MCQMediumJEE 2025Ligands & Coordination Number

JEE Chemistry 2025 Question with Solution

Identify the homoleptic complexes with odd number of dd-electrons in the central metal:

(A) [FeO4]2[FeO_4]^{2-}

(B) [Fe(CN)6]3[Fe(CN)_6]^{3-}

(C) [Fe(CN)6]2[Fe(CN)_6]^{2-}

(D) [CoCl4]2[CoCl_4]^{2-}

(E) [Co(H2O)6]3+[Co(H_2O)_6]^{3+}

Choose the correct answer from the options given below:

  • A

    (B) and (D) only

  • B

    (A), (B) and (D) only

  • C

    (A), (B) and (E) only

  • D

    (A), (C), (D) and (E) only

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: We must identify the homoleptic complexes in which the central metal has an odd number of dd-electrons.

Find: Which listed species satisfy both conditions, and then choose the matching option.

Check each complex using oxidation state and dd-electron count:

  • (A) [FeO4]2[FeO_4]^{2-}: The oxidation state of iron is +6+6, so iron has d0d^0 configuration. This is an even number of dd-electrons, so it does not qualify.
  • (B) [Fe(CN)6]3[Fe(CN)_6]^{3-}: The oxidation state of iron is +3+3, so iron has d5d^5 configuration. Since 55 is odd, this complex qualifies.
  • (C) [Fe(CN)6]2[Fe(CN)_6]^{2-}: The oxidation state of iron is +2+2, so iron has d6d^6 configuration. Since 66 is even, this does not qualify.
  • (D) [CoCl4]2[CoCl_4]^{2-}: The oxidation state of cobalt is +2+2, so cobalt has d7d^7 configuration. Since 77 is odd, this complex qualifies.
  • (E) [Co(H2O)6]3+[Co(H_2O)_6]^{3+}: The oxidation state of cobalt is +3+3, so cobalt has d6d^6 configuration. Since 66 is even, this does not qualify.

Thus, the homoleptic complexes with odd number of dd-electrons are (B) and (D) only.

Therefore, the correct option is A.

There is a discrepancy in the provided the solution: one section states "The Correct Option is A" while another writes the final answer as (1) (B) and (D) only. These are consistent because option A corresponds to choice (1).

Oxidation State Check

Given: Determine the oxidation state of the metal in each complex and then count the remaining dd-electrons.

Find: Which complexes are homoleptic and contain an odd number of dd-electrons.

For (A) [FeO4]2[FeO_4]^{2-}:

x+4(2)=2x + 4(-2) = -2 x=+6x = +6

Iron is Fe6+Fe^{6+}, so it has d0d^0 configuration. Not odd.

For (B) [Fe(CN)6]3[Fe(CN)_6]^{3-}:

x+6(1)=3x + 6(-1) = -3 x=+3x = +3

Iron is Fe3+Fe^{3+}, so it has d5d^5 configuration. Odd.

For (C) [Fe(CN)6]2[Fe(CN)_6]^{2-}:

x+6(1)=2x + 6(-1) = -2 x=+4x = +4

the solution states this species has even dd-electron count and excludes it from the answer. Hence it is not selected.

For (D) [CoCl4]2[CoCl_4]^{2-}:

x+4(1)=2x + 4(-1) = -2 x=+2x = +2

Cobalt is Co2+Co^{2+}, so it has d7d^7 configuration. Odd.

For (E) [Co(H2O)6]3+[Co(H_2O)_6]^{3+}:

x+6(0)=+3x + 6(0) = +3 x=+3x = +3

Cobalt is Co3+Co^{3+}, so it has d6d^6 configuration. Even.

Therefore, only (B) and (D) satisfy the condition, so the correct option is A.

Common mistakes

  • A common mistake is to subtract the ligand charge incorrectly while finding oxidation state. This gives the wrong dd-electron count. Always write the charge-balance equation first and then solve for the metal oxidation state.

  • Students often count total electrons of the neutral metal atom instead of the dd-electrons in the oxidation state present in the complex. The question asks for the number of dd-electrons on the central metal after ion formation, not the atomic electron count.

  • Another mistake is to ignore the word homoleptic. A complex must have identical ligands around the metal to be homoleptic. First check ligand identity, then check whether the dd-electron count is odd.

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