Coordination Compounds Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Coordination Compounds with step-by-step solutions covering crystal field theory, CFSE, spin-only magnetic moment, isomerism, spectrochemical series, and metal carbonyl bonding.
Practice the most recent JEE Main 2026 Coordination Compounds questions with clean, step-by-step worked solutions.
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Solution
Water is a weak-field ligand, so each ion is high spin. Use $\mu = \sqrt{n(n+2)}$ BM where $n$ is the number of unpaired electrons.
$$\text{A: } \text{Cr}^{2+} = d^4 \Rightarrow n=4,\ \mu=\sqrt{4\cdot 6}=\sqrt{24}=4.90 \Rightarrow \text{III}$$$$\text{B: } \text{Co}^{2+} = d^7 \Rightarrow n=3,\ \mu=\sqrt{3\cdot 5}=\sqrt{15}=3.87 \Rightarrow \text{I}$$$$\text{C: } \text{Cu}^{2+} = d^9 \Rightarrow n=1,\ \mu=\sqrt{1\cdot 3}=\sqrt{3}=1.73 \Rightarrow \text{IV}$$$$\text{D: } \text{Mn}^{2+} = d^5 \Rightarrow n=5,\ \mu=\sqrt{5\cdot 7}=\sqrt{35}=5.92 \Rightarrow \text{II}$$So A-III, B-I, C-IV, D-II.
Answer: D (A-III, B-I, C-IV, D-II)
Solution
In a tetrahedral field the $e$ orbitals lie lower ($-0.6\,\Delta_t$ each) and the $t_2$ orbitals higher ($+0.4\,\Delta_t$ each). Tetrahedral complexes are high spin, and the $e$ set (2 orbitals) fills first, then $t_2$ (3 orbitals).
$$\text{A: } d^2 = e^2 \Rightarrow 2(-0.6) = -1.2\,\Delta_t \Rightarrow \text{III}$$$$\text{B: } d^4 = e^2 t_2^2 \Rightarrow 2(-0.6)+2(+0.4) = -0.4\,\Delta_t \Rightarrow \text{IV}$$$$\text{C: } d^6 = e^3 t_2^3 \Rightarrow 3(-0.6)+3(+0.4) = -0.6\,\Delta_t \Rightarrow \text{I}$$$$\text{D: } d^8 = e^4 t_2^4 \Rightarrow 4(-0.6)+4(+0.4) = -0.8\,\Delta_t \Rightarrow \text{II}$$So A-III, B-IV, C-I, D-II.
Answer: C (A-III, B-IV, C-I, D-II)
Solution
In a tetrahedral field the splitting is inverted relative to octahedral: the $t_2$ set $(d_{xy}, d_{yz}, d_{xz})$ is higher in energy, while the $e$ set $(d_{x^2-y^2}, d_{z^2})$ is lower. Within each set the orbitals are degenerate.
$$e\text{-set: } d_{x^2-y^2} = d_{z^2} \quad (\text{lower}), \qquad t_2\text{-set: } d_{xy} = d_{yz} = d_{xz} \quad (\text{higher})$$- A. $d_{xy} = d_{xz}\ (t_2) > d_{x^2-y^2}\ (e)$ — TRUE.
- B. $d_{xy} = d_{yz}\ (t_2) > d_{z^2}\ (e)$ — TRUE.
- C. claims $d_{x^2-y^2} > d_{z^2}$, but these are degenerate ($=$), so FALSE.
- D. $d_{x^2-y^2} = d_{z^2}\ (e) < d_{xz}\ (t_2)$ — TRUE.
So A, B and D are correct.
Answer: A (A, B and D only)
Solution
For the same metal ion, $\Delta_0$ increases with ligand field strength along the spectrochemical series:
$$F^- < H_2O < en < CN^-$$Ordering the given $\Delta_0$ values with the ligands from weakest to strongest field:
$$\text{B }[CrF_6]^{3-} = 15{,}060 \Rightarrow \text{I}$$$$\text{C }[Cr(H_2O)_6]^{3+} = 17{,}400 \Rightarrow \text{II}$$$$\text{D }[Cr(en)_3]^{3+} = 22{,}300 \Rightarrow \text{III}$$$$\text{A }[Cr(CN)_6]^{3-} = 26{,}600 \Rightarrow \text{IV}$$So A-IV, B-I, C-II, D-III.
Answer: D (A-IV, B-I, C-II, D-III)
Solution
A 1 : 3 electrolyte means all three chlorides are ionizable: the complex is $[Cr(H_2O)_6]Cl_3$.
Molar mass of CrCl$_3\cdot$6H$_2$O:
$$M = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5\ \text{g mol}^{-1}$$Moles of complex:
$$n_{\text{complex}} = \frac{5.33}{266.5} = 0.02\ \text{mol}$$Passing through a cation exchanger retains the cation $[Cr(H_2O)_6]^{3+}$; the eluate carries the $3\,Cl^-$ (as HCl). These precipitate as AgCl ($M = 108 + 35.5 = 143.5$):
$$n_{\text{AgCl}} = \frac{8.61}{143.5} = 0.06\ \text{mol}$$Required ratio:
$$\frac{n_{\text{complex}}}{n_{\text{AgCl}}} = \frac{0.02}{0.06} = 0.333 = 33.3 \times 10^{-2}$$Nearest integer $= 33$.
