D- and F-Block Elements MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: The d-block elements occupy the central region of the periodic table, extending from Group 3 to Group 12. Their characteristic electronic filling involves the d subshell of the penultimate principal energy level. The general relationship is represented by \((n-1)d\), where \(n\) denotes the period number. For the fourth period, \(n=4\), so the relevant subshell is \(3d\). The first d-block series therefore extends from scandium (\(Z=21\)) to zinc (\(Z=30\)). The \(4d\) subshell is associated with the fifth-period d-block series, while \(5d\) characterises the sixth-period series. The \(4p\) subshell belongs to the p-block portion of the fourth period. This distinction between the period number and the principal quantum number of the filling d subshell is essential for interpreting transition-element configurations.

Explanation: A transition element possesses an incompletely filled d subshell in its atom or in at least one of its oxidation states. Scandium has the ground-state configuration \([Ar]\,3d^14s^2\), so its neutral atom contains an incompletely filled d subshell. When scandium forms \(\mathrm{Sc^{3+}}\), it loses two \(4s\) electrons and one \(3d\) electron, producing the configuration \([Ar]\). The resulting ion is \(3d^0\), but this does not invalidate scandium's transition-element classification because the neutral atom satisfies the defining condition. Zinc has the configuration \([Ar]\,3d^{10}4s^2\), and its common ion, \(\mathrm{Zn^{2+}}\), has the configuration \([Ar]\,3d^{10}\). Thus, the d subshell remains completely filled in both the zinc atom and its common oxidation state. Zinc is consequently a d-block element but is not classified as a transition element. Membership of the d-block and qualification as a transition element must therefore be evaluated separately.

Explanation: Copper has the exceptional neutral ground-state configuration \([Ar]\,3d^{10}4s^1\). On forming \(\mathrm{Cu^+}\), it first loses the \(4s\) electron and retains the completely filled \(3d^{10}\) subshell. Further oxidation to \(\mathrm{Cu^{2+}}\) removes one electron from \(3d\), giving \(3d^9\). This ion contains an incompletely filled d subshell and therefore establishes copper's transition-element character. Zinc has the neutral configuration \([Ar]\,3d^{10}4s^2\). The formation of its common ion, \(\mathrm{Zn^{2+}}\), removes both \(4s\) electrons without disturbing the filled \(3d^{10}\) subshell. The distinction is therefore not whether the neutral atom possesses ten d electrons, since both atoms do. It is whether an incompletely filled d subshell occurs in the atom or in a relevant oxidation state. Copper satisfies this condition through \(\mathrm{Cu^{2+}}\), whereas zinc does not satisfy it in its ordinary chemical states.

Explanation: The usual outer electronic configuration of d-block elements is expressed in terms of \((n-1)d\) and \(ns\) orbitals. However, this pattern has exceptions because the two subshells have relatively close energies. Palladium provides an important example, since its neutral ground-state configuration is \([Kr]\,4d^{10}5s^0\), rather than a configuration containing one or two \(5s\) electrons. Its position in the fifth-period d-block series is not altered by this exceptional occupancy. The transition-element definition must then be applied independently of the usual configuration pattern. Palladium forms \(\mathrm{Pd^{2+}}\), which contains eight \(4d\) electrons. As \(4d^8\) is incompletely filled, the ion satisfies the electronic criterion for a transition element. Therefore, a completely filled d subshell in a neutral atom does not necessarily exclude transition-element classification. The comparison also shows why the general d-block configuration should not be treated as an exception-free requirement.

Explanation: Chromium has atomic number 24, so six electrons are distributed between its \(3d\) and \(4s\) subshells beyond the argon core. A simple filling prediction would suggest \(3d^44s^2\), but the observed ground-state configuration is \(3d^54s^1\). This arrangement gives chromium a half-filled d subshell, which benefits from additional electronic stabilisation. Copper has atomic number 29 and eleven electrons beyond the argon core. Its simple predicted configuration would be \(3d^94s^2\), but the actual configuration is \(3d^{10}4s^1\). Here, the completely filled d subshell provides an important stabilising contribution. The small energy difference between the \(3d\) and \(4s\) subshells allows these arrangements to be energetically favourable. The exceptions illustrate the importance of electron distribution and exchange stabilisation rather than a rigid application of the simple filling sequence. However, this principle does not justify transferring electrons arbitrarily in every d-block element; the established ground-state configuration of each element must be used.

