Coordination Compounds JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Coordination Compounds, free to read — no sign-in needed. The full chapter has 53 questions; sign in to attempt the remaining 48 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Consider that the coordinating atoms of the ligands in $cis$-[Co(NH$_3$)$_4$Cl$_2$]Cl and $mer$-[Co(NH$_3$)$_3$Cl$_3$] octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangular faces in both the complexes having one N atom and two Cl atoms at their corners is ____.
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   Answer: 6

   In an octahedral complex, the six coordinating atoms are located at the vertices of an octahedron, which has 8 triangular faces. Each face is formed by 3 vertices, and any two adjacent (cis) vertices form an edge that is shared by exactly two faces. For the $cis$-[Co(NH$_3$)$_4$Cl$_2$]$^+$ complex: The two Cl atoms are at adjacent (cis) positions, forming exactly one Cl-Cl edge. This edge is shared by two triangular faces. Since the remaining four vertices are occupied by N atoms, the third vertex of each of these two faces must be an N atom. Therefore, there are 2 faces with exactly two Cl atoms and one N atom. For the $mer$-[Co(NH$_3$)$_3$Cl$_3$] complex: The three Cl atoms are in a meridional arrangement, meaning two Cl atoms are trans to each other and the third is cis to both. This arrangement forms exactly two Cl-Cl edges. Since the three Cl atoms do not occupy the vertices of the same face (unlike the $fac$ isomer), no face contains three Cl atoms. Each of the two Cl-Cl edges is shared by two faces, yielding $2 \times 2 = 4$ distinct faces that contain exactly two Cl atoms. The third vertex in all these 4 faces is an N atom. Therefore, there are 4 faces with exactly two Cl atoms and one N atom. The sum of the total number of such faces in both complexes is $2 + 4 = 6$. Answer: $6$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   The total number of all possible isomers for the square planar complex with formula $\text{K}[\text{M(NCS)(NO}_2\text{)(gly)}]$ is ____. ($M$ = metal ion and gly = $\text{NH}_2\text{CH}_2\text{COO}^-$)
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   Answer: 8

   The given complex is $\text{K}[\text{M(NCS)(NO}_2\text{)(gly)}]$. The complex anion is $[\text{M(NCS)(NO}_2\text{)(gly)}]^{-}$. The ligands present in the complex are: 1. $\text{gly}^-$ (glycinate): An unsymmetrical bidentate ligand coordinating through nitrogen (N) and oxygen (O). Let this be represented as $(AB)$. 2. $\text{NCS}^-$: An ambidentate monodentate ligand that can coordinate through nitrogen ($\text{-NCS}$) or sulfur ($\text{-SCN}$). 3. $\text{NO}_2^-$: An ambidentate monodentate ligand that can coordinate through nitrogen ($\text{-NO}_2$) or oxygen ($\text{-ONO}$). First, we find the number of linkage isomers due to the ambidentate ligands: $\text{NCS}^-$ has 2 linkage modes (N or S). $\text{NO}_2^-$ has 2 linkage modes (N or O). Number of linkage combinations = $2 \times 2 = 4$. For each linkage combination, the complex is of the type $[\text{M}(a)(b)(AB)]$, where $a$ and $b$ are monodentate ligands and $AB$ is an unsymmetrical bidentate ligand. In a square planar geometry, the bidentate ligand $(AB)$ must occupy adjacent (cis) positions. The monodentate ligands $a$ and $b$ will occupy the remaining two positions. This gives rise to 2 geometrical isomers for each linkage combination: 1. Ligand $a$ is trans to A (and $b$ is trans to B). 2. Ligand $a$ is trans to B (and $b$ is trans to A). Since square planar complexes of this type possess a molecular plane of symmetry, they do not exhibit optical isomerism. Total number of isomers = (Number of linkage combinations) $\times$ (Number of geometrical isomers per combination) Total isomers = $4 \times 2 = 8$. Answer: $8$
3. Q3JEE Advanced Adv 2026 (Paper 1)
   Reaction of $\text{PtF}_6$ with oxygen ($\text{O}_2$) gas results in the formation of an ionic compound, $\textbf{X}^+\textbf{Y}^-$. Correct statement(s) is(are)
   1. A.The bond order of $\textbf{X}^+$ is $1.5$.
   2. B.Valence $d$-orbitals of the metal ion in $\textbf{X}^+\textbf{Y}^-$ has $5$ electrons.
   3. C.$\text{PtF}_6$ acts as an oxidant in this reaction.
   4. D.$\text{PtF}_6$ acts as a fluorinating agent in this reaction.
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   Answer: B,C

   The reaction between $\text{O}_2$ and $\text{PtF}_6$ is: $\text{O}_2 + \text{PtF}_6 \rightarrow \text{O}_2^+[\text{PtF}_6]^-$ Here, $X^+$ is $\text{O}_2^+$ and $Y^-$ is $[\text{PtF}_6]^-$. For $\text{O}_2^+$, the total number of electrons is $15$. The molecular orbital configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1}$. The bond order is $\dfrac{10 - 5}{2} = 2.5$. Thus, the bond order is not $1.5$. In the complex ion $[\text{PtF}_6]^-$, the oxidation state of $\text{Pt}$ is $+5$. The ground state electronic configuration of $\text{Pt}$ is $[\text{Xe}] 4f^{14} 5d^9 6s^1$. The configuration of $\text{Pt}^{5+}$ is $[\text{Xe}] 4f^{14} 5d^5$. Therefore, the valence $d$-orbitals of the metal ion contain $5$ electrons. During the reaction, $\text{O}_2$ loses an electron to form $\text{O}_2^+$ and $\text{PtF}_6$ gains an electron to form $[\text{PtF}_6]^-$. Since $\text{PtF}_6$ accepts an electron, it undergoes reduction and acts as an oxidant. Since no fluorine atoms are transferred to $\text{O}_2$, $\text{PtF}_6$ does not act as a fluorinating agent. Answer: Valence $d$-orbitals of the metal ion in $\textbf{X}^+\textbf{Y}^-$ has $5$ electrons.; $\text{PtF}_6$ acts as an oxidant in this reaction.
4. Q4JEE Advanced Adv 2025 (Paper 1)
   The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is
   1. A.$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5(\mathrm{Cl})\right]^{2+}$
   2. B.$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5(\mathrm{Cl})\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$
   3. C.$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5(\mathrm{Cl})\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
   4. D.$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5(\mathrm{Cl})\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+}$
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   Answer: (A)

   $\Delta_{\mathrm{O}} \propto \frac{1}{\lambda}$ $\begin{aligned} &\therefore \text { The absorb wave length order is }\\ &\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} \end{aligned}$
5. Q5JEE Advanced Adv 2024 (Paper 2)
   Among $\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-},\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right], \mathrm{XeF}_4,\left[\mathrm{PCl}_4\right]^{+},\left[\mathrm{PdCl}_4\right]^{2-},\left[\mathrm{ICl}_4\right]^{-},\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$ and $\mathrm{P}_4$ the total number of species with tetrahedral geometry is _______
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   Answer: 5

   $\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-},\left[\mathrm{Co}(\mathrm{CO})_3 \mathrm{NO}\right],\left[\mathrm{PCl}_4\right]^{+},\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-} ~\&~ \mathrm{P}_4$ are with tetrahedral geometry.