states. Therefore, neither the electron counts nor the shared change of ionic charge establish identical reduction potentials. The comparison illustrates how electronic configurations can help explain oxidation-state exceptions while remaining insufficient for an absolute prediction about every chemical environment. It is the differing effect on the half-filled f subshell, not merely the fact that both ions lose one electron, that provides the key chemical distinction.

Explanation: Statement I is correct because ionic charge is only one factor determining ionic size and coordination behaviour. Fe3+, Cr3+ and Gd3+ have different nuclear charges, electron configurations and electron-shell structures. Therefore, they need not possess equal radii or form equally stable complexes with every ligand. Statement II is correct because iron(III) and chromium(III) involve partially occupied 3d subshells, while gadolinium(III) contains a half-filled 4f subshell. The 4f orbitals are more deeply buried relative to the outer parts of the ion than the 3d orbitals involved in characteristic transition-metal interactions. Statement III is correct because this relative shielding reduces the direct influence of the surrounding chemical environment on many lanthanoid f-electron transitions. Consequently, lanthanoid absorption bands can be comparatively narrow, whereas transition-metal d-level transition energies commonly depend strongly on ligand environment. Statement IV is incorrect because equal charge does not imply equal electron occupancy. Cr3+ has three unpaired d electrons in its elementary free-ion description. Fe3+ has five unpaired d electrons in the high-spin description, while Gd3+ has seven unpaired f electrons. Their optical behaviour also cannot be inferred merely from ionic charge or unpaired-electron count. The electronic origins and energies of absorption differ between d-block and f-block species. The comparison therefore demonstrates why ionic charge, magnetic electron count, complex formation and colour must be considered as related but distinct properties.

Explanation: \(\textbf{Evaluate zinc displacement of iron:}\) Experiment P proposes oxidation of zinc metal and reduction of iron(II). The corresponding reaction is \[ \mathrm{Zn(s)}+\mathrm{Fe^{2+}(aq)} \longrightarrow \mathrm{Zn^{2+}(aq)}+\mathrm{Fe(s)} \] \(\textbf{Calculate the standard potential for P:}\) \[ E^\circ_P=E^\circ(\mathrm{Fe^{2+}/Fe})-E^\circ(\mathrm{Zn^{2+}/Zn}) \] \[ E^\circ_P=-0.44-(-0.76) \] \[ E^\circ_P=+0.32\,\mathrm{V} \] The positive value favours the proposed displacement under standard conditions. \(\textbf{Evaluate iron displacement of copper:}\) Experiment Q proposes oxidation of iron metal and reduction of copper(II). \[ \mathrm{Fe(s)}+\mathrm{Cu^{2+}(aq)} \longrightarrow \mathrm{Fe^{2+}(aq)}+\mathrm{Cu(s)} \] \(\textbf{Calculate the standard potential for Q:}\) \[ E^\circ_Q=E^\circ(\mathrm{Cu^{2+}/Cu})-E^\circ(\mathrm{Fe^{2+}/Fe}) \] \[ E^\circ_Q=0.34-(-0.44) \] \[ E^\circ_Q=+0.78\,\mathrm{V} \] This reaction is also thermodynamically favourable. \(\textbf{Interpret the electronic configurations:}\) Zn2+ has the filled \(3d^{10}\) configuration, whereas Fe2+ has \(3d^6\). Copper(II) has \(3d^9\), while copper metal has its characteristic metallic electronic structure. These electronic differences influence the energetic properties of the chemical species involved. However, no single d-subshell occupancy rule independently establishes both displacement directions. \(\textbf{Distinguish classification from electrochemical behaviour:}\) Zinc is not classified as a transition element under the usual electronic definition. Nevertheless, zinc metal can act as a reducing agent and displace another metal from an appropriate salt solution. Transition-element classification is therefore not a prerequisite for metal-displacement behaviour. \(\textbf{Final answer:}\) \[ \boxed{E^\circ_P=+0.32\,\mathrm{V},\quad E^\circ_Q=+0.78\,\mathrm{V}} \]

