Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 2.
Q185 (2025)
Identify the diamagnetic octahedral complex ions from below;
A. $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$
B. $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$
C. $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$
D. $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$
Choose the correct answer from the options given below :
- B and D Only
- A and D Only
- A and C Only
- B and C Only
### Related Formula
According to Crystal Field Theory (CFT):
- A complex is diamagnetic if all electrons are paired up (unpaired electrons, $n=0$).
- Strong field ligands (like $\mathrm{CN}^-$, $\mathrm{NH}_3$ with $Co^{3+}$) cause pairing of electrons if $\Delta_o > P$.
{{SOLUTION_IMG}}
### Core Logic
Analyze each complex:
- **A. $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$**:
- $Mn^{3+}$ has $d^4$ configuration.
- Strong field ligand $\mathrm{CN}^-$ causes pairing in $t_{2g}$ orbitals: $t_{2g}^4 e_g^0$.
- There are 2 unpaired electrons $\rightarrow$ *Paramagnetic*.
- **B. $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$**:
- $Co^{3+}$ has $d^6$ configuration.
- $\mathrm{NH}_3$ acts as strong field ligand with $Co^{3+}$, causing complete pairing: $t_{2g}^6 e_g^0$.
- No unpaired electrons $\rightarrow$ *Diamagnetic*.
### Step 1: Analyze complexes C and D
- **C. $[\mathrm{Fe}(mathrm{CN})_6]^{4-}$**:
- $Fe^{2+}$ has $d^6$ configuration.
- Strong field ligand $\mathrm{CN}^-$ causes complete pairing: $t_{2g}^6 e_g^0$.
- No unpaired electrons $\rightarrow$ *Diamagnetic*.
- **D. $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$**:
- $Co^{3+}$ has $d^6$ configuration.
- Weak field ligands ($\mathrm{F}^-$, $\mathrm{H}_2\mathrm{O}$) do not cause pairing: $t_{2g}^4 e_g^2$.
- There are 4 unpaired electrons $\rightarrow$ *Paramagnetic*.
### Step 2: Conclusion
Only complexes B and C are diamagnetic, matching Option (4).
### Pattern Recognition
Octahedral $d^6$ ions (such as $Co^{3+}$ or $Fe^{2+}$) coupled with strong-field ligands are exceptionally stable and always form low-spin, fully paired, diamagnetic complexes ($t_{2g}^6 e_g^0$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q44 (2025)
An octahedral complex having molecular composition $\mathrm{Co \cdot 5NH_3 \cdot Cl \cdot SO_4}$ 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 a white precipitate with $\mathrm{BaCl_2}$ solution. The type of isomerism exhibited by the complex is:
- $\text{Coordination isomerism}$
- $\text{Linkage isomerism}$
- $\text{Ionisation isomerism}$
- $\text{Geometrical isomerism}$
### Core Logic
The complex molecular composition is $\mathrm{Co \cdot 5NH_3 \cdot Cl \cdot SO_4}$. Let's formulate the formulas for the two isomers:
1. **Isomer A**: Gives a white precipitate of $\mathrm{AgCl}$ when reacted with $\mathrm{AgNO_3}$. This means free chloride ions ($\mathrm{Cl}^-$) are present in the outer ionization sphere:
$$[\mathrm{Co(NH_3)_5(SO_4)}]\mathrm{Cl}$$
2. **Isomer B**: Gives a white precipitate of $\mathrm{BaSO_4}$ when reacted with $\mathrm{BaCl_2}$. This means free sulphate ions ($\mathrm{SO}_4^{2-}$) are present in the outer ionization sphere:
$$[\mathrm{Co(NH_3)_5Cl}]\mathrm{SO_4}$$
Since these two isomers yield different ions in solution due to exchange of ligands between the coordination sphere and the ionization sphere, they exhibit **Ionisation isomerism**.
### Pattern Recognition
Test for ions:
- $\mathrm{AgNO_3}$ PPT $\rightarrow$ free halide ion in outer sphere.
- $\mathrm{BaCl_2}$ PPT $\rightarrow$ free sulphate ion in outer sphere.
