Q.Using valence bond theory, explain the following in relation to the complexes given below:
$[Mn(CN)_6]^{3-}$, $[Co(NH_3)_6]^{3+}$, $[Cr(H_2O)_6]^{3+}$, $[FeCl_6]^{4-}$

Showing the 12 most recent of 45 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Given below are two statements Statement-I: In the conversion of $O_2^+$ to $O_2^{2+}$ bond length increases Statement-II: In the conversion of $O_2^+$ to $O_2^{2+}$ magnetic property changes (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

   ›Reveal solutionSolution

   $\text{O}_2^{+}$ (bond order 2.5) $\to$ $\text{O}_2^{2+}$ (bond order 3.0): bond order rises, so bond length decreases — Statement I is wrong. The species go from paramagnetic (1 unpaired $e^-$) to diamagnetic (0 unpaired), so the magnetic property does change — Statement II is correct. Option (C).

   Concept

   Using MO theory, $\text{bond order} = \tfrac{1}{2}(\text{bonding} - \text{antibonding electrons})$. Neutral $\text{O}_2$ has 16 electrons with the two highest in $\pi^*$ orbitals (singly occupied, hence paramagnetic). Removing electrons comes off these $\pi^*$ antibonding orbitals, which raises the bond order and shortens the bond.

   Solution

   1. $\text{O}_2^{+}$ (15 $e^-$): antibonding $\pi^*$ holds 1 electron.

   $\text{B.O.} = \frac{8-3}{2} = 2.5, \qquad \text{one unpaired } e^- \Rightarrow \text{paramagnetic}.$

   2. $\text{O}_2^{2+}$ (14 $e^-$): both $\pi^*$ electrons removed. …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Given below are two statements Statement-I: The percent compositions of Ni$^{2+}$ and Ni$^{3+}$ in Ni$_{0.98}$O is 96% and 4% respectively Statement-II: The fraction of Fe$^{3+}$ and Fe$^{2+}$ ions in 1 mole of Fe$_{0.93}$O is 0.14 and 0.79 respectively (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

   ›Reveal solutionSolution

   The problem tests the ability to compute the fraction of different oxidation states in a non-stoichiometric oxide using charge balance. For Ni₀.₉₈O, the given percentages (96% Ni²⁺, 4% Ni³⁺) are correct; for Fe₀.₉₃O, the given fractions (0.14 Fe³⁺, 0.79 Fe²⁺) are also correct. Hence both statements are true, and the correct option is (A).

   ---

   Concept and Intuition

   In a perfect ionic oxide like NiO or FeO, the metal ion is in the +2 state and the oxide ion is O²⁻, so the formula is exactly 1:1. But real crystals often have non-stoichiometry — missing some metal ions (cation vacancies). To keep the overall crystal electrically neutral, some of the remaining metal ions must adopt a higher oxidation state (+3 instead of +2) to compensate for the missing positive charge.

   The key idea:

   • Let the formula be $M_{1-x}O$.
   • There are $1-x$ metal ions per O²⁻.
   • If a fraction $f$ of those metal ions are M³⁺ and the rest (1−$f$) are M²⁺, then the total positive charge must exactly balance the −2 charge from O²⁻.

   That gives one equation, which we solve for the fraction of M³⁺ (and hence M²⁺). This is a classic solid-state chemistry charge-balance problem.

   ---

   Step-by-step solution

   1. Set up the charge balance for Ni₀.₉₈O

   The formula means: per oxide ion O²⁻, there are 0.98 nickel ions.

   Let $x$ = fraction of Ni ions that are Ni³⁺. Then the fraction that are Ni²⁺ is $1-x$.

   Total positive charge from nickel ions =

   $0.98 \times [\, 3x + 2(1-x) \,] = 0.98 \times (2 + x)$

   This must equal the negative charge from one O²⁻, which is 2.

