Q.Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In which of the following, ratio of $t_{2g}$ and $e_g$ electrons is $3:2$? (A) $[FeF_6]^{3-}$ (B) $[CoF_6]^{3-}$ (C) $[Fe(CN)_6]^{3-}$ (D) $[Co(NH_3)_6]^{3+}$

   ›Reveal solutionSolution

   This tests crystal field electron configurations for octahedral complexes; the answer is $[FeF_6]^{3-}$, the only complex with $t_{2g}:e_g = 3:2$.

   Concept and Intuition

   The distribution of d-electrons between $t_{2g}$ and $e_g$ orbitals in an octahedral complex depends on both the metal's d-electron count and whether the ligand is weak-field (high spin, obeying Hund's rule across all 5 orbitals before pairing) or strong-field (low spin, filling $t_{2g}$ completely before touching $e_g$).

   Step-by-Step Solution

   1. $[FeF_6]^{3-}$: $Fe^{3+}$ is $d^5$. $F^-$ is a weak-field ligand ⇒ high spin: electrons fill all 5 orbitals singly first — $t_{2g}^3 e_g^2$. Ratio = 3:2. Matches.
   2. $[CoF_6]^{3-}$: $Co^{3+}$ is $d^6$. $F^-$ weak field ⇒ high spin: $t_{2g}^4 e_g^2$. Ratio = 4:2 = 2:1. Does not match.
   3. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $d^5$. $CN^-$ is a strong-field ligand ⇒ low spin: $t_{2g}^5 e_g^0$. Ratio = 5:0. Does not match. …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Which of the following has maximum number of electrons in $t_{2g}$ orbitals? (A) $[Fe(CN)_6]^{3-}$ (B) $[FeCl_6]^{3-}$ (C) $[CoCl_6]^{3-}$ (D) $[Co(NH_3)_6]^{3+}$

   ›Reveal solutionSolution

   This tests crystal field theory: assign the metal oxidation state, its $d^n$ configuration, and whether the ligand is strong- or weak-field, then fill $t_{2g}/e_g$. $[Co(NH_3)_6]^{3+}$, a low-spin $d^6$ complex, has the maximum of 6 $t_{2g}$ electrons.

   Concept and Intuition

   In an octahedral field, $d$-orbitals split into the lower-energy $t_{2g}$ set (3 orbitals) and higher-energy $e_g$ set (2 orbitals). Whether electrons pair up in $t_{2g}$ first (low spin) or spread out with parallel spins across both sets first (high spin, Hund's rule) depends on whether the ligand is strong-field (large $\Delta_o$, causes pairing) or weak-field (small $\Delta_o$, favours unpaired spread). $CN^-$ and $NH_3$ are strong-field ligands; $Cl^-$ is a weak-field ligand.

   Step-by-Step Solution

   1. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is strong field $\Rightarrow$ low spin: all 5 electrons pack into $t_{2g}$ first $\Rightarrow t_{2g}^5 e_g^0$. $t_{2g}$ count = 5.
   2. $[FeCl_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $Cl^-$ is weak field $\Rightarrow$ high spin: electrons fill all 5 $d$-orbitals singly first $\Rightarrow t_{2g}^3 e_g^2$. $t_{2g}$ count = 3. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Match the following. List – I (Complex ion) : (A) $[\text{MnCl}_6]^{3-}$ (B) $[\text{Mn(CN)}_6]^{3-}$ (C) $[\text{Fe(CN)}_6]^{4-}$ (D) $[\text{CoF}_6]^{3-}$ List – II (d-Electronic configuration) : (I) $t_{2g}^6\,e_g^0$ (II) $t_{2g}^4\,e_g^2$ (III) $t_{2g}^3\,e_g^1$ (IV) $t_{2g}^4\,e_g^0$ The correct answer is: (A) A-III, B-IV, C-I, D-II (B) A-III, B-IV, C-II, D-I (C) A-II, B-III, C-IV, D-I (D) A-II, B-I, C-IV, D-III

   ›Reveal solutionSolution

   Determining the metal's oxidation state and d-electron count, then applying strong-field (low spin) vs weak-field (high spin) splitting for each ligand, matches A-III, B-IV, C-I, D-II.

   Concept and Intuition

   For an octahedral complex, the d-electron filling pattern depends on (i) how many d-electrons the metal ion has, and (ii) whether the ligand is strong-field (causes pairing → low spin, maximises $t_{2g}$ filling before using $e_g$) or weak-field (no forced pairing → high spin, electrons spread out following Hund's rule across both $t_{2g}$ and $e_g$ first). CN⁻ is a strong field ligand; Cl⁻ and F⁻ are weak field ligands.

