Coordination Compounds JEE Main previous year questions with solutions

5 solved JEE Main questions on Coordination Compounds, free to read — no sign-in needed. The full chapter has 373 questions; sign in to attempt the remaining 368 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Crystal Field Theory
   Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex? A. $d_{xy} = d_{xz} \gt d_{x^2-y^2}$ B. $d_{xy} = d_{yz} \gt d_{z^2}$ C. $d_{x^2-y^2} \gt d_{z^2} \gt d_{xz}$ D. $d_{x^2-y^2} = d_{z^2} \lt d_{xz}$ Choose the correct answer from the given below:
   1. A.A, B and D only
   2. B.A and B only
   3. C.B and D only
   4. D.B, C and D only
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   Answer: (A)

   In a tetrahedral complex, the five d-orbitals split into two sets due to the crystal field created by the ligands. The ligands approach the central metal ion between the axes. The $d_{xy}$, $d_{yz}$, and $d_{xz}$ orbitals (forming the $t_2$ set) point between the axes, experiencing greater repulsion from the ligands, and thus have higher energy. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (forming the $e$ set) point along the axes, experiencing lesser repulsion, and thus have lower energy. Therefore, the energy order is $d_{xy} = d_{yz} = d_{xz} \gt d_{x^2-y^2} = d_{z^2}$. Evaluating the given statements: A. $d_{xy} = d_{xz} \gt d_{x^2-y^2}$ is true. B. $d_{xy} = d_{yz} \gt d_{z^2}$ is true. C. $d_{x^2-y^2} \gt d_{z^2} \gt d_{xz}$ is false. D. $d_{x^2-y^2} = d_{z^2} \lt d_{xz}$ is true. Hence, statements A, B, and D are true. Answer: A, B and D only
2. Q2JEE Main 2025 (07 Apr, Shift 1)Isomerism of coordination compounds
   An octahedral complex having molecular composition Co. $5 \mathrm{NH}_3. \mathrm{Cl}. \mathrm{SO}_4$ has two isomers A and B. The solution of A gives a white precipitate with $\mathrm{AgNO}_3$ solution and the solution of B gives white precipitate with $\mathrm{BaCl}_2$ solution. The type of isomerism exhibited by the complex is,
   1. A.Co-ordinate isomerism
   2. B.Linkage isomerism
   3. C.Ionisation isomerism
   4. D.Geometrical isomerism
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   Answer: (C)

   (A) complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{Cl}$ (B) complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$ Both $(\mathrm{A})$ and $(\mathrm{B})$ are Ionisation isomers.
3. Q3JEE Main 2024 (06 Apr, Shift 2)Nomenclature of coordination compounds
   The correct IUPAC name of $\left[\mathrm{PtBr}_2\left(\mathrm{PMe}_3\right)_2\right]$ is :
   1. A.dibromodi(trimethylphosphine)platinum(II)
   2. B.bis(trimethylphosphine)dibromoplatinum(II)
   3. C.dibromobis(trimethylphosphine)platinum(II)
   4. D.bis[bromo(trimethylphosphine)]platinum(II)
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   Answer: (C)

   Dibromo bis(trimethylphosphine) platinum (II)
4. Q4JEE Main 2023 (15 Apr, Shift 1)Terminology of coordination compounds
   The homoleptic and octahedral complex of ${Co}^{2+}$ and ${H}_{2}O$ has ______ unpaired electron(s) in the ${t}_{2g}$ set of orbitals.
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   Answer: 1

   The electronic configuration of ${Co}^{2+}$ is $3{d}^{7}$. In an octahedral complex of ${Co}^{2+}$ with ${H}_{2}O$ ligands, the d orbitals split into two sets of energy levels: ${t}_{2g}$ (lower energy) and ${e}_{g}$ (higher energy). In this case, since the complex is octahedral, we have six ligands surrounding the central cobalt ion. Each ligand donates a pair of electrons to the cobalt ion, occupying the available d orbitals. So, ${\left[Co{\left({H}_{2}O\right)}_{6}\right]}^{2+}={d}^{7}\left(Highspin\right)={t}_{2g}^{5}{e}_{g}^{2}$ In the ${t}_{2g}$set of orbitals, there will be $1$ unpaired electron
5. Q5JEE Main 2022 (27 Jul, Shift 2)Carbonyl Compounds
   Low oxidation state of metals in their complexes are common when ligands
   1. A.have good $\pi$-accepting character
   2. B.have good $\sigma$-donor character
   3. C.are havind good $\pi$-donating ability
   4. D.are havind poor $\sigma$-donating ability
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   Answer: (A)

   When metal has low oxidation state, it has more electron density in d orbital. So it has more tendency to back donate electrons. Thus compound must have good $\pi$ acceptor ligand. π backbonding, also called π backdonation, is when electrons move from an atomic orbital on one atom to an appropriate symmetry antibonding orbital on a π-acceptor ligand.It is especially common in the organometallic chemistry of transition metals with multi-atomic ligands such as carbon monoxide, ethylene or the nitrosonium cation. Electrons from the metal are used to bond to the ligand, in the process relieving the metal of excess negative charge. Compounds where π backbonding occurs include $Ni(CO{)}_{4}$ and Zeise's salt. The metal to ligand bonding creates a synergic effect which strengthens the bond between ligand and the metal