201. The following standard reduction potentials are supplied for specified aqueous conditions:
\[
E^\circ(\mathrm{Ce^{4+}/Ce^{3+}})=+1.61\,\mathrm{V}
\]
\[
E^\circ(\mathrm{Pu^{4+}/Pu^{3+}})=+0.98\,\mathrm{V}
\]
\[
E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=+0.77\,\mathrm{V}
\]
Assume that the stated species remain soluble and that no competing chemical reactions occur.
Which statement correctly predicts the possible redox reactions and the formal 5f-electron change associated with plutonium oxidation?
ⓐ. Fe3+ can oxidise Pu3+, but Ce4+ cannot; Pu3+ changes from \(5f^4\) to \(5f^5\).
ⓑ. Both Ce4+ and Fe3+ can oxidise Pu3+; Pu3+ changes from \(5f^5\) to \(5f^4\).
ⓒ. Ce4+ can oxidise Pu3+, but Fe3+ cannot under standard conditions; Pu3+ changes from \(5f^5\) to \(5f^4\).
ⓓ. Neither Ce4+ nor Fe3+ can oxidise Pu3+ because Pu3+ has a partially filled 5f subshell.
Correct Answer: Ce4+ can oxidise Pu3+, but Fe3+ cannot under standard conditions; Pu3+ changes from \(5f^5\) to \(5f^4\).
Explanation: \(\textbf{Identify the plutonium electron-transfer process:}\)
Oxidation of Pu(III) produces Pu(IV).
\[
\mathrm{Pu^{3+}}\longrightarrow\mathrm{Pu^{4+}}+e^-
\]
The formal configurations are
\[
\mathrm{Pu^{3+}}:5f^5
\]
\[
\mathrm{Pu^{4+}}:5f^4
\]
Thus, plutonium loses one 5f electron.
\(\textbf{Evaluate cerium(IV) as the oxidising agent:}\)
Cerium(IV) is reduced according to
\[
\mathrm{Ce^{4+}}+e^-\longrightarrow\mathrm{Ce^{3+}}
\]
Its standard reduction potential is greater than that of the Pu(IV)/Pu(III) couple.
\[
E^\circ_{\mathrm{Ce/Pu}}=1.61-0.98
\]
\[
E^\circ_{\mathrm{Ce/Pu}}=+0.63\,\mathrm{V}
\]
The positive value favours oxidation of Pu(III) by Ce(IV).
\(\textbf{Evaluate iron(III) as the oxidising agent:}\)
Iron(III) accepts one electron to form iron(II).
For the proposed oxidation of Pu(III),
\[
E^\circ_{\mathrm{Fe/Pu}}=0.77-0.98
\]
\[
E^\circ_{\mathrm{Fe/Pu}}=-0.21\,\mathrm{V}
\]
The negative value does not favour the proposed forward reaction under standard conditions.
\(\textbf{Compare the outcomes:}\)
The formal charge on both Ce(IV) and Pu(IV) is +4, but this does not give their redox couples equal reduction potentials.
The different nuclear and electronic structures contribute to different electron-accepting tendencies.
Similarly, the +3 charge of Pu(III) does not by itself establish whether another ion can oxidise it.
The reaction direction follows from the relative potentials of the specified couples.
\(\textbf{Final answer:}\) Ce(IV), but not Fe(III), can oxidise Pu(III) under the stated standard conditions; the plutonium configuration changes from \(5f^5\) to \(5f^4\).
202. Trivalent and tetravalent actinoid ions can undergo hydrolysis in aqueous solution. Consider an actinoid M that forms both \(\mathrm{M^{3+}}\) and \(\mathrm{M^{4+}}\).
The following equilibria are proposed as simplified representations:
\[
\mathrm{M^{3+}}+\mathrm{H_2O}
\rightleftharpoons
\mathrm{MOH^{2+}}+\mathrm{H^+}
\]
\[
\mathrm{M^{4+}}+\mathrm{H_2O}
\rightleftharpoons
\mathrm{MOH^{3+}}+\mathrm{H^+}
\]
Assume that these are the only hydrolysis steps being considered.
Which statement correctly distinguishes hydrolysis from a redox transformation?
ⓐ. Both reactions reduce the metal by one oxidation-state unit because one hydrogen atom leaves the coordinated water.
ⓑ. Both reactions involve hydrolysis without changing the metal oxidation state; changing the acidity can influence the equilibria.
ⓒ. The first reaction is hydrolysis, but the second must involve oxidation because the metal initially has charge +4.
ⓓ. Both reactions oxidise the metal by one unit because the product contains one additional oxygen-containing group.
Correct Answer: Both reactions involve hydrolysis without changing the metal oxidation state; changing the acidity can influence the equilibria.
Explanation: Hydrolysis of a metal ion involves interaction with water and can produce a metal-bound hydroxide group together with a hydrogen ion. In the first proposed reaction, the starting metal ion has charge +3. The hydroxide group carries an oxidation-number contribution of \(-1\), and the hydrolysed complex has charge +2. The metal therefore retains oxidation state +3. In the second reaction, the starting ion has charge +4. The product \(\mathrm{MOH^{3+}}\) also contains the metal in oxidation state +4. Consequently, neither hydrolysis reaction requires electron transfer at the metal centre. These processes differ from oxidation, in which the metal oxidation state increases, and reduction, in which it decreases. The release of hydrogen ions also makes the equilibria sensitive to solution acidity. Increasing hydrogen-ion concentration generally opposes the forward hydrolysis processes under the simplified conditions. Conversely, conditions that remove hydrogen ions can favour hydrolysis. The distinction is particularly important for actinoid chemistry because the +3 and +4 oxidation states both occur and their ions can undergo hydrolysis. A change in coordination or aqueous speciation must not automatically be treated as evidence of an oxidation-state change. Identifying the metal oxidation state in each species resolves the misconception.
203. The ionisation enthalpies of early actinoids are generally lower than those of the corresponding early lanthanoids, although exact actinoid values are difficult to establish experimentally.
Consider the following Assertion and Reason.
