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Chemistry Question 60 – JEE-MAIN 2025

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The correct order of [FeF6]3−, [CoF6]3−, [Ni(CO)4] and [Ni(CN)4]2− complex species based on the number of unpaired electrons present is :

First, find the oxidation state of the central metal atom in each complex and its corresponding d-electron configuration.

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Ninja Strategy•Identify Diamagnetic Complexes First

Recognize that [Ni(CO)4] and [Ni(CN)4]2− are both diamagnetic (0 unpaired electrons), meaning they must be equal in the final order, which immediately points to option 3.

Step 1: Determine Oxidation State and d-electron Configuration✦ Active

Calculate the oxidation state of the central metal ion and its d-electron configuration for each complex:

[FeF6]3−:Fe3+(d5) [CoF6]3−:Co3+(d6) [Ni(CO)4]:Ni(0)(d10 from 3d84s2) [Ni(CN)4]2−:Ni2+(d8)
💡 Teacher's Secret Hint

Remember that CO is a neutral ligand and CN- is a -1 ligand.

Step 2: Determine Number of Unpaired Electrons○ Expand

Apply Crystal Field Theory considering ligand strength and geometry:

[FeF6]3−:F− is a weak field ligand, octahedral. d5 high spin⇒n=5 [CoF6]3−:F− is a weak field ligand, octahedral. d6 high spin⇒n=4 [Ni(CO)4]:CO is a strong field ligand, tetrahedral. Ni(0) is d10⇒n=0 [Ni(CN)4]2−:CN− is a strong field ligand, square planar. d8 low spin⇒n=0
💡 Teacher's Secret Hint

Recall the spectrochemical series to identify strong and weak field ligands. For d8 with strong field ligands, square planar geometry is common and leads to diamagnetism.

Step 3: Order the Complexes○ Expand

Arrange the complexes in decreasing order of unpaired electrons:

n([FeF6]3−)=5 n([CoF6]3−)=4 n([Ni(CN)4]2−)=0 n([Ni(CO)4])=0 Order: [FeF6]3−(5)>[CoF6]3−(4)>[Ni(CN)4]2−(0)=[Ni(CO)4](0)
💡 Teacher's Secret Hint

Carefully compare the number of unpaired electrons for each complex to establish the correct order.

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