Coordination Compounds NEET PYQs (2005–2026) with Solutions

10 solved Coordination Compounds PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumOptical isomerism in octahedral complexes

   Given below are two statements: Statement-I : $\mathrm{[Fe(ox)_3]^{3-}}$ is chiral. Statement-II : trans – $\mathrm{[Cr(H_2O)_2(ox)_2]^{-}}$ is chiral. (Given : $\mathrm{oxH_2 = HOOC-COOH}$) In light of the above statements, choose the most appropriate answer from the options given below:

   1. AOption A: Both Statement-I and Statement-II are correct
   2. BOption B: Both Statement-I and Statement-II are incorrect
   3. COption C: Statement-I is correct but Statement-II is incorrect
   4. DOption D: Statement-I is incorrect but Statement-II is correct

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   Correct answer: (C) Statement-I is correct but Statement-II is incorrect

   Explanation

   Statement-I: $\mathrm{[Fe(ox)_3]^{3-}}$ is an octahedral complex of the type $\mathrm{[M(AA)_3]}$ with three bidentate oxalate ligands. It has no plane of symmetry and exists as two non-superimposable mirror images (d and l forms). So it is chiral, and Statement-I is correct. Statement-II: in trans-$\mathrm{[Cr(H_2O)_2(ox)_2]^{-}}$ the two oxalate ligands lie in one plane and the two water molecules are on opposite axial positions. The plane containing the oxalate ligands is a mirror plane, so the trans isomer is superimposable on its mirror image and is achiral. Only the cis isomer is optically active. Statement-II is incorrect. So the answer is option (C).

2. Question 2 (NEET 2026)

   NEET 2026EasyDouble salts: potash alum

   In potash alum, the ratio of $\mathrm{K^+}$ and $\mathrm{SO_4^{2-}}$ ions is

   1. AOption A: $1 : 2$
   2. BOption B: $2 : 1$
   3. COption C: $2 : 3$
   4. DOption D: $3 : 2$

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   Correct answer: (A) $1 : 2$

   Explanation

   Potash alum is the double salt $\mathrm{KAl(SO_4)_2 \cdot 12H_2O}$, also written $\mathrm{K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O}$. In water it dissociates completely into simple ions: $$\mathrm{KAl(SO_4)_2 \rightarrow K^+ + Al^{3+} + 2SO_4^{2-}}$$ Each formula unit gives 1 $\mathrm{K^+}$for every 2$\mathrm{SO_4^{2-}}$ions, so$\mathrm{K^+ : SO_4^{2-}} = 1 : 2$, option (A).

3. Question 3 (NEET 2026)

   NEET 2026MediumIsomerism

   Match List I with List II: Choose the correct answer from the options given below :

   1. AOption A: A–III, B–I, C–II, D–IV
   2. BOption B: A–I, B–III, C–II, D–IV
   3. COption C: A–II, B–IV, C–III, D–I
   4. DOption D: A–III, B–I, C–IV, D–II

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   Correct answer: (D) A–III, B–I, C–IV, D–II

   Explanation

   [Pt(NH₃)₂Cl₂] shows geometrical isomerism; [Co(en)₃]³⁺ shows optical isomerism; [Co(NH₃)₅NO₂]Cl₂ shows linkage isomerism; [Cr(H₂O)₆]Cl₃ shows solvate isomerism. Hence option (D) is correct.

4. Question 4 (NEET 2026)

   NEET 2026EasySpectrochemical series of ligands

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

   1. AOption A: $\mathrm{CO > NH_3 > H_2O > Cl^-}$
   2. BOption B: $\mathrm{CO > H_2O > NH_3 > Cl^-}$
   3. COption C: $\mathrm{Cl^- > H_2O > NH_3 > CO}$
   4. DOption D: $\mathrm{Cl^- > NH_3 > H_2O > CO}$

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   Correct answer: (A) $\mathrm{CO > NH_3 > H_2O > Cl^-}$

   Explanation

   The spectrochemical series (increasing field strength) from NCERT is: $$\mathrm{I^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < edta^{4-} < NH_3 < en < CN^- < CO}$$ Taking the four given ligands in decreasing order of field strength: $\mathrm{CO > NH_3 > H_2O > Cl^-}$. So the answer is option (A).