Assertion (A): The greater availability of some actinoid valence electrons for chemical bonding is compatible with comparatively low ionisation enthalpies among early actinoids.
Reason (R): Early actinoid 5f orbitals penetrate less deeply into the inner electronic core than corresponding lanthanoid 4f orbitals, making the 5f electrons relatively more shielded from nuclear attraction.
Choose the correct option.
ⓐ. Both A and R are true, but R does not explain A.
ⓑ. A is true, but R is false.
ⓒ. A is false, but R is true.
ⓓ. Both A and R are true, and R correctly explains A.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: The Assertion is true because the outer electrons of early actinoids can be relatively available for bonding. This contributes to their characteristic chemical behaviour and the accessibility of multiple oxidation states. The Reason is also true because the 5f orbitals of early actinoids differ from the 4f orbitals of lanthanoids in their penetration towards the inner core. The 5f electrons penetrate less deeply and are consequently more effectively shielded from nuclear attraction. This can make electron removal comparatively easier in early actinoids. NCERT connects this difference with the observation that their ionisation enthalpies are lower than those of the corresponding early lanthanoids. The comparison does not imply that all individual actinoid ionisation enthalpies are accurately known. Nor does it establish that every actinoid has a lower ionisation enthalpy than every lanthanoid. The statement concerns the characteristic comparison between the early members of the two series. The reduced effective attraction also helps rationalise the greater availability of actinoid valence electrons for bonding. This reasoning is consistent with the broader observation that the 5f, 6d and 7s orbitals are relatively close in energy. The comparable orbital energies and the less deeply buried 5f electrons contribute to the oxidation-state variability of actinoids. Thus, the Reason provides an electronic explanation for the Assertion rather than merely repeating the observed trend.
204. A series of actinoid compounds is examined to compare the stability of two oxidation states, +3 and +4. The observations are:
| Case | Observation |
|---|
| P | An actinoid(III) ion is converted into actinoid(IV) by a suitable oxidising agent. |
| Q | The corresponding actinoid(IV) species is reduced back to actinoid(III) by a sufficiently strong reducing agent. |
| R | A solid compound stabilises actinoid(IV), whereas the trivalent state is favoured in a different aqueous environment. |
Which conclusion correctly reconciles all three observations?
ⓐ. The +4 oxidation state is intrinsically impossible because +3 is common among actinoids.
ⓑ. The +3 state must be thermodynamically more stable than +4 in every compound and solvent.
ⓒ. Both +3 and +4 can be accessible, while their relative stability depends on the chemical environment and the available redox partners.
ⓓ. The oxidation state of the metal remains unchanged in P and Q because the element itself is unchanged.
Correct Answer: Both +3 and +4 can be accessible, while their relative stability depends on the chemical environment and the available redox partners.
Explanation: The +3 oxidation state is important throughout actinoid chemistry, but it is not necessarily the only accessible state. Many actinoids can exhibit +4 in appropriate chemical species. In Case P, oxidation increases the formal metal oxidation state from +3 to +4. The actinoid centre consequently loses one electron. In Case Q, reduction reverses this change and restores the +3 state. These two observations demonstrate that different redox partners can favour opposite electron-transfer directions. Case R introduces a further distinction between chemical environments. Stabilisation of +4 in a solid compound does not require free tetravalent aqueous ions to be favoured under every condition. The bonding environment, surrounding ligands and solvent can contribute differently to the free energies of the oxidation states. Consequently, an oxidation state may be accessible or stable in one chemical form without being predominant in another. The presence of 5f electrons is relevant, but f-electron occupancy alone cannot determine every redox equilibrium. The three observations are therefore compatible rather than contradictory. A correct interpretation must distinguish an accessible oxidation state, the most stable state under specified conditions and the direction of a particular redox reaction.
205. The following ions are considered in their formal electronic descriptions:
\[
\mathrm{Pa^{4+}}:[Rn]\,5f^1
\]
\[
\mathrm{U^{5+}}:[Rn]\,5f^1
\]
\[
\mathrm{Np^{6+}}:[Rn]\,5f^1
\]
Each ion is converted separately into the next higher oxidation state by removal of one electron.
Assume that the resulting metal centres retain the radon core and that the electron is removed from the 5f subshell.
Which statement is correct?
ⓐ. Each \(5f^1\) starting ion forms \(5f^0\) upon oxidation, but these changes need not have identical redox potentials.
ⓑ. The three starting ions possess different 5f-electron counts because their oxidation states differ.
ⓒ. Only Pa4+ can form a \(5f^0\) product because protactinium has the smallest nuclear charge.
ⓓ. All three conversions must have identical standard reduction potentials because each involves one electron.
Correct Answer: Each \(5f^1\) starting ion forms \(5f^0\) upon oxidation, but these changes need not have identical redox potentials.
Explanation: \(\textbf{Identify the common starting configuration:}\)
Each ion is explicitly represented by the configuration
\[
[Rn]\,5f^1
\]
Therefore, each contains one electron in the 5f subshell.
However, their nuclear charges and oxidation states differ.
\(\textbf{Track the specified electron removal:}\)
Each ion loses one electron.
\[
\mathrm{Pa^{4+}}\longrightarrow\mathrm{Pa^{5+}}+e^-
\]
\[
\mathrm{U^{5+}}\longrightarrow\mathrm{U^{6+}}+e^-
\]
\[
\mathrm{Np^{6+}}\longrightarrow\mathrm{Np^{7+}}+e^-
\]
Under the stated assumptions, the common electronic change is
\[
5f^1\longrightarrow5f^0
\]
\(\textbf{Compare final oxidation states:}\)
Protactinium reaches +5.
Uranium reaches +6.
Neptunium reaches +7.
These are representative high oxidation states accessible to the corresponding early actinoids.
\(\textbf{Distinguish electron count from redox tendency:}\)
Each transformation removes one electron and produces an empty formal 5f subshell.
Nevertheless, the complete thermodynamic cost of electron removal need not be identical.
The ions have different nuclear charges, initial charges and chemical environments.
