Q.Explain $[Co(NH_3)_6]^{3+}$ is an inner orbital complex whereas $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

Showing the 12 most recent of 17 on this concept.

1. KCET 2026Set D31 markMCQ
   Q.How many ions per molecule are produced from the complex $[Co(NH_3)_6]Cl_3$ in solution? (A) 6 (B) 4 (C) 3 (D) 2

   ›Reveal solutionSolution

   The number of ions per formula unit equals the complex ion plus the free counter-ions written outside the coordination sphere.

   Step 1 — Identify the coordination sphere

   In $[Co(NH_3)_6]Cl_3$, all six $NH_3$ ligands are bound directly to cobalt inside the square brackets, so they do not ionize; only what is written outside the brackets ionizes in solution.

   Step 2 — Write the dissociation …

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2. KCET 2025Set D-41 markMCQ
   Q.A ligand which has two different donor atoms and either of the two ligates with the central metal atom/ion in the complex is called (A) Chelate ligand (B) Unidentate ligand (C) Polydentate ligand (D) Ambidentate ligand

   ›Reveal solutionSolution

   The defining phrase is "two different donor atoms, either of the two ligates" — one binding site used at a time, chosen from two candidates — which is exactly the definition of an ambidentate ligand.

   Step 1 — Parse the definition given in the stem

   The stem specifies three things:

   1. the ligand has two different donor atoms;
   2. either one of them can bond to the metal;
   3. (implicitly) only one at a time actually coordinates — hence "either", not "both".

   So the ligand occupies only one coordination site, but it has a choice of which of its own atoms to use.

   Step 2 — Run through the four terms and see which fits

   (A) Chelate ligand — a di- or polydentate ligand that grips the metal through two or more donor atoms simultaneously, forming a ring (the "claw"). E.g. ethylenediamine (en) binds through both N atoms at once; oxalate through both O's. ✗ — this binds through both, not either.

   (B) Unidentate (monodentate) ligand — donates through exactly one donor atom, and it has only that one available. E.g. $\ce{NH3}$ (N only), $\ce{Cl-}$, $\ce{H2O}$. ✗ — it uses one site, yes, but there is no choice of two different donor atoms. That is precisely the feature the stem adds.

   (C) Polydentate ligand — binds through several donor atoms at once. E.g. EDTA⁴⁻ is hexadentate (2 N + 4 O, all six coordinating). ✗ — again, all sites used together.

   (D) Ambidentate ligand — from Latin ambi = "both / either of two". It has two different donor atoms but coordinates through only one of them at a time. ✓ This is exactly the stem.

   Step 3 — The standard examples (worth knowing)

   Ligand	Donor used	Name of complex prefix	Example
   $\ce{NO2-}$	N	nitro ($\ce{-NO2}$)	$[\ce{Co(NH3)5(NO2)}]^{2+}$
   $\ce{NO2-}$	O	nitrito ($\ce{-ONO}$)	$[\ce{Co(NH3)5(ONO)}]^{2+}$

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3. KCET 2025Set D-41 markMCQ
   Q.In the complex ion $\left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_3\right]^{3-}$, the co-ordination number of Fe is (A) 4 (B) 5 (C) 6 (D) 3

   ›Reveal solutionSolution

   Coordination number counts donor atoms, not ligand molecules — oxalate is bidentate, so three oxalates supply $3 \times 2 = 6$ donor atoms.

   Step 1 — The concept: coordination number ≠ number of ligands

   The coordination number (CN) of the central metal is the number of donor atoms (i.e. the number of sigma bonds / ligand bonds) directly attached to it — not the number of ligand molecules or ions.

   These two counts agree only when every ligand is unidentate. Whenever a chelating ligand is present, you must multiply by its denticity:

   $\text{CN} = \sum_{\text{ligands}} (\text{number of that ligand}) \times (\text{its denticity})$

   Step 2 — What is the denticity of oxalate?

   The oxalate ion, $\ce{C2O4^{2-}}$ (abbreviated ox), is the conjugate base of oxalic acid:

   $\begin{array}{c} \ce{^-O2C-CO2^-} \end{array}$

   It carries two carboxylate groups, and one oxygen from each group coordinates to the metal. Both donor atoms bind simultaneously, wrapping around the metal to close a stable five-membered chelate ring (M–O–C–C–O).