   So:

   $0.98 (2 + x) = 2$

   $2 + x = \frac{2}{0.98} \approx 2.040816$

   $x \approx 0.040816$

   Thus fraction of Ni³⁺ ≈ 0.0408 → 4.08%, and fraction of Ni²⁺ ≈ 0.9592 → 95.92%.

   Statement-I says 96% and 4% — these are rounded values, perfectly acceptable. So Statement-I is correct.

   2. Set up the charge balance for Fe₀.₉₃O

   Per O²⁻, there are 0.93 iron ions.

   Let $y$ = fraction of Fe ions that are Fe³⁺. Then fraction Fe²⁺ = $1-y$.

   Total positive charge:

   $0.93 \times [\, 3y + 2(1-y) \,] = 0.93 \times (2 + y)$

   Set equal to 2:

   $0.93 (2 + y) = 2$

   $2 + y = \frac{2}{0.93} \approx 2.150538$

   $y \approx 0.150538$

   So fraction Fe³⁺ ≈ 0.1505, fraction Fe²⁺ ≈ 0.8495. …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The colour and magnetic nature of the compound formed, when MnO$_2$ is fused with a mixture of KOH and KNO$_3$ are respectively (A) Green, paramagnetic (B) Blue, paramagnetic (C) Green, diamagnetic (D) Violet, diamagnetic

   ›Reveal solutionSolution

   Fusing MnO₂ with KOH and KNO₃ oxidises Mn(IV) to Mn(VI), forming the green manganate ion MnO₄²⁻, which has one unpaired electron and is paramagnetic — so the answer is (A).

   The key here is recognising that the reaction is an oxidation in a strongly alkaline melt. MnO₂ (manganese in +4 oxidation state) is fused with KOH (base) and KNO₃ (a powerful oxidising agent). Under these conditions, Mn(IV) is oxidised further — not to Mn(VII) (purple permanganate), but to Mn(VI), because the alkaline melt stabilises the manganate ion, MnO₄²⁻. This ion is famously green in colour. Its electronic configuration (d¹) means it has one unpaired electron, making it paramagnetic.

   Let’s walk through the reasoning step by step.

   1. Identify the reaction type. Fusing MnO₂ with KOH and KNO₃ is a classic laboratory preparation of potassium manganate. The KNO₃ acts as the oxidising agent, converting Mn(IV) to Mn(VI). The balanced equation is:

   $\text{MnO}_2 + 2\text{KOH} + \text{KNO}_3 \rightarrow \text{K}_2\text{MnO}_4 + \text{KNO}_2 + \text{H}_2\text{O}$

   The product is potassium manganate, which contains the manganate ion, $\text{MnO}_4^{2-}$.

   2. Determine the colour.

      The manganate ion $\text{MnO}_4^{2-}$ is well-known to be green in solution (and in the solid state). This is a distinctive property — permanganate ($\text{MnO}_4^-$) is purple, while manganate is green. So the colour is green.

   3. Determine the magnetic nature.

      In $\text{MnO}_4^{2-}$, manganese is in the +6 oxidation state. The electronic configuration of Mn is $[Ar]\,3d^5 4s^2$. Removing six electrons (to get Mn⁶⁺) leaves a $3d^1$ configuration.

      • A d¹ ion has one unpaired electron in the d-orbital.
      • Any species with unpaired electrons is paramagnetic (attracted to a magnetic field). Therefore, the compound is paramagnetic.

   4. Eliminate other options. …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.In Lassaigne’s test, violet colour is formed when sodium fusion extract of an organic compound is treated with solution of X. What is X? (A) $\mathrm{Pb(CH_3COO)_2}$ (B) $\mathrm{Na_2[Fe(CN)_5\ NO]}$ (C) $\mathrm{Na_4[Fe(CN)_6]}$ (D) $\mathrm{(NH_4)_2MoO_4}$

   ›Reveal solutionSolution

   The violet colour in Lassaigne’s test for nitrogen comes from the formation of Prussian blue when the sodium fusion extract is treated with sodium nitroprusside — the correct reagent is $\mathrm{Na_2[Fe(CN)_5\ NO]}$, option (B).

