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MCQMediumJEE 2025Valence Bond Theory

JEE Chemistry 2025 Question with Solution

Match List - I with List - II: List - I: (A) [MnBr4]2−\text{[MnBr}_4]^{2-} (B) [FeF6]3−\text{[FeF}_6]^{3-} (C) [Co(C2O4)3]3−\text{[Co(C}_2\text{O}_4)_3]^{3-} (D) [Ni(CO)4]\text{[Ni(CO)}_4] List - II: (I) d2sp3d^2sp^3 & diamagnetic (II) sp2d2sp^2d^2 & paramagnetic (III) sp3sp^3 & diamagnetic (IV) sp3sp^3 & paramagnetic Choose the correct answer from the options given below:

  • A

    (A)-(III),(B)-(II),(C)-(I),(D)-(IV)\text{(A)-(III)}, \text{(B)-(II)}, \text{(C)-(I)}, \text{(D)-(IV)}

  • B

    (A)-(IV),(B)-(I),(C)-(II),(D)-(III)\text{(A)-(IV)}, \text{(B)-(I)}, \text{(C)-(II)}, \text{(D)-(III)}

  • C

    (A)-(I),(B)-(II),(C)-(III),(D)-(IV)\text{(A)-(I)}, \text{(B)-(II)}, \text{(C)-(III)}, \text{(D)-(IV)}

  • D

    (A)-(IV),(B)-(I),(C)-(III),(D)-(II)\text{(A)-(IV)}, \text{(B)-(I)}, \text{(C)-(III)}, \text{(D)-(II)}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The complexes are [MnBr4]2−\text{[MnBr}_4]^{2-}, [FeF6]3−\text{[FeF}_6]^{3-}, [Co(C2O4)3]3−\text{[Co(C}_2\text{O}_4)_3]^{3-} and [Ni(CO)4]\text{[Ni(CO)}_4].

Find: Match each complex with its hybridization and magnetic property.

To solve the problem of matching List-I with List-II, we need to determine the hybridization of each complex and their magnetic properties:

  1. For [MnBr4]2−[ \text{MnBr}_4]^{2-}: Manganese in [MnBr4]2−[ \text{MnBr}_4]^{2-} has a coordination number of 44 and contains no unpaired electrons; thus, it is diamagnetic. The hybridization of this complex is sp3sp^3.

  2. For [FeF6]3−[ \text{FeF}_6]^{3-}: The complex [FeF6]3−[ \text{FeF}_6]^{3-} involves Fe3+\text{Fe}^{3+}, which has a coordination number of 66. Fluoride is a weak field ligand, resulting in high spin, thus with unpaired electrons and is paramagnetic. The hybridization is sp2d2sp^2d^2.

  3. For [Co(C2O4)3]3−[ \text{Co(C}_2\text{O}_4)_3]^{3-}: In this case, Co3+\text{Co}^{3+} forms low-spin complexes with oxalate as a ligand. It has no unpaired electrons hence is diamagnetic and features d2sp3d^2sp^3 hybridization.

  4. For [Ni(CO)4][ \text{Ni(CO)}_4]: This complex contains Ni0\text{Ni}^{0} which forms a tetrahedral complex with carbonyl as a ligand, resulting in unpaired electrons. Therefore, it is paramagnetic with sp3sp^3 hybridization.

Thus, the correct matches are:

  • (A) - (III) = sp3sp^3 diamagnetic
  • (B) - (II) = sp2d2sp^2d^2 paramagnetic
  • (C) - (I) = d2sp3d^2sp^3 diamagnetic
  • (D) - (IV) = sp3sp^3 paramagnetic

Therefore, the correct option is A.

Match Each Complex Carefully

Given: A matching problem on hybridization and magnetic behavior of coordination compounds.

Find: The correct correspondence between List - I and List - II.

Use oxidation state, ligand strength, coordination number, and geometry for each complex.

[MnBr4]2−→coordination number 4→sp3→diamagnetic[FeF6]3−→coordination number 6, weak field ligand→sp2d2→paramagnetic[Co(C2O4)3]3−→low spin→d2sp3→diamagnetic[Ni(CO)4]→tetrahedral→sp3→paramagnetic\begin{aligned} \text{[MnBr}_4]^{2-} &\rightarrow \text{coordination number } 4 \rightarrow sp^3 \rightarrow \text{diamagnetic} \\ \text{[FeF}_6]^{3-} &\rightarrow \text{coordination number } 6, \text{ weak field ligand} \rightarrow sp^2d^2 \rightarrow \text{paramagnetic} \\ \text{[Co(C}_2\text{O}_4)_3]^{3-} &\rightarrow \text{low spin} \rightarrow d^2sp^3 \rightarrow \text{diamagnetic} \\ \text{[Ni(CO)}_4] &\rightarrow \text{tetrahedral} \rightarrow sp^3 \rightarrow \text{paramagnetic} \end{aligned}

So the final matching is:

(A)→(III)(B)→(II)(C)→(I)(D)→(IV)\begin{aligned} (A) &\rightarrow (III) \\ (B) &\rightarrow (II) \\ (C) &\rightarrow (I) \\ (D) &\rightarrow (IV) \end{aligned}

Hence, the correct option is A.

Common mistakes

  • Assuming every tetrahedral complex is paramagnetic. That is incorrect because magnetic behavior depends on the electron configuration and pairing, not only on geometry. First determine the metal oxidation state and electron count, then decide whether unpaired electrons are present.

  • Ignoring ligand field strength for octahedral complexes. This leads to wrong hybridization and spin-state assignment. For species like [FeF6]3−\text{[FeF}_6]^{3-}, identify fluoride as a weak field ligand before deciding high-spin or low-spin behavior.

  • Matching hybridization directly from coordination number without checking inner-orbital versus outer-orbital character. A coordination number of 66 does not by itself fix the hybridization. You must examine the metal ion and ligand nature to distinguish d2sp3d^2sp^3 from sp2d2sp^2d^2.

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