Their corresponding standard reduction potentials therefore cannot be determined from the shared \(5f^1\rightarrow5f^0\) change alone.
\(\textbf{Chemical interpretation:}\)
The comparison links the early-actinoid oxidation-state sequence with a common formal electronic endpoint.
It also illustrates why favourable electronic configurations do not by themselves provide a complete ranking of oxidising or reducing powers.
The actual redox tendencies require information about the specified couples and their environments.
206. A neutral actinoid atom is described by the configuration \([Rn]\,5f^76d^17s^2\). A hypothetical change in the relative orbital energies makes the configuration \([Rn]\,5f^87s^2\) energetically preferable, without changing the atomic number.
Consider the following statements.
Statement I: Both configurations contain the same total number of electrons beyond the radon core.
Statement II: The original configuration retains a half-filled 5f subshell, whereas the hypothetical configuration contains a paired electron in one 5f orbital.
Statement III: The hypothetical change necessarily increases the atomic number by one.
Statement IV: A change in the relative energies of 5f and 6d orbitals can change their occupancies without changing the identity of the element.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. I, II and IV only
ⓒ. I, III and IV only
ⓓ. II, III and IV only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because the first configuration contains seven 5f electrons, one 6d electron and two 7s electrons. Its total electron count beyond the radon core is therefore ten. The hypothetical configuration contains eight 5f electrons and two 7s electrons, again giving ten electrons beyond the radon core. Statement II is correct because \(5f^7\) represents a half-filled f subshell. The seven electrons can occupy seven different f orbitals with parallel spins under Hund's rule. In the hypothetical \(5f^8\) arrangement, the eighth electron must pair with an electron in one of those orbitals. Statement III is incorrect because no electron has been added to the atom. One electron has simply been redistributed from a 6d orbital into a 5f orbital. The total electron count and nuclear charge remain unchanged. Statement IV is correct because the relative energies of the 5f and 6d orbitals influence how electrons are distributed among them. The actual ground-state configuration results from the energetic balance between these possible occupations. The half-filled \(5f^7\) arrangement provides an important stabilising contribution to the original configuration. However, the question explicitly changes the relative orbital energies, so a different distribution is allowed as a hypothetical outcome. The comparison illustrates why the close energies of 5f and 6d orbitals produce electronic-configuration irregularities among actinoids. Such irregularities cannot be explained by a rigid rule that electrons must always enter only the 5f subshell.
207. A comparison is made between two hypothetical f-block systems having the same number of valence electrons and the same formal oxidation state.
System P has its partially occupied f orbitals relatively deeply buried beneath outer electronic shells.
System Q has its corresponding f orbitals less deeply buried and closer in energy to accessible d orbitals.
Which prediction is most consistent with the characteristic differences between lanthanoids and actinoids?
ⓐ. System P should necessarily exhibit a wider range of oxidation states because buried f electrons always form stronger chemical bonds.
ⓑ. Both systems must exhibit identical oxidation-state behaviour because their formal charges and valence-electron counts are equal.
ⓒ. System Q should necessarily possess only one oxidation state because close f and d orbital energies prevent electron participation in bonding.
ⓓ. System Q may show greater f-electron participation in bonding and a wider range of accessible oxidation states, although the exact states depend on the chemical environment.
Correct Answer: System Q may show greater f-electron participation in bonding and a wider range of accessible oxidation states, although the exact states depend on the chemical environment.
Explanation: The first distinction concerns the spatial availability of f electrons for bonding. In System P, the f orbitals are relatively deeply buried beneath outer electronic shells. This reduces their direct participation in ordinary chemical bonding. Such behaviour resembles the relatively shielded 4f electrons of lanthanoids. In System Q, the f orbitals are less deeply buried and can interact more strongly with the surrounding chemical environment. Their proximity in energy to accessible d orbitals also allows a greater variety of valence-electron distributions. These features resemble important aspects of actinoid electronic structure. Greater orbital participation can help stabilise several different oxidation states. Nevertheless, the exact oxidation states accessible to a particular element still depend on the relative free energies of the compounds formed. Therefore, one cannot infer a universal oxidation-state sequence solely from orbital depth or the number of valence electrons. Equal formal charge does not require identical chemical behaviour when the orbital energies and bonding contributions differ. The question thus requires connecting orbital location, relative orbital energies and oxidation-state variability. The most defensible prediction is that System Q may exhibit greater chemical flexibility, not that any particular oxidation state is guaranteed.
208. Consider the following first-row transition-metal data:
| Metal | Enthalpy of atomisation (\(\mathrm{kJ\,mol^{-1}}\)) | First ionisation enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| V | 515 | 650 |
| Cr | 398 | 653 |
| Mn | 279 | 717 |
A student claims that the metal with the greatest first ionisation enthalpy must also have the strongest metallic cohesion.
Which statement correctly evaluates this prediction?
ⓐ. The prediction is correct because manganese has the largest first ionisation enthalpy and therefore the largest atomisation enthalpy.
ⓑ. The prediction is correct because metallic cohesion depends only on the energy required to remove one electron from a gaseous atom.
ⓒ. The prediction is incorrect because manganese has the greatest first ionisation enthalpy among the three, whereas vanadium has the greatest atomisation enthalpy.
ⓓ. The prediction is incorrect because all three metals possess identical first ionisation enthalpies despite different atomisation enthalpies.
Correct Answer: The prediction is incorrect because manganese has the greatest first ionisation enthalpy among the three, whereas vanadium has the greatest atomisation enthalpy.