   The Lassaigne’s test is a classic qualitative analysis method used to detect elements like nitrogen, sulphur, and halogens in organic compounds. The key idea is that when an organic compound is fused with sodium metal, the elements present are converted into water-soluble ionic salts. For nitrogen, the fusion produces sodium cyanide ($\mathrm{NaCN}$). The test then relies on a specific chemical reaction that gives a distinctive coloured product — in this case, a deep violet or blue colour.

   The violet colour is not the final Prussian blue itself but an intermediate complex that forms when the cyanide ion reacts with a particular iron-containing reagent. The reagent must supply iron in a form that can combine with cyanide to produce the coloured complex. Among the options, only sodium nitroprusside ($\mathrm{Na_2[Fe(CN)_5\ NO]}$) contains iron in a suitable oxidation state and ligand environment to react with $\mathrm{CN^-}$ from the fusion extract, yielding the characteristic violet colour.

   Let’s go through the reasoning step by step.

   1. What happens in the sodium fusion? The organic compound is heated with sodium metal. If nitrogen is present, it forms sodium cyanide:

   $\mathrm{Na + C + N \rightarrow NaCN}$

   This $\mathrm{NaCN}$ dissolves in water when the fusion product is extracted, giving a solution containing cyanide ions.

   2. The test for cyanide using sodium nitroprusside:

      When the sodium fusion extract (containing $\mathrm{CN^-}$) is treated with sodium nitroprusside ($\mathrm{Na_2[Fe(CN)_5\ NO]}$), a reaction occurs that produces a violet-coloured complex. The exact complex formed is $\mathrm{Na_3[Fe(CN)_5\ NO\ CN]}$ — a substitution product where the nitroso group ($\mathrm{NO}$) is replaced by cyanide. This is the characteristic positive test for nitrogen.

   3. Why the other options are wrong:

      • (A) $\mathrm{Pb(CH_3COO)_2}$: This is lead acetate, used to test for sulphur (black precipitate of $\mathrm{PbS}$), not nitrogen. …

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5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The pairs of ions which exhibit same colour in aquated state are I. $\mathrm{Fe^{2+},\ Ni^{2+}}$ II. $\mathrm{V^{2+},\ Cr^{2+}}$ III. $\mathrm{Cr^{2+},\ Cu^{2+}}$ The correct answer is Options : (A) I, II only (B) II, III only (C) I, II, III (D) I, III only

   ›Reveal solutionSolution

   The colour of an aquated transition-metal ion depends on the number of d-electrons and the crystal-field splitting. Ions with the same d-electron count often show similar colours. Here, Fe²⁺ (d⁶) and Ni²⁺ (d⁸) differ; V²⁺ (d³) and Cr²⁺ (d⁴) differ; Cr²⁺ (d⁴) and Cu²⁺ (d⁹) are both Jahn–Teller active and appear blue-green. Only pair III matches.

   Concept & Intuition

   In aqueous solution, transition-metal ions are surrounded by water ligands, creating an octahedral crystal field. The d-orbitals split into two sets: lower-energy $t_{2g}$ and higher-energy $e_g$. When visible light is absorbed, electrons jump from $t_{2g}$ to $e_g$, and the complementary colour is observed. The energy gap $\Delta_o$ depends on the metal ion’s charge and its position in the periodic table, but a key factor is the number of d-electrons. Ions with the same d-count often have similar absorption spectra and thus similar colours — but exceptions occur due to Jahn–Teller distortions or different oxidation states.

   Let’s examine each pair.

   1. Pair I: Fe²⁺ and Ni²⁺

      • Fe²⁺ has the electron configuration $[Ar]\,3d^6$. In an octahedral field, it is high-spin: $t_{2g}^4 e_g^2$.
      • Ni²⁺ has $[Ar]\,3d^8$, giving $t_{2g}^6 e_g^2$.
      • These have different d-electron counts (d⁶ vs d⁸), so the crystal-field splitting and the number of d–d transitions differ. Fe²⁺ solutions are pale green, while Ni²⁺ solutions are bright green. They are not the same colour. → Pair I is incorrect.