   $\therefore\ \text{oxalate is } \textbf{BIDENTATE} \quad (\text{denticity} = 2)$

   This is why oxalate is a chelating ligand and why $[\ce{Fe(C2O4)3}]^{3-}$ is unusually stable (the chelate effect).

   Step 3 — Count the donor atoms

   The complex is $\left[\ce{Fe(C2O4)3}\right]^{3-}$ — three oxalate ligands, each bidentate:

   $\text{CN} = 3 \;\text{ligands} \times 2 \;\frac{\text{donor atoms}}{\text{ligand}} = \boxed{6}$

   All six donor atoms are oxygens, arranged octahedrally about the iron.

   Step 4 — The trap …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.An aqueous solution of $\mathrm{CrCl}_3 .6 \mathrm{H}_2 \mathrm{O}$ (Molar mass $=266.5 \mathrm{~g} / \mathrm{mol}$ ) containing 2.665 g of the solute after processing, when treated with excess of $\mathrm{AgNO}_3$ gave 2.87 g of AgCl (Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$ ) Choose the correct formula of the compound which give these results. (A) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \cdot 2 \mathrm{H}_2 \mathrm{O}$ (B) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right] \cdot 3 \mathrm{H}_2 \mathrm{O}$ (C) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$ (D) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$

   ›Reveal solutionSolution

   The AgCl formed shows $2$ free chloride ions per formula unit, which fixes the structure as $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$ — option (C).

   Concept

   In a coordination compound, chloride ions bound to the metal (written inside the square brackets) stay put, while chloride ions outside the coordination sphere are free ions that precipitate with $\text{AgNO}_3$ as AgCl. So counting the AgCl tells us the number of ionizable chlorides.

   Solution

   1. Moles of complex: $\dfrac{2.665}{266.5} = 0.0100\,\text{mol}$.
   2. Moles of AgCl: $\dfrac{2.87}{143.5} = 0.0200\,\text{mol}$.
   3. Free chlorides per formula unit: $\dfrac{0.0200}{0.0100} = 2$.
   4. Test the options for the number of chlorides outside the bracket:
      • (A) $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl}\cdot 2\text{H}_2\text{O}$: 1 free Cl.
      • (B) $[\text{Cr}(\text{H}_2\text{O})_3\text{Cl}_3]\cdot 3\text{H}_2\text{O}$: 0 free Cl.
      • (C) $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$: 2 free Cl. …

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5. KCET 2024Set B-21 markMCQ
   Q.On treating 100 mL of 0.1 M aqueous solution of the complex $\mathrm{CrCl_3.6H_2O}$ with excess of $\mathrm{AgNO_3}$, 2.86 g of AgCl was obtained. The complex is : (A) $\left[\mathrm{Cr(H_2O)_3Cl_3}\right].3\mathrm{H_2O}$ (B) $\left[\mathrm{Cr(H_2O)_4Cl_2}\right]\mathrm{Cl}.2\mathrm{H_2O}$ (C) $\left[\mathrm{Cr(H_2O)_5Cl}\right]\mathrm{Cl}_2.\mathrm{H_2O}$ (D) $\left[\mathrm{Cr(H_2O)_6}\right]\mathrm{Cl}_3$

   ›Reveal solutionSolution

   The number of moles of AgCl precipitated tells us how many chloride ions are free (outside the coordination sphere). From the given data, 0.02 mol of AgCl forms from 0.01 mol of complex, meaning 2 Cl⁻ are free — so the complex has one Cl inside and two outside. The correct option is (C).

   The key idea is that only chloride ions that are outside the coordination sphere — the counter-ions — will react with AgNO₃ to give AgCl. Chloride ions that are coordinated to the metal (inside the square brackets) do not precipitate. So the mass of AgCl tells us exactly how many free Cl⁻ ions each formula unit releases.

   1. Find moles of AgCl formed.

      Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol.

