D- and F-Block Elements MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Establish the electron counts:}\) Vanadium has atomic number 23. \[ N_e(\mathrm{V^{2+}})=23-2=21 \] Chromium has atomic number 24. \[ N_e(\mathrm{Cr^{3+}})=24-3=21 \] Manganese has atomic number 25. \[ N_e(\mathrm{Mn^{4+}})=25-4=21 \] Thus, all three ions contain 21 electrons. \(\textbf{Determine their d-electron configurations:}\) Each ion possesses the electronic configuration \[ [Ar]\,3d^3 \] The three d electrons occupy separate d orbitals according to Hund's rule. Therefore, every ion contains three unpaired d electrons. \(\textbf{Compare spin-only magnetic moments:}\) \[ \mu=\sqrt{n(n+2)}\,\mathrm{BM} \] For each ion, \[ n=3 \] Hence, \[ \mu=\sqrt{3(3+2)} \] \[ \mu=\sqrt{15}\approx3.87\,\mathrm{BM} \] Their calculated spin-only magnetic moments are equal. \(\textbf{Determine the ionic-radius order:}\) Although the ions contain the same number of electrons, their nuclear charges differ. The nuclear charges increase in the sequence \[ Z_{\mathrm V}=23,\quad Z_{\mathrm{Cr}}=24,\quad Z_{\mathrm{Mn}}=25 \] Because the electron populations are identical, increasing nuclear charge produces greater attraction towards the nucleus. The electronic cloud becomes more strongly contracted. Thus, the ionic-radius order is \[ r(\mathrm{V^{2+}})\gt r(\mathrm{Cr^{3+}})\gt r(\mathrm{Mn^{4+}}) \] \(\textbf{Connected inference:}\) Equal electron counts do not imply equal ionic radii when nuclear charges differ. However, equal free-ion d-electron configurations produce equal spin-only unpaired-electron counts. This explains how the ions can possess different sizes but identical calculated spin-only magnetic moments. \(\textbf{Final answer:}\) The ionic-radius order is \(\mathrm{V^{2+}\gt Cr^{3+}\gt Mn^{4+}}\), and all three ions have calculated spin-only moments of approximately \(3.87\,\mathrm{BM}\).

Explanation: \(\textbf{Identify the spontaneous redox direction:}\) Iron(III) has the more positive reduction potential. It therefore acts as the electron acceptor. \[ \mathrm{Fe^{3+}}+e^-\longrightarrow\mathrm{Fe^{2+}} \] Copper metal supplies electrons. \[ \mathrm{Cu}\longrightarrow\mathrm{Cu^{2+}}+2e^- \] \(\textbf{Check the standard driving potential:}\) \[ E^\circ=0.77-0.34 \] \[ E^\circ=+0.43\,\mathrm{V} \] The forward oxidation of copper by iron(III) is thermodynamically favourable under standard conditions. \(\textbf{Construct the balanced ionic reaction:}\) Two iron(III) ions are required to accept the two electrons released by one copper atom. \[ \mathrm{Cu}+2\mathrm{Fe^{3+}} \longrightarrow \mathrm{Cu^{2+}}+2\mathrm{Fe^{2+}} \] \(\textbf{Calculate the initial copper amount:}\) \[ n(\mathrm{Cu})=\frac{6.35}{63.5} \] \[ n(\mathrm{Cu})=0.100\,\mathrm{mol} \] \(\textbf{Identify the limiting reactant:}\) Complete oxidation of \(0.100\,\mathrm{mol}\) copper would require \[ n(\mathrm{Fe^{3+}})_{\mathrm{required}}=2(0.100) \] \[ n(\mathrm{Fe^{3+}})_{\mathrm{required}}=0.200\,\mathrm{mol} \] Only \(0.160\,\mathrm{mol}\) iron(III) is available. Therefore, iron(III) is the limiting reactant. \(\textbf{Calculate copper oxidation:}\) \[ n(\mathrm{Cu})_{\mathrm{reacted}}=\frac12n(\mathrm{Fe^{3+}}) \] \[ n(\mathrm{Cu})_{\mathrm{reacted}}=\frac12(0.160) \] \[ n(\mathrm{Cu})_{\mathrm{reacted}}=0.080\,\mathrm{mol} \] The same amount of copper(II) is formed. \[ n(\mathrm{Cu^{2+}})=0.080\,\mathrm{mol} \] \(\textbf{Calculate the unreacted copper:}\) \[ n(\mathrm{Cu})_{\mathrm{remaining}}=0.100-0.080 \] \[ n(\mathrm{Cu})_{\mathrm{remaining}}=0.020\,\mathrm{mol} \] \[ m(\mathrm{Cu})_{\mathrm{remaining}}=0.020\times63.5 \] \[ m(\mathrm{Cu})_{\mathrm{remaining}}=1.27\,\mathrm{g} \] \(\textbf{Chemical significance:}\) Copper cannot ordinarily reduce hydrogen ions to hydrogen gas in a dilute non-oxidising acid. However, the more positive reduction potential of the iron(III)/iron(II) couple permits iron(III) to oxidise copper. Thus, the behaviour of a metal depends on the oxidising species present rather than merely on whether the solution is acidic. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Cu^{2+}})=0.080\,\mathrm{mol},\quad m(\mathrm{Cu\ remaining})=1.27\,\mathrm{g}} \]

