Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry).
Q27 (2025)
The d-orbital electronic configuration of the complex among $[\mathrm{Co(en)_3}]^{3+}$, $[\mathrm{CoF_6}]^{3-}$, $[\mathrm{Mn(H_2O)_6}]^{2+}$ and $[\mathrm{Zn(H_2O)_6}]^{2+}$ that has the highest CFSE is:
- $\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$
- $\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{4}$
- $\mathrm{t}_{2\mathrm{g}}^{3}\mathrm{e}_{\mathrm{g}}^{2}$
- $\mathrm{t}_{2\mathrm{g}}^{4}\mathrm{e}_{\mathrm{g}}^{2}$
### Related Formula
$$\text{CFSE} = \left( -0.4 n_{\mathrm{t}_{2\mathrm{g}}} + 0.6 n_{\mathrm{e}_{\mathrm{g}}} \right) \Delta_{\mathrm{o}} + n_{\mathrm{p}} P$$
### Core Logic
Crystal Field Stabilization Energy (CFSE) is maximized (becomes most negative) when electrons populate lower-energy $\mathrm{t_{2g}}$ orbitals and stay out of higher-energy $\mathrm{e_g}$ orbitals. This is favored by strong-field ligands (SFL) that induce large $\Delta_o$ splitting, leading to low-spin configurations.
### Step 1: Analyze Ligand Strength and Configuration
Let us check each of the given complexes:
1. $[\mathrm{Co(en)_3}]^{3+}$: Here $\mathrm{Co^{3+}}$ has a $3\mathrm{d^6}$ configuration. Since ethylenediamine ($\mathrm{en}$) is a strong-field ligand, it causes pairing of all $6$ electrons in the $\mathrm{t_{2g}}$ subshell. The configuration is $\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.
{{SOL_IMG1}}
2. $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}$ is $3\mathrm{d^6}$. Since $\mathrm{F^-}$ is a weak-field ligand (WFL), no pairing occurs. The configuration is $\mathrm{t}_{2\mathrm{g}}^{4}\mathrm{e}_{\mathrm{g}}^{2}$.
3. $[\mathrm{Mn(H_2O)_6}]^{2+}$: $\mathrm{Mn^{2+}}$ is $3\mathrm{d^5}$. Since $\mathrm{H_2O}$ is a weak-field ligand, the configuration is high-spin: $\mathrm{t}_{2\mathrm{g}}^{3}\mathrm{e}_{\mathrm{g}}^{2}$.
4. $[\mathrm{Zn(H_2O)_6}]^{2+}$: $\mathrm{Zn^{2+}}$ is $3\mathrm{d^{10}}$. The d-subshell is fully filled, yielding $\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{4}$.
### Step 2: Compare CFSE Values
Calculating CFSE (neglecting pairing energy term for simplicity):
- For $[\mathrm{Co(en)_3}]^{3+}$: $\text{CFSE} = 6 \times (-0.4 \Delta_o) = -2.4 \Delta_o$
- For $[\mathrm{CoF_6}]^{3-}$: $\text{CFSE} = [4(-0.4) + 2(0.6)] \Delta_o = -0.4 \Delta_o$
- For $[\mathrm{Mn(H_2O)_6}]^{2+}$: $\text{CFSE} = [3(-0.4) + 2(0.6)] \Delta_o = 0$
- For $[\mathrm{Zn(H_2O)_6}]^{2+}$: $\text{CFSE} = [6(-0.4) + 4(0.6)] \Delta_o = 0$
Hence, $[\mathrm{Co(en)_3}]^{3+}$ has the highest crystal field stabilization energy, corresponding to the d-orbital electronic configuration $\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.
### Pattern Recognition
For octahedral complexes of $\mathrm{d^6}$ metals, a low-spin configuration ($\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$) achieves the theoretical maximum orbital stabilization since the $\mathrm{e_g}$ levels are completely empty and $\mathrm{t_{2g}}$ is fully filled.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q33 (2025)
The type of hybridization and the magnetic property of $[\mathrm{MnCl}_6]^{3-}$ are:
- $\mathrm{d}^2\mathrm{sp}^3\text{, paramagnetic with four unpaired electrons}$
- $\mathrm{sp}^3\mathrm{d}^2\text{, paramagnetic with four unpaired electrons}$
- $\mathrm{d}^2\mathrm{sp}^3\text{, paramagnetic with two unpaired electrons}$
- $\mathrm{sp}^3\mathrm{d}^2\text{, paramagnetic with two unpaired electrons}$
### Related Formula
$$\mu_{\text{spin-only}} = \sqrt{n(n+2)}~\mathrm{B.M.}$$
### Core Logic
Let's find the oxidation state of Manganese in the complex $[\mathrm{MnCl}_6]^{3-}$:
$$x + 6(-1) = -3 \implies x = +3$$
Thus, manganese is in the $+3$ oxidation state: $\mathrm{Mn^{3+}} = [\mathrm{Ar}]3\mathrm{d}^4$.