Explanation: \(\textbf{Initial electronic configuration:}\) Chromium has the exceptional ground-state configuration \[ \mathrm{Cr}:[Ar]\,3d^54s^1 \] The formation of \(\mathrm{Cr^{2+}}\) requires removal of two electrons. The \(4s\) electron is removed before a \(3d\) electron during ion formation. \(\textbf{Successive electron removal:}\) \[ \mathrm{Cr}\longrightarrow\mathrm{Cr^+}+e^- \] \[ [Ar]\,3d^54s^1\longrightarrow[Ar]\,3d^5 \] The second electron must therefore be removed from the \(3d\) subshell. \[ \mathrm{Cr^+}\longrightarrow\mathrm{Cr^{2+}}+e^- \] \[ [Ar]\,3d^5\longrightarrow[Ar]\,3d^4 \] \(\textbf{Applying Hund's rule:}\) The five d orbitals are degenerate in a free ion, and electrons occupy different orbitals singly with parallel spins before pairing occurs. Consequently, the four \(3d\) electrons occupy four different orbitals. There are four unpaired electrons in the free \(\mathrm{Cr^{2+}}\) ion. \(\textbf{Final answer:}\) The ion has the configuration \(3d^4\), with four unpaired electrons. The crucial distinction is that chromium's neutral \(3d^54s^1\) configuration must be used before applying the electron-removal rule.

Explanation: The two given configurations reveal that the second ionisation removes an electron from a half-filled \(3d^5\) subshell. The monovalent ion contains no \(4s\) electron, so the neutral atom must be identified using the actual ground-state configurations of the first-row elements. Chromium has the neutral configuration \([Ar]\,3d^54s^1\). Removal of its single \(4s\) electron produces \(\mathrm{Cr^+}\) with \(3d^5\). Removal of one additional \(3d\) electron produces \(\mathrm{Cr^{2+}}\) with \(3d^4\). Manganese, by comparison, has the neutral configuration \([Ar]\,3d^54s^2\), so its monovalent ion retains a \(4s\) electron. Its divalent ion is \(3d^5\), not \(3d^4\). Vanadium and iron have different d-electron occupancies and cannot produce the stated pair through ordinary successive ionisation. Interpreting two consecutive ionic states rather than only one electron count identifies chromium. Its exceptional ground-state configuration determines the successive electron-removal sequence.

Explanation: \(\textbf{Electron counting:}\) For a positive ion, the number of electrons equals the atomic number minus the positive ionic charge. Chromium has atomic number 24 and forms \(\mathrm{Cr^+}\). \[ N_e(\mathrm{Cr^+})=24-1=23 \] Manganese has atomic number 25 and forms \(\mathrm{Mn^{2+}}\). \[ N_e(\mathrm{Mn^{2+}})=25-2=23 \] Iron has atomic number 26 and forms \(\mathrm{Fe^{3+}}\). \[ N_e(\mathrm{Fe^{3+}})=26-3=23 \] \(\textbf{Electronic configuration comparison:}\) Each ion possesses the configuration \([Ar]\,3d^5\), so all three ions are isoelectronic. However, their nuclear charges are \(+24e\), \(+25e\) and \(+26e\), respectively. Isoelectronic species contain the same number of electrons but need not have equal nuclear charge, ionic size or chemical behaviour. \(\textbf{Final inference:}\) The common electron count is 23, while the three nuclear charges remain different. This distinction prevents the incorrect assumption that identical d-electron configurations automatically imply identical chemical properties.