Answer: 33
Solution
In all three, nickel is $\text{Ni}^{2+} = d^8$.
- A $[Ni(en)_3]^{2+}$: octahedral $d^8$ gives $t_{2g}^6 e_g^2$ regardless of field strength $\Rightarrow$ 2 unpaired.
- B $[NiCl_4]^{2-}$: tetrahedral $d^8$ gives $e^4 t_2^4 \Rightarrow$ 2 unpaired.
- C $[Ni(NH_3)_6]^{2+}$: octahedral $d^8 \Rightarrow$ 2 unpaired.
So the count is 2, 2, 2.
Frequency of absorption $\propto \Delta$. Octahedral $\Delta_o > \Delta_t$, and field strength follows $en > NH_3 > Cl^-$:
$$\Delta:\quad [Ni(en)_3]^{2+} > [Ni(NH_3)_6]^{2+} > [NiCl_4]^{2-}$$Hence frequency order $(A) > (C) > (B)$.
Answer: A (2, 2, 2 and (A) > (C) > (B))
Solution
Statement I — third ionization energy ($IE_3$) removes an electron from the $M^{2+}$ ion:
- Sc: $[Ar]3d^1 4s^2$; $IE_3$ removes the lone $3d^1$ electron to give the stable noble-gas $Sc^{3+}$ ($[Ar]$) $\Rightarrow$ lowest $IE_3$.
- Zn: $[Ar]3d^{10} 4s^2$; $IE_3$ must break the stable, fully filled $3d^{10}$ shell $\Rightarrow$ highest $IE_3$.
Statement I is TRUE.
Statement II — comparing CFSE magnitudes:
$$[Co(H_2O)_6]^{2+}:\ Co^{2+}(d^7),\ \text{weak field, high spin (small }\Delta)$$$$[Co(H_2O)_6]^{3+}:\ Co^{3+}(d^6),\ \text{higher charge} \Rightarrow \text{larger }\Delta$$$$[Co(en)_3]^{3+}:\ Co^{3+}(d^6),\ en\ \text{stronger field} \Rightarrow \text{largest }\Delta,\ \text{low spin } t_{2g}^6$$CFSE increases in exactly this order, so Statement II is TRUE.
Answer: A (Both Statement I and Statement II are true)
Solution
Coordination isomerism arises when both the cation and the anion are complex ions and the ligands can be interchanged between the two different metal centres to give a genuinely different compound.
- A. $[Ag(NH_3)_2][Ag(CN)_2]$: same metal (Ag/Ag) $\Rightarrow$ interchange gives an identical compound $\Rightarrow$ NO.
- B. $[Co(NH_3)_6][Cr(CN)_6]$: Co and Cr differ $\Rightarrow$ isomer $[Cr(NH_3)_6][Co(CN)_6]$ $\Rightarrow$ YES.
- C. $[Co(NH_3)_6][Co(CN)_6]$: same metal in the same oxidation state (Co(III)/Co(III)) $\Rightarrow$ interchange gives the same compound $\Rightarrow$ NO.
- D. $[Fe(NH_3)_6][Co(CN)_6]$: Fe and Co differ $\Rightarrow$ YES.
- E. $[Co(NH_3)_6][Fe(CN)_6]$: Co and Fe differ $\Rightarrow$ YES.
So B, D and E only.
Answer: B (B, D and E Only)
Solution
- A. $[NiCl_4]^{2-}$: $Ni^{2+}(d^8)$, $Cl^-$ weak field $\Rightarrow sp^3$, tetrahedral, 2 unpaired $\Rightarrow$ paramagnetic. CORRECT.
- B. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}(d^8)$ octahedral $\Rightarrow sp^3d^2$, 2 unpaired $\Rightarrow$ paramagnetic. CORRECT.
- C. $[Ni(CO)_4]$: $Ni(0)(d^{10})$, $sp^3$, tetrahedral, all electrons paired $\Rightarrow$ diamagnetic, not paramagnetic. INCORRECT.
- D. $[Ni(CN)_4]^{2-}$: $Ni^{2+}(d^8)$, $CN^-$ strong field $\Rightarrow$ pairing, $dsp^2$, square planar, 0 unpaired $\Rightarrow$ diamagnetic. CORRECT.
So A, B and D only.
Answer: D (A, B and D only)
Solution
The complex is $[Cr(H_2O)_4Cl_2]Cl$. Two chlorides are inside the coordination sphere and do not ionize; only the one chloride outside the sphere is free and precipitable by $Ag^+$.
Moles of complex:
$$n = 0.100\ \text{L} \times 0.05\ \text{M} = 5 \times 10^{-3}\ \text{mol}$$Each formula unit gives 1 free $Cl^-$, so:
$$n_{\text{AgCl}} = 5 \times 10^{-3}\ \text{mol}$$Hence the value is $5$.
Answer: 5
Solution
- A. The M–C bond has both $\sigma$ (ligand $\to$ metal) and $\pi$ (metal $\to$ ligand) character. TRUE.