Explanation: First ionisation enthalpy measures the energy needed to remove one electron from an isolated gaseous atom. Enthalpy of atomisation measures the energy needed to separate the atoms of one mole of metallic solid into isolated gaseous atoms. These properties describe different physical processes. The supplied data show that manganese has the greatest first ionisation enthalpy among the three metals. Its value is \(717\,\mathrm{kJ\,mol^{-1}}\), compared with \(653\,\mathrm{kJ\,mol^{-1}}\) for chromium and \(650\,\mathrm{kJ\,mol^{-1}}\) for vanadium. However, manganese has the smallest atomisation enthalpy of the three. Vanadium has the largest supplied atomisation enthalpy, \(515\,\mathrm{kJ\,mol^{-1}}\). Thus, the element requiring the greatest energy to remove its first electron does not possess the greatest metallic cohesive energy. First ionisation depends strongly on the electronic configuration of the isolated atom and the resulting ion. Metallic cohesion depends on interactions among atoms in the solid, including the participation of valence electrons in metallic bonding. The half-filled \(3d^5\) configuration of manganese contributes to its electronic behaviour but does not guarantee exceptionally strong metallic bonding. The data therefore demonstrate why different energetic properties of transition metals cannot be ranked using one simple electron-counting rule. Their trends share electronic influences but need not be identical.
209. Two first-row transition-metal samples are compared under the following conditions:
Sample P consists of isolated gaseous chromium atoms in their ground states.
Sample Q consists of chromium atoms incorporated into a bulk metallic chromium sample.
A researcher proposes that the six unpaired electrons of an isolated gaseous chromium atom must remain six independent, localised unpaired electrons on every chromium atom in the metallic solid.
Which conclusion is justified?
ⓐ. The proposal is necessarily correct because metallic bonding does not alter the electronic description of constituent atoms.
ⓑ. The proposal is unjustified: electronic states in a metal reflect delocalisation and interactions absent from an isolated-atom orbital diagram.
ⓒ. The proposal is necessarily correct because chromium is a transition metal and transition metals do not form energy bands.
ⓓ. The proposal is necessarily false because metallic chromium must contain only \(3d^{10}\) closed-shell atoms.
Correct Answer: The proposal is unjustified: electronic states in a metal reflect delocalisation and interactions absent from an isolated-atom orbital diagram.
Explanation: An isolated gaseous chromium atom has the ground-state configuration \([Ar]\,3d^54s^1\). In a simple isolated-atom orbital description, its five 3d electrons occupy separate orbitals and the 4s electron is also unpaired. This gives six unpaired electrons for the separated ground-state atom. However, metallic chromium is not merely an assembly of electronically independent gaseous chromium atoms. Metallic bonding involves interactions among the valence orbitals of neighbouring atoms. The resulting electronic states extend through the solid and differ from the discrete atomic orbitals used to describe isolated gaseous atoms. Consequently, an isolated-atom unpaired-electron count cannot be transferred directly into an exact description of the magnetic behaviour of the bulk metal. This does not mean that the original chromium atom must somehow become a closed-shell \(3d^{10}\) species. Instead, the relevant electronic description changes because the atoms are interacting in a metallic lattice. The distinction is important when interpreting metallic properties such as cohesion, electrical conductivity and magnetic behaviour. Isolated-atom configurations help explain general tendencies, but they do not independently determine every property of a metallic solid. Therefore, the proposal overextends a useful atomic model beyond the conditions in which it directly applies.
210. The following first-row transition-metal data are supplied:
| Metal | Melting point (K) | Enthalpy of atomisation (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Fe | 1811 | 418 |
| Co | 1768 | 427 |
| Ni | 1728 | 431 |
Which conclusion is best supported by these data?
ⓐ. From Fe to Ni, both melting point and atomisation enthalpy decrease, so the two measurements follow an identical trend.
ⓑ. From Fe to Ni, both properties increase, so nickel has the greatest melting point and the greatest atomisation enthalpy.
ⓒ. From Fe to Ni, both properties stay effectively constant, so the three metals must have identical metallic bonding.
ⓓ. From Fe to Ni, atomisation enthalpy rises while melting point falls, so the two measured properties need not show the same trend.
Correct Answer: From Fe to Ni, atomisation enthalpy rises while melting point falls, so the two measured properties need not show the same trend.
Explanation: The melting point is the temperature at which a solid and its liquid phase coexist in equilibrium at a specified pressure. Enthalpy of atomisation is the energy required to convert one mole of metallic solid into isolated gaseous atoms. Both properties are influenced by bonding within the metal, but they are not identical thermodynamic quantities. The supplied melting points decrease in the sequence iron, cobalt and nickel. Iron melts at \(1811\,\mathrm{K}\), cobalt at \(1768\,\mathrm{K}\), and nickel at \(1728\,\mathrm{K}\). The enthalpies of atomisation show the opposite direction across these three data points. Iron has \(418\,\mathrm{kJ\,mol^{-1}}\), cobalt has \(427\,\mathrm{kJ\,mol^{-1}}\), and nickel has \(431\,\mathrm{kJ\,mol^{-1}}\). Therefore, nickel has the greatest supplied atomisation enthalpy but the lowest melting point among the three. The observations are not contradictory because melting involves conversion from solid to liquid, whereas atomisation involves separation into gaseous atoms. The complete energetic and structural changes involved are different. Close atomisation enthalpies also do not establish identical metallic bonding or crystal structures. The data thus warn against assuming that a single metallic property provides an exact ranking of every other property. A correct interpretation must use the measured quantities relevant to the process being considered.
211. The first and second ionisation enthalpies of selected first-row transition metals are:
| Element | \(IE_1\) (\(\mathrm{kJ\,mol^{-1}}\)) | \(IE_2\) (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Cr | 653 | 1592 |
| Mn | 717 | 1509 |
| Fe | 762 | 1561 |
An energy comparison concerns conversion of isolated gaseous neutral atoms into isolated gaseous divalent ions.
Which conclusion correctly identifies the element requiring the greatest total ionisation energy per mole for this conversion and relates the result to the appropriate electronic configurations?
ⓐ. Chromium requires the greatest total energy because its first ionisation enthalpy is the smallest.
ⓑ. Manganese requires the greatest total energy because formation of Mn2+ destroys a half-filled \(3d^5\) subshell.
ⓒ. Iron has the greatest combined first-and-second ionisation enthalpy; the contrasting Cr and Mn values reflect \(3d^5\) stability effects.
ⓓ. All three require the same total energy because each neutral atom loses two electrons.
Correct Answer: Iron has the greatest combined first-and-second ionisation enthalpy; the contrasting Cr and Mn values reflect \(3d^5\) stability effects.