Explanation: \(\textbf{Second ionisation enthalpy comparison:}\) Chromium has the neutral configuration \([Ar]\,3d^54s^1\). Its monovalent ion, \(\mathrm{Cr^+}\), possesses the stable half-filled \(3d^5\) configuration. The second ionisation of chromium removes an electron from this half-filled subshell. \[ \mathrm{Cr^+}:[Ar]\,3d^5\longrightarrow\mathrm{Cr^{2+}}:[Ar]\,3d^4 \] Manganese has the neutral configuration \([Ar]\,3d^54s^2\). After the first ionisation, \(\mathrm{Mn^+}\) retains the configuration \([Ar]\,3d^54s^1\). Its second ionisation removes the remaining \(4s\) electron and produces the favourable \(3d^5\) configuration. \[ \mathrm{Mn^+}:[Ar]\,3d^54s^1\longrightarrow\mathrm{Mn^{2+}}:[Ar]\,3d^5 \] Consequently, the second ionisation enthalpy of chromium exceeds that of manganese. \(\textbf{Third ionisation enthalpy comparison:}\) Manganese(II) already has a stable half-filled d subshell. Its third ionisation removes an electron from this arrangement. \[ \mathrm{Mn^{2+}}:3d^5\longrightarrow\mathrm{Mn^{3+}}:3d^4 \] Iron(II), in contrast, possesses \(3d^6\). Its third ionisation produces the favourable half-filled configuration. \[ \mathrm{Fe^{2+}}:3d^6\longrightarrow\mathrm{Fe^{3+}}:3d^5 \] Therefore, the third ionisation enthalpy of manganese exceeds that of iron. \(\textbf{Consistency with NCERT values:}\) \[ \Delta_iH_2(\mathrm{Cr})=1592\,\mathrm{kJ\,mol^{-1}}\gt 1509\,\mathrm{kJ\,mol^{-1}}=\Delta_iH_2(\mathrm{Mn}) \] \[ \Delta_iH_3(\mathrm{Mn})=3260\,\mathrm{kJ\,mol^{-1}}\gt 2962\,\mathrm{kJ\,mol^{-1}}=\Delta_iH_3(\mathrm{Fe}) \] The two irregularities arise at different ionisation stages, but both illustrate how half-filled d-subshell stability contributes to the observed energetics.

Explanation: \(\textbf{Recovering the atomic numbers:}\) Each ion contains the argon core of 18 electrons and three additional d electrons. \[ N_e=18+3=21 \] A positive ion has fewer electrons than its neutral parent atom. Therefore, the atomic number of the parent element is obtained by adding the positive ionic charge to the electron count. For \(P^{2+}\), \[ Z_P=21+2=23 \] The element with atomic number 23 is vanadium. For \(Q^{3+}\), \[ Z_Q=21+3=24 \] The element with atomic number 24 is chromium. For \(R^{4+}\), \[ Z_R=21+4=25 \] The element with atomic number 25 is manganese. \(\textbf{Configuration verification:}\) Vanadium has the neutral configuration \([Ar]\,3d^34s^2\), so \(\mathrm{V^{2+}}\) is \(3d^3\). Chromium has the exceptional neutral configuration \([Ar]\,3d^54s^1\), so \(\mathrm{Cr^{3+}}\) is \(3d^3\). Manganese has the neutral configuration \([Ar]\,3d^54s^2\), so \(\mathrm{Mn^{4+}}\) is \(3d^3\). All three assignments satisfy both the electron-count and oxidation-state conditions. The essential principle is that equal ionic electronic configurations do not imply identical parent elements. Different nuclear charges and oxidation states can produce isoelectronic transition-metal ions.

Explanation: The Assertion is true because identical d-electron configurations do not guarantee identical thermodynamic stability in different chemical environments. Manganese(II) and iron(III) both possess five d electrons, and the half-filled \(d^5\) subshell has additional electronic stabilisation. However, the two ions belong to different elements and have different nuclear charges and ionic charges. Their interactions with surrounding water molecules are consequently not identical. The Reason is also true because the stability of an aqueous oxidation state involves several energetic factors. Ionisation enthalpy represents the energy needed to remove electrons from gaseous atoms or ions. Hydration enthalpy contributes when gaseous ions interact with water to form hydrated species. Other energetic contributions can also influence the relative stability of oxidation states and the direction of redox reactions. Therefore, the energetic preference suggested by a favourable \(d^5\) configuration is only one part of the overall thermodynamic picture. A configuration-based argument cannot independently establish the relative aqueous stability of two different metal ions. The Reason directly explains why the Assertion is correct. This distinction is particularly important when using electronic configurations to interpret transition-metal oxidation states.