Explanation: \(\textbf{Determine the initial manganese oxidation state:}\) Permanganate has the formula \(\mathrm{MnO_4^-}\). Oxygen contributes a total oxidation number of \(-8\). Since the ion carries charge \(-1\), manganese has oxidation state +7. \(\textbf{Reduction in acidic medium:}\) The manganese product is \(\mathrm{Mn^{2+}}\). Therefore, manganese changes from +7 to +2. \[ \Delta x=7-2=5 \] Each permanganate ion accepts five electrons. The acidic reduction half-reaction is \[ \mathrm{MnO_4^-}+8\mathrm{H^+}+5e^- \longrightarrow \mathrm{Mn^{2+}}+4\mathrm{H_2O} \] \(\textbf{Electron amount in the acidic sample:}\) \[ n(e^-)_{\mathrm{acidic}}=5n(\mathrm{MnO_4^-}) \] \[ n(e^-)_{\mathrm{acidic}}=5(0.040) \] \[ n(e^-)_{\mathrm{acidic}}=0.200\,\mathrm{mol} \] \(\textbf{Reduction under neutral conditions:}\) The manganese product is \(\mathrm{MnO_2}\). Manganese has oxidation state +4 in this oxide. Therefore, manganese changes from +7 to +4. \[ \Delta x=7-4=3 \] Each permanganate ion accepts three electrons. A suitable reduction half-reaction is \[ \mathrm{MnO_4^-}+2\mathrm{H_2O}+3e^- \longrightarrow \mathrm{MnO_2}+4\mathrm{OH^-} \] \(\textbf{Electron amount in the neutral sample:}\) \[ n(e^-)_{\mathrm{neutral}}=3n(\mathrm{MnO_4^-}) \] \[ n(e^-)_{\mathrm{neutral}}=3(0.040) \] \[ n(e^-)_{\mathrm{neutral}}=0.120\,\mathrm{mol} \] \(\textbf{Compare the capacities:}\) \[ 0.200-0.120=0.080\,\mathrm{mol} \] Thus, the acidic sample accepts \(0.080\,\mathrm{mol}\) more electrons than the neutral sample. \(\textbf{Chemical interpretation:}\) The initial oxidation state of manganese is +7 in both samples. However, the final oxidation states differ because the reduction products depend on the reaction medium. Consequently, the electron-acceptance capacity per mole of permanganate is not identical under all conditions. This distinction is essential when applying redox stoichiometry to reactions involving manganese(VII). \(\textbf{Final answer:}\) \[ \boxed{n(e^-)_{\mathrm{acidic}}=0.200\,\mathrm{mol},\quad n(e^-)_{\mathrm{neutral}}=0.120\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the electronic comparison:}\) The third ionisation of manganese converts Mn(II) into Mn(III). The third ionisation of iron converts Fe(II) into Fe(III). Each process removes one electron from a divalent gaseous ion. However, the initial d-electron configurations are different. \(\textbf{Exchange pairs in manganese(II):}\) Manganese(II) has the \(3d^5\) configuration. Hund's rule gives five parallel-spin electrons occupying five different d orbitals. The number of distinct parallel-spin pairs is \[ N_{\mathrm{pairs}}=\frac{n(n-1)}2 \] For five parallel-spin electrons, \[ N_{\mathrm{pairs}}(d^5)=\frac{5(5-1)}2 \] \[ N_{\mathrm{pairs}}(d^5)=10 \] \(\textbf{After manganese oxidation:}\) Removal of one d electron gives \(3d^4\). Four parallel-spin electrons occupy four different d orbitals. \[ N_{\mathrm{pairs}}(d^4)=\frac{4(4-1)}2 \] \[ N_{\mathrm{pairs}}(d^4)=6 \] Thus, the loss of favourable parallel-spin pairs is \[ 10-6=4 \] \(\textbf{Exchange pairs in iron(II):}\) Iron(II) has the \(3d^6\) configuration. Five electrons have parallel spins in different d orbitals. The sixth electron has the opposite spin and pairs in one orbital. In the simplified model, the five parallel-spin electrons provide \[ N_{\mathrm{pairs}}(d^6)=\frac{5(5-1)}2 \] \[ N_{\mathrm{pairs}}(d^6)=10 \] \(\textbf{After iron oxidation:}\) Removal of the oppositely oriented electron produces \(3d^5\). Five parallel-spin electrons remain. \[ N_{\mathrm{pairs}}(d^5)=10 \] Thus, the loss of favourable parallel-spin pairs is \[ 10-10=0 \] \(\textbf{Connect exchange stabilisation with ionisation:}\) The manganese ionisation disrupts a favourable half-filled d arrangement and reduces the number of parallel-spin pairs. The iron ionisation produces a half-filled d subshell without reducing the number of these pairs in the simplified model. This difference contributes to the greater third ionisation enthalpy of manganese compared with iron. The exchange-pair model identifies an important electronic contribution rather than replacing the complete energetic description of ionisation. \(\textbf{Final answer:}\) The respective losses of favourable parallel-spin pairs are four for Mn(II) and zero for Fe(II).