Explanation: Statement I is correct because magnetite contains one Fe(II) and two Fe(III) centres per formula unit. The supplied hydrochloric-acid reaction produces one FeCl2 and two FeCl3 units. Iron consequently retains its respective initial oxidation states during this transformation. The process involves acid-assisted dissolution without a net oxidation-state change at iron. Statement II is correct because Co3O4 contains one Co(II) and two Co(III) centres in the stated formal description. The product CoCl2 contains only Co(II). Therefore, both Co(III) centres gain one electron each. Two chloride ions supply these electrons when they are oxidised to molecular chlorine. Statement III is incorrect because identical starting distributions do not require identical final oxidation states. Reaction P involves no net electron transfer, whereas Reaction Q transfers two electrons from chloride to cobalt(III). Statement IV is correct because the actual reaction depends on the redox behaviour of the metal centres, not merely on the oxide formula. The two compounds share the general M3O4 composition but have different reactions with hydrochloric acid. This comparison also shows why the use of an acid must not automatically be classified as either exclusively acid–base or necessarily redox. The products must be examined before assigning the chemical process.

Explanation: Lanthanoid contraction describes the overall decrease in size associated with increasing nuclear charge and incomplete shielding by 4f electrons. The effect is particularly clear when comparable trivalent ions are examined. The supplied trivalent radii decrease from samarium through europium to gadolinium. They also decrease from ytterbium to lutetium. Thus, the ionic data are consistent with lanthanoid contraction. The metallic atomic radii show a different pattern because europium and ytterbium have unusually large values relative to neighbouring elements. Their characteristic divalent configurations contribute to differences in metallic bonding compared with lanthanoids showing predominantly trivalent metallic behaviour. Europium(II) possesses the favourable half-filled \(4f^7\) subshell, while ytterbium(II) has the filled \(4f^{14}\) subshell. These electronic features help explain their anomalous behaviour. However, the neutral metallic atoms and isolated trivalent ions are not chemically identical systems. Their radius trends therefore need not show the same local irregularities. The anomalies do not indicate that nuclear charge decreases at either element. The correct interpretation distinguishes a general periodic contraction from local changes associated with oxidation-state preference and metallic bonding.

Explanation: Observation P concerns magnetism arising from unpaired electrons. Iron(III) has the formal \(3d^5\) configuration and possesses unpaired electrons in its ordinary high-spin aqueous description. Therefore, P matches Explanation 2. Observation Q concerns heterogeneous catalysis on a metallic surface. Reactant molecules can interact with accessible nickel surface sites, enabling an alternative reaction pathway. Thus, Q matches Explanation 3. Observation R concerns interstitial compound formation. Small atoms such as carbon can occupy suitable positions within a metallic structure, contributing to high hardness without necessarily eliminating metallic conductivity. Therefore, R matches Explanation 1. Observation S concerns optical behaviour. Chromium(VI) in chromate has the formal \(d^0\) configuration, so its colour cannot arise from an ordinary transition between occupied metal d orbitals. Charge-transfer processes provide the appropriate explanation, giving S–4. The student's proposal fails because these observations arise from different electronic or structural features. Unpaired electrons are relevant to magnetism but do not independently explain interstitial hardness. Surface interactions can contribute to catalysis without being the origin of chromate's colour. Similarly, the presence of a transition metal does not guarantee a single universal mechanism for all its properties. The correct matching therefore distinguishes four characteristic but chemically different aspects of transition-metal behaviour.

