Coordination Compounds Previous Year Questions — JEE Main Chemistry

68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 14.

Q73 (2026)

X is the number of geometrical isomers exhibited by $[Pt(NH_{3})(H_{2}O)BrCl]$. Y is the number of optically inactive isomer(s) exhibited by $[CrCl_{2}(ox)_{2}]^{3-}$ Z is the number of geometrical isomers exhibited by $[Co(NH_{3})_{3}(NO_{2})_{3}]$ The value of $X + Y + Z$ is _____.
### Step 1: Evaluate X $[Pt(NH_3)(H_2O)BrCl]$ is a square planar complex of the type $[Mabcd]$. Such complexes exhibit exactly 3 geometrical isomers (by fixing one ligand and placing the other three opposite to it, creating two cis-like and one trans-like variations relative to the fixed ligand).\nThus, $X = 3$. ### Step 2: Evaluate Y $[CrCl_2(ox)_2]^{3-}$ is an octahedral complex of the type $[M(AA)_2b_2]$. It has two geometrical isomers: cis and trans.\n$\bullet$ The cis-isomer lacks a plane of symmetry and is optically active (exists as a pair of enantiomers).\n$\bullet$ The trans-isomer has a plane of symmetry and is optically inactive (meso).\nThe question asks for the number of optically inactive isomers. Thus, $Y = 1$. ### Step 3: Evaluate Z $[Co(NH_3)_3(NO_2)_3]$ is an octahedral complex of the type $[Ma_3b_3]$. It exhibits exactly 2 geometrical isomers: Facial (fac) and Meridional (mer).\nThus, $Z = 2$. ### Final Conclusion $X + Y + Z = 3 + 1 + 2 = 6$ ### Pattern Recognition $[Mabcd]$ planar = 3 G.I. $[Ma_3b_3]$ octahedral = 2 G.I. (fac/mer). $[M(AA)_2b_2]$ octahedral trans isomer = ALWAYS optically inactive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q69 (2026)

The correct increasing order of spin-only magnetic moment values of the complex ions $[MnBr_{4}]^{2-}$ (A), $[Cu(H_{2}O)_{6}]^{2+}$ (B), $[Ni(CN_{4})]^{2-}$ (C) and $[Ni(H_{2}O)_{6}]^{2+}$ (D) is:
  1. $(1)\text{ } A = B < C < D$
  2. $(2)\text{ } A = B < D < C$
  3. $(3)\text{ } C = D < B < A$
  4. $(4)\text{ } C < B < D < A$
### Related Formula $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$ where $n$ is the number of unpaired electrons. ### Core Logic (A) $[MnBr_{4}]^{2-}$: $Mn^{2+}$ has $3d^5$ configuration. $Br^-$ is a weak field ligand (WFL). It forms a tetrahedral complex with 5 unpaired electrons ($n=5$). (B) $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ has $3d^9$ configuration. It has exactly 1 unpaired electron ($n=1$). (C) $[Ni(CN_{4})]^{2-}$: $Ni^{2+}$ has $3d^8$ configuration. $CN^-$ is a strong field ligand (SFL). It forms a square planar complex ($dsp^2$) forcing pairing. Unpaired electrons $n=0$. (D) $[Ni(H_{2}O)_{6}]^{2+}$: $Ni^{2+}$ has $3d^8$ configuration. $H_2O$ is a weak field ligand (WFL). In an octahedral field, it leaves 2 unpaired electrons ($n=2$). ### Step 1: Final Conclusion Unpaired electrons (n): (C) = 0 (B) = 1 (D) = 2 (A) = 5 Increasing order of magnetic moment: C < B < D < A. ### Pattern Recognition Nickel + Strong Ligand ($CN^-$) = Square Planar, Diamagnetic ($n=0$). Nickel + Weak Ligand ($H_2O$) = Octahedral, Paramagnetic ($n=2$). Copper (II) is always $d^9$ ($n=1$). Manganese (II) with WFL is high spin $d^5$ ($n=5$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q75 (2026)

The number of isoelectronic species among $Sc^{3+}$, $Cr^{2+}$, $Mn^{3+}$, $Co^{3+}$ and $Fe^{3+}$ is 'n'. If 'n' moles of AgCl is formed during the reaction of complex with formula $CoCl_{3}(en)_{2}NH_{3}$ with excess of $AgNO_{3}$ solution, then the number of electrons present in the $t_{2g}$ orbital of the complex is ____.
### Core Logic Step 1: Find 'n' (number of isoelectronic species) Electrons count: $Sc^{3+} = 21 - 3 = 18$ $Cr^{2+} = 24 - 2 = 22$ $Mn^{3+} = 25 - 3 = 22$ $Co^{3+} = 27 - 3 = 24$ $Fe^{3+} = 26 - 3 = 23$ The isoelectronic species are $Cr^{2+}$ and $Mn^{3+}$ (both have 22 electrons). So, $n = 2$. Step 2: Complex formulation. Reaction produces 'n' moles of AgCl, so 2 moles of $Cl^-$ are ionizable outside the coordination sphere. Complex formula is $[Co(en)_2(NH_3)Cl]Cl_2$. Step 3: Crystal field splitting. The central metal ion is $Co^{3+}$ ($3d^6$ configuration). The ligands are $en$ (strong field) and $NH_3$ (strong field), which will cause complete pairing of electrons. $Cl^-$ inside the sphere is weak, but the overwhelming presence of $en$ and $NH_3$ combined with $+3$ oxidation state makes it a strong field (low spin) complex. $Co^{3+}$ ($3d^6$) low spin splitting: $t_{2g}^{6} \; e_g^{0}$. ### Step 1: Final Conclusion The number of electrons present in the $t_{2g}$ orbital of the complex is 6. ### Pattern Recognition $Co^{3+}$ in predominantly N-donor environment forms low spin octahedral complexes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Structure of Atom
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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