- Outer-inner ion exchanges are always called **Ionisation isomerism**.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q50 (2025)
The number of paramagnetic complexes among $[\mathrm{FeF}_{6}]^{3-}$, $[\mathrm{Fe(CN)}_{6}]^{3-}$, $[\mathrm{Mn(CN)}_{6}]^{3-}$, $[\mathrm{Co(C}_{2}\mathrm{O}_{4})_{3}]^{3-}$, $[\mathrm{MnCl}_{6}]^{3-}$ and $[\mathrm{CoF}_{6}]^{3-}$, which involve $\mathrm{d}^{2}\mathrm{sp}^{3}$ hybridization is ______.
### Core Logic
Let's systematically analyze the coordination, ligand strength, hybridization, and magnetic behavior of each complex:
1. **$[\mathrm{FeF}_{6}]^{3-}$**:
- $\mathrm{Fe}^{3+}$ ($3\mathrm{d}^5$). $\mathrm{F}^-$ is a weak-field ligand (WFL). No pairing occurs.
- Outer-orbital complex: $\mathrm{sp}^3\mathrm{d}^2$.
- Paramagnetic ($5$ unpaired electrons).
2. **$[\mathrm{Fe(CN)}_{6}]^{3-}$**:
- $\mathrm{Fe}^{3+}$ ($3\mathrm{d}^5$). $\mathrm{CN}^-$ is a strong-field ligand (SFL). Pairing occurs.
- Config: $\mathrm{t}_{2\mathrm{g}}^5\ \mathrm{e}_{\mathrm{g}}^0$ (one unpaired electron remains $\implies$ **Paramagnetic**).
- Inner-orbital complex: **$\mathrm{d}^2\mathrm{sp}^3$**.
3. **$[\mathrm{Mn(CN)}_{6}]^{3-}$**:
- $\mathrm{Mn}^{3+}$ ($3\mathrm{d}^4$). $\mathrm{CN}^-$ is an SFL. Pairing occurs.
- Config: $\mathrm{t}_{2\mathrm{g}}^4\ \mathrm{e}_{\mathrm{g}}^0$ (two unpaired electrons remain $\implies$ **Paramagnetic**).
- Inner-orbital complex: **$\mathrm{d}^2\mathrm{sp}^3$**.
4. **$[\mathrm{Co(C}_2\mathrm{O}_4)_3]^{3-}$**:
- $\mathrm{Co}^{3+}$ ($3\mathrm{d}^6$). Oxalate is a chelating SFL here. Full pairing occurs.
- Config: $\mathrm{t}_{2\mathrm{g}}^6\ \mathrm{e}_{\mathrm{g}}^0$ (zero unpaired electrons $\implies$ Diamagnetic).
- Inner-orbital complex: $\mathrm{d}^2\mathrm{sp}^3$.
5. **$[\mathrm{MnCl}_{6}]^{3-}$**:
- $\mathrm{Mn}^{3+}$ ($3\mathrm{d}^4$). $\mathrm{Cl}^-$ is a WFL. No pairing occurs.
- Outer-orbital complex: $\mathrm{sp}^3\mathrm{d}^2$.
- Paramagnetic ($4$ unpaired electrons).
6. **$[\mathrm{CoF}_{6}]^{3-}$**:
- $\mathrm{Co}^{3+}$ ($3\mathrm{d}^6$). $\mathrm{F}^-$ is a WFL. No pairing occurs.
- Outer-orbital complex: $\mathrm{sp}^3\mathrm{d}^2$.
- Paramagnetic ($4$ unpaired electrons).
Thus, only $[\mathrm{Fe(CN)}_{6}]^{3-}$ and $[\mathrm{Mn(CN)}_{6}]^{3-}$ are both **paramagnetic** and involve **$\mathrm{d}^2\mathrm{sp}^3$** hybridization.
### Pattern Recognition
VBT shortcut:
- Strong-field ligand complexes with $d^4\text{--}d^6$ central ions form inner-orbital $\mathrm{d}^2\mathrm{sp}^3$ complexes.
- Of those, check the number of electrons: $d^6$ is completely paired (diamagnetic), but $d^5$ ($[Fe(CN)_6]^{3-}$) and $d^4$ ($[Mn(CN)_6]^{3-}$) both leave unpaired electrons in the $t_{2g}$ orbitals (paramagnetic).
