Coordination Compounds Previous Year Questions — JEE Main Chemistry
68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 13.
Q58 (2026)
Consider a mixture 'X' which is made by dissolving $0.4 \text{ mol}$ of $[Co(NH_3)_5SO_4]Br$ and $0.4 \text{ mol}$ of $[Co(NH_3)_5Br]SO_4$ in water to make $4 \text{ L}$ of solution.
When $2 \text{ L}$ of mixture 'X' is allowed to react with excess of $AgNO_3$, it forms precipitate 'Y'. The rest $2 \text{ L}$ of mixture 'X' reacts with excess $BaCl_2$ to form precipitate 'Z'. Which of the following statements is CORRECT.
- $0.2 \text{ mol of 'Z' is formed}$
- $\text{'Y' is } BaSO_4 \text{ and 'Z' is } AgBr$
- $0.4 \text{ mol of 'Z' is formed}$
- $0.1 \text{ mol of 'Y' is formed}$
### Core Logic
The $4 \text{ L}$ solution contains:
$0.4 \text{ mol}$ of $[Co(NH_3)_5SO_4]Br$ $\implies$ Gives $0.4 \text{ mol } Br^-$ in solution.
$0.4 \text{ mol}$ of $[Co(NH_3)_5Br]SO_4$ $\implies$ Gives $0.4 \text{ mol } SO_4^{2-}$ in solution.
The solution is divided into two $2 \text{ L}$ portions. Each $2 \text{ L}$ portion will contain exactly half the moles of each species:
Moles of $Br^-$ in $2 \text{ L} = 0.2 \text{ mol}$
Moles of $SO_4^{2-}$ in $2 \text{ L} = 0.2 \text{ mol}$
### Step 1: Reaction with Silver Nitrate
When the first $2 \text{ L}$ of the mixture reacts with excess $AgNO_3$:
$$Ag^+ + Br^- \rightarrow AgBr \downarrow (\text{Precipitate 'Y'})$$
Since there are $0.2 \text{ mol}$ of free $Br^-$ ions, $0.2 \text{ mol}$ of $AgBr$ (Y) will be formed.
### Step 2: Reaction with Barium Chloride
When the remaining $2 \text{ L}$ of the mixture reacts with excess $BaCl_2$:
$$Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \downarrow (\text{Precipitate 'Z'})$$
Since there are $0.2 \text{ mol}$ of free $SO_4^{2-}$ ions, $0.2 \text{ mol}$ of $BaSO_4$ (Z) will be formed.
### Step 3: Final Conclusion
Checking the options:
(1) $0.2 \text{ mol}$ of 'Z' is formed $\implies$ True.
(2) 'Y' is $BaSO_4$ and 'Z' is $AgBr$ $\implies$ False, they are reversed.
(3) $0.4 \text{ mol}$ of 'Z' is formed $\implies$ False, it's $0.2 \text{ mol}$.
(4) $0.1 \text{ mol}$ of 'Y' is formed $\implies$ False, it's $0.2 \text{ mol}$.
### Pattern Recognition
Only the counter-ions (outside the coordination sphere) ionize in water and participate in precipitation reactions. Always account for volume fractions when a solution is split.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q59 (2026)
Given below are two statements
Statements-I: The number of paramagnetic species among $[CoF_6]^{3-}$, $[TiF_6]^{3-}$, $V_2O_5$ and $[Fe(CN)_6]^{3-}$ is 3.
Statement-II: $K_4[Fe(CN)_6] < K_3[Fe(CN)_6] < [Fe(H_2O)_6]SO_4 \cdot H_2O < [Fe(H_2O)_6]Cl_3$ is the correct order in terms of number of unpaired electron(s) in the complexes.
In the light of the above statements, choose the correct answer from the options given below.
- $\text{Both statement-I and statement-II are true}$
- $\text{Both statement-I and statement-II are false}$
- $\text{Statement-I is true but statement-II is false}$
- $\text{Statement-I is false but statement-II is true}$
### Core Logic
Analyzing Statement I:
$[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. F is WFL $\implies$ high spin, $4$ unpaired electrons (paramagnetic).
$[TiF_6]^{3-}$: $Ti^{3+}$ is $3d^1$. $1$ unpaired electron (paramagnetic).
$V_2O_5$: $V^{5+}$ is $3d^0$. $0$ unpaired electrons (diamagnetic).
$[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. CN is SFL $\implies$ low spin, $1$ unpaired electron (paramagnetic).
Total paramagnetic species = 3. Statement I is true.
Analyzing Statement II:
$K_4[Fe(CN)_6]$: $Fe^{2+}$ is $3d^6$. SFL $\implies t_{2g}^6 e_g^0$, $0$ unpaired electrons.
$K_3[Fe(CN)_6]$: $Fe^{3+}$ is $3d^5$. SFL $\implies t_{2g}^5 e_g^0$, $1$ unpaired electron.
$[Fe(H_2O)_6]SO_4 \cdot H_2O$: $Fe^{2+}$ is $3d^6$. $H_2O$ is WFL $\implies t_{2g}^4 e_g^2$, $4$ unpaired electrons.
$[Fe(H_2O)_6]Cl_3$: $Fe^{3+}$ is $3d^5$. $H_2O$ is WFL $\implies t_{2g}^3 e_g^2$, $5$ unpaired electrons.
The order $0 < 1 < 4 < 5$ is correct. Statement II is true.
### Step 1: Final Conclusion
Both Statement I and Statement II are true.
