Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 6.
Q49 (2025)
The number of paramagnetic metal complex species among $[\text{Co}(\text{NH}_3)_6]^{3+}$, $[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$, $[\text{MnCl}_6]^{3-}$, $[\text{Mn}(\text{CN})_6]^{3-}$, $[\text{CoF}_6]^{3-}$, $[\text{Fe}(\text{CN})_6]^{3-}$ and $[\text{FeF}_6]^{3-}$ with same number of unpaired electrons is $\dots$.
### Related Formula
$$\text{Paramagnetic species: Complexes with unpaired electron count } (n) > 0$$
### Core Logic
Let's perform electron tracking across every entry using CFT parameters:
1. $[\text{Co}(\text{NH}_3)_6]^{3+}$: $\text{Co}^{3+}$ ($3d^6$), $\text{NH}_3$ is SFL $\implies t{2g}^6 e_g^0$, unpaired electrons $= 0$ (Diamagnetic).
2. $[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$: $\text{Co}^{3+}$ ($3d^6$), Oxalate acts as SFL here $\implies t_{2g}^6 e_g^0$, unpaired electrons $= 0$ (Diamagnetic).
3. $[\text{MnCl}_6]^{3-}$: $\text{Mn}^{3+}$ ($3d^4$), $\text{Cl}^-$ is WFL $\implies t_{2g}^3 e_g^1$, unpaired electrons $= 4$.
4. $[\text{Mn}(\text{CN})_6]^{3-}$: $\text{Mn}^{3+}$ ($3d^4$), $\text{CN}^-$ is SFL $\implies t_{2g}^4 e_g^0$, unpaired electrons $= 2$.
5. $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+}$ ($3d^6$), $\text{F}^-$ is WFL $\implies t_{2g}^4 e_g^2$, unpaired electrons $= 4$.
6. $[\text{Fe}(\text{CN})_6]^{3-}$: $\text{Fe}^{3+}$ ($3d^5$), $\text{CN}^-$ is SFL $\implies t_{2g}^5 e_g^0$, unpaired electrons $= 1$.
7. $[\text{FeF}_6]^{3-}$: $\text{Fe}^{3+}$ ($3d^5$), $\text{F}^-$ is WFL $\implies t_{2g}^3 e_g^2$, unpaired electrons $= 5$.
### Step 1: Finding Common Electronic Counts
Reviewing unpaired counts among paramagnetic entities:
- $n=1$: 1 complex ($[\text{Fe}(\text{CN})_6]^{3-}$)
- $n=2$: 1 complex ($[\text{Mn}(\text{CN})_6]^{3-}$)
- $n=4$: 2 complexes ($[\text{MnCl}_6]^{3-}$ and $[\text{CoF}_6]^{3-}$)
- $n=5$: 1 complex ($[\text{FeF}_6]^{3-}$)
The highest matching sub-group frequency has a count of 2.
### Pattern Recognition
CFT Shortcut tracking: For $3d^4$ weak field and $3d^6$ weak field systems, the unpaired counts identically match ($n=4$). Spotting that $\text{Mn}^{3+}\text{/WFL}$ and $\text{Co}^{3+}\text{/WFL}$ both leave 4 electrons unpaired immediately provides the pair answer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q33 (2025)
Find the compound 'A' from the following reaction sequences.
$$\mathrm{A} \xrightarrow{\text{aqua-regia}} \mathrm{B} \xrightarrow{\text{(1) } \mathrm{KNO}{2} | \mathrm{NH}{4}\mathrm{OH}, \text{ (2) } \mathrm{AcOH}} \text{yellow ppt}$$
- \text{ZnS}
- \text{CoS}
- \text{MnS}
- \text{NiS}
### Core Logic
This pathway corresponds to the standard confirmatory test for cobalt ($\mathrm{Co}^{2+}$) ions in qualitative inorganic analysis:
1. $\mathrm{CoS}$ dissolves in aqua regia to yield cobalt chloride ($\mathrm{CoCl}_2$):
$$\mathrm{CoS} + \text{aqua regia} \rightarrow \mathrm{CoCl}_{2}$$
2. Treating this solution with potassium nitrite ($\mathrm{KNO}_{2}$) in the presence of acetic acid ($\mathrm{AcOH}$) oxidizes $\mathrm{Co}^{2+}$ to $\mathrm{Co}^{3+}$, precipitating potassium cobaltinitrite as a characteristic yellow solid:
$$\mathrm{CoCl}_{2} + 7\mathrm{KNO}_{2} + 2\mathrm{CH}_{3}\mathrm{COOH} \rightarrow \mathrm{K}_{3}[\mathrm{Co}(\mathrm{NO}_{2})_{6}]\downarrow (\text{yellow}) + 2\mathrm{NaCl} + \mathrm{NO} + 2\mathrm{CH}_{3}\mathrm{COOK} + \mathrm{H}_{2}\mathrm{O}$$
### Pattern Recognition
A yellow precipitate formed specifically upon adding $\mathrm{KNO}_2$ and acetic acid is a definitive signature of potassium cobaltinitrite, $\mathrm{K}_3[\mathrm{Co}(\mathrm{NO}_2)_6]$. This confirms the starting sulfide was $\mathrm{CoS}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Qualitative Analysis
Q37 (2025)
When Ethane-1, 2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:
- \text{Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Green }
ightarrow \text{ Violet}
- \text{Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Violet }
ightarrow \text{ Green}
- \text{Green }
ightarrow \text{ Pale Blue }
ightarrow \text{ Blue }