### Step 1: Determine Ligand Splitting and Orbitals
$\mathrm{Cl^-}$ is a weak-field ligand (WFL). Consequently, crystal field splitting is small ($\Delta_o < P$), and no pairing of the $3\mathrm{d}$ electrons occurs.
The distribution of the $4$ electrons in the $3\mathrm{d}$ orbitals remains high-spin:
$$\text{Unpaired electrons (n)} = 4$$
Because the inner $3\mathrm{d}$ orbitals are not empty (since they contain 4 singly occupied orbitals), the complex must utilize the outer $4\mathrm{d}$ orbitals for hybridization.
### Step 2: Assign Hybridization
The vacant outer orbitals used for bonding are one $4\mathrm{s}$, three $4\mathrm{p}$, and two $4\mathrm{d}$ orbitals, which hybridize to form six **$\mathrm{sp^3d^2}$** hybrid orbitals.
Since there are four unpaired electrons, the complex is **paramagnetic** with four unpaired electrons.
### Pattern Recognition
Whenever WFL (like halides $\mathrm{Cl^-}$, $\mathrm{F^-}$) are present with octahedral transition metal complexes with $\mathrm{d^4}$ to $\mathrm{d^7}$ configuration, they always yield high-spin, outer orbital complexes with $\mathrm{sp^3d^2}$ hybridization.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q50 (2025)
The spin-only magnetic moment value of $\mathbf{M}^{\mathrm{n+}}$ ion formed among Ni, Zn Mn and Cu that has the least enthalpy of atomisation is (in nearest integer)
Here $\mathrm{n}$ is equal to the number of diamagnetic complexes among $\mathrm{K}_2\mathrm{[NiCl_4]}$, $\mathrm{[Zn(H_2O)_6]Cl_2}$, $\mathrm{K}_3[\mathrm{Mn}(\mathrm{CN})_6]$ and $[\mathrm{Cu}(\mathrm{PPh}_3)_3\mathrm{I}]$
### Related Formula
$$\mu = \sqrt{n_{\text{unpaired}}(n_{\text{unpaired}}+2)}~\mathrm{B.M.}$$
### Core Logic
This question requires three distinct sequential conceptual steps:
1. Determine the count $n$ of diamagnetic complexes.
2. Identify which of the metal ions (Ni, Zn, Mn, Cu) has the lowest enthalpy of atomisation.
3. Compute the spin-only magnetic moment of that metal in its $+n$ state.
### Step 1: Count Diamagnetic Complexes to find n
- $\mathrm{K_2[NiCl_4]}$: $\mathrm{Ni^{2+}} = 3\mathrm{d^8}$. Weak field chloride ligand leads to 2 unpaired electrons $\implies$ **Paramagnetic**.
- $\mathrm{[Zn(H_2O)_6]Cl_2}$: $\mathrm{Zn^{2+}} = 3\mathrm{d^{10}}$. Completely filled subshell $\implies$ **Diamagnetic**.
- $\mathrm{K_3[Mn(CN)_6]}$: $\mathrm{Mn^{3+}} = 3\mathrm{d^4}$. Strong field cyanide ligand gives low-spin state with 2 unpaired electrons $\implies$ **Paramagnetic**.
- $\mathrm{[Cu(PPh_3)_3I]}$: $\mathrm{Cu^+}$ = $3\mathrm{d^{10}}$. Completely filled subshell $\implies$ **Diamagnetic**.
Thus, there are exactly 2 diamagnetic complexes: **$n = 2$**.
### Step 2: Identify Metal with Lowest Enthalpy of Atomisation
Among the transition metals of the 3d series (Ni, Zn, Mn, Cu), **Zinc (Zn)** has the lowest enthalpy of atomisation ($126~\mathrm{kJ~mol^{-1}}$). This is because Zinc has a fully occupied d-subshell ($3\mathrm{d^{10}}4\mathrm{s^2}$) and lacks any unpaired d-electrons to participate in metallic bonding.
### Step 3: Calculate Spin-only Magnetic Moment of M^n+
With $M = \text{Zinc}$ and $n = 2$, the ion is $\mathrm{Zn^{2+}}$.