Explanation: \(\textbf{Iron oxidation:}\) Iron has the neutral ground-state configuration \([Ar]\,3d^64s^2\). Removal of the two \(4s\) electrons gives \[ \mathrm{Fe^{2+}}:[Ar]\,3d^6 \] According to Hund's rule, the five d orbitals first receive one electron each with parallel spins. The sixth electron pairs with one of the existing electrons. Consequently, \(\mathrm{Fe^{2+}}\) has four unpaired d electrons. Oxidation to \(\mathrm{Fe^{3+}}\) removes one d electron. \[ \mathrm{Fe^{3+}}:[Ar]\,3d^5 \] The resulting half-filled subshell contains five unpaired electrons. Thus, the unpaired-electron count increases from four to five. \(\textbf{Chromium oxidation:}\) Chromium has the neutral configuration \([Ar]\,3d^54s^1\). Its divalent ion contains four d electrons. \[ \mathrm{Cr^{2+}}:[Ar]\,3d^4 \] These occupy four separate d orbitals, giving four unpaired electrons. Further oxidation produces \[ \mathrm{Cr^{3+}}:[Ar]\,3d^3 \] The trivalent ion therefore contains three unpaired d electrons. The unpaired-electron count decreases from four to three. \(\textbf{Electronic consequence:}\) Although both processes involve the removal of one electron, their effects on unpaired-electron populations are opposite. The result follows from the initial d-electron occupancy and Hund's rule, not merely from the change in ionic charge.

Explanation: Copper has the neutral ground-state configuration \([Ar]\,3d^{10}4s^1\). Removal of the \(4s\) electron gives \(\mathrm{Cu^+}\), which possesses a completely filled \(3d^{10}\) subshell. All five d orbitals contain paired electrons, so the ion has no unpaired d electrons. A filled d subshell also lacks an available d-orbital vacancy for an ordinary d–d electronic excitation. Oxidation of \(\mathrm{Cu^+}\) to \(\mathrm{Cu^{2+}}\) removes one electron from the d subshell, producing \(3d^9\). The nine electrons occupy the five d orbitals with one electron remaining unpaired. The ion therefore becomes paramagnetic as a consequence of its unpaired electron. Because the d subshell is now incompletely filled, electronic excitation between appropriately split d levels becomes possible. Such excitation can contribute to colour when the energy separation corresponds to absorption in the visible region. The actual colour depends on the coordination environment and the energies of the electronic transitions. The important conclusion is that changing the oxidation state from +1 to +2 modifies both the electron-pairing arrangement and the possibilities for d–d excitation. The electronic criterion for transition-element classification remains satisfied for copper because its divalent state contains a partially filled d subshell.

Explanation: \(\textbf{Given:}\) The mineral formula is \(\mathrm{FeCr_2O_4}\). Iron remains in oxidation state +2. Oxygen has oxidation state \(-2\). Each formula unit contains two chromium atoms. \(\textbf{Oxidation state of chromium in chromite:}\) Let the oxidation state of chromium be \(x\). Charge neutrality requires \[ (+2)+2x+4(-2)=0 \] \[ 2+2x-8=0 \] \[ 2x=6 \] \[ x=+3 \] Thus, each chromium atom in chromite is initially in oxidation state +3. \(\textbf{Oxidation state in dichromate:}\) The dichromate ion is \(\mathrm{Cr_2O_7^{2-}}\). Let the oxidation state of chromium in dichromate be \(y\). \[ 2y+7(-2)=-2 \] \[ 2y-14=-2 \] \[ 2y=12 \] \[ y=+6 \] Chromium is therefore oxidised from +3 to +6. \(\textbf{Electron-transfer calculation:}\) The increase in oxidation state per chromium atom is \[ \Delta x=(+6)-(+3)=3 \] Each chromium atom loses three electrons. Since one formula unit of chromite contains two chromium atoms, \[ n(e^-)=2\times3=6 \] Therefore, one mole of chromite releases six moles of electrons through chromium oxidation. \(\textbf{Electron-balance check:}\) The corresponding oxidation half-reaction in acidic medium is \[ 2\mathrm{Cr^{3+}}+7\mathrm{H_2O}\longrightarrow\mathrm{Cr_2O_7^{2-}}+14\mathrm{H^+}+6e^- \] The chromium atoms, oxygen atoms, hydrogen atoms and total charge are balanced. Iron contributes no electrons because its oxidation state is specified to remain unchanged. \(\textbf{Final answer:}\) \[ \boxed{6\,\mathrm{mol}\ e^-\text{ per mol of }\mathrm{FeCr_2O_4}} \]