   2. Pair II: V²⁺ and Cr²⁺

      • V²⁺ is d³: $t_{2g}^3$.
      • Cr²⁺ is d⁴: in water it is high-spin, $t_{2g}^3 e_g^1$.
      • Different d-counts again. V²⁺ solutions are violet, Cr²⁺ solutions are blue. They are not the same colour. → Pair II is incorrect.

   3. Pair III: Cr²⁺ and Cu²⁺

      • Cr²⁺ is d⁴ (high-spin, $t_{2g}^3 e_g^1$). …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Identify the incorrect statement from the following? (A) $m_l$ designates the orientation of the orbital (B) The probability density of electron is expressed by $|\psi|^3$ (C) The total information about electron in atom is stored in its $\psi$ (D) Total number of orbitals in a sub level is equal to $(2l+1)$

   ›Reveal solutionSolution

   The question asks to identify the incorrect statement among four options about quantum numbers and orbitals. The wrong one is (B) because probability density is $|\psi|^2$, not $|\psi|^3$.

   The key concept here is the Born interpretation of the wavefunction in quantum mechanics. The wavefunction $\psi$ itself is not directly observable; it is a mathematical object that contains all information about the electron's state. However, the physical meaning comes from $|\psi|^2$, which gives the probability density of finding the electron at a given point in space. Any statement that misstates this exponent is automatically false.

   Let’s examine each option step by step.

   1. Option (A): "$m_l$ designates the orientation of the orbital."

      This is correct. The magnetic quantum number $m_l$ can take integer values from $-l$ to $+l$, and each value corresponds to a specific orientation of the orbital in space (e.g., $p_x$, $p_y$, $p_z$ for $l=1$).

   2. Option (B): "The probability density of electron is expressed by $|\psi|^3$."

      This is incorrect. According to Max Born’s statistical interpretation, the probability density is $|\psi|^2$ (the square of the magnitude of the wavefunction). The cube has no physical meaning in standard quantum mechanics. This is the classic trap — students sometimes misremember the exponent.

   3. Option (C): "The total information about electron in atom is stored in its $\psi$." …

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7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The incorrect statement about crystals with Schottky defect is (A) It is due to missing of equal number of cations and anions from lattice points (B) On the whole crystal is electrically neutral (C) It is shown by ionic compounds in which cation and anion are of almost same size (D) Density of the crystal increases

   ›Reveal solutionSolution

   Schottky defect involves missing equal numbers of cations and anions, preserving neutrality and decreasing density; the incorrect statement is that density increases — so the answer is (D).

   Concept and Intuition

   Schottky defect is a type of point defect in ionic crystals. Imagine a perfect crystal lattice where every cation and anion sits in its assigned spot. In a Schottky defect, a pair of ions — one cation and one anion — simply vanish from their lattice sites, leaving behind vacancies. Because the numbers of missing positive and negative ions are equal, the crystal remains electrically neutral overall. This defect is common in ionic compounds where the cation and anion are similar in size (like NaCl), because the lattice can “afford” to lose both without collapsing. A key consequence: the crystal loses mass but its volume stays roughly the same, so its density decreases — not increases. That’s the trap in option (D).

   ---

   Step-by-Step Reasoning

   1. What is Schottky defect?

      It is a vacancy defect where an equal number of cations and anions are missing from their lattice sites. This preserves the stoichiometry and electrical neutrality of the crystal.

      Example: In NaCl, one Na⁺ and one Cl⁻ leave, creating two vacancies.

   2. Check option (A):

      “It is due to missing of equal number of cations and anions from lattice points.”

      This is exactly the definition. So (A) is correct.

   3. Check option (B):

      “On the whole crystal is electrically neutral.”