Explanation: \(\textbf{Identify manganese(III)'s two possible changes:}\) Manganese(III) has the formal \(3d^4\) configuration. During disproportionation, one manganese centre accepts an electron and becomes Mn(II). The other loses an electron and becomes Mn(IV) in manganese dioxide. \(\textbf{Reduction of manganese(III):}\) \[ \mathrm{Mn^{3+}}+e^-\longrightarrow\mathrm{Mn^{2+}} \] The supplied reduction potential is \[ E^\circ_{\mathrm{reduction}}=+1.51\,\mathrm{V} \] \(\textbf{Oxidation of manganese(III):}\) Reverse the supplied manganese dioxide reduction half-reaction. \[ \mathrm{Mn^{3+}}+2\mathrm{H_2O} \longrightarrow \mathrm{MnO_2}+4\mathrm{H^+}+e^- \] Its corresponding reduction potential, used in the cell-potential subtraction, is \(+0.95\,\mathrm{V}\). \(\textbf{Determine the disproportionation potential:}\) \[ E^\circ_{\mathrm{cell}}=1.51-0.95 \] \[ E^\circ_{\mathrm{cell}}=+0.56\,\mathrm{V} \] The positive result favours the stated disproportionation under the supplied standard conditions. \(\textbf{Combine the half-reactions:}\) \[ 2\mathrm{Mn^{3+}}+2\mathrm{H_2O} \longrightarrow \mathrm{Mn^{2+}}+\mathrm{MnO_2}+4\mathrm{H^+} \] \(\textbf{Track the d-electron populations:}\) For the reduced manganese centre, \[ d^4\longrightarrow d^5 \] For the oxidised manganese centre, \[ d^4\longrightarrow d^3 \] The reduced centre attains the favourable half-filled configuration. \(\textbf{Interpret the reaction:}\) The same initial oxidation state produces two different final oxidation states. This is the defining feature of disproportionation. The supplied potentials establish the standard thermodynamic tendency, while actual reaction rates and speciation can depend on the chemical environment. \(\textbf{Final answer:}\) \[ \boxed{E^\circ_{\mathrm{cell}}=+0.56\,\mathrm{V}} \]

Explanation: Observation P specifies a divalent ion with four unpaired d electrons. High-spin Fe2+ has the formal \(3d^6\) configuration and contains four unpaired electrons. Chromium(II), with \(d^4\), also contains four unpaired electrons, so P alone cannot distinguish chromium from iron. Observation Q introduces the crucial additional condition. Oxidation of Fe2+ to Fe3+ removes an electron from the \(d^6\) configuration and produces \(d^5\). The number of unpaired electrons consequently increases from four to five. In contrast, oxidation of chromium(II) from \(d^4\) to \(d^3\) decreases its unpaired-electron count. Observation R provides an independent consistency check because Fe3+ has the favourable half-filled \(3d^5\) configuration. Cobalt(II) has \(d^7\) and ordinarily contains three unpaired electrons in the high-spin description. Manganese(II) has \(d^5\) with five unpaired electrons, so neither matches the first observation. Iron also forms the oxides FeO and Fe2O3, containing Fe(II) and Fe(III), respectively. Thus, the oxide formulas, magnetic observations and electronic stability point to the same element. The question illustrates why one measured property may be insufficient for identification, whereas several connected electronic consequences can distinguish otherwise plausible candidates.

Explanation: \(\textbf{Identify the proposed electron transfer:}\) Chromium(II) donates an electron to iron(III). \[ \mathrm{Cr^{2+}}+\mathrm{Fe^{3+}} \longrightarrow \mathrm{Cr^{3+}}+\mathrm{Fe^{2+}} \] Iron(III) is reduced, while chromium(II) is oxidised. \(\textbf{Evaluate the initial environment:}\) The standard driving potential is \[ E^\circ_{\mathrm{initial}}=0.77-(-0.41) \] \[ E^\circ_{\mathrm{initial}}=+1.18\,\mathrm{V} \] The positive value favours electron transfer under the supplied standard conditions. \(\textbf{Apply the ligand-modified potentials:}\) The question explicitly supplies effective reduction potentials appropriate to the ligand-containing environment. For iron, \[ E^\circ_L(\mathrm{Fe^{3+}/Fe^{2+}})=+0.35\,\mathrm{V} \] For chromium, \[ E^\circ_L(\mathrm{Cr^{3+}/Cr^{2+}})=+0.10\,\mathrm{V} \] \(\textbf{Calculate the modified driving potential:}\) \[ E^\circ_{L,\mathrm{cell}}=0.35-0.10 \] \[ E^\circ_{L,\mathrm{cell}}=+0.25\,\mathrm{V} \] The reaction remains favourable. \(\textbf{Compare the two environments:}\) \[ \Delta E^\circ=1.18-0.25 \] \[ \Delta E^\circ=0.93\,\mathrm{V} \] Thus, the specified ligand reduces the standard driving potential without reversing its sign. \(\textbf{Interpret the chemistry:}\) Preferential stabilisation of an oxidation state can alter the relative free energies of the oxidised and reduced metal species. Consequently, complex formation can influence redox tendencies. However, changing a reduction potential does not automatically reverse the direction of a reaction. The numerical effective potentials must be compared before making that inference. \(\textbf{Final answer:}\) \[ \boxed{E^\circ_{\mathrm{initial}}=+1.18\,\mathrm{V},\quad E^\circ_{L,\mathrm{cell}}=+0.25\,\mathrm{V}} \]