### Pattern Recognition
Identify the oxidation state, count $d$-electrons, apply spectrochemical series to determine pairing, then count unpaired electrons to establish diamagnetic ($n=0$) vs paramagnetic ($n>0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 12 Chemistry: The d- and f-Block Elements
Q57 (2026)
The wavelength of light absorbed for the following complexes are in the order.
I: $[Co(NH_{3})_{6}]^{3+}$
II: $[Co(H_{2}O)_{6}]^{3+}$
III: $[Co(CN)_{6}]^{3-}$
IV: $[Co(NH_{3})_{5}(H_{2}O)]^{3+}$
V: $[CoF_{6}]^{3-}$
- III < I < II < IV < V
- III < I < IV < V < II
- III < IV < I < II < V
- III < I < IV < II < V
### Related Formula
$$E = \frac{hc}{\lambda} \implies \lambda \propto \frac{1}{\Delta_o}$$
### Core Logic
Wavelength of light absorbed increases as the Crystal Field Splitting Energy (C.F.S.E, $\Delta_o$) of the complex decreases.
According to the spectrochemical series, the ligand field strength follows the order:
$CN^- > NH_3 > H_2O > F^-$
Based on this:
Ligand field strength $\uparrow$ ; C.F.S.E $\uparrow$ ; Absorbed wavelength ($\lambda$) $\downarrow$
$[Co(CN)_{6}]^{3-}$ (III) has the strongest ligand, so maximum CFSE and minimum $\lambda$.
$[CoF_{6}]^{3-}$ (V) has the weakest ligand, so least CFSE and maximum $\lambda$.
Comparing $(NH_3)_6$ (I) vs $(NH_3)_5(H_2O)$ (IV) vs $(H_2O)_6$ (II):
Since $NH_3 > H_2O$, the field strength decreases as we replace $NH_3$ with $H_2O$.
Thus, field strength order: III > I > IV > II > V
Correct absorbed wavelength ($\lambda$) order will be the reverse:
III < I < IV < II < V
### Pattern Recognition
Strong field ligands cause high splitting (large $\Delta$), meaning they absorb higher energy photons, which correspond to shorter wavelengths. Remember: $\lambda \propto 1/\text{field strength}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q73 (2026)
A chromium complex with a formula $CrCl_{3}\cdot6H_{2}O$ has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to 1:2 electrolyte. 2.75 g of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of $AgNO_{3}$ . The amount of AgCl formed in the above process is ____ g. (Nearest integer)
[Given : Molar mass in g mol $^{-1}$ Cr : 52; Cl : 35.5, Ag : 108, O : 16, H : 1]
### Core Logic
Since the complex behaves as a 1:2 electrolyte, it must dissociate into 3 ions: 1 complex cation and 2 counter anions.
This means two chloride ions ($Cl^-$) must reside outside the coordination sphere.
The structural formula is: $[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}$.
Reaction with silver nitrate:
$[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O} + 2\mathrm{AgNO}_{3} \rightarrow 2\mathrm{AgCl} \downarrow + \text{complex nitrate}$
### Step 1: Calculate Moles of Complex
Molar mass of $CrCl_3\cdot 6H_2O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 \text{ g/mol}$.
Moles of the complex = $\frac{2.75}{266.5} \simeq 0.01031 \text{ moles}$.
### Step 2: Calculate Mass of Precipitate
Since each mole of complex gives 2 moles of $AgCl$:
Moles of $AgCl$ formed = $2 \times 0.01031 = 0.02063 \text{ moles}$.
Molar mass of $AgCl = 108 + 35.5 = 143.5 \text{ g/mol}$.
Mass of $AgCl = 0.02063 \times 143.5 \simeq 2.96 \text{ gm}$.
Rounding to the nearest integer yields 3.
### Pattern Recognition
A 1:2 electrolyte contains 3 total ions. For $CrCl_3 \cdot 6H_2O$, this strictly dictates $2 Cl^-$ are outside the coordination sphere. Hence $n$ factor for chloride precipitation is exactly 2.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q70 (2026)
The correct statement among the following is :
- $[Ni(CN)_{4}]^{2-}$ and $[NiCl_{4}]^{2-}$ are diamagnetic and $Ni(CO)_{4}$ is paramagnetic.
- $Ni(CO)_{4}$ and $[NiCl_{4}]^{2-}$ are diamagnetic and $[Ni(CN)_{4}]^{2-}$ is paramagnetic.
- $Ni(CO)_{4}$ and $[Ni(CN)_{4}]^{2-}$ are diamagnetic and $[NiCl_{4}]^{2-}$ is paramagnetic.
- $Ni(CO)_{4}$ is diamagnetic and $[NiCl_{4}]^{2-}$ and $[Ni(CN)_{4}]^{2-}$ are paramagnetic.
### Core Logic
Analyze the oxidation state, electronic configuration, and ligand field strength for each Nickel complex:
$\bullet$ $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand. It causes pairing of electrons, resulting in $dsp^2$ hybridization and 0 unpaired electrons (Diamagnetic).
$\bullet$ $Ni(CO)_4$: $Ni^0$ is $3d^8 4s^2$. $CO$ is a strong field ligand, forcing the $4s$ electrons into the $3d$ subshell, resulting in a $3d^{10}$ configuration. It undergoes $sp^3$ hybridization with 0 unpaired electrons (Diamagnetic).
$\bullet$ $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand. No pairing occurs, resulting in $sp^3$ hybridization with 2 unpaired electrons (Paramagnetic).
### Final Conclusion
$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic, while $[NiCl_4]^{2-}$ is paramagnetic.
### Pattern Recognition
Ni with strong ligands ($CN^-, CO$) collapses into paired diamagnetic states. Ni with weak halogens ($Cl^-$) stays paramagnetic and tetrahedral.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