      Since equal numbers of positive and negative ions are removed, the net charge remains zero. So (B) is correct.

   4. Check option (C):

      “It is shown by ionic compounds in which cation and anion are of almost same size.” …

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Noble gas ‘X’ is used as a diluent for oxygen in modern diving apparatus and noble gas ‘Y’ is used mainly to provide an inert atmosphere in high temperature metallurgical processes. ‘Y’ and ‘X’ are respectively? (A) Ar, Kr (B) He, Kr (C) He, Ar (D) Ar, He

   ›Reveal solutionSolution

   Helium (He) is the diluent mixed with oxygen in diving apparatus, and argon (Ar) provides the inert atmosphere in high-temperature metallurgy. Since the question asks for 'Y and X' in that order — metallurgy gas first, then diving gas — the answer is Ar, He, option (D).

   Concept & Intuition

   Noble gases are prized for their chemical inertness, but each has distinct physical properties suited to particular applications.

   • Helium (He) is extremely light and has very low solubility in blood, making it ideal for mixing with oxygen in deep-sea diving — it reduces the risk of decompression sickness ("the bends") and avoids nitrogen narcosis.
   • Argon (Ar) is denser than air, cheap, and abundant (about 0.93% of the atmosphere). It forms a stable, non-reactive blanket that protects hot metals from oxidation during welding, casting, and other high-temperature metallurgical processes.

   The question asks for "Y and X respectively" — so Y is the metallurgy gas and X is the diving diluent.

   Step-by-step reasoning

   1. Identify the diving diluent (X)

      Modern diving apparatus mixes oxygen with a diluent to prevent oxygen toxicity and nitrogen narcosis. Helium is the standard choice — it is far less soluble in blood than nitrogen and does not cause narcosis. Argon and krypton are unsuitable (argon itself causes narcosis at depth; krypton is far too costly).

      → X = He

   2. Identify the metallurgy gas (Y)

      High-temperature metallurgical processes need an inert blanket to prevent oxidation. Argon is the standard choice for its low cost, abundance, and complete inertness. Krypton is far too expensive for this bulk use.

      → Y = Ar

   3. Match to the options

      "Y and X" means Y first, then X — so we need (Ar, He).

      • (A) Ar, Kr → wrong (Kr is not the diving diluent) …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The incorrect statement about crystals with Schottky defect is (A) On the whole crystal is electrically neutral (B) It is due to missing of equal number of cations and anions from lattice points (C) It is shown by ionic compounds in which cation and anion are of almost same size (D) Density of the crystal increases

   ›Reveal solutionSolution

   Schottky defect involves missing equal numbers of cations and anions, keeping the crystal neutral but decreasing its density. The incorrect statement is the one claiming density increases.

   The key concept here is Schottky defect — a type of point defect in ionic crystals where a pair of one cation and one anion is missing from their lattice sites, creating vacancies. This defect preserves electrical neutrality because the missing ions are equal in number and opposite in charge. However, because atoms are removed, the mass of the crystal decreases while its volume remains nearly the same, so the density decreases — not increases. The question asks for the incorrect statement, so we need to spot the one that contradicts this behavior.

   Let’s examine each option step by step:

   1. Option (A): "On the whole crystal is electrically neutral"

      This is correct. In Schottky defect, equal numbers of cations and anions are missing, so the net charge remains zero. The crystal as a whole stays electrically neutral.

   2. Option (B): "It is due to missing of equal number of cations and anions from lattice points"

      This is the very definition of Schottky defect. It is correct.

   3. Option (C): "It is shown by ionic compounds in which cation and anion are of almost same size"

      This is also correct. Schottky defect is common in compounds like NaCl, KCl, CsCl, where the cation and anion have similar ionic radii. This similarity allows the lattice to tolerate vacancies without collapsing. …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.A metal (M), crystallizes in fcc lattice with edge length of 4.242 Å. What is the radius of M atom (in Å)? (A) 1.25 (B) 1.75 (C) 1.5 (D) 1.0

    ›Reveal solutionSolution

    For an fcc lattice, the face diagonal equals four atomic radii; using the given edge length, the atomic radius is calculated as $r = \frac{a}{2\sqrt{2}} \approx 1.5 \, \text{Å}$, so the correct option is (C).

