Coordination Compounds Previous Year Questions — JEE Main Chemistry

68 past-year Coordination Compounds questions from JEE Main (Chemistry) — page 8.

Q71 (2024)

Which of the following complex is homoleptic?
  1. $[Ni(CN)_4]^{2-}$
  2. $[Ni(NH_3)_2Cl_2]$
  3. $[Fe(NH_3)_4Cl_2]^+$
  4. $[Co(NH_3)_4Cl_2]^+$
### Core Logic A homoleptic complex is one in which the central metal atom/ion is bound to only one kind of donor group (ligand). A heteroleptic complex is one in which the central metal atom/ion is bound to more than one kind of donor group. ### Step 1: Analyze Options (1) $[Ni(CN)_4]^{2-}$: Only one type of ligand ($CN^-$). Homoleptic. (2) $[Ni(NH_3)_2Cl_2]$: Two types of ligands ($NH_3$ and $Cl^-$). Heteroleptic. (3) $[Fe(NH_3)_4Cl_2]^+$: Two types of ligands. Heteroleptic. (4) $[Co(NH_3)_4Cl_2]^+$: Two types of ligands. Heteroleptic. ### Pattern Recognition Homo = same, leptic = ligands. Look for the formula bracket containing only one symbol type after the central metal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q1476 (2024)

The correct IUPAC name of $\mathrm{K}_2\mathrm{MnO}_4$ is
  1. Potassium tetraoxopermanganate (VI)
  2. Potassium tetraoxidomanganate (VI)
  3. Dipotassium tetraoxidomanganate (VII)
  4. Potassium tetraoxidomanganese (VI)
### Related Formula $$\text{Oxidation State of Mn Evaluation: } 2(+1) + x + 4(-2) = 0$$ ### Core Logic Solving for $x$: $$2 + x - 8 = 0 \implies x = +6$$ Since the complex is anionic, the metal name ends with the suffix '-ate', making it 'manganate(VI)'. The ligands are oxygen atoms, designated systematically as 'tetraoxido' or 'tetraoxo' according to newer recommendations. ### Step 1: Assembly Combining parts systematically yields the correct IUPAC string: Potassium tetraoxidomanganate(VI). ### Pattern Recognition Anionic metal centers must include the trailing '-ate' modifier followed immediately by their absolute Roman oxidation state indicators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q1492 (2024)

The oxidation number of iron in the compound formed during brown ring test for $\mathrm{NO}_3^-$ ion is ________.
### Related Formula $$\text{Formula of Brown Ring Complex: } [\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_5(\mathrm{NO})]^{2+}$$ ### Core Logic In this specific coordination complex, charge transfer occurs where nitric oxide transfers an electron to the iron center. As a result, $\text{NO}$ exists as a positive $\text{NO}^+$ ligand, and iron drops to an unusual oxidation state: $$x + 5(0) + 1(+1) = +2$$ $$x + 1 = 2 \implies x = +1$$ ### Step 1: Final Value Assignment Solving this charge balance confirms that the oxidation number of iron in the brown ring complex is $+1$. ### Pattern Recognition The brown ring test features a rare $+1$ oxidation state for iron because $\text{NO}$ coordinates as the positive nitrosonium cation ($\text{NO}^+$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q68 (2024)

Consider the following complex ions: $P=[FeF_6]^{3-}$ $Q=[V(H_2O)_6]^{2+}$ $R=[Fe(H_2O)_6]^{2+}$ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
  1. $R < Q < P$
  2. $R < P < Q$
  3. $Q < R < P$
  4. $Q < P < R$
### Step 1: Evaluation of $P=[FeF_6]^{3-}$ $$\text{Fe}^{3+} \rightarrow 3\text{d}^5$$ Since $\text{F}^-$ is a weak field ligand, no pairing occurs. Number of unpaired electrons ($n$) = 5. $$\mu = \sqrt{5(5+2)} = \sqrt{35}\text{ BM}$$ ### Step 2: Evaluation of $Q=[V(H_2O)_6]^{2+}$ $$\text{V}^{2+} \rightarrow 3\text{d}^3$$ Number of unpaired electrons ($n$) = 3. $$\mu = \sqrt{3(3+2)} = \sqrt{15}\text{ BM}$$ ### Step 3: Evaluation of $R=[Fe(H_2O)_6]^{2+}$ $$\text{Fe}^{2+} \rightarrow 3\text{d}^6$$ Since $\text{H}_2\text{O}$ is a weak field ligand, configuration is $\text{t}_{2\text{g}}^4 \text{e}_{\text{g}}^2$. Number of unpaired electrons ($n$) = 4. $$\mu = \sqrt{4(4+2)} = \sqrt{24}\text{ BM}$$ ### Step 4: Comparison Comparing values: $$\mu(Q) < \mu(R) < \mu(P) \implies Q < R < P$$ ### Pattern Recognition Count unpaired electrons strictly accounting for weak field vs strong field rules. Order scales monotonically with $n$. ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q1665 (2024)

In which one of the following metal carbonyls, CO forms a bridge between metal atoms?
  1. $\left[\mathrm{Co}_{2}(\mathrm{CO})_{8}\right]$
  2. $[\mathrm{Mn}_2(\mathrm{CO})_{10}]$
  3. $[\mathrm{Os}_3(\mathrm{CO})_{12}]$
  4. $[\mathrm{Ru}_3(\mathrm{CO})_{12}]$
### Core Logic Let's examine the structure of each polynuclear metal carbonyl: 1. **$[Co_2(CO)_8]$**: Solid octacarbonyldicobalt(0) exists in a bridged structure. It has a $Co-Co$ bond and two bridging $CO$ ligands, along with three terminal $CO$ ligands on each cobalt atom. So, it contains bridging CO groups. 2. **$[Mn_2(CO)_{10}]$**: Decacarbonyldimanganese(0) has a single $Mn-Mn$ bond and all 10 CO ligands are terminal. It does NOT have any bridging carbonyls. 3. **$[Os_3(CO)_{12}]$**: Dodecacarbonyltriosmium(0) forms a triangular cluster of Os atoms with all 12 CO ligands being terminal. 4. **$[Ru_3(CO)_{12}]$**: Dodecacarbonyltriruthenium(0) similarly forms a triangular $Ru_3$ cluster where all 12 CO ligands are terminal. ### Step 1: Conclusion Only $[Co_2(CO)_8]$ features $CO$ forming a bridge between the metal atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
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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