Explanation: Statement I is correct because lanthanoid 4f electrons are relatively shielded by outer electronic shells. Their electronic transitions are therefore often less strongly affected by changes in the surrounding environment than transitions involving more exposed d orbitals. This contributes to comparatively narrow spectral bands in many lanthanoid compounds. Statement II is correct because actinoid 5f electrons can participate more extensively in chemical bonding. The relatively close energies of their valence subshells also contribute to the accessibility of several oxidation states, particularly among early actinoids. Statement III is incorrect because ligand replacement can alter the electronic environment of a transition-metal centre without transferring electrons to or from that centre. A change in ligand identity may influence colour or complex stability while leaving the formal oxidation state unchanged. Statement IV is correct because lanthanoid contraction affects the elements following the lanthanoid series. Hafnium has a radius remarkably close to that of zirconium, even though it belongs to the next transition series. The incomplete shielding of increasing nuclear charge by 4f electrons is responsible for much of this unexpected similarity. Thus, each correct statement connects a different electronic feature to an observable chemical consequence. The incorrect statement confuses a change in coordination environment with a redox process. The comparison demonstrates why electronic shielding, orbital participation and ligand effects must be interpreted separately rather than treated as one universal mechanism.

Explanation: \(\textbf{Identify chromium's reduction requirement:}\) Chromium is in oxidation state +6 in both chromate and dichromate. Reduction to chromium(III) requires three electrons per chromium atom. \[ \mathrm{Cr(VI)}+3e^-\longrightarrow\mathrm{Cr(III)} \] \(\textbf{Count chromium atoms in Sample P:}\) Each chromate ion contains one chromium atom. Each dichromate ion contains two chromium atoms. Therefore, \[ n(\mathrm{Cr})_P=0.120+2(0.040) \] \[ n(\mathrm{Cr})_P=0.200\,\mathrm{mol} \] \(\textbf{Determine P's electron capacity:}\) \[ n(e^-)_P=3(0.200) \] \[ n(e^-)_P=0.600\,\mathrm{mol} \] \(\textbf{Count chromium atoms in Sample Q:}\) \[ n(\mathrm{Cr})_Q=0.040+2(0.080) \] \[ n(\mathrm{Cr})_Q=0.200\,\mathrm{mol} \] \(\textbf{Determine Q's electron capacity:}\) \[ n(e^-)_Q=3(0.200) \] \[ n(e^-)_Q=0.600\,\mathrm{mol} \] \(\textbf{Compare the result:}\) \[ n(e^-)_P=n(e^-)_Q \] Both samples possess equal chromium(VI) amounts despite having different oxoanion distributions. \(\textbf{Interpret the invariant quantity:}\) Acidification changes the equilibrium between chromate and dichromate. It does not change chromium's oxidation state. Therefore, the total electron capacity for complete reduction to Cr(III) is determined by the total number of chromium(VI) atoms rather than by the number of chromate or dichromate ions separately. This remains true even if the solutions have different yellow and orange colour contributions. \(\textbf{Final answer:}\) \[ \boxed{n(e^-)_P=n(e^-)_Q=0.600\,\mathrm{mol}} \]

Explanation: An f subshell contains seven orbitals and can accommodate fourteen electrons. A configuration containing seven f electrons is half-filled. Under Hund's rule, these electrons occupy separate f orbitals with parallel spins before pairing occurs. This arrangement provides an important electronic stabilising contribution. The proposed Ln3+ ion therefore possesses a favourable \(4f^7\) subshell. However, the stability of a chemical oxidation state is not determined exclusively by the electronic occupancy of an isolated ion. Formation of Ln2+, Ln3+ or Ln4+ involves different ionisation and electron-transfer energetics. The resulting species may also experience different hydration or complex-formation effects. Consequently, their relative free energies can depend on the chemical environment. The presence of a half-filled f subshell does not make electron removal or addition impossible. Nor do the \(4f^8\) and \(4f^6\) configurations automatically possess identical thermodynamic stability merely because they lie on opposite sides of \(4f^7\). Electronic stabilisation is a useful explanatory factor but not a universal guarantee of oxidation-state predominance. The student's initial observation is therefore reasonable, while the conclusion that +3 must dominate in every environment is unjustified.