    Concept & Intuition

    In a face-centered cubic (fcc) lattice, atoms touch along the face diagonal — not along the edge. The face diagonal spans from one corner atom to the opposite corner atom on the same square face. Along that diagonal, there are two half-atoms at the corners and one full atom at the face center, making a total of two full atomic diameters. So the face diagonal length equals $4r$ (where $r$ is the atomic radius). Since the face diagonal of a cube of edge $a$ is $a\sqrt{2}$, we get the key relation:

    $a\sqrt{2} = 4r \quad \Rightarrow \quad r = \frac{a}{2\sqrt{2}}.$

    Step-by-step solution

    1. Identify the geometry

       The metal crystallizes in an fcc lattice. In fcc, atoms are located at each corner and at the center of each face. The closest contact occurs between a corner atom and a face-centered atom — this distance is half the face diagonal.

    2. Write the relation between edge length and radius

       For fcc:

    $\text{Face diagonal} = a\sqrt{2} = 4r.$

    Therefore,

    $r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}.$

    3. Plug in the given edge length Edge length $a = 4.242 \, \text{Å}$. $r = \frac{4.242}{2\sqrt{2}}.$ …

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11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.The pair of ions with paramagnetic nature and same number of electrons is (A) $\mathrm{Lu}^{3+},\ \mathrm{Yb}^{2+}$ (B) $\mathrm{Eu}^{3+},\ \mathrm{Pm}^{2+}$ (C) $\mathrm{Eu}^{2+},\ \mathrm{Gd}^{3+}$ (D) $\mathrm{La}^{3+},\ \mathrm{Ce}^{4+}$

    ›Reveal solutionSolution

    The key is to compare the electron configurations of lanthanide ions: paramagnetic ions have unpaired electrons, and the pair must also have the same number of electrons (be isoelectronic). Only one option satisfies both conditions.

    Concept & Intuition

    Paramagnetism arises from unpaired electrons. In lanthanides, the 4f subshell is being filled. When ions form, they typically lose the two 6s electrons and sometimes one 4f electron. To check paramagnetism, we count unpaired 4f electrons. To check “same number of electrons,” we compare total electron count (atomic number minus charge). The trick: many lanthanide ions have the same number of electrons but different configurations — only some are paramagnetic.

    Step-by-step reasoning

    1. Determine electron counts for each ion

       • Lu³⁺: Lu (Z=71) → 71 – 3 = 68 electrons.
       • Yb²⁺: Yb (Z=70) → 70 – 2 = 68 electrons.
       • Eu³⁺: Eu (Z=63) → 63 – 3 = 60 electrons.
       • Pm²⁺: Pm (Z=61) → 61 – 2 = 59 electrons.
       • Eu²⁺: Eu (Z=63) → 63 – 2 = 61 electrons.
       • Gd³⁺: Gd (Z=64) → 64 – 3 = 61 electrons.
       • La³⁺: La (Z=57) → 57 – 3 = 54 electrons.
       • Ce⁴⁺: Ce (Z=58) → 58 – 4 = 54 electrons.

    2. Identify which pairs are isoelectronic (same electron count)

       • (A) Lu³⁺ (68) and Yb²⁺ (68) → same
       • (B) Eu³⁺ (60) and Pm²⁺ (59) → different
       • (C) Eu²⁺ (61) and Gd³⁺ (61) → same
       • (D) La³⁺ (54) and Ce⁴⁺ (54) → same

       So (B) is out.

