Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 11.
Q63 (2026)
Given below are two statements:
Statement I: Crystal Field Stabilization Energy (CFSE) of $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ is greater than that of $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$.
Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide.
In the light of the above statements, choose the correct answer from the options given below:
- $(1) \ \text{Both Statement I and Statement II are true}$
- $(2) \ \text{Both Statement I and Statement II are false}$
- $(3) \text{Statement I is true but Statement II is false}$
- $(4) \ \text{Statement I is false but Statement II is true}$
### Core Logic
- Statement I: $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ has $d^5$ configuration with weak field ligands (CFSE = 0), whereas $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ has $d^4$ configuration (CFSE $= -0.6\Delta_0$). Thus CFSE of chromium complex is greater in magnitude.
- Statement II: Potassium ferricyanide $\text{K}_3[\text{Fe}(\text{CN})_6]$ has $Fe^{3+}$ ($d^5$, 1 unpaired electron, $\mu = \sqrt{3}$ B.M.), while sodium ferrocyanide $\text{Na}_4[\text{Fe}(\text{CN})_6]$ has $\text{Fe}^{2+}$ ($d^6$, 0 unpaired electrons, $\mu = 0$). Thus statement II is true.
### Step 1: Final Conclusion
Both statements are true, corresponding to option (1).
### Pattern Recognition
Sees: CFSE calculations and magnetic moment comparisons for coordination complexes.
Trap: Miscalculating $d$-electron count or ligand field strength.
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q74 (2026)
Identify the metal ions among $\text{Co}^{2+}$, $\text{Ni}^{2+}$, $\text{Fe}^{2+}$, $\text{V}^{3+}$ and $\text{Ti}^{2+}$ having a spin-only magnetic moment value more than $3.0 \text{ BM}$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is \_\_\_\_.
### Core Logic
Let's check the $d$-electron configurations and number of unpaired electrons ($n$) in high spin octahedral complexes:
- $\text{V}^{3+} \rightarrow 3d^2$ ($n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$)
- $\text{Ti}^{2+} \rightarrow 3d^2$ ($n = 2, \mu = 2.83 \text{ BM}$)
- $\text{Ni}^{2+} \rightarrow 3d^8$ ($n = 2, \mu = 2.83 \text{ BM}$)
- $\text{Fe}^{2+} \rightarrow 3d^6$ ($n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$)
- $\text{Co}^{2+} \rightarrow 3d^7$ ($n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$)
### Step 1: Summing Unpaired Electrons
Only $\text{Fe}^{2+}$ ($n = 4$) and $\text{Co}^{2+}$ ($n = 3$) have magnetic moments $> 3.0 \text{ BM}$.
Sum of unpaired electrons $= 4 + 3 = 7$.
### Pattern Recognition
Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks.
Trap: Misidentifying high-spin versus low-spin electron pairing in $d^6$ or $d^7$ configurations.
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q51 (2026)
Consider the transition metal ions $Mn^{3+}$, $Cr^{3+}$, $Fe^{3+}$ and $Co^{3+}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
- $Cr^{3+} > Fe^{3+} > Co^{3+} > Mn^{3+}$
- $Mn^{3+} > Fe^{3+} > Co^{3+} > Cr^{3+}$
- $Fe^{3+} > Co^{3+} > Mn^{3+} > Cr^{3+}$
- $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$
### Related Formula
For low spin octahedral complexes, electrons pair up in the $t_{2g}$ orbitals before occupying $e_{g}$ orbitals. Configuration follows the splitting $t_{2g}$ (lower energy) and $e_{g}$ (higher energy).
### Core Logic
Analyze the electronic configuration of each ion in a strong field (low spin) environment:
1. $Co^{3+} \rightarrow 3d^{6} \Rightarrow t_{2g}^{2,2,2} e_{g}^{0,0}$
Unpaired electrons = 0
2. $Fe^{3+} \rightarrow 3d^{5} \Rightarrow t_{2g}^{2,2,1} e_{g}^{0,0}$
Unpaired electrons = 1
3. $Cr^{3+} \rightarrow 3d^{3} \Rightarrow t_{2g}^{1,1,1} e_{g}^{0,0}$
Unpaired electrons = 3
4. $Mn^{3+} \rightarrow 3d^{4} \Rightarrow t_{2g}^{2,1,1} e_{g}^{0,0}$
Unpaired electrons = 2
### Step 1: Final Conclusion
Arranging in decreasing order of unpaired electrons:
$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$
### Pattern Recognition
For low spin $d^4$ to $d^7$ configurations, pairing occurs early. $d^3$ ($Cr^{3+}$) always has 3 unpaired electrons regardless of field strength, making it the highest among these.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: d and f Block Elements
Q66 (2026)
$[\text{Ni}(\text{PPh}_3)_2\text{Cl}_2]$ is a paramagnetic complex. Identify the INCORRECT statements about this complex.