Explanation: \(\textbf{Identify the required energetic process:}\)
The conversion is
\[
\mathrm{M(g)}\longrightarrow\mathrm{M^{2+}(g)}+2e^-
\]
Two successive electrons are removed from each isolated gaseous neutral atom.
Therefore, both the first and second ionisation enthalpies must be included.
\(\textbf{Governing relation:}\)
\[
\Delta H=IE_1+IE_2
\]
\(\textbf{Calculate the chromium total:}\)
\[
\Delta H_{\mathrm{Cr}}=653+1592
\]
\[
\Delta H_{\mathrm{Cr}}=2245\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the manganese total:}\)
\[
\Delta H_{\mathrm{Mn}}=717+1509
\]
\[
\Delta H_{\mathrm{Mn}}=2226\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the iron total:}\)
\[
\Delta H_{\mathrm{Fe}}=762+1561
\]
\[
\Delta H_{\mathrm{Fe}}=2323\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the calculated energies:}\)
\[
2323\gt 2245\gt 2226
\]
Iron therefore requires the greatest total ionisation energy for the stated gaseous conversion.
\(\textbf{Interpret chromium's electronic behaviour:}\)
Neutral chromium has the configuration \(3d^54s^1\).
Its first ionisation removes the 4s electron, producing a \(3d^5\) ion.
The second ionisation disrupts that half-filled arrangement.
This contributes to chromium's relatively high second ionisation enthalpy.
\(\textbf{Interpret manganese's electronic behaviour:}\)
Neutral manganese has the configuration \(3d^54s^2\).
Removal of two 4s electrons produces Mn2+ with the favourable \(3d^5\) configuration.
Thus, its second ionisation does not remove a d electron from the half-filled subshell.
\(\textbf{Evaluate the final ranking:}\)
These electronic effects help explain differences between chromium and manganese.
However, the greatest total energy in the supplied data belongs to iron.
The correct ranking must therefore be established from the actual sum rather than from one stabilisation argument alone.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{Fe}}=2323\,\mathrm{kJ\,mol^{-1}}}
\]
212. A hypothetical first-row transition-metal atom M has the ground-state configuration \([Ar]\,3d^54s^2\).
The first two ionisations remove the two 4s electrons. A third ionisation then removes one 3d electron.
Consider the following statements.
Statement I: The divalent ion possesses a half-filled d subshell.
Statement II: The third ionisation converts a \(d^5\) ion into a \(d^4\) ion.
Statement III: The third ionisation is expected to be particularly easy because it creates a completely filled d subshell.
Statement IV: The electronic stability of the \(d^5\) divalent ion contributes to the relatively high energy needed for the third electron removal.
Which combination is correct?
ⓐ. I, II and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: The neutral configuration \([Ar]\,3d^54s^2\) corresponds to manganese. Formation of the divalent ion removes the two 4s electrons. The resulting ion has the configuration \([Ar]\,3d^5\). Statement I is therefore correct because the five d electrons occupy separate orbitals in the usual high-spin free-ion description. Statement II is also correct because the third ionisation removes one electron from the \(3d^5\) subshell. This produces the \(3d^4\) configuration of manganese(III). Statement III is incorrect because the third ionisation does not create a completely filled d subshell. Instead, it disrupts the favourable half-filled \(d^5\) arrangement. Statement IV is correct because half-filled subshell stability contributes to the energetic resistance to removing a d electron from Mn2+. Nevertheless, this electronic contribution is not the only factor affecting the numerical ionisation enthalpy. Nuclear attraction, shielding and the energies of the initial and final gaseous ions must also be considered. The important distinction is between removal of 4s electrons during formation of Mn2+ and subsequent removal of a 3d electron during formation of Mn3+. These two stages have different electronic consequences. Therefore, the configuration sequence provides a sound explanation of the third-ionisation behaviour without implying that electron removal is governed by half-filled stability alone.
213. Consider three first-row transition-metal ions in the gaseous state:
\[
\mathrm{Cr^{2+}}:3d^4,\qquad
\mathrm{Mn^{2+}}:3d^5,\qquad
\mathrm{Fe^{2+}}:3d^6
\]
Each ion loses one electron to form its trivalent ion.
Assume the usual free-ion high-spin configurations.
Which statement correctly compares the changes in unpaired-electron counts?
ⓐ. All three ions lose one unpaired electron because all three undergo oxidation.
ⓑ. Chromium(II) gains one unpaired electron, manganese(II) loses one, and iron(II) gains one.
ⓒ. Chromium(II) loses one unpaired electron, manganese(II) gains one, and iron(II) loses one.
ⓓ. Chromium(II) loses one unpaired electron, manganese(II) loses one, and iron(II) gains one.
Correct Answer: Chromium(II) loses one unpaired electron, manganese(II) loses one, and iron(II) gains one.
Explanation: \(\textbf{Chromium(II) oxidation:}\)
Chromium(II) has the \(3d^4\) configuration.
Its four d electrons occupy four separate orbitals.
Therefore, Cr2+ has four unpaired electrons.
Oxidation removes one d electron.
\[
3d^4\longrightarrow3d^3
\]
The resulting Cr3+ has three unpaired electrons.
Thus, the number of unpaired electrons decreases by one.
\(\textbf{Manganese(II) oxidation:}\)
Manganese(II) possesses the half-filled \(3d^5\) configuration.
All five d electrons are unpaired in the free-ion high-spin description.
Oxidation removes one electron.
\[
3d^5\longrightarrow3d^4
\]
The resulting Mn3+ has four unpaired electrons.
Again, the unpaired-electron count decreases by one.
\(\textbf{Iron(II) oxidation:}\)
Iron(II) possesses the \(3d^6\) configuration.
Five electrons occupy separate d orbitals, while the sixth is paired with one of them.
Therefore, Fe2+ has four unpaired electrons.
Oxidation removes one electron and produces Fe3+.
\[
3d^6\longrightarrow3d^5
\]
The resulting ion has five unpaired electrons.