Explanation: \(\textbf{Oxidation-state cycle:}\) The active vanadium(V) oxide catalyst participates in an idealised redox cycle during the oxidation of sulfur dioxide to sulfur trioxide. \[ \mathrm{V_2O_5}+\mathrm{SO_2}\longrightarrow\mathrm{V_2O_4}+\mathrm{SO_3} \] In this step, vanadium is reduced from +5 to +4, whereas sulfur is oxidised from +4 to +6. \(\textbf{Regeneration:}\) \[ \mathrm{V_2O_4}+\tfrac12\mathrm{O_2}\longrightarrow\mathrm{V_2O_5} \] Vanadium returns from +4 to +5, and the original oxide is regenerated. Adding the two steps gives the net reaction \[ \mathrm{SO_2}+\tfrac12\mathrm{O_2}\longrightarrow\mathrm{SO_3} \] The supplied energy diagram is consistent with an alternative pathway having a lower energy barrier while the initial and final energies remain the same. Catalyst regeneration means there is no net consumption of \(\mathrm{V_2O_5}\), and catalysis does not change the overall reaction enthalpy. \(\textbf{Final answer:}\) Vanadium cycles from V(V) to V(IV) and back to V(V); the catalyst is regenerated without changing the overall reaction enthalpy.

Explanation: Manganese exhibits several oxidation states because different numbers of its valence electrons can participate in chemical bonding. In \(\mathrm{MnO}\), oxygen has oxidation state \(-2\). Since the compound is neutral, manganese must have oxidation state +2. Manganese(II) possesses the formal \(3d^5\) configuration. In the usual high-spin ionic description, five unpaired d electrons contribute to its paramagnetic character. Manganese(II) oxide is predominantly basic, consistent with the general behaviour of a relatively low-oxidation-state transition-metal oxide. In \(\mathrm{Mn_2O_7}\), seven oxygen atoms contribute a total oxidation number of \(-14\). The two manganese atoms therefore contribute +14, giving an oxidation state of +7 for each manganese centre. Manganese(VII) has a formal \(3d^0\) configuration. Its metal centres consequently have no unpaired d electrons in this formal description. Manganese(VII) oxide exhibits acidic character and is associated with permanganic acid. The higher oxidation state is stabilised by strong metal–oxygen bonding, including significant covalent character. Thus, increasing the oxidation state of manganese changes both the metal-centre electronic configuration and the acid–base character of its oxides. The comparison should not be interpreted as a general rule that every oxide must be purely ionic or that electron count alone completely determines acidity. Rather, it illustrates the combined influence of oxidation state, bonding and electronic structure. The two oxides therefore provide contrasting examples of low-oxidation-state basic behaviour and high-oxidation-state acidic behaviour within the chemistry of a single transition element.

Explanation: Statement I is correct because the relatively high melting points of many transition metals are associated with strong interatomic metallic bonding. Electrons belonging to the outer s subshell and the underlying d subshell can participate in these interactions. The availability of several suitable valence electrons contributes to strong cohesion within many transition-metal solids. Statement II is incorrect because a half-filled d subshell does not guarantee the highest melting point of a particular metallic series. Although melting points generally rise towards the middle of a transition series, manganese and technetium show important anomalies. Manganese possesses a half-filled \(3d^5\) subshell in its isolated neutral atom but has a comparatively low enthalpy of atomisation. The stability of an isolated-atom configuration cannot therefore be used as an unconditional predictor of metallic cohesion. Statement III is correct because Zn, Cd and Hg possess completely filled d subshells in their ground-state atoms. Their filled d subshells contribute relatively little to the strong metallic bonding characteristic of many transition metals. Consequently, these elements exhibit comparatively low melting points. Statement IV is correct because the electronic structure of the metallic solid involves collective interactions among atoms. Metallic cohesion cannot always be inferred by counting unpaired electrons in separated gaseous atoms. The general trend reflects electronic contributions to bonding, while its exceptions demonstrate that the detailed energetic environment remains important.

