Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 10.
Q61 (2024)
Match List-I with 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 ion)</th><th style="border: 1px solid #888; padding: 8px;">List - II (Electronic Configuration)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. $[Cr(H_2O)_6]^{3+}$</td><td style="border: 1px solid #888; padding: 8px;">I. $t_{2g}^2e_g^0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. $[Fe(H_2O)_6]^{3+}$</td><td style="border: 1px solid #888; padding: 8px;">II. $t_{2g}^3e_g^0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. $[Ni(H_2O)_6]^{2+}$</td><td style="border: 1px solid #888; padding: 8px;">III. $t_{2g}^3e_g^2$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. $[V(H_2O)_6]^{3+}$</td><td style="border: 1px solid #888; padding: 8px;">IV. $t_{2g}^6e_g^2$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-III, B-II, C-IV, D-I}$
- $\text{A-IV, B-I, C-II, D-III}$
- $\text{A-IV, B-III, C-I, D-II}$
- $\text{A-II, B-III, C-IV, D-I}$
### Core Logic
Identify the central metal ion, its oxidation state, and outer electronic configuration:
1) $[Cr(H_2O)_6]^{3+}$ contains $Cr^{3+} : [Ar] 3d^3$. In an octahedral field with weak field ligands ($H_2O$), the configuration is $t_{2g}^3 e_g^0$.
2) $[Fe(H_2O)_6]^{3+}$ contains $Fe^{3+} : [Ar] 3d^5$. With weak field ligand ($H_2O$), it forms a high spin complex: $t_{2g}^3 e_g^2$.
3) $[Ni(H_2O)_6]^{2+}$ contains $Ni^{2+} : [Ar] 3d^8$. In an octahedral field, the configuration is $t_{2g}^6 e_g^2$.
4) $[V(H_2O)_6]^{3+}$ contains $V^{3+} : [Ar] 3d^2$. In an octahedral field, the configuration is $t_{2g}^2 e_g^0$.
### Step 1: Final Mapping
A $\rightarrow$ II ($t_{2g}^3e_g^0$)
B $\rightarrow$ III ($t_{2g}^3e_g^2$)
C $\rightarrow$ IV ($t_{2g}^6e_g^2$)
D $\rightarrow$ I ($t_{2g}^2e_g^0$)
This matches option (4).
### Pattern Recognition
Shortcut: Vanadium (V) in $+3$ state has $2$ electrons, so $D \rightarrow I$. Only options (1) and (4) have $D \rightarrow I$. Nickel (Ni) in $+2$ state has $8$ electrons, so $C \rightarrow IV$. Only option (4) has $C \rightarrow IV$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q68 (2024)
Select the option with correct property -
- $\text{(1) }[Ni(CO)_4]\text{ and }[NiCl_4]^{2-}\text{ both diamagnetic}$
- $\text{(2) }[Ni(CO)_4]\text{ and }[NiCl_4]^{2-}\text{ both paramagnetic}$
- $\text{(3) }[NiCl_4]^{2-}\text{ diamagnetic, }[Ni(CO)_4]\text{ paramagnetic}$
- $\text{(4) }[Ni(CO)_4]\text{ diamagnetic, }[NiCl_4]^{2-}\text{ paramagnetic}$
### Core Logic
For $[Ni(CO)_4]$:
Nickel is in $0$ oxidation state: $Ni^0 = [Ar] 3d^8 4s^2$. $CO$ is a strong field ligand, forcing pairing of electrons. The $4s$ electrons move to the $3d$ orbital, making the configuration $3d^{10}$. With no unpaired electrons, the hybridization is $sp^3$ and it is diamagnetic.
For $[NiCl_4]^{2-}$:
Nickel is in $+2$ oxidation state: $Ni^{2+} = [Ar] 3d^8$. $Cl^-$ is a weak field ligand, so pairing does not occur against Hund's rule. The configuration remains $t_{2g}^6 e_g^2$ (or simply two unpaired electrons in the tetrahedral $d$ splitting). The hybridization is $sp^3$ and with 2 unpaired electrons, it is paramagnetic.
### Step 1: Final Conclusion
$[Ni(CO)_4]$ is diamagnetic, while $[NiCl_4]^{2-}$ is paramagnetic. This correctly corresponds to option (4).
