Given below are two statements: Statement I: The number of paramagnetic species among [CoF6]3−, [TiF6]3−, V2O5 and [Fe(CN)6]3− is 3. Statement II: K4[Fe(CN)6] < K3[Fe(CN)6] < [Fe(H2O)6]Cl3 is the correct order in terms of number of unpaired electron(s) present in the complexes. In the light of the above statements, choose the correct answer from the options given below.

Given below are two statements about paramagnetism and unpaired electrons in coordination compounds

Q. Given below are two statements:

Statement I: The number of paramagnetic species among [CoF₆]³⁻, [TiF₆]³⁻, V₂O₅ and [Fe(CN)₆]³⁻ is 3.

Statement II: K₄[Fe(CN)₆] < K₃[Fe(CN)₆] < [Fe(H₂O)₆]Cl₃ is the correct order in terms of number of unpaired electron(s).

(A) Statement I is true but Statement II is false

(B) Both Statement I and Statement II are false

(D) Both Statement I and Statement II are true

Correct Answer: (D)

Explanation

Statement I:

[CoF₆]³⁻: Co³⁺ → d⁶, F⁻ is weak field → high spin → paramagnetic

[TiF₆]³⁻: Ti³⁺ → d¹ → paramagnetic

V₂O₅: Vanadium in +5 oxidation state → d⁰ → diamagnetic

[Fe(CN)₆]³⁻: Fe³⁺ → d⁵, CN⁻ strong field → low spin → 1 unpaired electron → paramagnetic

Hence, total paramagnetic species = 3 → Statement I is true.

Statement II:

K₄[Fe(CN)₆]: Fe²⁺ (d⁶), CN⁻ strong field → low spin → 0 unpaired

K₃[Fe(CN)₆]: Fe³⁺ (d⁵), CN⁻ strong field → low spin → 1 unpaired

[Fe(H₂O)₆]Cl₃: Fe³⁺ (d⁵), H₂O weak field → high spin → 5 unpaired

Therefore: K₄[Fe(CN)₆] < K₃[Fe(CN)₆] < [Fe(H₂O)₆]Cl₃

Statement II is also true.

Explanation

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