Explanation: Chromium, molybdenum and tungsten possess the same group relationship, but the relative stabilities of their oxidation states are not identical. Chromium(VI) occurs in important compounds such as chromate and dichromate. Acidified dichromate is a strong oxidising agent because chromium(VI) can be reduced to more stable chromium(III) species under appropriate conditions. Molybdenum(VI) and tungsten(VI), in contrast, are comparatively stable in their corresponding high-oxidation-state compounds. Their oxides \(\mathrm{MoO_3}\) and \(\mathrm{WO_3}\) do not show the same strong oxidising behaviour as acidified dichromate under ordinary comparable conditions. This reflects an important trend in the transition series: higher oxidation states often become more stable when moving from the lighter to the heavier members of a group. The tendency is particularly relevant to Groups 4–10. In many heavier p-block elements, a contrasting tendency occurs because the outer \(ns^2\) electrons may participate less readily in bonding. This can stabilise lower oxidation states through the inert-pair effect. The transition-metal trend should therefore not be predicted by mechanically applying the p-block inert-pair argument. Nevertheless, greater stability of the +6 oxidation state does not mean that molybdenum or tungsten cannot exhibit lower oxidation states. The observed chemistry depends on metal–ligand bonding and the relative energies of the compounds involved. The correct comparison is consequently one of relative oxidation-state stability, not an absolute prohibition on other oxidation states.

Explanation: \(\textbf{Identify the oxidation of chloride:}\) Chloride ions must lose electrons to form chlorine. \[ 2\mathrm{Cl^-}\longrightarrow\mathrm{Cl_2}+2e^- \] The corresponding standard reduction potential is \[ E^\circ(\mathrm{Cl_2/Cl^-})=+1.36\,\mathrm{V} \] A trivalent metal ion can oxidise chloride under standard conditions when its reduction potential is greater than this value. \(\textbf{Test cobalt(III):}\) \[ E^\circ_{\mathrm{Co}}=1.97-1.36 \] \[ E^\circ_{\mathrm{Co}}=+0.61\,\mathrm{V} \] The positive value favours reduction of cobalt(III) to cobalt(II), accompanied by chloride oxidation. \(\textbf{Test manganese(III):}\) \[ E^\circ_{\mathrm{Mn}}=1.57-1.36 \] \[ E^\circ_{\mathrm{Mn}}=+0.21\,\mathrm{V} \] This value is also positive. Manganese(III) can therefore oxidise chloride under the stipulated conditions. \(\textbf{Test iron(III):}\) \[ E^\circ_{\mathrm{Fe}}=0.77-1.36 \] \[ E^\circ_{\mathrm{Fe}}=-0.59\,\mathrm{V} \] The negative value does not favour oxidation of chloride by iron(III) under standard conditions. Thus, only the cobalt(III) and manganese(III) samples contribute to chlorine formation. \(\textbf{Balance the electron transfer:}\) Each trivalent ion accepts one electron. \[ \mathrm{M^{3+}}+e^-\longrightarrow\mathrm{M^{2+}} \] Formation of one chlorine molecule releases two electrons. Consequently, \[ 2\mathrm{M^{3+}}+2\mathrm{Cl^-} \longrightarrow 2\mathrm{M^{2+}}+\mathrm{Cl_2} \] \(\textbf{Calculate chlorine from cobalt(III):}\) \[ n(\mathrm{Cl_2})_{\mathrm{Co}}=\frac{0.100}{2} \] \[ n(\mathrm{Cl_2})_{\mathrm{Co}}=0.050\,\mathrm{mol} \] \(\textbf{Calculate chlorine from manganese(III):}\) \[ n(\mathrm{Cl_2})_{\mathrm{Mn}}=\frac{0.100}{2} \] \[ n(\mathrm{Cl_2})_{\mathrm{Mn}}=0.050\,\mathrm{mol} \] \(\textbf{Total chlorine formation:}\) \[ n(\mathrm{Cl_2})_{\mathrm{total}}=0.050+0.050 \] \[ n(\mathrm{Cl_2})_{\mathrm{total}}=0.100\,\mathrm{mol} \] \(\textbf{Chemical interpretation:}\) The three metal ions have the same formal positive charge but different tendencies to accept electrons. Their oxidising powers cannot therefore be inferred from ionic charge alone. The relative standard reduction potentials determine which ions can oxidise chloride under the stated conditions. \(\textbf{Final answer:}\) \[ n(\mathrm{Cl_2})_{\mathrm{total}}=0.100\,\mathrm{mol} \] Only cobalt(III) and manganese(III) contribute to chlorine formation.

