Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 12.
Q63 (2026)
The statements that are incorrect about the nickel (II) complex of dimethylglyoxime are:
A. It is red in colour
B. It has a high solubility in water at $pH = 9$
C. The Ni ion has two unpaired d-electrons
D. The N-Ni-N bond angle is almost close to $90^{\circ}$
E. The complex contains four five-membered metallacycles (metal containing rings)
Choose the correct answer from the options given below :
- $\text{C and E only}$
- $\text{A, D and B only}$
- $\text{B, C and E only}$
- $\text{C and D only}$
### Core Logic
Analyze the structural and electronic properties of the $[Ni(dmg)_2]$ complex.
{{SOL_IMG1}}
### Step 1: Evaluating Each Statement
(A) It is a rosy red precipitate. (True)
(B) It forms a precipitate in a basic medium (like ammonium hydroxide), indicating it is insoluble in water at $pH=9$. (False)
(C) The $Ni^{2+}$ ion is $3d^8$. DMG is a strong field ligand in a square planar geometry, causing electron pairing ($dsp^2$ hybridization). Number of unpaired electrons = 0. (False)
(D) Square planar geometry ensures N-Ni-N bond angles are close to $90^\circ$. (True)
(E) The complex contains two 5-membered rings and two 6-membered rings formed by hydrogen bonding. (False)
### Step 2: Identifying Incorrect Statements
Statements B, C, and E are incorrect.
### Pattern Recognition
$Ni-DMG$ complex = Rosy red ppt, Square Planar ($dsp^2$), Diamagnetic ($n=0$), Hydrogen bonded (extra stability).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q75 (2026)
The crystal field splitting energy of $[Co(\text{oxalate})_{3}]^{3-}$ complex is '$n$' times that of the $[Cr(\text{oxalate})_{3}]^{3-}$ complex. Here '$n$' is ____. [Assume $\Delta_{0} \gg P$]
### Related Formula
$$\text{CFSE} = \left( -0.4 \times n_{t_{2g}} + 0.6 \times n_{eg} \right) \Delta_0$$
*(Note: Pairing energy is neglected with respect to $\Delta_0$ based on assumption)*
### Core Logic
Since $\Delta_0 \gg P$ (Strong field logic / low spin complexes), electrons will pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
Identify the oxidation state and $d$-electron count for the central metal in both complexes.
### Step 1: Cobalt Complex Analysis
Complex: $[Co(\text{ox})_{3}]^{3-}$
Cobalt oxidation state = $+3$.
Electronic configuration of $Co^{3+}$: $[Ar] 3d^6$.
Under $\Delta_0 \gg P$, the $d^6$ configuration is $t_{2g}^{2,2,2} e_{g}^{0,0}$ (i.e., $t_{2g}^6 e_g^0$).
$\text{CFSE}_{Co^{3+}} = 6 \times (-0.4 \Delta_0) = -2.4 \Delta_0$
### Step 2: Chromium Complex Analysis
Complex: $[Cr(\text{ox})_{3}]^{3-}$
Chromium oxidation state = $+3$.
Electronic configuration of $Cr^{3+}$: $[Ar] 3d^3$.
Configuration is $t_{2g}^{1,1,1} e_{g}^{0,0}$ (i.e., $t_{2g}^3 e_g^0$).
$\text{CFSE}_{Cr^{3+}} = 3 \times (-0.4 \Delta_0) = -1.2 \Delta_0$
### Step 3: Calculating n
Ratio $n = \frac{\text{CFSE of } Co^{3+}}{\text{CFSE of } Cr^{3+}}$
$n = \frac{|-2.4 \Delta_0|}{|-1.2 \Delta_0|} = 2$
{{SOL_IMG1}}
### Pattern Recognition
$d^6$ low-spin always gives max CFSE for octahedral (-2.4 $\Delta_0$). $d^3$ is strictly half of that (-1.2 $\Delta_0$). Ratio is always 2.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q55 (2026)
Identify the CORRECT set of details from the following:
A. $[Co(NH_{3})_{6}]^{3+}$ : Inner orbital complex; $d^{2}sp^{3}$ hybridized
B. $[MnCl_{6}]^{3-}$ : Outer orbital complex; $sp^{3}d^{2}$ hybridized
C. $[CoF_{6}]^{3-}$ : Outer orbital complex; $d^{2}sp^{3}$ hybridized
D. $[FeF_{6}]^{3-}$ : Outer orbital complex; $sp^{3}d^{2}$ hybridized
E. $[Ni(CN)_{4}]^{2-}$ : Inner orbital complex; $sp^{3}$ hybridized
Choose the correct answer from the options given below:
- $\text{C & D only}$
- $\text{A, B & D only}$
- $\text{A, C & E only}$
- $\text{A, B, C, D & E}$
### Related Formula
null
### Core Logic
Evaluate each complex individually using Valence Bond Theory:
(A) $[Co(NH_{3})_{6}]^{3+}$: Central metal is $Co^{3+}$ ($3d^{6}$). $NH_3$ acts as a strong field ligand (SFL) for $Co^{3+}$, causing pairing. This leads to $d^2sp^3$ hybridization, forming an inner orbital complex. (Correct)
(B) $[MnCl_{6}]^{3-}$: Central metal is $Mn^{3+}$ ($3d^{4}$). $Cl^{-}$ is a weak field ligand (WFL), so no pairing occurs. It utilizes outer $4d$ orbitals for hybridization ($sp^3d^2$), forming an outer orbital complex. (Correct)
(C) $[CoF_{6}]^{3-}$: Central metal is $Co^{3+}$ ($3d^{6}$). $F^{-}$ is a weak field ligand (WFL), causing no pairing. It undergoes $sp^3d^2$ hybridization (outer orbital complex). The statement says it is $d^2sp^3$ hybridized, which is incorrect. (Incorrect)
(D) $[FeF_{6}]^{3-}$: Central metal is $Fe^{3+}$ ($3d^{5}$). $F^{-}$ is a weak field ligand (WFL), leading to no pairing. It undergoes $sp^3d^2$ hybridization, making it an outer orbital complex. (Correct)
(E) $[Ni(CN)_{4}]^{2-}$: Central metal is $Ni^{2+}$ ($3d^{8}$). $CN^{-}$ is a strong field ligand (SFL), causing pairing. It undergoes $dsp^2$ hybridization (inner orbital complex/square planar), not $sp^3$. (Incorrect)
### Step 1: Final Conclusion
Only statements A, B, and D are correct.