- B. Synergic bonding strengthens the M–C bond (and weakens the C–O bond) because back-donation reinforces M–C. So “becomes weak” is FALSE.
- C. The $\sigma$ bond forms by donation of the carbon lone pair into a vacant metal orbital. TRUE.
- D. The $\pi$ bond forms by back-donation from a filled metal $d$-orbital into the vacant $\pi^*$ of CO. TRUE.
Correct statements: A, C and D.
Answer: B (A, C and D Only)
Solution
Statement I — In an octahedral field, relative to the barycentre, each $e_g$ electron raises the energy by $+0.6\,\Delta_o$ and each $t_{2g}$ electron lowers it by $-0.4\,\Delta_o$. This is the standard CFSE convention. TRUE.
Statement II — In a free (gaseous) metal atom/ion the five $d$-orbitals are degenerate. On complex formation the ligand field lifts (destroys) this degeneracy, splitting them into $t_{2g}$ and $e_g$. TRUE.
Both statements are correct.
Answer: A (Both Statement I and Statement II are correct)
Solution
- A. Oxalate ($C_2O_4^{2-}$) is a bidentate chelating ligand; the chelate effect makes $[Fe(C_2O_4)_3]^{3-}$ the most stable of the three. TRUE.
- B. $en$ is a chelating bidentate ligand, so $[Cu(en)_2]^{2+}$ is more stable than $[Cu(NH_3)_4]^{2+}$. The statement is reversed $\Rightarrow$ FALSE.
- C. In $K_4[Fe(CN)_6]$, $Fe^{2+}(d^6)$ with strong-field $CN^-$ is low spin $\Rightarrow d^2sp^3$. TRUE.
- D. $NO_2^-$ is ambidentate (binds through N or O), so $[Fe(NO_2)_3Cl_3]^{3-}$ shows linkage isomerism. TRUE.
- E. Both $NO_2^-$ and $SCN^-$ are ambidentate, so this statement is FALSE.
Correct statements: A, C and D.
Answer: C (A, C and D only)
Solution
Fill each configuration in the octahedral $t_{2g}/e_g$ scheme:
$$d^3:\ t_{2g}^3 \Rightarrow 3\ \text{unpaired}$$$$d^4\ (\text{low spin}):\ t_{2g}^4 \Rightarrow 2\ \text{unpaired}$$$$d^5\ (\text{high spin}):\ t_{2g}^3 e_g^2 \Rightarrow 5\ \text{unpaired}$$$$d^6\ (\text{high spin}):\ t_{2g}^4 e_g^2 \Rightarrow 4\ \text{unpaired}$$$$d^7\ (\text{low spin}):\ t_{2g}^6 e_g^1 \Rightarrow 1\ \text{unpaired}$$Total:
$$3 + 2 + 5 + 4 + 1 = 15$$Answer: 15
Solution
A complex is paramagnetic if it has one or more unpaired electrons.
$$[MnBr_4]^{2-}:\ Mn^{2+}(d^5),\ \text{tetrahedral high spin} \Rightarrow 5\ \text{unpaired} \Rightarrow \text{para}$$$$[NiCl_4]^{2-}:\ Ni^{2+}(d^8),\ \text{tetrahedral} \Rightarrow 2\ \text{unpaired} \Rightarrow \text{para}$$$$[Ni(CN)_4]^{2-}:\ Ni^{2+}(d^8),\ \text{square planar} \Rightarrow 0 \Rightarrow \text{dia}$$$$[Ni(CO)_4]:\ Ni(0)(d^{10}) \Rightarrow 0 \Rightarrow \text{dia}$$$$[CoF_6]^{3-}:\ Co^{3+}(d^6),\ F^-\ \text{weak, high spin} \Rightarrow 4\ \text{unpaired} \Rightarrow \text{para}$$$$[Fe(CN)_6]^{4-}:\ Fe^{2+}(d^6),\ CN^-\ \text{low spin} \Rightarrow 0 \Rightarrow \text{dia}$$$$[Mn(CN)_6]^{3-}:\ Mn^{3+}(d^4),\ CN^-\ \text{low spin }t_{2g}^4 \Rightarrow 2\ \text{unpaired} \Rightarrow \text{para}$$$$[Ti(CN)_6]^{3-}:\ Ti^{3+}(d^1) \Rightarrow 1\ \text{unpaired} \Rightarrow \text{para}$$$$[Cu(H_2O)_6]^{2+}:\ Cu^{2+}(d^9) \Rightarrow 1\ \text{unpaired} \Rightarrow \text{para}$$$$[Co(C_2O_4)_3]^{3-}:\ Co^{3+}(d^6),\ \text{low spin} \Rightarrow 0 \Rightarrow \text{dia}$$Paramagnetic ones: $[MnBr_4]^{2-},\ [NiCl_4]^{2-},\ [CoF_6]^{3-},\ [Mn(CN)_6]^{3-},\ [Ti(CN)_6]^{3-},\ [Cu(H_2O)_6]^{2+}$ $\Rightarrow$ 6.
Answer: 6