Explanation: Statement I is correct because carbon monoxide is a neutral ligand in the specified metal carbonyls. In neutral \(\mathrm{Ni(CO)_4}\), the sum of the ligand charges is zero, so nickel has formal oxidation state zero. The same reasoning gives iron an oxidation state of zero in \(\mathrm{Fe(CO)_5}\). Statement II is also correct because carbon monoxide can stabilise low oxidation states through suitable bonding interactions with transition-metal centres. Its ability to accept electron density contributes to the stability of several low-oxidation-state metal carbonyl compounds. A detailed orbital treatment of this bonding is not necessary to recognise its significance. Statement III is incorrect because formation of a zero-oxidation-state carbonyl does not prevent the metal from forming compounds in positive oxidation states. Nickel commonly forms nickel(II) compounds, while iron commonly exhibits the +2 and +3 oxidation states. These oxidation states are stabilised in different chemical environments. Statement IV is therefore correct because oxidation state belongs to a metal within a specified chemical species. It is not a fixed property of the elemental metal independent of its surroundings. The carbonyl examples illustrate how ligand identity can influence the range of oxidation states accessible to transition metals. This is consistent with the broader importance of variable oxidation states and complex formation in transition-metal chemistry. However, the formal oxidation-state comparison should not be confused with a detailed prediction of magnetic behaviour or molecular structure, which requires additional information about bonding.

Explanation: Vanadium exhibits several oxidation states, and the acid–base character of its oxides changes with the oxidation state. In \(\mathrm{V_2O_3}\), vanadium has oxidation state +3. This oxide exhibits basic character. In \(\mathrm{V_2O_4}\), vanadium has oxidation state +4, and the oxide is less basic than \(\mathrm{V_2O_3}\). In \(\mathrm{V_2O_5}\), vanadium has oxidation state +5. Vanadium(V) oxide is amphoteric, although its acidic character predominates. Reaction P illustrates its reaction with hydrogen ions to produce an oxygen-containing vanadium(V) cation. In \(\mathrm{VO_2^+}\), the two oxygen atoms contribute \(-4\), while the ionic charge is +1. Therefore, vanadium remains in oxidation state +5. Reaction Q illustrates its reaction with alkali to form the vanadate ion \(\mathrm{VO_4^{3-}}\). In this ion, four oxygen atoms contribute \(-8\), while the overall charge is \(-3\), again giving vanadium oxidation state +5. Neither reaction changes the oxidation state of vanadium. They consequently demonstrate acid–base behaviour rather than electron-transfer reactions. The two equations are also balanced in atoms and electrical charge. The comparison establishes that increasing oxidation state in transition-metal oxides generally favours increasing acidic character. Nevertheless, an oxide such as \(\mathrm{V_2O_5}\) may remain amphoteric even when its acidic behaviour is predominant.

Explanation: \(\textbf{Identify the first acid-forming reaction:}\) Chromium(VI) oxide reacts with water according to \[ \mathrm{CrO_3}+\mathrm{H_2O}\longrightarrow\mathrm{H_2CrO_4} \] The oxide-to-acid mole ratio is \[ n(\mathrm{CrO_3}):n(\mathrm{H_2CrO_4})=1:1 \] \(\textbf{Calculate chromic-acid formation:}\) \[ n(\mathrm{H_2CrO_4})=n(\mathrm{CrO_3}) \] \[ n(\mathrm{H_2CrO_4})=0.120\,\mathrm{mol} \] \(\textbf{Identify the second acid-forming reaction:}\) Manganese(VII) oxide is an acidic oxide associated with permanganic acid. \[ \mathrm{Mn_2O_7}+\mathrm{H_2O}\longrightarrow2\mathrm{HMnO_4} \] One mole of manganese(VII) oxide contains two manganese atoms. It therefore produces two moles of mononuclear permanganic acid in the stated idealised reaction. \(\textbf{Calculate permanganic-acid formation:}\) \[ n(\mathrm{HMnO_4})=2n(\mathrm{Mn_2O_7}) \] \[ n(\mathrm{HMnO_4})=2(0.080) \] \[ n(\mathrm{HMnO_4})=0.160\,\mathrm{mol} \] \(\textbf{Check the chromium oxidation state:}\) Chromium has oxidation state +6 in \(\mathrm{CrO_3}\). It also has oxidation state +6 in \(\mathrm{H_2CrO_4}\). Thus, the oxide-to-acid conversion does not involve chromium oxidation or reduction. \(\textbf{Check the manganese oxidation state:}\) Manganese has oxidation state +7 in \(\mathrm{Mn_2O_7}\). It retains oxidation state +7 in \(\mathrm{HMnO_4}\). Consequently, the second reaction is also non-redox. \(\textbf{Chemical interpretation:}\) The different acid amounts result from the stoichiometric formulas of the two oxides. Their high oxidation states and strong metal–oxygen bonding are associated with predominantly acidic character. The reactions do not require electron transfer because the metal oxidation states remain unchanged. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{H_2CrO_4})=0.120\,\mathrm{mol},\quad n(\mathrm{HMnO_4})=0.160\,\mathrm{mol}} \]