      Moles of AgCl = $\frac{2.86}{143.5} \approx 0.02$ mol.

   2. Find moles of the complex taken.

      Volume = 100 mL = 0.1 L, concentration = 0.1 M.

      Moles of complex = $0.1 \times 0.1 = 0.01$ mol.

   3. Find the ratio of free Cl⁻ per complex molecule.

      $\frac{\text{moles AgCl}}{\text{moles complex}} = \frac{0.02}{0.01} = 2$.

      So each formula unit of the complex gives 2 chloride ions that precipitate with Ag⁺.

   4. Interpret the ratio.

      The complex has the formula $\mathrm{CrCl_3 \cdot 6H_2O}$, so there are 3 Cl atoms total. If 2 are free (outside the coordination sphere), then only 1 Cl is inside the coordination sphere (bonded to Cr). The water molecules also distribute between inside and outside.

   5. Match with the options.

      • (A) $\mathrm{[Cr(H_2O)_3Cl_3] \cdot 3H_2O}$ — all 3 Cl are inside → 0 free Cl⁻. Wrong. …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.The molar conductivity of the complex $\mathrm{CoCl}_3 \cdot 4 \mathrm{NH}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is found to be the same as that of a $1: 3$ electrolyte. The structural formula of the compound is : (A) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_3$ (B) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\left(\mathrm{H}_2 \mathrm{O}\right)\right] \mathrm{Cl}_2 \mathrm{H}_2 \mathrm{O}$ (C) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right) \mathrm{Cl}_2\right] \mathrm{Cl} \mathrm{H}_2 \mathrm{O}$ (D) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}\right] \mathrm{Cl}_2$

   ›Reveal solutionSolution

   Behaving as a $1:3$ electrolyte (4 ions) requires all three $Cl^-$ outside the coordination sphere: $[\mathrm{Co}(\mathrm{NH_3})_4(\mathrm{H_2O})_2]\mathrm{Cl}_3$.

   A $1:3$ electrolyte dissociates into 4 ions total: one complex cation and three anions. So all three chloride ions must be outside the coordination sphere (ionisable), and both water molecules coordinate to cobalt. The complex cation is

   $[\mathrm{Co}(\mathrm{NH_3})_4(\mathrm{H_2O})_2]^{3+}$

   (giving $\mathrm{Co}$ its $+3$ oxidation state, coordination number 6), balanced by $3\,\mathrm{Cl}^-$: …

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7. COMEDK 2024Set 2024-M1 markMCQ
   Q.A Coordination compound is represented by the formula $[\mathrm{CoBr}_3(e n) x]$. This compound required one mole of $\mathrm{AgNO}_3$ to form a pale yellow precipitate of $\mathrm{AgBr}$. What is the value of $\mathrm{x}$ in the compound? (A) 2 (B) 1 (C) 4 (D) 3

   ›Reveal solutionSolution

   The key is that one mole of AgNO₃ precipitates one mole of free Br⁻ ions; the complex has only one ionic bromide, so the coordination sphere must contain the other two bromides and the ethylenediamine (en) ligands, giving x = 2.

   Concept & Intuition

   Coordination compounds can have some ligands tightly bound inside the coordination sphere (non‑ionic) and others outside as counter‑ions (ionic). When you add AgNO₃, only the free, ionic bromide ions (Br⁻) react to form pale yellow AgBr precipitate. The problem says exactly one mole of AgNO₃ is needed per mole of the compound, meaning only one Br⁻ is ionic. The rest must be inside the coordination sphere. Since the formula is written as $[\mathrm{CoBr}_3(en)_x]$, the square brackets indicate the coordination sphere. The total number of bromides is 3, but only one is outside the brackets (ionic), so the other two must be inside. That forces the charge balance and tells us how many neutral en ligands are needed.

   Step‑by‑step reasoning

   1. Identify the ionic bromide count

      One mole of AgNO₃ gives one mole of AgBr precipitate. This means the compound releases exactly one mole of free Br⁻ ions per mole of complex. So there is one ionic bromide outside the coordination sphere.