   Step-by-Step Solution

   1. A: $[\text{MnCl}_6]^{3-}$ — Cl is $-1$ each ($6 \times -1 = -6$); overall charge $-3$ ⟹ Mn is $+3$. Mn ($Z=25$): $[\text{Ar}]3d^5 4s^2$; $\text{Mn}^{3+}$ removes the two $4s$ electrons and one $3d$ electron ⟹ $d^4$. Cl⁻ is weak field ⟹ high spin: $t_{2g}^3 e_g^1$ — matches (III).
   2. B: $[\text{Mn(CN)}_6]^{3-}$ — same $\text{Mn}^{3+}$, $d^4$, but CN⁻ is strong field ⟹ low spin: electrons pair up in $t_{2g}$ before occupying $e_g$: $t_{2g}^4 e_g^0$ — matches (IV).
   3. C: $[\text{Fe(CN)}_6]^{4-}$ — CN is $-1$ each ($-6$); overall charge $-4$ ⟹ Fe is $+2$. Fe ($Z=26$): $[\text{Ar}]3d^6 4s^2$; $\text{Fe}^{2+}$ removes $4s^2$ ⟹ $d^6$. CN⁻ strong field ⟹ low spin: $t_{2g}^6 e_g^0$ — matches (I). …

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4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.In which of the following sets, given species are not only diamagnetic in nature but also inner orbital complexes? I. $[Ni(CN)_4]^{2-}, [Fe(CN)_6]^{4-}$ II. $[Mn(CN)_6]^{3-}, [Fe(CN)_6]^{3-}$ III. $[Co(en)_3]^{3+}, [Ni(CN)_4]^{2-}$ The correct answer is (only = only) (A) I, II only (B) I, III only (C) II, III only (D) I, II, III

   ›Reveal solutionSolution

   "Diamagnetic + inner orbital" needs a strong-field ligand forcing all d-electrons to pair using (n−1)d orbitals. Sets I and III both satisfy this; set II fails because its complexes are paramagnetic.

   Concept and Intuition

   An inner orbital complex uses an (n−1)d orbital in its hybridisation ($dsp^2$ for 4-coordinate square planar, or $d^2sp^3$ for 6-coordinate octahedral), which happens only with strong-field ligands (like $CN^-$, en) that force pairing of the metal's d-electrons (low spin). Diamagnetism additionally requires that ALL electrons end up paired — so we must check the d-electron count of the actual oxidation state, not just "is the ligand strong field".

   Step-by-Step Solution

   1. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. With strong-field $CN^-$ in a 4-coordinate geometry, $Ni^{2+}$ adopts square planar geometry via $dsp^2$ hybridisation (using one 3d orbital) — this is the textbook example of an inner-orbital, diamagnetic complex (all 8 d-electrons pair up in the four lower-lying d-orbitals, leaving the $d_{x^2-y^2}$ empty for the hybrid set).
   2. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. Strong field $CN^-$ gives low-spin $d^2sp^3$ (inner orbital), $t_{2g}^6 e_g^0$ — all 6 electrons paired → diamagnetic. Set I: both diamagnetic + inner orbital. ✓
   3. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. Even low-spin (strong field, $d^2sp^3$, inner orbital), $t_{2g}^4$ across 3 orbitals means 3 are singly occupied first, and the 4th electron must pair into one of them: net 2 unpaired electrons → paramagnetic. This alone disqualifies set II (even without checking the second species). …

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5. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Which of the following orders correctly represent the strength of ligands in the spectrochemical series? I. $I^- < Br^- < S^{2-} < SCN^-$ II. $H_2O < NCS^- < NH_3 < en$ III. $Cl^- < F^- < N^{3-} < OH^-$ Correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   Orders I and II match the standard spectrochemical series; order III has $F^-$ and $N^{3-}$ swapped, so it is incorrect.

   Concept and Intuition

   The spectrochemical series ranks ligands by the crystal field splitting energy ($\Delta$) they produce, from weak field to strong field. It is a memorised empirical order (from spectroscopic data) rather than something derivable from first principles, so each proposed order must be checked against the standard sequence.