    3. Check paramagnetism (unpaired 4f electrons)

       • Lu³⁺: Lu is [Xe]4f¹⁴6s² → Lu³⁺ loses 6s² and one 4f? No, it loses 6s² and one 4f? Actually Lu³⁺ = [Xe]4f¹⁴ (full 4f, all paired) → diamagnetic.

       • Yb²⁺: Yb is [Xe]4f¹⁴6s² → Yb²⁺ loses 6s² → [Xe]4f¹⁴ → diamagnetic.

         → Pair (A) is diamagnetic-diamagnetic → not paramagnetic.

       • Eu²⁺: Eu is [Xe]4f⁷6s² → Eu²⁺ loses 6s² → [Xe]4f⁷ (half-filled, 7 unpaired) → paramagnetic.

       • Gd³⁺: Gd is [Xe]4f⁷5d¹6s² → Gd³⁺ loses 6s² and 5d¹ → [Xe]4f⁷ → paramagnetic.

         → Pair (C) is paramagnetic-paramagnetic ✓ …

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12. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.The pair of ions with paramagnetic nature and same number of electrons is (A) $\mathrm{La^{3+},\ Ce^{4+}}$ (B) $\mathrm{Eu^{2+},\ Gd^{3+}}$ (C) $\mathrm{Eu^{3+},\ Pm^{2+}}$ (D) $\mathrm{Lu^{3+},\ Yb^{2+}}$

    ›Reveal solutionSolution

    The key is to compare the electron configurations of the lanthanide ions, checking for unpaired electrons (paramagnetism) and equal total electron count. The pair that satisfies both is Eu²⁺ and Gd³⁺, option (B).

    Concept & Intuition

    Paramagnetism arises from unpaired electrons. In lanthanide ions, the 4f subshell is gradually filled. To find a pair that is both paramagnetic and has the same number of electrons, we need to:

    • Determine the atomic number of each element.
    • Subtract the charge to get the number of electrons in the ion.
    • Write the electron configuration (focus on the 4f subshell).
    • Check for unpaired electrons (Hund’s rule: half-filled or partially filled f-orbitals are paramagnetic; empty, full, or exactly half-filled f⁷ are special cases).

    Step-by-step reasoning

    1. Identify atomic numbers

       La (57), Ce (58), Eu (63), Gd (64), Pm (61), Lu (71), Yb (70).

    2. Calculate electron counts for each ion

       • La³⁺: 57 − 3 = 54 electrons
       • Ce⁴⁺: 58 − 4 = 54 electrons
       • Eu²⁺: 63 − 2 = 61 electrons
       • Gd³⁺: 64 − 3 = 61 electrons
       • Eu³⁺: 63 − 3 = 60 electrons
       • Pm²⁺: 61 − 2 = 59 electrons
       • Lu³⁺: 71 − 3 = 68 electrons
       • Yb²⁺: 70 − 2 = 68 electrons

    3. Write the 4f configurations (the core is [Xe] 4fⁿ)

       • La³⁺: [Xe] → 4f⁰ (no unpaired electrons, diamagnetic)
       • Ce⁴⁺: [Xe] → 4f⁰ (diamagnetic)
       • Eu²⁺: [Xe] 4f⁷ (half-filled, all 7 electrons unpaired → paramagnetic)
       • Gd³⁺: [Xe] 4f⁷ (half-filled, paramagnetic)
       • Eu³⁺: [Xe] 4f⁶ (4 unpaired electrons, paramagnetic)
       • Pm²⁺: [Xe] 4f⁵ (5 unpaired electrons, paramagnetic)
       • Lu³⁺: [Xe] 4f¹⁴ (full, diamagnetic)
       • Yb²⁺: [Xe] 4f¹⁴ (full, diamagnetic)

    4. Check each option

       • (A) La³⁺ (diamagnetic) and Ce⁴⁺ (diamagnetic) → both diamagnetic, so not paramagnetic.
       • (B) Eu²⁺ (paramagnetic, 61 e⁻) and Gd³⁺ (paramagnetic, 61 e⁻) → both paramagnetic and same electron count. …

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