   2. Determine the composition of the coordination sphere

      The formula is written as $[\mathrm{CoBr}_3(en)_x]$. The square brackets enclose the coordination sphere. Since there are three bromine atoms total and only one is ionic, the other two bromine atoms must be inside the brackets as ligands. So the coordination sphere contains Co, two Br ligands, and x molecules of ethylenediamine (en).

   3. Balance the charge

      Cobalt in this complex is typically in the +3 oxidation state (common for Co(III) ammine/amine complexes). Each bromide ligand inside the sphere carries a –1 charge, so two bromides contribute –2. The en ligand is neutral. Therefore the charge on the complex ion is:

   $\text{Charge} = (+3) + 2(-1) = +1$

   So the complex ion is $[\mathrm{CoBr}_2(en)_x]^+$. To balance this, there must be one counter‑ion — and that counter‑ion is the one ionic Br⁻ we already identified. This matches perfectly: the compound is $[\mathrm{CoBr}_2(en)_x]\mathrm{Br}$.

   4. Determine x from coordination number …

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8. KCET 2023Set D-21 markMCQ
   Q.The number of protons, neutrons and electrons in the ion ${}^{32}_{16}\mathrm{S}^{2-}$ respectively are (A) $16, 18, 16$ (B) $16, 16, 18$ (C) $18, 16, 16$ (D) $16, 16, 16$

   ›Reveal solutionSolution

   The atomic number gives protons and electrons in a neutral atom; the mass number gives neutrons; the charge tells how many extra electrons are present. For ${}^{32}_{16}\mathrm{S}^{2-}$, the counts are 16 protons, 16 neutrons, and 18 electrons — option (B).

   The key is to read the notation ${}^{32}_{16}\mathrm{S}^{2-}$ piece by piece. The subscript (16) is the atomic number $Z$ — that’s the number of protons. In any atom or ion, the number of protons never changes; it defines the element. So protons = 16.

   The superscript (32) is the mass number $A$, which is the sum of protons and neutrons. Neutrons = $A - Z = 32 - 16 = 16$.

   Now the charge: $2-$ means the ion has gained two extra electrons compared to the neutral atom. A neutral sulfur atom has as many electrons as protons — 16. With a $2-$ charge, it has $16 + 2 = 18$ electrons.

   Let’s check each option:

   1. Protons: always 16. That eliminates (A) and (C) immediately, since they show 18 or 16 but in the wrong order.
   2. Neutrons: 16. Both remaining options (B) and (D) have 16 neutrons — so far so good. …

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9. COMEDK 2023Set 2023-E1 markMCQ
   Q.Match the Coordination compounds given in Column I with their characteristic features listed in Column II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} S.No. Coordination compounds S.No. Characteristic features W $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_3$ P $\text { Oxidation state }=+3 \text { Configuration }=\mathrm{d}^5 \quad \mu=5.92 \mathrm{BM}$ X $\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right.$ Q $\text { Oxidation state }=+3 \text { Configuration }=\mathrm{d}^3 \quad \mu=3.87 \mathrm{BM}$ Y $\left[\mathrm{CrCl}_3(\mathrm{py})_3\right]$ R $\text { Oxidation state }=+3 \text { Configuration }=d^6 \quad \mu=0 \mathrm{BM}$ Z $\mathrm{Cs}\left[\mathrm{FeCl}_4\right]$ S $\text { Oxidation state }=+2 \text { Configuration }=\mathrm{d}^5 \quad \mu=1.732 \mathrm{BM}$ (A) $\mathrm{W}=\mathrm{S} \quad \mathrm{X}=\mathrm{R} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$ (B) $\mathrm{W}=\mathrm{S} \quad \mathrm{X}=\mathrm{P} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{R}$ (C) $\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{P} \quad \mathrm{Z}=\mathrm{Q}$ (D) $\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$

   ›Reveal solutionSolution

   [!TLDR]

   Work out oxidation state, d-count and spin state for each complex; the spin-only moments identify W=R (d⁶, μ=0), X=S (d⁵ LS, μ=1.73), Y=Q (d³, μ=3.87), Z=P (d⁵ HS, μ=5.92).

