Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 3.
Q44 (2025)
Match the coordination complexes listed in LIST-I with their geometric shape and magnetic moment characteristics described in LIST-II:
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">LIST-I (Complex/Species)</th><th style="border: 1px solid #888; padding: 8px;">LIST-II (Shape & magnetic moment)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. $[\text{Ni(CO)}_4]$</td><td style="border: 1px solid #888; padding: 8px;">I. Tetrahedral, 2.8 BM</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. $[\text{Ni(CN)}_4]^{2-}$</td><td style="border: 1px solid #888; padding: 8px;">II. Square planar, 0 BM</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. $[\text{NiCl}_4]^{2-}$</td><td style="border: 1px solid #888; padding: 8px;">III. Tetrahedral, 0 BM</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. $[\text{MnBr}_4]^{2-}$</td><td style="border: 1px solid #888; padding: 8px;">IV. Tetrahedral, 5.9 BM</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-III, B-IV, C-II, D-I}$
- $\text{A-I, B-II, C-III, D-IV}$
- $\text{A-III, B-II, C-I, D-IV}$
- $\text{A-IV, B-I, C-III, D-II}$
### Core Logic
Let us apply Valence Bond Theory (VBT) and crystal field rules to evaluate each coordination complex:
* **A. $[\text{Ni(CO)}_4]$**: Nickel is in the $0$ oxidation state ($3d^8 4s^2$). Carbon monoxide ($\text{CO}$) is a strong field ligand, forcing the $4s$ electrons into the $3d$ shell to produce a fully paired $3d^{10}$ configuration. The vacant $4s$ and three $4p$ orbitals hybridize into an **$sp^3$ tetrahedral** geometry. All spins are paired, so $\mu = 0 \text{ BM}$. Thus, $\text{A} \rightarrow \text{III}$. {{SOL_IMG1}}
* **B. $[\text{Ni(CN)}_4]^{2-}$**: Nickel is in the $+2$ state ($3d^8$). Cyanide ($\text{CN}^-$) is a strong field ligand, forcing the pairing of the two unpaired $3d$ electrons. This leaves one internal $3d$ orbital vacant, leading to **$dsp^2$ square planar** hybridization with zero unpaired electrons ($\mu = 0 \text{ BM}$). Thus, $\text{B} \rightarrow \text{II}$. {{SOL_IMG2}}
* **C. $[\text{NiCl}_4]^{2-}$**: Nickel is in the $+2$ state ($3d^8$). Chloride ($\text{Cl}^-$) is a weak field ligand, leaving the two $3d$ electrons unpaired ($n = 2$). The system adopts **$sp^3$ tetrahedral** hybridization with a spin-only moment of $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 \text{ BM}$. Thus, $\text{C} \rightarrow \text{I}$. {{SOL_IMG4}}
* **D. $[\text{MnBr}_4]^{2-}$**: Manganese is in the $+2$ state ($3d^5$). Bromide ($\text{Br}^-$) is a weak field ligand, preserving five unpaired parallel spins ($n = 5$). The geometry is **$sp^3$ tetrahedral** with a maximum spin-only moment of $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM}$. Thus, $\text{D} \rightarrow \text{IV}$. {{SOL_IMG6}}
### Step 1: Alignment Summary
Consolidating our results:
$$\text{A-III, B-II, C-I, D-IV}$$
This matches Option (3).
### Pattern Recognition
Nickel complexes provide classic benchmarks: Nickel zero tetracarbonyl is always tetrahedral diamagnetic. Nickel $+2$ tetracyanide is square planar diamagnetic due to strong ligand field pairing. Spotting these properties cuts down the problem solving time significantly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q26 (2025)
The calculated spin-only magnetic moments of $K_{3}[Fe(OH)_{6}]$ and $K_{4}[Fe(OH)_{6}]$ respectively are:
(1) 4.90 and 4.90 B.M.
(2) 5.92 and 4.90 B.M.
(3) 3.87 and 4.90 B.M.
(4) 4.90 and 5.92 B.M.
- 4.90 and 4.90 B.M.
- 5.92 and 4.90 B.M.
- 3.87 and 4.90 B.M.
- 4.90 and 5.92 B.M.
### Related Formula
$$mu = sqrt{n(n+2)}\text{ B.M.}$$
### Core Logic
In $K_{3}[Fe(OH)_{6}]$, iron is in the $+3$ oxidation state ($Fe^{3+} = 3d^{5}$).
Since $OH^{-}$ is a weak field ligand, no pairing of electrons takes place. The number of unpaired electrons ($n$) is $5$.
$$mu = sqrt{5(5+2)} = sqrt{35} approx 5.92\text{ B.M.}$$
In $K_{4}[Fe(OH)_{6}]$, iron is in the $+2$ oxidation state ($Fe^{2+} = 3d^{6}$).