Electronic configuration of $\mathrm{Zn^{2+}}$ is $[\mathrm{Ar}]3\mathrm{d^{10}}$, which contains zero unpaired electrons ($n_{\text{unpaired}} = 0$).
Therefore, the spin-only magnetic moment is:
$$\mu = 0~\mathrm{B.M.}$$
### Pattern Recognition
Zinc is always a unique outlier in the d-block. Because it has a completely filled $d^{10}$ shell in both its atomic and $+2$ oxidation states, it exhibits no d-orbital metallic bonding (leading to lowest melting point, boiling point, and atomisation enthalpy in the 3d series) and is always diamagnetic.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d- and f-Block Elements
Q471 (2025)
A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential $(\mathrm{M}^{3+} / \mathrm{M}^{2+})$. It forms a metal complex of the type $[\mathrm{M}(\mathrm{CN})_6]^{4-}$. The number of electrons present in the $\mathbf{e}_{\mathrm{g}}$ orbital of the complex is
### Related Formula
Crystal Field Splitting configuration rule for strong field ligands:
$$\Delta_0 > P \implies \text{Electrons fill } \mathrm{t_{2g}} \text{ completely before entering } \mathrm{e_g}$$
### Core Logic
Let's isolate properties step-by-step:
1. Among the given first-row elements ($\mathrm{Mn, Cr, Co, Fe}$), Cobalt ($\mathrm{Co}$) possesses the highest standard electrode potential value for the $3+/2+$ couple:
$$E^\circ(\mathrm{Co^{3+}/Co^{2+}}) = +1.81\mathrm{~V}$$
2. The oxidation state configuration within complex $[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ is $\mathrm{Co^{2+}}$, which has a $\mathrm{d^7}$ valence configuration.
3. Cyanide ($\mathrm{CN^-}$) acts as a strong field ligand, forcing maximum electron pairing within the lower energy levels.
### Step 1: Subshell Filling Matrix
Distribute 7 electrons across the split crystal field levels:
* First 6 electrons fill the lower $\mathrm{t_{2g}}$ levels completely, forming paired tracks.
* The 7th electron has no choice but to step up to the higher level.
This structural splitting is visualized below:
{{SOL_IMG1}}
Therefore, the number of electrons present in the $\mathrm{e_g}$ orbital block is exactly 1.
### Pattern Recognition
Always double check the specific oxidation state value: $[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ gives $\mathrm{Co^{2+}}$ ($\mathrm{d^7}$), whereas $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ would be $\mathrm{Co^{3+}}$ ($\mathrm{d^6}$), which has zero electrons in its $\mathrm{e_g}$ level.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d- and f-Block Elements
Q40 (2025)
Given below are two statements :
Statement (I): In octahedral complexes, when $\Delta_0 < P$ high spin complexes are formed. When $\Delta_0 > P$ low spin complexes are formed.
Statement (II) : In tetrahedral complexes because of $\Delta_{\mathrm{t}} < \mathrm{P}$, low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $(1)\ \text{Statement I is correct but Statement II is incorrect.}$
- $(2)\ \text{Both Statement I and Statement II are incorrect}$
- $(3)\ \text{Statement I is incorrect but Statement II is correct}$
- $(4)\ \text{Both Statement I and Statement II are correct}$
### Related Formula
Crystal field splitting values relation for matching configuration choices:
$$\Delta_{\mathrm{t}} = \frac{4}{9}\Delta_0$$
### Core Logic
Let's verify both rules based on Crystal Field Theory principles:
* **Statement I**: In octahedral configurations, if pairing penalty energy $P$ exceeds field split magnitude $\Delta_0$, electrons prefer moving to upper sub-shells, creating high-spin states. Conversely, if $\Delta_0 > P$, forced pairing occurs, creating low-spin complexes. (Statement I is accurate).
* **Statement II**: Because tetrahedral configurations separate by an extremely narrow gap magnitude $\Delta_{\mathrm{t}}$ (about half of octahedral field splits), the value almost never exceeds standard pairing energy $P$. Electrons consistently choose higher sub-levels rather than pairing up, meaning low-spin arrangements are extremely rare. (Statement II is accurate).
### Step 1: Verdict
Therefore, both Statement I and Statement II are correct.
### Pattern Recognition
Tetrahedral configurations are systematically assumed to be high-spin unless special structural properties dictate otherwise, due to the $\Delta_{\mathrm{t}} < P$ constraint.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
