NEET Coordination Compounds PYQs & Practice Questions

10 free NEET Chemistry questions on Coordination Compounds, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Magnetic Properties · Magnetic Properties of Complexes · Crystal Field Splitting / Spectrochemical Series · Ionic vs Non-ionizable Ligands / Conductance · Wavelength Absorption Order from Crystal Field Splitting · Determining Electrolyte Type from AgCl Precipitation · Determining a Complex's Formula from AgCl Precipitation · IUPAC Name of [Pt(NH3)2Cl2] · Type of Isomerism Shown by Two Complexes Differing Only in Counter-Ion · Homoleptic and Heteroleptic Complexes · Stability of Complexes (Chelate Effect) · Spectrochemical Series - Crystal Field Theory

1. Q1NEET 2026Spectrochemical Series - Crystal Field Theory

   According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is:

   • A. CO > H2O > NH3 > Cl⁻
   • B. Cl⁻ > H2O > NH3 > CO
   • C. Cl⁻ > NH3 > H2O > CO
   • D. CO > NH3 > H2O > Cl⁻
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   Answer: C — Cl⁻ > NH3 > H2O > CO

   According to the spectrochemical series, ligand field strength decreases in the order: CO > CN⁻ > NH3 > H2O > OH⁻ > Cl⁻ (among others). So among the given ligands: CO > NH3 > H2O > Cl⁻.

2. Q2NEET 2026Optical Isomerism in Coordination Complexes

   Given below are two statements: Statement-I: [Fe(ox)3]3− is chiral. Statement-II: trans-[Cr(H2O)2(ox)2]⁻ is chiral. (Given: oxH2 = HOOC–COOH, i.e., ox = oxalate, a bidentate ligand) In light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both Statement-I and Statement-II are incorrect.
   • B. Statement-I is correct but Statement-II is incorrect.
   • C. Statement-I is incorrect but Statement-II is correct.
   • D. Both Statement-I and Statement-II are correct.
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   Answer: A — Both Statement-I and Statement-II are incorrect.

   [Fe(ox)3]3− is a tris-chelate octahedral complex with three bidentate oxalate ligands; such tris-bidentate complexes lack any plane of symmetry and exist as non-superimposable mirror-image (Δ/Λ) enantiomers, so they are chiral -- Statement-I is correct. In trans-[Cr(H2O)2(ox)2]⁻, the two water ligands are trans to each other, and this arrangement possesses a plane of symmetry passing through the two water ligands and bisecting the two oxalate rings, making the trans isomer achiral (it is the cis isomer that is chiral) -- Statement-II is incorrect.

3. Q3NEET 2026Spin-Only Magnetic Moment of Complexes

   Among the species given below, the spin-only magnetic moment is highest for: (Given: Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

   • A. [Fe(CN)6]3−
   • B. [Co(NH3)6]3+
   • C. [Ti(H2O)6]3+
   • D. [Mn(CN)6]3−
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   Answer: C — [Ti(H2O)6]3+

   [Fe(CN)6]3−: Fe3+ is d5; CN⁻ is strong field, low spin, giving t2g5eg0 with 1 unpaired electron, μ=√3≈1.73 BM. [Co(NH3)6]3+: Co3+ is d6, low spin (common for Co3+), t2g6eg0, 0 unpaired electrons, μ=0. [Ti(H2O)6]3+: Ti3+ is d1, 1 unpaired electron, μ=√3≈1.73 BM. [Mn(CN)6]3−: Mn3+ is d4; even with strong-field CN⁻, low-spin d4 gives t2g4eg0 with 2 unpaired electrons, μ=√(2×4)=√8≈2.83 BM, which is the highest among these four.

4. Q4NEET 2026Facial and Meridional Isomerism

   The complex which has facial and meridional isomers is: (Given: py = pyridine and en = H2N–CH2–CH2–NH2)

   • A. [Cr(H2O)6]3+
   • B. [Co(NH3)4(H2O)2]3+
   • C. [Ni(en)2(H2O)2]2+
   • D. [Cr(py)3Cl3]
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   Answer: C — [Ni(en)2(H2O)2]2+

   Facial (fac) and meridional (mer) isomerism occurs specifically in octahedral complexes of the type MA3B3, with three identical ligands of each of two kinds. [Cr(py)3Cl3] is of this type (three pyridine and three chloride ligands) and shows fac/mer isomerism. [Cr(H2O)6]3+ has only one type of ligand (no isomerism of this kind). [Co(NH3)4(H2O)2]3+ is an MA4B2 type, which shows cis/trans isomerism, not fac/mer. [Ni(en)2(H2O)2]2+ involves a bidentate ligand and doesn't fit the MA3B3 pattern either.

