Coordination Compounds JEE Main PYQs (2020–2026)

Solved Coordination Compounds PYQs

3 recent questions with answers. The full set of 274 is free to practise.

1. JEE Main 2026EasyCoordination Entities, Ligands, Coordination Number & IUPAC Nomenclature
   Q1

   An excess of ${AgNO}_3$ is added to 100 mL of a 0.05 M solution of tetraaquadichloridochromium (III) chloride. The number of moles of AgCl precipitated will be $\_\_\_\_\times 10^{-3}$ .
   
   (Nearest integer)

   Numerical answer type

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   Correct answer: 5

   The given complex is $[Cr(H_{2}O{)}_4{Cl}_2]Cl$ .
   
   When an excess of ${AgNO}_3$ is added, the $C{l}^{-}$ ion outside the coordination sphere reacts with $A{g}^{+}$ to form a white precipitate of silver chloride ( $AgCl$ ).
   
   The reaction is:
   
   $$[Cr(H_{2}O{)}_4{Cl}_2]Cl+{AgNO}_3$$
   
   $$\to [Cr(H_{2}O{)}_4{Cl}_2]{NO}_3+AgCl\downarrow$$
   
   From the reaction, 1 mole of the complex gives 1 mole of $AgCl$ .
   
   Now, the number of moles of the given complex is calculated as:
   
   $$\text{Moles of complex}=M\times V=0.05\times 100\times 10^{-3}=5\times 10^{-3}\text{moles}$$
   
   Hence, moles of $AgCl$ formed $=5\times 10^{-3}$ moles.
   
   Therefore, the required value of $x=5$ .
2. JEE Main 2026EasyWerner's Theory of Coordination Compounds
   Q2

   5.33 g of $CrC{l}_3\cdot 6H_{2}O$, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with AgNO3 results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is ________ ×10^-2. (Nearest integer)
   
   [Molar mass in g mol^–1 Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

   Numerical answer type

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   Correct answer: 33

   Let the compound be of the type
   
   $$[\text{Cr}({\text{H}}_2\text{O}{)}_x{\text{Cl}}_y]{\text{Cl}}_{3-y}\cdot n{\text{H}}_2\text{O}$$
   
   Since it is a $1:3$ electrolyte, it gives a total of $4$ ions in solution.
   
   So, the complex must be:
   
   $$[\text{Cr}({\text{H}}_2\text{O}{)}_6]{\text{Cl}}_3$$
   
   because this dissociates as
   
   $$[\text{Cr}({\text{H}}_2\text{O}{)}_6{]}^{3+}+3{\text{Cl}}^{-}$$
   
   Now calculate moles of the complex.
   
   Molar mass of ${\text{CrCl}}_3\cdot 6{\text{H}}_2\text{O}$ :
   
   $$=52+3(35.5)+6(18)$$
   
   $$=52+106.5+108=266.5{\text{ g mol}}^{-1}$$
   
   Moles of complex taken:
   
   $$\frac{5.33}{266.5}=0.02\text{ mol}$$
   
   Now, mass of AgCl formed $=8.61\text{ g}$
   
   Molar mass of AgCl:
   
   $$108+35.5=143.5{\text{ g mol}}^{-1}$$
   
   Moles of AgCl formed:
   
   $$\frac{8.61}{143.5}=0.06\text{ mol}$$
   
   Required ratio:
   
   $$\frac{\text{moles of complex reacted}}{\text{moles of AgCl formed}}=\frac{0.02}{0.06}=\frac{1}{3}=0.3333$$
   
   Now express this as
   
   $$x\times 10^{-2}$$
   
   So,
   
   $$0.3333=33.33\times 10^{-2}$$
   
   Nearest integer $=33$ .
   
   Final answer:
   
   $$\boxed{33}$$
3. JEE Main 2026MediumCrystal Field Theory (CFT): Octahedral & Tetrahedral Splitting
   Q3

   The number of isoelectronic species among $S{c}^{3+},C{r}^{2+},M{n}^{3+},C{o}^{3+}$ and $F{e}^{3+}$ is ‘n’. If ‘n’ moles of AgCl is formed during the reaction of complex with formula $CoC{l}_3(en{)}_{2}N{H}_3$ with excess of $AgN{O}_3$ solution, then the number of electrons present in the $t_{2g}$ orbital of the complex is ________.

   Numerical answer type

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   Correct answer: 6

   $$\begin{array}{cc}{Sc}^{+3}& 18\\ {Cr}^{+2}& 22\\ {Mn}^{+3}& 22\\ {Co}^{+3}& 24\\ {Fe}^{+3}& 23\\ & \end{array}$$
   
   In the table, the number written for each ion is its total number of electrons.
   
   ${Cr}^{2+}$ and ${Mn}^{3+}$ both have $22$ electrons, so they are isoelectronic (same number of electrons).
   
   So, the number of isoelectronic species is
   
   $$n=2$$
   
   Now, for the complex $CoC{l}_3(en{)}_{2}N{H}_3$:
   
   $en$ and $N{H}_3$ are neutral ligands, so they do not change the charge.
   
   In the given formula, total $Cl$ are $3$. If $n=2$ moles of $AgCl$ are formed with excess $AgN{O}_3$, it means $2$ chloride ions are outside the coordination sphere (they are ionisable and precipitate as $AgCl$).
   
   So the complex must be:
   
   Complex is : $\left[Co(en{)}_2{NH}_{3}Cl\right]{Cl}_2$
   
   Here, $2$ Cl${}^{-}$ are outside, so the complex ion has charge $+2$.
   
   Let oxidation state of Co be $x$.
   
   Inside the bracket: $en$ and $N{H}_3$ contribute $0$, and one coordinated Cl${}^{-}$ contributes $-1$.
   
   So, $x-1=+2\Rightarrow x=+3$.
   
   Thus cobalt is $C{o}^{3+}$, and its $d$-electron configuration is:
   
   $$\Rightarrow {Co}^{3+}3d^6{t}_{2g}^{2,2,2}{e}_g^{0,0}$$
   
   So, total electrons in $t_{2g}$ orbitals $=2+2+2=6$.

Topic-wise question count

Where JEE Main has asked from Coordination Compounds, most frequent first.

• Valence Bond Theory (VBT) & Magnetic Properties94
• Crystal Field Theory (CFT): Octahedral & Tetrahedral Splitting53
• Coordination Entities, Ligands, Coordination Number & IUPAC Nomenclature42
• Structural and Stereoisomerism in Coordination Compounds35
• Colour, Magnetic Moments & Stability of Complexes29
• Werner's Theory of Coordination Compounds12
• Organometallic Compounds & Metal Carbonyls9

Year-wise Coordination Compounds questions in JEE Main

Number of questions from this chapter in each year's paper(s).