5. Question 5 (NEET 2026)

   NEET 2026EasyWerner Theory & Basic Concepts

   Which one of the following is an ambidentate ligand?

   1. AOption A: Ethane-1,2-diamine
   2. BOption B: Ethylenediaminetetraacetate ion
   3. COption C: Thiocyanate
   4. DOption D: Oxalate

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   Correct answer: (C) Thiocyanate

   Explanation

   Ambidentate ligands can coordinate through two different donor atoms. Thiocyanate (SCN⁻) can bind through S or N, making it ambidentate. Hence option (C) is correct.

6. Question 6 (NEET 2026)

   NEET 2026MediumCrystal field splitting and spin-only magnetic moment

   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)

   1. AOption A: $\mathrm{[Mn(CN)_6]^{3-}}$
   2. BOption B: $\mathrm{[Fe(CN)_6]^{3-}}$
   3. COption C: $\mathrm{[Co(NH_3)_6]^{3+}}$
   4. DOption D: $\mathrm{[Ti(H_2O)_6]^{3+}}$

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   Correct answer: (A) $\mathrm{[Mn(CN)_6]^{3-}}$

   Explanation

   Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}\ \mathrm{BM}$, where $n$ is the number of unpaired electrons. (A) $\mathrm{Mn^{3+}}$ is $3d^4$. $\mathrm{CN^-}$ is a strong-field ligand, so it is low spin, $t_{2g}^4$: $n = 2$, $\mu = \sqrt{8} \approx 2.83\ \mathrm{BM}$. (B) $\mathrm{Fe^{3+}}$ is $3d^5$, low spin with $\mathrm{CN^-}$, $t_{2g}^5$: $n = 1$, $\mu \approx 1.73\ \mathrm{BM}$. (C) $\mathrm{Co^{3+}}$ is $3d^6$, low spin with $\mathrm{NH_3}$, $t_{2g}^6$: $n = 0$, diamagnetic. (D) $\mathrm{Ti^{3+}}$ is $3d^1$: $n = 1$, $\mu \approx 1.73\ \mathrm{BM}$. The highest value is for $\mathrm{[Mn(CN)_6]^{3-}}$, option (A).

7. Question 7 (NEET 2026)

   NEET 2026MediumBonding & Crystal Field Theory

   Match List I with List II: Choose the correct answer from the options given below :

   1. AOption A: A–III, B–IV, C–I, D–II
   2. BOption B: A–III, B–I, C–IV, D–II
   3. COption C: A–IV, B–I, C–III, D–II
   4. DOption D: A–I, B–III, C–IV, D–II

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   Correct answer: (B) A–III, B–I, C–IV, D–II

   Explanation

   [PtCl₂(NH₃)₂] → square planar; [Co(NH₃)₆]³⁺ → octahedral; [NiCl₄]²⁻ → tetrahedral; [Fe(CO)₅] → trigonal bipyramidal. Hence option (B) is correct.

8. Question 8 (NEET 2026)

   NEET 2026EasyIUPAC nomenclature and formulas of complexes

   The formula of tetraammineaquachloridocobalt(III) chloride is

   1. AOption A: $\mathrm{[Co(NH_3)_4Cl_2] \times H_2O}$
   2. BOption B: $\mathrm{[Co(NH_3)_4]Cl_3 \times H_2O}$
   3. COption C: $\mathrm{[Co(NH_3)_4(H_2O)Cl]Cl}$
   4. DOption D: $\mathrm{[Co(NH_3)_4(H_2O)Cl]Cl_2}$