Explanation: \(\textbf{Determine iron's initial oxidation state:}\) Ferrate has the formula \(\mathrm{FeO_4^{2-}}\). Four oxygen atoms contribute a total oxidation number of \(-8\). Let the oxidation state of iron be \(x\). \[ x-8=-2 \] \[ x=+6 \] Thus, ferrate contains iron(VI). \(\textbf{Determine iron's final oxidation state:}\) In \(\mathrm{Fe_2O_3}\), three oxygen atoms contribute \(-6\). The two iron atoms must together contribute +6. \[ 2x-6=0 \] \[ x=+3 \] Therefore, iron changes from +6 to +3. \(\textbf{Count electrons gained by iron:}\) Each iron centre undergoes a decrease of three oxidation-state units. \[ \Delta x=6-3=3 \] Thus, each Fe(VI) centre accepts three electrons. For four ferrate ions, the electron acceptance is \[ 4\times3=12 \] \(\textbf{Identify the compensating oxidation:}\) In the ferrate ion and water, oxygen has oxidation state \(-2\). In molecular oxygen, it has oxidation state zero. Each oxygen atom converted into \(\mathrm{O_2}\) therefore loses two electrons. Three oxygen molecules contain six oxygen atoms. \[ 6\times2=12 \] Hence, the total electron loss by oxygen balances the twelve electrons gained by iron. \(\textbf{Apply the balanced reaction:}\) \[ 4\mathrm{FeO_4^{2-}}+4\mathrm{H_2O} \longrightarrow 2\mathrm{Fe_2O_3}+3\mathrm{O_2}+8\mathrm{OH^-} \] Four moles of ferrate produce three moles of oxygen. \(\textbf{Calculate oxygen evolution:}\) \[ n(\mathrm{O_2})=\frac34n(\mathrm{FeO_4^{2-}}) \] \[ n(\mathrm{O_2})=\frac34(0.080) \] \[ n(\mathrm{O_2})=0.060\,\mathrm{mol} \] \(\textbf{Check the redox nature:}\) Iron undergoes reduction, while part of the oxygen undergoes oxidation. The same reacting system therefore provides both the electron-accepting and electron-donating transformations. This explains how oxygen can be evolved during decomposition of a high-oxidation-state metal oxoanion. The calculation is consistent with the instability of ferrate(VI) towards decomposition under suitable conditions. \(\textbf{Final answer:}\) Oxygen evolved: \[ n(\mathrm{O_2})=0.060\,\mathrm{mol} \] Each Fe(VI) centre accepts three electrons.

Explanation: Statement I is correct because fluorine has oxidation state \(-1\) in \(\mathrm{MnF_4}\). The four fluorine atoms contribute a total oxidation number of \(-4\), so manganese must have oxidation state +4. Statement II is correct because both oxygen and fluorine must be considered when determining the oxidation state in \(\mathrm{MnO_3F}\). The three oxygen atoms contribute \(-6\), while fluorine contributes \(-1\). The total negative contribution is \(-7\), giving manganese an oxidation state of +7. In \(\mathrm{Mn_2O_7}\), the seven oxygen atoms contribute \(-14\). Each of the two manganese atoms consequently has oxidation state +7. Statement III is correct because manganese has atomic number 25 and neutral configuration \([Ar]\,3d^54s^2\). Formal removal of seven electrons gives the \(3d^0\) configuration for Mn(VII). Statement IV is incorrect because the ability of manganese to form an oxyfluoride in oxidation state +7 does not establish the existence of a corresponding simple heptafluoride. Oxygen can stabilise high oxidation states through favourable metal–oxygen bonding, including significant multiple-bonding contributions. The highest well-established simple manganese fluoride is \(\mathrm{MnF_4}\), whereas manganese reaches +7 in suitable oxygen-containing compounds. Therefore, the maximum oxidation state found in an oxyfluoride need not be attainable in a simple fluoride. The comparison demonstrates why oxidation-state accessibility depends on the complete chemical environment and cannot be predicted solely from the formal charge assigned to the metal.

