Coordination Compounds JEE Main previous year questions with solutions

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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 (08 Apr, Shift 1)Back Bonding
   Given below are two statements: Statement I: $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ can act as ligands to form transition metal complexes. Statement II: As N and P are from same group, the nature of bonding of $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ is always same with transition metals. In the light of the above statements, choose the most appropriate answer from the options given below:
   1. A.Statement I is correct but Statement II is incorrect.
   2. B.Statement I is incorrect but Statement II is correct.
   3. C.Both Statement I and Statement II are correct.
   4. D.Both Statement I and Statement II are incorrect.
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   Answer: (A)

   $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ both are Lewis base and acts as ligand, However, $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ has a $\pi$-acceptor character.
4. Q4JEE Main 2023 (06 Apr, Shift 2)Nomenclature of coordination compounds
   The IUPAC name of ${K}_{3}[Co({C}_{2}{O}_{4}{)}_{3}]$ is:
   1. A.Potassium tris(oxalato)cobaltate$\left(III\right)$
   2. B.Potassium tris(oxalato)cobalt$(III)$
   3. C.Potassium trioxalatocobalt$(III)$
   4. D.Potassium trioxalatocobaltate$\left(III\right)$
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   Answer: (D)

   Ligands that include a numerical prefix in the name use the prefixes bis for $2$, tris for $3$, or tetrakis for $4$ to indicate their number. If the complex ion is an anion, we drop the ending of the metal name and add –ate. In the given coordination compound, $Co$ is the cobalt metal, ${C}_{2}{O}_{4}^{2-}$ are the ligands named as oxalato. The charge on $[Co({C}_{2}{O}_{4}{)}_{3}]$ is $-3$ and oxalate is having charge of $-2$. $X+3(-2)=-3 X=+3$ Hence, the IUPAC name of the compound ${K}_{3}[Co({C}_{2}{O}_{4}{)}_{3}]$ is Potassium trioxalatocobaltate$(III)$.
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