Thus, the number of unpaired electrons increases by one.
\(\textbf{Interpret the contrasting outcomes:}\)
All three ions undergo oxidation by losing one electron.
However, the change in unpaired-electron count depends on whether the removed electron is unpaired or belongs to an occupied pair.
Chromium(II) and manganese(II) lie at or below the half-filled configuration, whereas iron(II) lies beyond it.
Consequently, identical one-electron oxidation processes can produce opposite changes in the spin-only magnetic contribution.
The result must be inferred from the electronic configurations rather than from the oxidation direction alone.
214. Two first-row transition-metal ions, P2+ and Q3+, have the electronic configurations:
\[
\mathrm{P^{2+}}:[Ar]\,3d^6
\]
\[
\mathrm{Q^{3+}}:[Ar]\,3d^6
\]
Their respective neutral atoms belong to the first transition series.
Which statement correctly identifies the two elements and predicts the effect of reducing Q3+ to Q2+ on its free-ion spin-only magnetic moment?
ⓐ. P is iron and Q is cobalt; reduction of Q3+ to Q2+ decreases the number of unpaired electrons from four to three.
ⓑ. P is manganese and Q is iron; reduction of Q3+ to Q2+ decreases the number of unpaired electrons from four to three.
ⓒ. P is iron and Q is cobalt; reduction of Q3+ to Q2+ increases the number of unpaired electrons from four to five.
ⓓ. P is cobalt and Q is nickel; reduction of Q3+ to Q2+ increases the number of unpaired electrons from three to four.
Correct Answer: P is iron and Q is cobalt; reduction of Q3+ to Q2+ decreases the number of unpaired electrons from four to three.
Explanation: \(\textbf{Identify P from the divalent configuration:}\)
The ion P2+ contains eighteen electrons in the argon core and six electrons in the 3d subshell.
Therefore, its total electron count is
\[
N_e(\mathrm{P^{2+}})=18+6
\]
\[
N_e(\mathrm{P^{2+}})=24
\]
The neutral atom has two additional electrons.
\[
Z_P=24+2=26
\]
Atomic number 26 corresponds to iron.
\(\textbf{Identify Q from the trivalent configuration:}\)
The ion Q3+ also contains 24 electrons.
Its neutral atom therefore has atomic number
\[
Z_Q=24+3
\]
\[
Z_Q=27
\]
Atomic number 27 corresponds to cobalt.
\(\textbf{Determine the reduction product:}\)
Cobalt(III) has the free-ion configuration \(3d^6\).
Reduction adds one electron.
\[
\mathrm{Co^{3+}}+e^-\longrightarrow\mathrm{Co^{2+}}
\]
\[
3d^6\longrightarrow3d^7
\]
\(\textbf{Compare unpaired-electron counts:}\)
Under Hund's rule for free ions, a \(d^6\) configuration contains four unpaired electrons.
A \(d^7\) configuration contains three unpaired electrons.
Thus, the number of unpaired electrons decreases during reduction.
\(\textbf{Calculate the initial spin-only moment:}\)
\[
\mu=\sqrt{n(n+2)}\,\mathrm{BM}
\]
\[
\mu_{\mathrm{Co^{3+}}}=\sqrt{4(4+2)}
\]
\[
\mu_{\mathrm{Co^{3+}}}=\sqrt{24}\approx4.90\,\mathrm{BM}
\]
\(\textbf{Calculate the final spin-only moment:}\)
\[
\mu_{\mathrm{Co^{2+}}}=\sqrt{3(3+2)}
\]
\[
\mu_{\mathrm{Co^{2+}}}=\sqrt{15}\approx3.87\,\mathrm{BM}
\]
\(\textbf{Interpret the result:}\)
Equal d-electron configurations do not imply that two ions belong to the same element.
The ionic charges must also be used to recover the atomic numbers.
Furthermore, reduction can decrease the number of unpaired electrons when an additional electron enters an already more-than-half-filled d subshell.
\(\textbf{Final answer:}\) P is iron, Q is cobalt, and reduction of Q3+ decreases its free-ion spin-only magnetic moment from approximately \(4.90\,\mathrm{BM}\) to \(3.87\,\mathrm{BM}\).
215. An imaginary first-row transition-metal ion M2+ is placed separately in two chemical environments.
In Environment P, it remains as a hydrated divalent ion.
In Environment Q, a suitable ligand forms a particularly stable complex with M3+ but interacts much less favourably with M2+.
The electronic configurations of the free ions are \(M^{2+}:3d^6\) and \(M^{3+}:3d^5\).
Which conclusion is most defensible?
ⓐ. Environment Q must favour M2+ because conversion to M3+ reduces the total number of d electrons.
ⓑ. The \(d^5\) configuration alone proves that M3+ is the predominant species in every conceivable chemical environment.
ⓒ. The metal must remain in oxidation state +2 in both environments because a change in ligand cannot influence oxidation-state stability.
ⓓ. A ligand that stabilises M(III) more strongly can shift the +2/+3 balance despite M(III)'s \(d^5\) configuration.
Correct Answer: A ligand that stabilises M(III) more strongly can shift the +2/+3 balance despite M(III)'s \(d^5\) configuration.
Explanation: The supplied configurations identify a divalent ion with six d electrons and a corresponding trivalent ion with five d electrons. Iron provides a familiar first-row example of this relationship. Oxidation from M2+ to M3+ removes one d electron and produces the half-filled \(d^5\) configuration. This configuration can provide an important electronic stabilising contribution. However, the stability of an oxidation state is not determined solely by the isolated-ion configuration. The surrounding chemical environment influences the energies of the species involved. In Environment P, the divalent ion is stabilised by its hydration interactions. In Environment Q, the ligand is specified to stabilise M3+ particularly strongly. This additional stabilisation can make formation of the trivalent state relatively more favourable. Therefore, changing the ligand environment can alter the thermodynamic preference between oxidation states. The fact that oxidation removes an electron does not require the product to be less stable in every chemical situation. Conversely, formation of a half-filled d subshell does not guarantee that the trivalent state predominates universally. The comparison requires considering both the electronic configuration and the differential stabilisation supplied by the ligand. This is an important reason why transition-metal oxidation-state behaviour can vary between free aqueous ions and coordinated species.