   Step-by-Step Solution

   1. Standard spectrochemical series (weak → strong): $I^- < Br^- < S^{2-} < SCN^- < Cl^- < N^{3-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < CH_3CN < py \approx NH_3 < en < \ldots < CN^- \approx CO$.
   2. I: $I^-<Br^-<S^{2-}<SCN^-$ — matches the series exactly in this stretch. Correct.
   3. II: $H_2O<NCS^-<NH_3<en$ — matches the series exactly (water is weaker than NCS⁻, which is weaker than ammonia, which is weaker than ethylenediamine). Correct. …

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6. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Match the following List-I (aquated ion) List-II (colour) A) $\mathrm{Ni^{2+}}$ I) violet B) $\mathrm{Fe^{3+}}$ II) blue C) $\mathrm{Mn^{3+}}$ III) yellow D) $\mathrm{V^{4+}}$ IV) red V) green Correct answer is (A) A-V, B-III, C-IV, D-II (B) A-IV, B-V, C-I, D-III (C) A-I, B-III, C-IV, D-V (D) A-V, B-III, C-I, D-II

   ›Reveal solutionSolution

   $\mathrm{Ni^{2+}}$-green, $\mathrm{Fe^{3+}}$-yellow, $\mathrm{Mn^{3+}}$-violet, $\mathrm{V^{4+}}$-blue → option (D).

   Concept and Intuition

   Colours of hydrated transition-metal ions arise from d-d electronic transitions and are tabulated in NCERT. Matching each ion to its standard aqueous colour resolves the question.

   Step-by-Step Solution

   1. $\mathrm{Ni^{2+}}$ (d$^8$): green → V.
   2. $\mathrm{Fe^{3+}}$ (d$^5$): yellow → III.
   3. $\mathrm{Mn^{3+}}$ (d$^4$): violet → I.
   4. $\mathrm{V^{4+}}$, present as the vanadyl $\mathrm{VO^{2+}}$ ion (d$^1$): blue → II. …

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7. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.In which of the following, complex ions are not in correct order with respect to their magnitude of crystal field splitting ? (A) $[\mathrm{Fe(H_2O)_6}]^{3+} > [\mathrm{FeF_6}]^{3-}$ (B) $[\mathrm{Fe(en)_3}]^{3+} > [\mathrm{Fe(NCS)_6}]^{3-}$ (C) $[\mathrm{Fe(CN)_6}]^{4-} > [\mathrm{Fe(H_2O)_6}]^{2+}$ (D) $[\mathrm{Fe(H_2O)_6}]^{2+} > [\mathrm{Fe(NH_3)_6}]^{2+}$

   ›Reveal solutionSolution

   Tests the spectrochemical series ordering of common ligands and its effect on crystal field splitting energy, Δ₀.

   Concept and Intuition

   Crystal field splitting energy Δ₀ for an octahedral complex depends chiefly on the field strength of the ligand (for a fixed metal ion/oxidation state). The spectrochemical series ranks ligands by this field strength, largely independent of the metal: $\mathrm{I^- < Br^- < S^{2-} < SCN^- < Cl^- < N_3^-,F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < py, NH_3 < en < NO_2^- < CN^- < CO}$.

   Step-by-Step Solution

   1. (A): $[\mathrm{Fe(H_2O)_6}]^{3+}$ vs $[\mathrm{FeF_6}]^{3-}$ — $\mathrm{H_2O}$ lies above $\mathrm{F^-}$ in the series, so $\mathrm{H_2O}$ gives greater splitting: order is correct.
   2. (B): $[\mathrm{Fe(en)_3}]^{3+}$ vs $[\mathrm{Fe(NCS)_6}]^{3-}$ — en (a strong bidentate N-donor) is well above NCS⁻: order is correct.
   3. (C): $[\mathrm{Fe(CN)_6}]^{4-}$ vs $[\mathrm{Fe(H_2O)_6}]^{2+}$ — CN⁻ is one of the strongest-field ligands, far above H₂O: order is correct. …

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8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.In which one of the following complexes the metal ion has $t_{2g}^3 e_g^2$ configuration ? (A) $\mathrm{[Mn(H_2O)_6]^{2+}}$ (B) $\mathrm{[Fe(H_2O)_6]^{2+}}$ (C) $\mathrm{[Co(NH_3)_6]^{3+}}$ (D) $\mathrm{[Ni(H_2O)_6]^{2+}}$

   ›Reveal solutionSolution

   This tests $d$-electron counting plus crystal field splitting: $t_{2g}^3 e_g^2$ is the high-spin $d^5$ pattern, and only $Mn^{2+}$ among the choices is $d^5$.