Explanation: \(\textbf{Identify the electron donor:}\) The reduction potentials of all three \(\mathrm{M^{3+}/M^{2+}}\) couples are negative relative to the hydrogen electrode. Therefore, oxidation of their divalent ions by hydrogen ions is thermodynamically favourable under standard conditions. The general oxidation half-reaction is \[ \mathrm{M^{2+}}\longrightarrow\mathrm{M^{3+}}+e^- \] \(\textbf{Hydrogen-ion reduction:}\) \[ 2\mathrm{H^+}+2e^-\longrightarrow\mathrm{H_2} \] Two electrons are required to produce one hydrogen molecule. \(\textbf{Check the three standard driving potentials:}\) For titanium(II), \[ E^\circ_{\mathrm{Ti}}=0.00-(-0.37)=+0.37\,\mathrm{V} \] For vanadium(II), \[ E^\circ_{\mathrm V}=0.00-(-0.26)=+0.26\,\mathrm{V} \] For chromium(II), \[ E^\circ_{\mathrm{Cr}}=0.00-(-0.41)=+0.41\,\mathrm{V} \] All three values are positive. \(\textbf{Combine each oxidation with hydrogen reduction:}\) The general balanced ionic equation is \[ 2\mathrm{M^{2+}}+2\mathrm{H^+} \longrightarrow 2\mathrm{M^{3+}}+\mathrm{H_2} \] Thus, two moles of divalent ions produce one mole of hydrogen. \(\textbf{Total amount of divalent ions:}\) \[ n(\mathrm{M^{2+}})_{\mathrm{total}}=0.020+0.030+0.050 \] \[ n(\mathrm{M^{2+}})_{\mathrm{total}}=0.100\,\mathrm{mol} \] \(\textbf{Electron amount released:}\) Each divalent ion loses one electron. \[ n(e^-)=0.100\times1 \] \[ n(e^-)=0.100\,\mathrm{mol} \] \(\textbf{Calculate hydrogen evolution:}\) \[ n(\mathrm{H_2})=\frac{n(e^-)}2 \] \[ n(\mathrm{H_2})=\frac{0.100}{2} \] \[ n(\mathrm{H_2})=0.050\,\mathrm{mol} \] \(\textbf{Chemical interpretation:}\) The reducing behaviour of these divalent ions is distinct from the reaction of their corresponding metals with acids. Here, the reacting species are already divalent ions, and their oxidation to trivalent ions supplies the electrons needed for hydrogen evolution. The total hydrogen amount depends on electron equivalents rather than on the number of different transition-metal ions present. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{H_2})=0.050\,\mathrm{mol}} \]

Explanation: Statement I is correct because fluorine is an electronegative element capable of stabilising relatively high metal oxidation states. In suitable ionic fluorides, favourable lattice-energy contributions can assist this stabilisation. In more covalent fluorides, favourable bond enthalpies may play an important role. Statement II is correct because the supplied reduction potential for the cobalt(III)/cobalt(II) couple is more positive than that for chlorine/chloride. Consequently, cobalt(III) has a greater standard tendency to accept electrons than molecular chlorine in the specified comparison. The corresponding electron transfer from chloride to cobalt(III) is thermodynamically favourable. Statement III is correct because cobalt has atomic number 27 and neutral configuration \([Ar]\,3d^74s^2\). Removal of three electrons gives the formal \(3d^6\) configuration for cobalt(III). Statement IV is incorrect because stability of an oxidation state in a crystalline compound does not require the corresponding free aqueous ion to be stable towards every reducing agent. The solid-state environment and aqueous environment provide different energetic contributions. A highly oxidising aqueous ion may nevertheless be present in a stable solid compound under appropriate conditions. This distinction is important when comparing halide compounds with hydrated metal ions. The existence of \(\mathrm{CoF_3}\) therefore cannot be interpreted as proof that aqueous cobalt(III) is chemically inert. The comparison illustrates how oxidation-state stability depends on ligand identity, bonding and reaction medium. It also demonstrates why standard reduction potentials must be interpreted for the specified redox couples rather than transferred unconditionally to every compound containing the metal.

Explanation: The acid–base character and bonding nature of transition-metal oxides often depend on the oxidation state of the metal. In a relatively low positive oxidation state, the metal centre generally has lower polarising power, and metal–oxygen bonding may possess appreciable ionic character. Such oxides commonly exhibit basic behaviour because their oxide-ion character favours reaction with acids. As the oxidation state increases, the greater effective positive charge associated with the metal centre generally enhances its ability to polarise electron density associated with oxygen. This favours a greater contribution from covalent bonding. High-oxidation-state oxides often exhibit acidic behaviour and can react with bases to form corresponding oxoanions. Therefore, Option A correctly relates the observed changes in oxidation state, covalency and acid–base character. Option B is incorrect because greater oxide-ion basicity would not explain the increased acidic character of Q. Option C is incorrect because stronger electrostatic lattice interactions do not independently establish greater covalent character or acidic behaviour. Lattice energy and bond covalency are distinct considerations and cannot be treated as interchangeable explanations. Option D is incorrect because the observed increase in covalency involves a greater degree of electron sharing or orbital interaction rather than its suppression. Chromium provides a representative comparison: chromium(II) oxide, \(\mathrm{CrO}\), is basic, whereas chromium(VI) oxide, \(\mathrm{CrO_3}\), is acidic. The latter also exhibits substantial covalent metal–oxygen bonding. These observations illustrate the general relationship between higher oxidation state, increased covalent character and increasing oxide acidity. The relationship is a chemical trend rather than an exception-free numerical rule for every transition-metal oxide.