Explanation: Complex formation is a characteristic chemical behaviour of many transition-metal ions, but it is not the defining criterion for transition elements. The formal classification depends on the presence of an incompletely filled d subshell in the atom or in a relevant oxidation state. Zinc provides an important counterexample because the neutral atom has the configuration \([Ar]\,3d^{10}4s^2\), while \(\mathrm{Zn^{2+}}\) has \([Ar]\,3d^{10}\). Both have completely filled d subshells, so zinc is not classified as a transition element. Nevertheless, zinc(II) can form complex ions with ligands such as ammonia. The formation of a complex involves interactions between the central metal ion and electron-pair-donating ligands. Such interactions do not universally require an incompletely filled d subshell or unpaired d electrons. An appreciable positive ionic charge and suitable metal–ligand interactions can favour coordination. Consequently, the ability to form a complex ion cannot independently establish transition-element character. This also illustrates why characteristic transition-metal properties should be distinguished from the electronic definition of the class. An element may exhibit one associated chemical behaviour without satisfying the defining electronic criterion.

Explanation: The colour of many transition-metal ions arises from electronic excitation between d orbitals having different energies in the surrounding chemical environment. This ordinary d–d mechanism requires suitable electronic occupancy, including an electron that can undergo the excitation and an available final state. Scandium(III) has the configuration \([Ar]\,3d^0\), so it contains no d electron capable of undergoing a d–d transition. Its ordinary aqueous solution is therefore colourless. Manganese in \(\mathrm{MnO_4^-}\) has oxidation state +7 because the four oxygen atoms contribute a total oxidation number of \(-8\), while the ion carries charge \(-1\). The manganese(VII) centre consequently has a \(3d^0\) configuration. Nevertheless, permanganate is intensely purple because its electronic absorption involves charge-transfer excitation rather than an ordinary metal-centred d–d transition. Electron density can be transferred electronically from orbitals associated with oxygen towards the manganese centre during absorption. This process does not require a pre-existing d electron on manganese. Therefore, the common statement that \(d^0\) ions are colourless must be understood in the context of ordinary d–d transitions and cannot be applied indiscriminately to every coloured compound. The difference also demonstrates why the observed colour of a compound cannot be used by itself to determine whether its metal centre possesses unpaired d electrons or whether the element qualifies as a transition element.

Explanation: Statement I is correct. Chromite has the formula \(\mathrm{FeCr_2O_4}\), in which iron is taken as Fe(II) and oxygen has oxidation state \(-2\). If the oxidation state of chromium is \(x\), charge neutrality gives \(2+2x-8=0\), so chromium has oxidation state +3. Statement II is also correct because manganese in neutral \(\mathrm{MnO_2}\) must balance two oxygen atoms, each having oxidation state \(-2\). Thus, manganese has oxidation state +4 in pyrolusite. Statement III is correct because chromium in chromate, \(\mathrm{CrO_4^{2-}}\), has oxidation state +6. Conversion of Cr(III) into Cr(VI) therefore involves loss of electrons and is an oxidation process. Statement IV is incorrect because manganese in permanganate, \(\mathrm{MnO_4^-}\), has oxidation state +7. Its conversion from Mn(IV) in pyrolusite to Mn(VII) involves an increase in oxidation state, not a decrease. Consequently, both prescribed mineral-to-oxoanion transformations require oxidation of their respective transition metals. The important distinction is that the initial oxidation state must be obtained from the mineral formula rather than assumed to be the same as that of the final compound.