   Concept and Intuition

   In an octahedral field the five $d$ orbitals split into the lower $t_{2g}$ (3 orbitals) and upper $e_g$ (2 orbitals) sets. The configuration $t_{2g}^3e_g^2$ has $3+2=5$ electrons distributed one-per-orbital across all five $d$ orbitals — this is only possible for a $d^5$ ion in the high-spin (weak field) case, where electrons singly occupy every orbital before any pairing (Hund's rule) because the ligand field splitting $\Delta_o$ is too small to force pairing.

   Step-by-Step Solution

   1. Count total electrons in the configuration: $t_{2g}^3 e_g^2 = 5$ electrons $\Rightarrow$ the metal ion is $d^5$.
   2. Find each ion's $d^n$ count:

   • $Mn^{2+}$: Mn (Z=25) is $[Ar]3d^54s^2$; losing the two $4s$ electrons gives $Mn^{2+} = 3d^5$.
   • $Fe^{2+}$: Fe (Z=26) is $[Ar]3d^64s^2$; $Fe^{2+} = 3d^6$.
   • $Co^{3+}$: Co (Z=27) is $[Ar]3d^74s^2$; removing 3 electrons gives $Co^{3+} = 3d^6$.
   • $Ni^{2+}$: Ni (Z=28) is $[Ar]3d^84s^2$; $Ni^{2+} = 3d^8$.

   3. Only $Mn^{2+}$ is $d^5$. …

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9. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.Arrange the following complexes in increasing order of number of unpaired electrons present in central metal ion I. $\mathrm{[Fe(CN)_6]^{3-}}$ II. $\mathrm{[Co(C_2O_4)_3]^{3-}}$ III. $\mathrm{[Mn(CN_6)]^{3-}}$ (A) II < III < I (B) III < I < II (C) II < I < III (D) I < II < III

   ›Reveal solutionSolution

   Tests counting unpaired $d$-electrons in low-spin octahedral complexes; the increasing order of

   unpaired electrons is II ($\mathrm{Co}$, 0) < I ($\mathrm{Fe}$, 1) < III ($\mathrm{Mn}$, 2).

   Concept and Intuition

   In an octahedral crystal field, $d$ orbitals split into a lower $t_{2g}$ set (3 orbitals) and a

   higher $e_g$ set (2 orbitals). Strong-field ligands (like $\mathrm{CN^-}$) and chelating ligands

   with high effective field strength for a high-oxidation-state metal (like oxalate on Co$^{3+}$)

   favour the low-spin arrangement, in which electrons fill and pair up in $t_{2g}$ before

   occupying $e_g$ at all.

   Step-by-Step Solution

   1. $[\mathrm{Fe(CN)_6}]^{3-}$: Fe is +3, so Fe$^{3+}$ is $d^5$. $\mathrm{CN^-}$ is a strong-field ligand → low spin: $t_{2g}^5 e_g^0$. Filling $t_{2g}$'s 3 orbitals with 5 electrons gives 2 fully paired orbitals + 1 singly occupied orbital → 1 unpaired electron.
   2. $[\mathrm{Co(C_2O_4)_3}]^{3-}$: Co is +3, so Co$^{3+}$ is $d^6$. Co$^{3+}$ complexes are strongly disposed to low spin even with moderately strong chelating ligands like oxalate → $t_{2g}^6 e_g^0$, all three $t_{2g}$ orbitals fully paired → 0 unpaired electrons.
   3. $[\mathrm{Mn(CN)_6}]^{3-}$: Mn is +3, so Mn$^{3+}$ is $d^4$. $\mathrm{CN^-}$ forces low spin: …

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10. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.$[Ni(H_2O)_6]^{2+}$ on reaction with ethane-1,2-diamine forms violet colour complex. Its formula and magnetic property respectively are (A) $[Ni(H_2O)_4\,en]^{2+}$, Paramagnetic nature (B) $[Ni(H_2O)_2\,en_2]^{2+}$, Diamagnetic nature (C) $[Ni(en)_3]^{2+}$, Paramagnetic nature (D) $[Ni(H_2O)_2\,en_2]^{2+}$, Paramagnetic nature

    ›Reveal solutionSolution

    $[Ni(H_2O)_6]^{2+}$ reacts with excess ethylenediamine to fully substitute all six water ligands (chelate effect), forming the octahedral, violet $[Ni(en)_3]^{2+}$ complex, which is paramagnetic since octahedral $d^8$ Ni(II) always retains 2 unpaired electrons.