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   Correct answer: (D) $\mathrm{[Co(NH_3)_4(H_2O)Cl]Cl_2}$

   Explanation

   The name means the coordination sphere has 4 ammine ($\mathrm{NH_3}$), 1 aqua ($\mathrm{H_2O}$) and 1 chlorido ($\mathrm{Cl^-}$) ligand around $\mathrm{Co^{3+}}$, and chloride is the counter ion. Charge on the complex ion $= +3 + 4(0) + 0 + (-1) = +2$, i.e. $\mathrm{[Co(NH_3)_4(H_2O)Cl]^{2+}}$. Two $\mathrm{Cl^-}$ ions are needed to balance it, so the formula is $\mathrm{[Co(NH_3)_4(H_2O)Cl]Cl_2}$, option (D). Option (C) has the wrong charge balance. Options (A) and (B) do not have water and a single chloride inside the coordination sphere.

9. Question 9 (NEET 2026)

   NEET 2026EasyGeometrical isomerism: fac and mer isomers

   The complex which has facial and meridional isomers is (Given : py = pyridine and en = $\mathrm{H_2N-CH_2-CH_2-NH_2}$)

   1. AOption A: $\mathrm{[Cr(py)_3(Cl)_3]}$
   2. BOption B: $\mathrm{[Cr(H_2O)_6]^{3+}}$
   3. COption C: $\mathrm{[Co(NH_3)_4(H_2O)_2]^{3+}}$
   4. DOption D: $\mathrm{[Ni(en)_2(H_2O)_2]^{2+}}$

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   Correct answer: (A) $\mathrm{[Cr(py)_3(Cl)_3]}$

   Explanation

   Facial (fac) and meridional (mer) isomers occur in octahedral complexes of the type $\mathrm{[Ma_3b_3]}$. In fac, the three identical ligands sit on one face of the octahedron; in mer, they lie around a meridian. (A) $\mathrm{[Cr(py)_3Cl_3]}$ is $\mathrm{Ma_3b_3}$, so it shows fac and mer isomers. (B) $\mathrm{[Cr(H_2O)_6]^{3+}}$ is $\mathrm{Ma_6}$ and has no geometrical isomers. (C) $\mathrm{[Co(NH_3)_4(H_2O)_2]^{3+}}$ is $\mathrm{Ma_4b_2}$ and shows only cis–trans isomerism. (D) $\mathrm{[Ni(en)_2(H_2O)_2]^{2+}}$ is $\mathrm{M(AA)_2b_2}$ and shows cis–trans isomerism (the cis form is also optically active), not fac–mer. So the answer is option (A).

10. Question 10 (NEET 2026)

    NEET 2026EasyIonisation of complexes in solution

    A 1 : 3 electrolyte in an aqueous solution is

    1. AOption A: $\mathrm{[CoCl_2(NH_3)_4]Cl}$
    2. BOption B: $\mathrm{[CoCl(NH_3)_5]Cl_2}$
    3. COption C: $\mathrm{[Co(NH_3)_6]Cl_3}$
    4. DOption D: $\mathrm{[Co(NH_3)_3(NO_2)_3]}$

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    Correct answer: (C) $\mathrm{[Co(NH_3)_6]Cl_3}$

    Explanation

    Only the ions outside the square brackets are released in solution. (A) $\mathrm{[CoCl_2(NH_3)_4]Cl \rightarrow [CoCl_2(NH_3)_4]^+ + Cl^-}$: 1 : 1 electrolyte. (B) $\mathrm{[CoCl(NH_3)_5]Cl_2 \rightarrow [CoCl(NH_3)_5]^{2+} + 2Cl^-}$: 1 : 2 electrolyte. (C) $\mathrm{[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-}$: 1 : 3 electrolyte. (D) $\mathrm{[Co(NH_3)_3(NO_2)_3]}$ is a neutral complex and a non-electrolyte. So the answer is option (C).