### Pattern Recognition
Spectrochemical series dictates SFL vs WFL. $Co^{3+}$ with $NH_3$ is a classic exception to memorize: $NH_3$ behaves as a SFL with $Co^{3+}$ (pairing occurs), whereas it acts as a WFL with many $+2$ ions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q74 (2026)
Total number of unpaired electrons present in the central metal atoms/ions of
$[Ni(CO)_{4}], [NiCl_{4}]^{2-}, [PtCl_{2}(NH_{3})_{2}], [Ni(CN_{4})]^{2-}$
and $[Pt(CN_{4})]^{2-}$ is __.
### Related Formula
null
### Core Logic
Let's determine the electronic configuration, oxidation state, and ligand nature for each complex:
1. $[Ni(CO)_4]$:
Oxidation state of Ni = 0. Configuration: $3d^8 4s^2$.
CO is a strong field ligand. The $4s$ electrons are pushed into the $3d$ orbital, making it a $3d^{10}$ configuration. It is $sp^3$ hybridized and diamagnetic.
Unpaired electrons = 0.
2. $[NiCl_4]^{2-}$:
Oxidation state of Ni = $+2$. Configuration: $3d^8$.
$Cl^-$ is a weak field ligand, so no pairing of the $d^8$ electrons occurs. The configuration is $t_{2g}^6 e_g^2$. It is $sp^3$ hybridized.
Unpaired electrons = 2.
3. $[PtCl_2(NH_3)_2]$:
Oxidation state of Pt = $+2$. Configuration: $5d^8$.
For 4d and 5d series metals, nearly all ligands act as strong field ligands. This causes pairing, leading to a $dsp^2$ hybridized square planar geometry.
Unpaired electrons = 0.
4. $[Ni(CN)_4]^{2-}$:
Oxidation state of Ni = $+2$. Configuration: $3d^8$.
$CN^-$ is a strong field ligand, forcing electron pairing. It becomes $dsp^2$ hybridized.
Unpaired electrons = 0.
5. $[Pt(CN)_4]^{2-}$:
Oxidation state of Pt = $+2$. Configuration: $5d^8$.
As established, 5d metals strictly form low spin complexes. $CN^-$ causes pairing ($dsp^2$).
Unpaired electrons = 0.
### Step 1: Total Sum
Summing all unpaired electrons across all listed complexes:
$0 + 2 + 0 + 0 + 0 = 2$.
### Pattern Recognition
Metals from 4d and 5d series (like Pd, Pt) ALWAYS form inner-orbital/low-spin complexes regardless of the ligand strength. So, $d^8$ configurations in $Pd^{2+}$ and $Pt^{2+}$ will always pair up to yield zero unpaired electrons.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q51 (2026)
Given below are two statements :
Statement-I Hybridisation, shape and spin only magnetic moment of $K_3[Co(CO_3)_3]$ is $sp^3d^2$, octahedral and $4.9 \text{ BM}$ respectively.
Statement-II Geometry, hybridisation and spin only magnetic moment values (BM) of the ions $[Ni(CN)_4]^{2-}$, $[MnBr_4]^{2-}$ and $[CoF_6]^{3-}$ respectively are square planar, tetrahedral, octahedral : $dsp^2$, $sp^3$, $sp^3d^2$ and $0$, $5.9$, $4.9$.
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
In $K_3[Co(CO_3)_3]$, the oxidation state of Co is $+3$ ($3d^6$ configuration). Since carbonate is a weak field ligand, electrons do not pair up.
Hybrdisation is $sp^3d^2$ (octahedral), with $4$ unpaired electrons.
$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ B.M.}$$
Thus, Statement I is true.
For Statement II:
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ ($3d^8$). $CN^-$ is a strong field ligand. Pairing occurs. Hybridisation is $dsp^2$ (square planar), $0$ unpaired electrons, $\mu = 0 \text{ B.M.}$
$[MnBr_4]^{2-}$: $Mn^{2+}$ ($3d^5$). $Br^-$ is a weak field ligand. $sp^3$ hybridised (tetrahedral), $5$ unpaired electrons, $\mu = 5.9 \text{ B.M.}$
$[CoF_6]^{3-}$: $Co^{3+}$ ($3d^6$). $F^-$ is a weak field ligand. $sp^3d^2$ hybridised (octahedral), $4$ unpaired electrons, $\mu = 4.9 \text{ B.M.}$
Thus, Statement II is also true.
### Step 1: Final Conclusion
Both Statement I and Statement II are true.
### Pattern Recognition
Check the ligand strength: $CO_3^{2-}$ and Halogens are WFL (outer orbital complexes, high spin), while $CN^-$ is SFL (inner orbital, low spin). Calculate unpaired electrons $n$, then use $\sqrt{n(n+2)}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