216. Consider a proposed catalytic cycle for the oxidation of sulphur dioxide to sulphur trioxide in the Contact process.
Step P:
\[
\mathrm{V_2O_5}+\mathrm{SO_2}
\longrightarrow
\mathrm{V_2O_4}+\mathrm{SO_3}
\]
Step Q:
\[
\mathrm{V_2O_4}+\frac12\mathrm{O_2}
\longrightarrow
\mathrm{V_2O_5}
\]
Which statement correctly identifies the oxidation-state changes and the overall role of the catalyst?
ⓐ. Vanadium changes from +5 to +4 in P and from +4 to +5 in Q; the net reaction is oxidation of SO2 by O2, with V2O5 regenerated.
ⓑ. Vanadium changes from +5 to +6 in P and from +6 to +5 in Q; V2O5 is permanently consumed.
ⓒ. Vanadium remains +5 throughout both stages, while SO2 is reduced from sulphur(VI) to sulphur(IV).
ⓓ. Vanadium changes from +5 to +4 in both stages; the catalyst becomes progressively reduced with each cycle.
Correct Answer: Vanadium changes from +5 to +4 in P and from +4 to +5 in Q; the net reaction is oxidation of SO2 by O2, with V2O5 regenerated.
Explanation: \(\textbf{Identify vanadium in the initial catalyst:}\)
In \(\mathrm{V_2O_5}\), oxygen contributes an oxidation number of \(-10\).
The two vanadium atoms consequently contribute +10.
Each vanadium centre therefore has oxidation state +5.
\(\textbf{Track vanadium during Step P:}\)
In \(\mathrm{V_2O_4}\), four oxygen atoms contribute \(-8\).
Each vanadium centre has oxidation state +4.
Thus, vanadium is reduced from +5 to +4.
\[
\mathrm{V(V)}+e^-\longrightarrow\mathrm{V(IV)}
\]
Two vanadium centres collectively accept two electrons.
\(\textbf{Track sulphur during Step P:}\)
Sulphur has oxidation state +4 in \(\mathrm{SO_2}\).
It has oxidation state +6 in \(\mathrm{SO_3}\).
Therefore, sulphur loses two electrons.
The electron loss by sulphur matches the total electron gain by vanadium.
\(\textbf{Regeneration during Step Q:}\)
Oxygen restores \(\mathrm{V_2O_4}\) to \(\mathrm{V_2O_5}\).
Vanadium returns from oxidation state +4 to +5.
Thus, the catalyst is regenerated.
\(\textbf{Obtain the overall reaction:}\)
Adding both steps gives
\[
\mathrm{V_2O_5}+\mathrm{SO_2}+\mathrm{V_2O_4}+\frac12\mathrm{O_2}
\longrightarrow
\mathrm{V_2O_4}+\mathrm{SO_3}+\mathrm{V_2O_5}
\]
The vanadium oxides appear on both sides and cancel.
Therefore, the net reaction is
\[
\mathrm{SO_2}+\frac12\mathrm{O_2}\longrightarrow\mathrm{SO_3}
\]
\(\textbf{Chemical interpretation:}\)
The proposed cycle illustrates how variable oxidation states can enable a transition-metal compound to transfer electrons between reactants.
The catalyst changes its oxidation state during intermediate stages but returns to its initial state after completing the cycle.
Consequently, an unchanged catalyst in the overall equation need not retain one oxidation state throughout every individual stage.
217. An aqueous solution contains an iron-containing species with formal \(3d^5\) occupancy. Two observations are made:
Observation P: Its measured magnetic behaviour is consistent with five unpaired electrons.
Observation Q: After adding a suitable ligand, the iron oxidation state remains unchanged, but the observed magnetic behaviour changes.
Which explanation best reconciles the observations without assuming a redox reaction?
ⓐ. Observation Q is impossible because the number of unpaired electrons depends only on the oxidation state.
ⓑ. The new ligand must reduce iron from +3 to +2 even if charge analysis shows no oxidation-state change.
ⓒ. Ligand changes can alter electron pairing and magnetism in iron complexes without changing the metal's oxidation state.
ⓓ. Observation Q proves that the total number of d electrons has increased from five to ten.
Correct Answer: Ligand changes can alter electron pairing and magnetism in iron complexes without changing the metal's oxidation state.
Explanation: An iron-containing species with a formal \(d^5\) configuration is consistent with iron(III). In a high-spin electronic arrangement, five d electrons occupy separate orbitals, giving five unpaired electrons. This accounts for Observation P. However, the electronic behaviour of a coordinated metal ion can depend on the nature of the surrounding ligands. A change in coordination environment can modify the relative energies of the metal's d orbitals. Under suitable circumstances, this can influence the pairing of electrons. Therefore, two complexes having the same metal oxidation state and the same formal d-electron count need not exhibit identical magnetic behaviour. The change in Observation Q does not independently establish that iron has undergone reduction or oxidation. Formal oxidation state must be determined from the chemical composition and charge of the species. Similarly, the total d-electron count cannot be inferred to have increased merely because magnetic behaviour has changed. Different ligand environments can alter the distribution of the same five d electrons between available orbital levels. Therefore, the free-ion spin-only description cannot be applied unconditionally to every coordinated species. The important distinction is between oxidation-state changes, which involve formal electron transfer, and changes in electron pairing caused by the coordination environment.
218. Four aqueous first-row metal-ion samples are labelled P, Q, R and S. Their experimentally assigned metal oxidation states, formal d-electron counts and observed properties are:
| Sample | Oxidation state | d-electron count | Observation |
|---|
| P | +3 | \(d^1\) | Purple and paramagnetic |
| Q | +2 | \(d^8\) | Green and paramagnetic |
| R | +2 | \(d^{10}\) | Colourless and diamagnetic |
| S | +2 | \(d^9\) | Blue and paramagnetic |
Assume these are the characteristic simple aqueous ions described in NCERT.