### Pattern Recognition
Strong field ligands (like $CO$, $CN^-$) usually lead to diamagnetism in $d^8$ metal ions by forming square planar ($dsp^2$) or forcing $d^{10}$ configurations (for $Ni^0$). Weak field halogens lead to paramagnetic $sp^3$ complexes for $Ni(II)$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q76 (2024)
The correct statements from following are:
A. The strength of anionic ligands can be explained by crystal field theory.
B. Valence bond theory does not give a quantitative interpretation of kinetic stability of coordination compounds.
C. The hybridization involved in formation of $[Ni(CN)_4]^{2-}$ complex is $dsp^2$
D. The number of possible isomer(s) of $cis-[PtCl_2(en)_2]^{2+}$ is one
Choose the correct answer from the options given below:
- $\text{A, D only}$
- $\text{A, C only}$
- $\text{B, D only}$
- $\text{B, C only}$
### Step 1: Analyzing Statement A
Crystal field theory considers ligands as point charges. Therefore, anionic ligands should exert the greatest splitting effect, but practically they are at the lower end of the spectrochemical series. CFT cannot explain this properly. Hence, Statement A is incorrect.
### Step 2: Analyzing Statement B
Valence bond theory (VBT) is qualitative and does not give a quantitative interpretation of either thermodynamic or kinetic stability. Hence, Statement B is correct.
### Step 3: Analyzing Statement C
In $[Ni(CN)_4]^{2-}$, $Ni$ is in $+2$ state ($3d^8$). Since $CN^-$ is a strong field ligand, pairing of electrons takes place leaving one inner 3d orbital empty. Thus, the hybridization is $dsp^2$ (square planar). Hence, Statement C is correct.
### Step 4: Analyzing Statement D
$cis-[PtCl_2(en)_2]^{2+}$ is an octahedral complex of type $M(AA)_2a_2$. The cis-isomer lacks a plane of symmetry and is optically active, thus it exists as a pair of enantiomers (d and l). Therefore, the number of possible isomers is 2, not 1. Statement D is incorrect.
### Final Conclusion
Statements B and C are correct.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q82 (2024)
The 'Spin only' Magnetic moment for $[Ni(NH_3)_6]^{2+}$ is ________ $\times 10^{-1}$ BM.
(given = Atomic number of Ni: 28)
### Related Formula
$$\mu = \sqrt{n(n + 2)} \text{ BM}$$
### Step 1: Electronic Configuration
Atomic number of $Ni = 28$.
$Ni^{2+} = [Ar] 3d^8$
Because the geometry is octahedral (coordination number 6) and $NH_3$ acts as a weak field ligand with $Ni^{2+}$ (or because $d^8$ configuration always has 2 unpaired electrons in an octahedral field regardless of ligand strength):
The configuration in the $t_{2g}$ and $e_g$ levels is $t_{2g}^6 e_g^2$.
### Step 2: Calculating Magnetic Moment
Number of unpaired electrons, $n = 2$.
$$\mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.828 \text{ BM}$$
Representing in $10^{-1}$ scale:
$$\mu \approx 28.28 \times 10^{-1} \text{ BM}$$
Rounding to the nearest integer as typically expected gives 28.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q65 (2026)
Given below are two statements:
Statement I: Among $[\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+}$, $[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}$, $[\mathrm{Ni}(\mathrm{NH}_{3})_{6}]^{2+}$ and $[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$, $[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$ has the maximum number of unpaired electrons.
Statement II: The number of pairs among $\{[NiCl_{4}]^{2-}, [Ni(CO)_{4}]\}$, $\{[NiCl_{4}]^{2-}, [Ni(CN)_{4}]^{2-}\}$ and $\{[Ni(CO)_{4}], [Ni(CN)_{4}]^{2-}\}$ that contain only diamagnetic species is two.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Statement I is false but Statement II is true}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is true but Statement II is false}$
### Core Logic
Evaluating Statement I:
- $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$, 1 unpaired electron.
- $[Ni(en)_3]^{2+}$: $Ni^{2+}$ is $3d^8$, in octahedral field, 2 unpaired electrons.
- $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$, 2 unpaired electrons.
- $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$, weak field ligand $H_2O$ leads to high spin, 5 unpaired electrons.
So $[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true.
Evaluating Statement II:
- $[Ni(CO)_4]$: $Ni(0)$ is $3d^8 4s^2$, strong field CO pairs electrons to $3d^{10}$, diamagnetic (0 unpaired).
- $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$, strong field $CN^-$ forces pairing $\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired).
- $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$, weak field $Cl^-$ does not pair $\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired).
The pairs containing ONLY diamagnetic species:
- $\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$ $\rightarrow$ 1 para, 1 dia (No)
- $\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$ $\rightarrow$ 1 para, 1 dia (No)
- $\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$ $\rightarrow$ Both dia (Yes)
The number of such pairs is exactly ONE. Statement II says two, so it is false.
### Step 1: Final Conclusion
Statement I is true, Statement II is false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