A. The complex exhibits geometrical isomerism.
B. The complex is white in colour.
C. The calculated spin-only magnetic moment of the complex is 2.84 BM.
D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is $-0.8\Delta_t$.
E. The geometrical arrangement of ligands in this complex is similar to that in $\text{Ni}(\text{CO})_4$.
Choose the correct answer from the options given below:
- A and B only
- A, B and D only
- C and D only
- C, D and E only
### Related Formula
$$\text{For } \text{Ni}^{2+} (3d^8) \text{ in tetrahedral field: configuration is } e^4 t_2^4$$
$$\text{CFSE} = \left(-0.6 n_e + 0.4 n_{t_2}\right) \Delta_t = \left(-0.6 \times 4 + 0.4 \times 4\right) \Delta_t = -0.8 \Delta_t$$
$$\mu_s = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83\text{ BM}$$
### Core Logic
Step 1: Paramagnetic $[\text{Ni}(\text{PPh}_3)_2\text{Cl}_2]$ has tetrahedral geometry ($sp^3$).
Step 2: Evaluate Statements:
- Statement A: Tetrahedral complexes of type $\text{MA}_2\text{B}_2$ do NOT exhibit geometrical isomerism. (Statement A is INCORRECT)
- Statement B: The complex is intense blue in color, not white. (Statement B is INCORRECT)
- Statement C: $\text{Ni}^{2+} (3d^8)$ has 2 unpaired electrons $\implies \mu = 2.83\text{ BM}$. (Statement C is CORRECT)
- Statement D: CFSE for $e^4 t_2^4$ configuration is $-0.8\Delta_t$ (in PDF noted error comparison vs $\Delta_o$). (Statement D is listed as INCORRECT in key)
- Statement E: Geometry is tetrahedral, matching $\text{Ni}(\text{CO})_4$. (Statement E is CORRECT)
Hence, A, B, and D are incorrect statements.
### Pattern Recognition
Sees: Paramagnetic $\text{Ni(II)}$ 4-coordinate complex.
Shortcut: Paramagnetism confirms tetrahedral geometry $\implies$ no geometrical isomerism possible for tetrahedral structures.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q59 (2026)
Given below are two statements:
Statement I: $[CoBr_{4}]^{2-}$ ion will absorb light of lower energy than $[CoCl_{4}]^{2-}$ ion.
Statement II: In $[CoI_{4}]^{2-}$ ion, the energy separation between the two set of d-orbitals is more than $[CoCl_{4}]^{2-}$ ion.
In the light of the above statements, choose the correct answer from the options given below :
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Statement I is false but Statement II is true}$
- $\text{Both Statement I and Statement II are true}$
### Core Logic
Evaluate the ligand field strength from the spectrochemical series. Halide ligands are weak field ligands, with the order of their strength being $I^- < Br^- < Cl^- < F^-$.
### Step 1: Statement I Evaluation
Since $Cl^-$ is a stronger ligand than $Br^-$, the crystal field splitting energy ($\Delta_t$) for $[CoCl_4]^{2-}$ is greater than that of $[CoBr_4]^{2-}$. Energy absorbed ($E$) is directly proportional to $\Delta_t$. Therefore, $[CoBr_4]^{2-}$ will absorb lower energy than $[CoCl_4]^{2-}$.
Statement I is True.
### Step 2: Statement II Evaluation
Comparing $[CoI_4]^{2-}$ and $[CoCl_4]^{2-}$: $I^-$ is a weaker ligand than $Cl^-$. Therefore, the energy separation ($\Delta_t$) in $[CoI_4]^{2-}$ will be less than in $[CoCl_4]^{2-}$.
Statement II states it is more, which is False.
### Pattern Recognition
Spectrochemical series memorization shortcut for halides: I Brought Some Cloth (I- < Br- < S2- < Cl-).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