   Concept

   Spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Strong-field ligands (CN⁻, NH₃) tend to pair electrons (low spin); weak-field ligands (Cl⁻) keep them unpaired (high spin) — core CBSE/NCERT coordination-chemistry ideas.

   Solution

   W = [Co(NH₃)₅Cl]³⁺ type, Co(III): $d^6$. With ammine ligands it is low spin, all electrons paired, $n=0$, $\mu=0$ → feature R ($d^6,\ \mu=0$).

   X = K₄[Mn(CN)₆], Mn(II): $d^5$. CN⁻ is strong field → low spin, one unpaired electron, $n=1$, $\mu=\sqrt{1\cdot3}=1.73$ BM → feature S (OS $+2$, $d^5$, $\mu=1.732$). …

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10. KCET 2022Set B-31 markMCQ
    Q.The complex hexamine platinum (IV) chloride will give ______ number of ions on ionization. (A) 3 (B) 2 (C) 5 (D) 4

    ›Reveal solutionSolution

    The key is to determine the correct formula of the complex and then count the ions produced when it dissociates in solution. Hexamine platinum(IV) chloride gives 5 ions on ionization.

    The question asks about "complex hexamine platinum (IV) chloride". The name tells you the coordination compound directly. "Hexamine" means six ammonia ($NH_3$) ligands are attached to the central metal. "Platinum (IV)" tells you the oxidation state of platinum is +4. "Chloride" at the end indicates that chlorine is present as a counter-ion (outside the coordination sphere), not as a ligand.

    So the coordination sphere is $[Pt(NH_3)_6]^{4+}$. To balance the +4 charge, you need four chloride ions, each with a -1 charge. The full formula is therefore $[Pt(NH_3)_6]Cl_4$.

    Now, when this compound is dissolved in water (ionization), the complex ion and the chloride ions separate. The coordination sphere itself does not break apart — the $NH_3$ ligands remain firmly attached to platinum. So the ionization is:

    $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$

    Count the ions produced: one complex cation and four chloride anions. That gives a total of 5 ions. …

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11. KCET 2021Set B-21 markMCQ
    Q.Homoleptic complexes among the following are (A) $\mathrm{K_3[Al(C_2O_4)_3]}$, (B) $\mathrm{[CoCl_2(en)_2]^+}$ (C) $\mathrm{K_2[Zn(OH)_4]}$ (A) A only (B) (A) and (B) only (C) (A) and (C) only (D) (C) only

    ›Reveal solutionSolution

    Homoleptic = one ligand type only; the oxalato-aluminate and the tetrahydroxozincate qualify, the mixed chloro/en cobalt complex does not.

    Step 1 — The definition.

    In a homoleptic complex the central metal ion is bonded to only one type of donor (ligand). If two or more different ligands are attached, the complex is heteroleptic.

    Step 2 — Examine each complex.

    1. $\mathrm{K_3[Al(C_2O_4)_3]}$ — the coordination entity is $[\mathrm{Al(C_2O_4)_3}]^{3-}$. The only ligand present is the oxalate ion $\mathrm{C_2O_4^{2-}}$ (three of them, each bidentate, giving CN = 6). One ligand type ⇒ homoleptic.

    2. $\mathrm{[CoCl_2(en)_2]^+}$ — here cobalt is bonded to two different ligands: two chloride ions $\mathrm{Cl^-}$ and two ethylenediamine molecules (en). Two ligand types ⇒ heteroleptic. …

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12. COMEDK 2021Set 20211 markMCQ
    Q.Which type of ligand is EDTA? (A) Monodentate (B) Hexadentate (C) Bidentate (D) Tridentate

    ›Reveal solutionSolution

    Total = 6 donor atoms, so EDTA is a HEXADENTATE chelating ligand; it wraps around an octahedral metal ion occupying all six coordination sites (e.g. [Ca(EDTA)]2-).

    Concept: Denticity of a ligand = number of donor atoms it uses to bind the metal.

    EDTA (ethylenediaminetetraacetate, EDTA4-) has:

    • 2 nitrogen donor atoms (the two amine N of the ethylenediamine backbone)
    • 4 oxygen donor atoms (one from each of the four carboxylate arms) …

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