    Concept and Intuition

    Ethane-1,2-diamine (en) is a bidentate ligand, and chelation with a multidentate ligand is entropically strongly favoured (the chelate effect) over monodentate water ligands. With enough en added, all six water molecules on $[Ni(H_2O)_6]^{2+}$ are displaced by three en molecules (each en occupying two coordination sites), giving the well-known tris(ethylenediamine)nickel(II) complex. For $Ni^{2+}$ ($d^8$) in an octahedral field, the electron configuration is $t_{2g}^6 e_g^2$ — the two $e_g$ electrons must occupy the two degenerate $e_g$ orbitals singly (Hund's rule), so an octahedral $d^8$ complex is always paramagnetic (2 unpaired electrons), regardless of the ligand's field strength; only a square-planar geometry (as with $CN^-$ or in $Pt^{2+}/Pd^{2+}$ complexes) can force pairing to diamagnetism for $d^8$.

    Step-by-Step Solution

    1. Reaction: $[Ni(H_2O)_6]^{2+} + 3\,en \to [Ni(en)_3]^{2+} + 6H_2O$ (complete displacement, driven by the chelate effect).
    2. Geometry: octahedral (6-coordinate, three bidentate en ligands). …

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11. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.The paramagnetic complex ion which has no unpaired electrons in $t_{2g}$ orbitals is (A) $\mathrm{[Fe(CN)_6]^{4-}}$ (B) $\mathrm{[Fe(CN)_6]^{3-}}$ (C) $\mathrm{[Zn(NH_3)_6]^{2+}}$ (D) $\mathrm{[Ni(NH_3)_6]^{2+}}$

    ›Reveal solutionSolution

    This tests crystal-field electron filling ($t_{2g}$ vs $e_g$) across four octahedral complexes; the one paramagnetic ion with an empty $t_{2g}$ subshell of unpaired spins is $[\mathrm{Ni(NH_3)_6}]^{2+}$.

    Concept and Intuition

    In an octahedral field the five $d$-orbitals split into the lower $t_{2g}$ set (3 orbitals) and upper $e_g$ set (2 orbitals). Whether electrons pair up in $t_{2g}$ before occupying $e_g$ depends on whether the ligand is strong field (low spin) or weak/medium field (high spin), and on the metal's $d^n$ count.

    Step-by-Step Solution

    1. $[\mathrm{Fe(CN)_6}]^{4-}$: $\mathrm{Fe}^{2+} = d^6$. $\mathrm{CN^-}$ is a strong-field ligand $\Rightarrow$ low spin: configuration $t_{2g}^6 e_g^0$. All electrons paired — diamagnetic, so it is not even paramagnetic; eliminated.
    2. $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe}^{3+} = d^5$, low spin with $\mathrm{CN^-}$: $t_{2g}^5 e_g^0$. This has 1 unpaired electron, but it sits IN $t_{2g}$ — fails the "no unpaired electrons in $t_{2g}$" condition.
    3. $[\mathrm{Zn(NH_3)_6}]^{2+}$: $\mathrm{Zn}^{2+} = d^{10}$, fully filled $t_{2g}^6 e_g^4$ regardless of field strength — diamagnetic, no unpaired electrons anywhere; eliminated. …

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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.How many of the following ligands are stronger than $\mathrm{H_2O}$? $\mathrm{S^{2-}, Br^-, C_2O_4^{2-}, CN^-, en, NH_3, CO, OH^-}$ (A) 5 (B) 3 (C) 4 (D) 6

    ›Reveal solutionSolution

    Checking each of the 8 ligands against the spectrochemical series relative to H₂O shows exactly 4 (CN⁻, en, NH₃, CO) are stronger field ligands than water.

    Concept and Intuition

    The spectrochemical series ranks ligands by their crystal-field splitting strength, from weak field (e.g. halides) to strong field (e.g. CO, CN⁻). Water sits roughly in the middle, so any given ligand must be individually checked against its position relative to H₂O.

    Step-by-Step Solution

    1. Standard spectrochemical series (weak → strong): $I^-<Br^-<S^{2-}<SCN^-<Cl^-<N_3^-<F^-<OH^-<C_2O_4^{2-}<H_2O<NCS^-<NH_3<en<NO_2^-<CN^-\approx CO$.
    2. Classify each of the 8 given ligands relative to $H_2O$:
       • $S^{2-}$ — weaker than $H_2O$.
       • $Br^-$ — weaker than $H_2O$.
       • $C_2O_4^{2-}$ — weaker than $H_2O$ (just below it in the series).
       • $CN^-$ — stronger than $H_2O$.
       • $en$ — stronger than $H_2O$.
       • $NH_3$ — stronger than $H_2O$.
       • $CO$ — stronger than $H_2O$.
       • $OH^-$ — weaker than $H_2O$. …

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