Since $OH^{-}$ is a weak field ligand, no pairing occurs. The number of unpaired electrons ($n$) is $4$.
$$mu = sqrt{4(4+2)} = sqrt{24} approx 4.90\text{ B.M.}$$
### Pattern Recognition
Identify the ligand field strength first. $OH^{-}$ is a weak field ligand in the spectrochemical series, so it does not cause pairing in either $Fe^{2+}$ or $Fe^{3+}$ configurations.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q37 (2025)
Identify the homoleptic complexes with odd number of d electrons in the central metal.
(A) $[FeO_{4}]^{2-}$
(B) $[Fe(CN)_{6}]^{3-}$
(C) $[Fe(CN)_{5}NO]^{2-}$
(D) $[CoCl_{4}]^{2-}$
(E) $[Co(H_{2}O)_{3}F_{3}]$
Choose the correct answer from the options given below:
- (B) and (D) only
- (C) and (E) only
- (A), (B) and (D) only
- (A), (C) and (E) only
### Core Logic
A complex is homoleptic if the metal is bound to only one kind of donor ligand group.
* (A) $[FeO_4]^{2-}$ is homoleptic, but $Fe^{+6}$ corresponds to a $3d^2$ (even) electronic configuration.
* (B) $[Fe(CN)_6]^{3-}$ is homoleptic. $Fe^{+3}$ corresponds to a $3d^5$ (odd) configuration.
* (C) $[Fe(CN)_5NO]^{2-}$ is heteroleptic (contains two types of ligands).
* (D) $[CoCl_4]^{2-}$ is homoleptic. $Co^{+2}$ corresponds to a $3d^7$ (odd) configuration.
* (E) $[Co(H_2O)_3F_3]$ is heteroleptic.
### Pattern Recognition
Filter by 'homoleptic' first to instantly eliminate multi-ligand mixed structures like options (C) and (E).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q46 (2025)
Consider the following low-spin complexes $K_{3}[Co(NO_{2})_{6}]$, $K_{4}[Fe(CN)_{6}]$, $K_{3}[Fe(CN)_{6}]$, $Cu_{2}[Fe(CN)_{6}]$ and $Zn_{2}[Fe(CN)_{6}]$.
The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer is nearest integer)
### Core Logic
From the given list, the complexes exhibiting a distinct yellow color are $K_{3}[Co(NO_{2})_{6}]$ and $K_{4}[Fe(CN)_{6}]$.
Let's calculate the spin-only magnetic moments for these low-spin configurations:
1) For $K_{3}[Co(NO_{2})_{6}]$, cobalt is in $+3$ oxidation state ($Co^{3+} = 3d^{6}$).
In the presence of the strong ligand field ($NO_2^-$), all six electrons pair up completely in the $t_{2g}$ orbitals:
$$t_{2g}^6 e_g^0 implies n = 0 \text{ unpaired electrons} implies mu = 0\text{ BM}$$
{{SOL_IMG46}}
2) For $K_{4}[Fe(CN)_{6}]$, iron is in $+2$ oxidation state ($Fe^{2+} = 3d^{6}$).
In the strong field of cyanide ligands ($CN^-$), pairing is complete:
$$t_{2g}^6 e_g^0 implies n = 0 \text{ unpaired electrons} implies mu = 0\text{ BM}$$
Therefore, the sum of their spin-only magnetic moments is $0 + 0 = 0$.
### Pattern Recognition
Low-spin $d^6$ octahedral complexes always yield a fully closed-shell $t_{2g}^6$ arrangement with zero unpaired electrons, leading deterministically to a magnetic moment of 0 BM.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q36 (2025)
The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is
- $\mathrm{Ti}^{3+}$
- $\mathrm{Ni}^{2+}$
- $\mathrm{Mn}^{2+}$
- $\mathrm{Cr}^{3+}$
### Core Logic
Nickel ($\mathrm{Ni}^{2+}$) exhibits a $d^8$ electronic profile. In regular octahedral complex splits:
$t_{2g}^6 e_g^2$
Because the lower $t_{2g}$ subshell is fully paired and the higher $e_g$ contains exactly 2 electrons matching Hund's rules, this orbital distribution remains configurationally identical under both strong-field and weak-field environments.
Additionally, $\mathrm{Ni}^{2+}$ compounds produce a characteristic violet bead during hot cycles in a non-luminous flame within the qualitative borax matrix.
### Pattern Recognition
Sees: Configuration invariant to ligand strength + qualitative test combination.
Shortcut: A $d^8$ structure in octahedral splitting always stays high-spin/low-spin identical, pointing strictly to $\mathrm{Ni}^{2+}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: The d-and f-Block Elements