### Evaluation Rubric / Model Answer
Detailed individual classification of each complex based on VBT/CFT to yield the correct count of $2$ inner-orbital paramagnetic complexes.
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q31 (2025)
Given below are two statements:
Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism.
Statement II: cis- and trans-platin are heteroleptic complexes of Pd.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Both Statement I and Statement II are false.}$
- $\text{Statement I is false but Statement II is true.}$
- $\text{Both Statement I and Statement II are true.}$
- $\text{Statement I is true but Statement II is false.}$
### Core Logic
Let us evaluate both statements individually:
* **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is $[Ma_6]$. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.** {{SOL_IMG1}}
* **Statement II**: Cis-platin and trans-platin have the chemical formula $[Pt(NH_3)_2Cl_2]$. While they are indeed heteroleptic complexes, they are coordination coordinates of **Platinum ($Pt$)**, not Palladium ($Pd$). **Statement II is false.** {{SOL_IMG2}}
### Pattern Recognition
Always read element symbols with immense focus in coordination chemistry. Changing a single letter from Pt to Pd creates a false assertion trap designed to test parsing alertness rather than chemical difficulty.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q41 (2025)
Determine the total number of chemical species from the list below that are specifically involved in an $sp^3d^2$ hybridization state:
$$[\text{Co(NH}_3)_6]^{3+}, \text{ SF}_6, \, [\text{CrF}_6]^{3-}, \, [\text{CoF}_6]^{3-}, \, [\text{Mn(CN)}_6]^{3-}, \text{ and } [\text{MnCl}_6]^{3-}$$
- 5
- 6
- 4
- 3
### Core Logic
Let us systematically determine the hybridization configuration of each species:
1. **$[\text{Co(NH}_3)_6]^{3+}$**: $\text{Co}^{3+}$ has a $3d^6$ configuration. Ammonia ($\text{NH}_3$) acts as a Strong Field Ligand (SFL), forcing the pairing of $3d$ electrons. This leaves two internal $3d$ orbitals vacant, leading to an **inner orbital** $d^2sp^3$ hybridization.
2. **$\text{SF}_6$**: Central sulfur has 6 valence electrons, forming 6 single bonds. Steric number $= 6$, resulting in a regular outer octahedral **$sp^3d^2$** hybridization.
3. **$[\text{CrF}_6]^{3-}$**: $\text{Cr}^{3+}$ has a $3d^3$ configuration. The $t_{2g}$ subshell holds 3 unpaired electrons, leaving the two $e_g$ orbitals empty regardless of ligand field strength. This results in a $d^2sp^3$ hybridization.
4. **$[\text{CoF}_6]^{3-}$**: $\text{Co}^{3+}$ has a $3d^6$ configuration. Fluoride ($F^-$) is a Weak Field Ligand (WFL) and cannot induce spin pairing. Thus, the complex utilizes outer shell $4d$ orbitals, yielding an **outer orbital** **$sp^3d^2$** configuration.
5. **$[\text{Mn(CN)}_6]^{3-}$**: $\text{Mn}^{3+}$ has a $3d^4$ configuration. Cyanide ($\text{CN}^-$) is a Strong Field Ligand (SFL), inducing pairing to leave two $3d$ slots vacant, giving a $d^2sp^3$ hybridization.
6. **$[\text{MnCl}_6]^{3-}$**: $\text{Mn}^{3+}$ has a $3d^4$ configuration. Chloride ($\text{Cl}^-$) is a Weak Field Ligand (WFL) and cannot cause pairing. It utilizes the outer $4d$ shell, resulting in an **outer orbital** **$sp^3d^2$** hybridization.
Counting the outer-orbital $sp^3d^2$ species: $\text{SF}_6$, $[\text{CoF}_6]^{3-}$, and $[\text{MnCl}_6]^{3-}$. Total count $= 3$.
### Pattern Recognition
Outer orbital complexes ($sp^3d^2$) require weak field ligands (like $F^-, Cl^-$) paired with metal configurations where internal $d$-orbitals cannot be cleared by pairing ($d^4, d^5, d^6$). $SF_6$ is a primary group molecule that always uses outer-shell d-orbitals. This brings our total to 3.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Chemical Bonding and Molecular Structure