Explanation: \(\textbf{Oxidation states in the starting mineral:}\) Chromite has the formula \(\mathrm{FeCr_2O_4}\). Iron is initially present as Fe(II), and chromium is present as Cr(III). One formula unit contains one iron atom and two chromium atoms. \(\textbf{Electron loss by iron:}\) Iron changes from oxidation state +2 to +3. \[ \mathrm{Fe^{2+}}\longrightarrow\mathrm{Fe^{3+}}+e^- \] Thus, one iron atom loses one electron. \(\textbf{Electron loss by chromium:}\) Each chromium atom changes from +3 to +6. \[ \mathrm{Cr^{3+}}\longrightarrow\mathrm{Cr^{6+}}+3e^- \] These ionic symbols represent oxidation-state electron accounting; the chromium(VI) product is present as an oxoanion. Since there are two chromium atoms, \[ n(e^-)_{\mathrm{Cr}}=2\times3=6 \] \(\textbf{Total oxidation electron equivalents:}\) \[ n(e^-)_{\mathrm{total}}=1+6=7 \] Therefore, one mole of chromite releases seven moles of electrons. \(\textbf{Oxygen reduction:}\) Oxygen changes from oxidation state zero in \(\mathrm{O_2}\) to \(-2\) in the products. Each oxygen atom accepts two electrons. One oxygen molecule therefore accepts four electrons. \[ \mathrm{O_2}+4e^-\longrightarrow2\mathrm{O^{2-}} \] This equation represents electron-equivalent accounting, not an assertion that free oxide ions remain as the final products. \(\textbf{Oxygen requirement:}\) \[ n(\mathrm{O_2})=\frac{n(e^-)_{\mathrm{released}}}{4} \] \[ n(\mathrm{O_2})=\frac{7}{4}=1.75\,\mathrm{mol} \] \(\textbf{Reaction consistency check:}\) In the presence of sodium carbonate, the overall balanced reaction is \[ 4\mathrm{FeCr_2O_4}+8\mathrm{Na_2CO_3}+7\mathrm{O_2} \longrightarrow8\mathrm{Na_2CrO_4}+2\mathrm{Fe_2O_3}+8\mathrm{CO_2} \] The equation confirms that four moles of chromite consume seven moles of oxygen. This differs from considering chromium oxidation alone because iron also loses electrons in the specified process. \(\textbf{Final answer:}\) \[ \boxed{\frac{7}{4}\,\mathrm{mol\ O_2\ per\ mol\ FeCr_2O_4}} \]

Explanation: The direction of a redox transformation is determined by the change in oxidation state of the particular element being followed. In Process P, manganese initially occurs in \(\mathrm{MnO_2}\), where its oxidation state is +4. In permanganate, \(\mathrm{MnO_4^-}\), manganese has oxidation state +7. The increase from +4 to +7 indicates the loss of three electrons per manganese atom. Consequently, Process P requires an oxidising agent capable of accepting those electrons. In Process Q, iron initially exists in oxidation state +2. Metallic iron has oxidation state zero because an element in its uncombined state is assigned oxidation number zero. Conversion of Fe(II) into Fe(0) therefore requires the gain of two electrons per iron atom. Process Q is consequently a reduction and requires an electron donor or reducing agent. Both transformations involve transition-metal chemistry, but their electron-transfer directions are opposite. The fact that a metal is obtained from an ore does not establish whether oxidation or reduction occurs until the initial and final oxidation states are compared. This comparison also explains why preparation of a high-oxidation-state oxoanion and extraction of a free metal generally involve different redox requirements.

Explanation: The enthalpy of atomisation of a metallic solid measures the energy required to convert one mole of the solid element into separated gaseous atoms. A larger atomisation enthalpy generally indicates stronger cohesive interactions in the metallic solid. Chromium and manganese both possess half-filled \(3d^5\) subshells in their neutral ground-state atoms. Their outer electronic configurations differ in the occupancy of the \(4s\) subshell. Chromium has \(4s^1\), whereas manganese has \(4s^2\). The tabulated atomisation enthalpies show that chromium has substantially stronger cohesive bonding than manganese. This observation demonstrates that simply counting \(4s\) electrons cannot reliably determine metallic bonding strength. Likewise, both elements have five unpaired atomic d electrons, but their atomisation enthalpies are not equal. Metallic bonding depends on the collective electronic structure of the solid, including the participation and interaction of suitable valence orbitals. The energy of isolated-atom subshell configurations cannot be transferred directly into a quantitative prediction of metallic cohesion. The relatively low atomisation enthalpy of manganese is an important irregularity within the first transition series. The correct conclusion is therefore a limitation on simple electron-counting rules, not a denial of the general relationship between valence-electron participation and metallic bonding.