Q.Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

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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