Explanation: Statement I is correct because a high-spin \(d^4\) arrangement has four singly occupied d orbitals. Removal of one electron produces \(d^3\) with three unpaired electrons. Statement II is correct because addition of one electron to the \(d^4\) configuration can produce the half-filled \(d^5\) arrangement. Under Hund's rule, the added electron occupies the remaining empty d orbital with parallel spin. The unpaired-electron count consequently increases from four to five. Statement III is incorrect because the two transformations have opposite effects on unpaired-electron number. The \(d^4\rightarrow d^3\) change decreases it, while \(d^4\rightarrow d^5\) increases it. Furthermore, the spin-only moment depends on the square root of \(n(n+2)\), so successive changes in the number of unpaired electrons do not produce a constant numerical change in magnetic moment. Statement IV is correct because oxidation and reduction specify electron removal and addition, respectively. They do not independently establish the resulting electron-pairing pattern for every starting configuration. The change must be determined from the actual initial and final orbital occupancies. In real coordination compounds, the ligand environment can additionally influence spin arrangement. The question's high-spin assumption therefore provides an essential qualification. The correct conclusion distinguishes the electron-transfer direction from the magnetic consequences of the resulting configuration.

Explanation: \(\textbf{Resolve the mixed-valence oxide:}\) Fe3O4 contains one Fe(II) and two Fe(III) centres per formula unit. The amount of oxide is \[ n(\mathrm{Fe_3O_4})=0.080\,\mathrm{mol} \] \(\textbf{Calculate iron(II) after dissolution:}\) \[ n(\mathrm{Fe^{2+}})_{\mathrm{initial}}=1(0.080) \] \[ n(\mathrm{Fe^{2+}})_{\mathrm{initial}}=0.080\,\mathrm{mol} \] \(\textbf{Calculate iron(III) after dissolution:}\) \[ n(\mathrm{Fe^{3+}})_{\mathrm{initial}}=2(0.080) \] \[ n(\mathrm{Fe^{3+}})_{\mathrm{initial}}=0.160\,\mathrm{mol} \] \(\textbf{Apply the cerium(IV) electron-transfer reaction:}\) \[ \mathrm{Fe^{2+}}+\mathrm{Ce^{4+}} \longrightarrow \mathrm{Fe^{3+}}+\mathrm{Ce^{3+}} \] One Ce4+ ion oxidises one Fe2+ ion. The supplied Ce4+ amount is \[ n(\mathrm{Ce^{4+}})=0.050\,\mathrm{mol} \] Since \(0.080\,\mathrm{mol}\) Fe2+ is available, Ce4+ is limiting. \(\textbf{Determine iron(II) remaining:}\) \[ n(\mathrm{Fe^{2+}})_{\mathrm{final}}=0.080-0.050 \] \[ n(\mathrm{Fe^{2+}})_{\mathrm{final}}=0.030\,\mathrm{mol} \] \(\textbf{Determine iron(III) after oxidation:}\) \[ n(\mathrm{Fe^{3+}})_{\mathrm{final}}=0.160+0.050 \] \[ n(\mathrm{Fe^{3+}})_{\mathrm{final}}=0.210\,\mathrm{mol} \] \(\textbf{Check iron conservation:}\) Initially, the oxide contains \[ n(\mathrm{Fe})=3(0.080)=0.240\,\mathrm{mol} \] Finally, \[ n(\mathrm{Fe})=0.030+0.210=0.240\,\mathrm{mol} \] The atom balance is satisfied. \(\textbf{Chemical interpretation:}\) The first stage dissolves a mixed-valence oxide without changing the initial iron oxidation-state distribution. The second stage selectively oxidises Fe2+ to Fe3+. Cerium(IV) is reduced to cerium(III), so its supplied amount controls how much Fe2+ can be converted. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Fe^{2+}})_{\mathrm{final}}=0.030\,\mathrm{mol},\quad n(\mathrm{Fe^{3+}})_{\mathrm{final}}=0.210\,\mathrm{mol}} \]