Which assignment correctly identifies all four metal ions and explains the colourlessness of R?
ⓐ. P = V3+, Q = Co2+, R = Cu2+, S = Zn2+; R is colourless because it has one unpaired electron.
ⓑ. P = Ti3+, Q = Ni2+, R = Zn2+, S = Cu2+; R lacks an ordinary d–d transition because its d subshell is completely filled.
ⓒ. P = Cr3+, Q = Ni2+, R = Sc3+, S = Cu2+; R is colourless because it possesses an empty d subshell.
ⓓ. P = Ti3+, Q = Fe2+, R = Zn2+, S = Ni2+; R lacks an ordinary d–d transition because its d subshell is half-filled.
Correct Answer: P = Ti3+, Q = Ni2+, R = Zn2+, S = Cu2+; R lacks an ordinary d–d transition because its d subshell is completely filled.
Explanation: The identification requires matching oxidation state, d-electron count and characteristic aqueous colour rather than relying on colour alone. Titanium(III) possesses the configuration \(3d^1\) and is associated with a purple aqueous solution. Therefore, Sample P corresponds to Ti3+. Nickel(II) possesses the configuration \(3d^8\), and its usual aqueous ion is green and paramagnetic. Thus, Q corresponds to Ni2+. Zinc(II) has the completely filled \(3d^{10}\) configuration. All ten d electrons are paired, so Zn2+ is diamagnetic. A completely filled d subshell does not permit the ordinary metal-centred d–d excitation associated with partially occupied d orbitals. This accounts for the characteristic colourlessness of the simple aqueous Zn2+ ion, making R zinc(II). Copper(II) has the configuration \(3d^9\) and commonly produces blue aqueous solutions. It contains one unpaired d electron, so S corresponds to Cu2+. Chromium(III), although coloured, has a \(d^3\) configuration and cannot represent P. Iron(II) has a \(d^6\) configuration and cannot represent Q. This combination of observations illustrates the importance of electronic configuration as a more reliable identifying criterion than a single property. The colour assignments also depend on the stated aqueous environment; a metal ion's colour need not remain unchanged in every complex.
219. Consider two different cobalt(II) complexes containing the same metal oxidation state but different ligands:
\[
\mathrm{[Co(H_2O)_6]^{2+}}
\]
\[
\mathrm{[CoCl_4]^{2-}}
\]
Suppose both complexes are prepared separately and their visible absorption maxima are measured. Their characteristic colours are different.
Which statement is the most defensible interpretation of the observation?
ⓐ. One complex must contain cobalt(III) because two complexes of cobalt(II) cannot have different colours.
ⓑ. Both complexes must exhibit identical absorption maxima because the central metal has the same oxidation state.
ⓒ. The colour difference proves that chloride has oxidised cobalt(II) into cobalt(IV).
ⓓ. Different ligands and coordination environments can change the energies of electronic transitions without changing the cobalt oxidation state.
Correct Answer: Different ligands and coordination environments can change the energies of electronic transitions without changing the cobalt oxidation state.
Explanation: Both displayed complexes contain cobalt in oxidation state +2. In the aqua complex, water is a neutral ligand, so the cobalt oxidation state equals the overall complex charge of +2. In the chloride complex, four chloride ligands contribute a total charge of \(-4\). The overall complex charge is \(-2\), so cobalt must again possess oxidation state +2. Therefore, no difference in cobalt oxidation state is required to account for their different colours. The attached ligands influence the electronic environment of the metal centre. Their identities and arrangements affect the energies associated with electronic excitation. Consequently, the wavelengths of visible light absorbed by the two complexes can differ. A difference in absorbed wavelength can produce a difference in observed colour. The existence of different colours therefore does not prove that one complex has undergone oxidation or reduction. Nor does equal formal d-electron count guarantee identical optical properties. This illustrates the limitation of treating the colour of a transition-metal ion as a fixed property independent of its ligands. The question concerns the general relationship between coordination environment and optical behaviour and does not require detailed crystal-field calculations. A correct interpretation must distinguish ligand exchange from electron-transfer chemistry.
220. The ions Fe2+ and Mn2+ are compared in their usual high-spin aqueous environments. A student argues that because Mn2+ has one more unpaired electron than Fe2+, every compound of Mn(II) must necessarily absorb visible light more strongly than every compound of Fe(II).
Which response correctly evaluates the claim?
ⓐ. The claim is correct because visible absorption intensity is exactly proportional to the number of unpaired electrons.
ⓑ. The claim is correct only if the Mn(II) compound contains oxygen ligands.
ⓒ. The claim is unjustified: unpaired electrons determine spin magnetism, while absorption intensity depends on allowed electronic transitions.
ⓓ. The claim is unjustified because Mn2+ is diamagnetic despite having a half-filled d subshell.
Correct Answer: The claim is unjustified: unpaired electrons determine spin magnetism, while absorption intensity depends on allowed electronic transitions.
Explanation: Iron(II) has the formal \(3d^6\) configuration, whereas manganese(II) has the formal \(3d^5\) configuration. In the usual high-spin description, Fe2+ contains four unpaired electrons and Mn2+ contains five. This difference is relevant to their spin-only magnetic moments. However, the number of unpaired electrons does not directly determine the intensity of a visible absorption band. Colour depends on the energies of available electronic transitions and on how strongly the corresponding transitions absorb light. These characteristics are influenced by electronic structure and the surrounding chemical environment. Manganese(II) commonly forms pale-pink aqueous solutions despite possessing five unpaired d electrons. Iron(II) commonly forms pale-green aqueous solutions with four unpaired electrons. Thus, a larger unpaired-electron count does not guarantee greater colour intensity. The general idea of d–d excitation helps explain why many transition-metal ions are coloured, but it is not an exact quantitative rule relating colour intensity to spin-only magnetic moment. The student's reasoning incorrectly combines two different electronic properties. Magnetic behaviour is associated with uncompensated electron spins, while optical absorption concerns transitions between electronic energy levels. Therefore, the correct assessment must keep these concepts distinct.