ightarrow \text{ Violet}
- \text{Violet }
ightarrow \text{ Blue }
ightarrow \text{ Pale Blue }
ightarrow \text{ Green}
### Core Logic
An aqueous nickel (II) chloride solution contains the green hexaquarickel(II) complex, $[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$. Ethane-1,2-diamine ('en') is a bidentate ligand that binds more strongly than water, shifting the crystal field splitting parameter ($\Delta_o$) to higher energies as it replaces water molecules:
1. Initial state:
$$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\text{ (Green)}$$
2. Adding 1 equivalent of 'en' forms a mono-en complex:
$$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+}\text{ (Pale Blue)} + 2\mathrm{H}_{2}\mathrm{O}$$
3. Adding a 2nd equivalent forms a bis-en complex:
$$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+}\text{ (Blue / Purple)} + 2\mathrm{H}_{2}\mathrm{O}$$
4. Adding a 3rd equivalent forms the tris-en complex:
$$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+} + \mathrm{en}
ightarrow [\mathrm{Ni}(\mathrm{en})_{3}]^{2+}\text{ (Violet)} + 2\mathrm{H}_{2}\mathrm{O}$$
This progressive ligand replacement shifts the absorption spectrum, changing the solution's visible color from Green $
ightarrow$ Pale Blue $
ightarrow$ Blue $
ightarrow$ Violet.
### Pattern Recognition
Replacing weak-field ligands (like $\mathrm{H}_2\mathrm{O}$) with stronger bidentate chelating ligands (like 'en') increases crystal field splitting. For $\mathrm{Ni}^{2+}$, this ligand substitution always follows the specific chromatic progression: Green $
ightarrow$ Pale Blue $
ightarrow$ Blue $
ightarrow$ Violet.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q38 (2025)
The conditions and consequence that favours the $t_{2g}^{3} e_{g}^{1}$ configuration in a metal complex are:
- \text{weak field ligand, high spin complex}
- \text{strong field ligand, high spin complex}
- \text{strong field ligand, low spin complex}
- \text{weak field ligand, low spin complex}
### Core Logic
Consider an octahedral coordination environment for a $d^4$ transition metal ion configuration:
* Weak Field Ligand (WFL):
The crystal field splitting energy is smaller than the pairing energy ($\Delta_o < P$). Consequently, electrons prefer to occupy the higher-energy $e_g$ orbitals rather than pair up in the lower-energy $t_{2g}$ orbitals. This leads to a high spin complex with the configuration:
$$t_{2g}^{3} e_{g}^{1}$$
* Strong Field Ligand (SFL):
The splitting energy is larger than the pairing energy ($\Delta_o > P$). Electrons pair up in the $t_{2g}$ orbitals before occupying the $e_g$ subshell, resulting in a low spin complex with the configuration:
$$t_{2g}^{4} e_{g}^{0}$$
### Pattern Recognition
An electron occupying an $e_g$ orbital before the $t_{2g}$ orbitals are fully paired requires a weak-field ligand. This configuration maximizes the number of unpaired electrons, which is the defining characteristic of a high-spin complex.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q28 (2025)
One mole of the octahedral complex compound $Co(NH_{3})_{5}Cl_{3}$ gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of $AgNO_{3}$ solution to yield two moles of $AgCl_{(s)}$. The structure of the complex is:
- $[Co(NH_{3})_{5}Cl]Cl_{2}$
- $[Co(NH_{3})_{4}Cl].Cl_{2}.NH_{3}$
- $[Co(NH_{3})_{4}Cl_{2}]Cl.NH_{3}$
- $[Co(NH_{3})_{3}Cl_{3}].2NH_{3}$
### Related Formula
$$\text{Moles of } AgCl \text{ precipitated} = \text{Moles of ionizable } Cl^{-} \text{ ions outside the coordination sphere}$$
### Core Logic
Since 1 mole of the complex yields 2 moles of $AgCl_{(s)}$, there must be exactly 2 chloride ions outside the coordination sphere to undergo precipitation:
$$[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+}(aq) + 2Cl^{-}(aq)$$
This dissociation produces a total of 3 moles of ions per mole of the complex, perfectly consistent with the problem constraints.
### Pattern Recognition
Number of precipitated $AgCl$ moles directly equates to the count of counter-anions located outside the square brackets.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