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5. Q5NEET 2026Electrolytic Nature of Coordination Compounds

   A 1:3 electrolyte in an aqueous solution is:

   • A. [CoCl(NH3)5]Cl2
   • B. [Co(NH3)6]Cl3
   • C. [Co(NH3)3(NO2)3]
   • D. [CoCl2(NH3)4]Cl
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   Answer: A — [CoCl(NH3)5]Cl2

   [Co(NH3)6]Cl3 dissociates in water into [Co(NH3)6]3+ and 3Cl⁻ ions, giving a 1:3 electrolyte (1 complex cation : 3 chloride anions). [CoCl(NH3)5]Cl2 is a 1:2 electrolyte, [Co(NH3)3(NO2)3] is a neutral, non-electrolyte complex (all ligands inside the coordination sphere), and [CoCl2(NH3)4]Cl is a 1:1 electrolyte.

6. Q6NEET 2026IUPAC Nomenclature of Coordination Compounds

   The formula of tetraammineaquachloridocobalt(III) chloride is:

   • A. [Co(NH3)4]Cl3 × H2O
   • B. [Co(NH3)4(H2O)Cl]Cl
   • C. [Co(NH3)4(H2O)Cl]Cl2
   • D. [Co(NH3)4Cl2] × H2O
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   Answer: B — [Co(NH3)4(H2O)Cl]Cl

   Tetraammineaquachloridocobalt(III) chloride has, inside the coordination sphere, 4 ammine (NH3), 1 aqua (H2O), and 1 chlorido (Cl⁻) ligands around Co(III): [Co(NH3)4(H2O)Cl]. The charge on this complex ion = +3 (Co) + 0 (NH3×4) + 0 (H2O) - 1 (Cl) = +2. To balance this +2 charge, 2 chloride ions are needed outside the coordination sphere, giving the formula [Co(NH3)4(H2O)Cl]Cl2.

7. Q7NEET 2026Matching complex geometry/shape

   Match List I with List II: List I (Complex/Ion) — A. [Pt(Cl2)(NH3)2], B. [Co(NH3)6]3+, C. [NiCl4]2-, D. [Fe(CO)5]. List II (Shape/geometry) — I. Octahedral, II. Trigonal bipyramidal, III. Square planar, IV. Tetrahedral.

   • A. A-I, B-III, C-IV, D-II
   • B. A-III, B-IV, C-I, D-II
   • C. A-III, B-I, C-IV, D-II
   • D. A-IV, B-I, C-III, D-II
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   Answer: B — A-III, B-IV, C-I, D-II

   [Pt(Cl2)(NH3)2] is square planar (III); [Co(NH3)6]3+ is octahedral (I); [NiCl4]2- is tetrahedral (IV); [Fe(CO)5] is trigonal bipyramidal (II). A-III, B-I, C-IV, D-II.

8. Q8NEET 2026Geometry and statements about ClF3

   Identify the correct statement about ClF3 from the following options:

   • A. It has T-shaped geometry with three lone pairs on Cl atom
   • B. It has T-shaped geometry with two lone pairs on Cl atom
   • C. It has a trigonal pyramidal geometry with two lone pairs on Cl atom
   • D. It has a planar trigonal geometry with two lone pairs on Cl atom
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   Answer: A — It has T-shaped geometry with three lone pairs on Cl atom

   ClF3 has sp3d hybridization with 2 lone pairs and 3 bond pairs, giving T-shaped geometry with two lone pairs on the Cl atom.

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9. Q9NEET 2026Matching complex to type of isomerism

   Match List I with List II: List I (Complex) — A. [Pt(NH3)2Cl2], B. [Co(en)3]3+, C. [Co(NH3)5NO2]Cl2, D. [Cr(H2O)6]Cl3. List II (Type of isomerism) — I. Optical, II. Solvate, III. Geometrical, IV. Linkage.

   • A. A-I, B-III, C-II, D-IV
   • B. A-II, B-IV, C-III, D-I
   • C. A-III, B-I, C-IV, D-II
   • D. A-III, B-I, C-II, D-IV
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   Answer: B — A-II, B-IV, C-III, D-I

   [Pt(NH3)2Cl2] shows geometrical isomerism (III); [Co(en)3]3+ shows optical isomerism (I); [Co(NH3)5NO2]Cl2 shows linkage isomerism (IV); [Cr(H2O)6]Cl3 shows solvate (hydrate) isomerism (II). A-III, B-I, C-IV, D-II.

10. Q10NEET 2026Incorrect statement among group 15 elements

    Identify the incorrect statement from the following: (1) Phosphorus, arsenic and antimony show catenation property. (2) P(C2H5)3 and As(C6H5)3 form dπ-dπ bond with transition metals. (3) Nitrogen can form dπ-pπ bond with oxygen. (4) Nitrogen can form pπ-pπ multiple bonds with itself.

    • A. Phosphorus, arsenic and antimony show catenation property
    • B. P(C2H5)3 and As(C6H5)3 form dπ-dπ bond with transition metals
    • C. Nitrogen can form dπ-pπ bond with oxygen
    • D. Nitrogen can form pπ-pπ multiple bonds with itself
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    Answer: B — P(C2H5)3 and As(C6H5)3 form dπ-dπ bond with transition metals

    Nitrogen, lacking d-orbitals, cannot form dπ-pπ bonds; it forms pπ-pπ bonds. Hence statement 3 is incorrect.