Explanation: Zirconium belongs to the second transition series, while hafnium belongs to the third transition series. Moving down a group would ordinarily be expected to increase atomic size because an additional principal electron shell is occupied. However, the transition from the 4d series to the 5d series is affected by the filling of the 4f subshell. The 4f electrons provide relatively ineffective shielding against the increasing nuclear charge. As the nuclear charge increases across the lanthanoid series, the incomplete shielding produces a progressive contraction of atomic and ionic sizes. This phenomenon is known as lanthanoid contraction. By the time the 5d transition series is reached, the contraction has substantially offset the size increase expected from occupying a higher principal shell. Consequently, corresponding elements of the 4d and 5d series can possess very similar metallic radii. Zirconium and hafnium are characteristic examples of this effect. The supplied radii differ by only \(1\,\mathrm{pm}\), despite the elements belonging to different periods. They do not have identical nuclear charges or the same number of occupied principal shells. Nor does the contraction require the radii of every corresponding pair to be mathematically identical. The similar sizes contribute to the pronounced resemblance between the physical and chemical properties of zirconium and hafnium. The explanation therefore requires recognising a competing electronic effect rather than applying the ordinary down-group radius trend without qualification.

Explanation: Interstitial compounds form when relatively small atoms become incorporated into suitable spaces within a metallic lattice. Carbon, hydrogen and nitrogen are common examples of small atoms that can occupy such positions in transition-metal structures. The question explicitly states that carbon occupies interstitial positions rather than replacing metal atoms. Therefore, the structural arrangement is consistent with an interstitial compound. Such compounds are frequently non-stoichiometric because the degree of occupation of interstitial sites can vary. Their formulas need not correspond to simple integer valence combinations expected for ordinary ionic compounds. The presence of carbon can modify the interactions within the metallic structure and contribute to increased hardness. Many transition-metal interstitial compounds also possess high melting points. Importantly, the incorporation of non-metal atoms does not necessarily eliminate the metallic electronic character of the parent framework. Consequently, appreciable electrical conductivity may be retained. The material therefore need not be classified as an electrical insulator merely because it contains carbon. A non-stoichiometric formula also does not establish that the lattice has been destroyed. The distinction between an interstitial compound and a substitutional alloy depends primarily on the sites occupied by the added atoms. The supplied combination of site occupancy, hardness, melting behaviour and conductivity is consistent with the characteristic properties of transition-metal interstitial compounds.

Explanation: \(\textbf{Determine the relevant comparison:}\) For a proposed metal-displacement reaction, the standard driving potential is obtained by subtracting the reduction potential of the metal being oxidised from that of the ion being reduced. \[ E^\circ=E^\circ_{\mathrm{cathodic\ reduction}}-E^\circ_{\mathrm{oxidised\ metal\ couple}} \] A positive value supports the proposed forward direction under standard conditions. \(\textbf{Reaction P — Zinc displaces nickel:}\) Nickel(II) undergoes reduction. \[ \mathrm{Ni^{2+}}+2e^-\longrightarrow\mathrm{Ni} \] Zinc undergoes oxidation. \[ \mathrm{Zn}\longrightarrow\mathrm{Zn^{2+}}+2e^- \] The standard driving potential is \[ E^\circ_P=-0.25-(-0.76) \] \[ E^\circ_P=+0.51\,\mathrm{V} \] Thus, P is thermodynamically favoured. \(\textbf{Reaction Q — Nickel displaces copper:}\) Copper(II) undergoes reduction, while nickel is oxidised. \[ E^\circ_Q=+0.34-(-0.25) \] \[ E^\circ_Q=+0.59\,\mathrm{V} \] Thus, Q is also thermodynamically favoured. \(\textbf{Reaction R — Copper displaces zinc:}\) The proposed reaction requires copper oxidation and zinc(II) reduction. \[ E^\circ_R=-0.76-(+0.34) \] \[ E^\circ_R=-1.10\,\mathrm{V} \] The negative value indicates that R is not favoured in the specified forward direction under standard conditions. \(\textbf{Compare the favourable reactions:}\) \[ E^\circ_Q=+0.59\,\mathrm{V}\gt E^\circ_P=+0.51\,\mathrm{V} \] Therefore, Q has the larger positive standard driving potential. \(\textbf{Chemical interpretation:}\) The metals and their ions follow the electron-transfer tendencies indicated by their relative reduction potentials. A more negative metal-ion reduction potential generally corresponds to a greater tendency of the metal to undergo oxidation in the specified comparisons. However, equal electron-transfer stoichiometry does not imply equal spontaneity. The sign of the standard driving potential must be checked for each reaction. \(\textbf{Final answer:}\) P and Q are favoured, R is not, and \[ \boxed{E^\circ_Q=+0.59\,\mathrm{V}\gt E^\circ_P=+0.51\,\mathrm{V}} \]