Coordination Compounds Hybridization and Magnetism

 

❓ Question

The number of paramagnetic complexes among:

$$[FeF_6]^{3-}, \quad [Fe(CN)_6]^{3-}, \quad [Mn(CN)_6]^{3-}, \quad [Co(C_2O_4)_3]^{3-}, \quad [MnCl_6]^{3-}, \quad [CoF_6]^{3-}$$

which involve:

$$d^2sp^3$$

hybridization is ________.

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🖼 Question Image

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✍️ Short Concept

This question combines:

👉 Oxidation state
👉 Strong vs weak ligands
👉 Hybridization
👉 Paramagnetism.

Main trick:

Only inner orbital octahedral complexes show:

$$d^2sp^3$$

hybridization.

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🔷 Step 1 — Identify Strong & Weak Ligands 💯

Strong field ligands:

$$CN^- ,\ C_2O_4^{2-}$$

Weak field ligands:

$$F^- ,\ Cl^-$$

Strong ligands → pairing possible → inner orbital complex

Weak ligands → no pairing → outer orbital complex.

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🔷 Step 2 — Check Each Complex

1️⃣ $[FeF_6]^{3-}$

Fe³⁺:

$$3d^5$$

$F^-$ is weak ligand.

High spin:

$$sp^3d^2$$

❌ Not $d^2sp^3$

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2️⃣ $[Fe(CN)_6]^{3-}$

Fe³⁺:

$$3d^5$$

$CN^-$strong ligand → pairing occurs.

Hybridization:

$$d^2sp^3$$

Unpaired electrons remain:

$$1$$

✅ Paramagnetic
✅ Counts

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3️⃣ $[Mn(CN)_6]^{3-}$

Mn³⁺:

$$3d^4$$

Strong ligand → pairing occurs.

Hybridization:

$$d^2sp^3$$

Still unpaired electrons present.

✅ Paramagnetic
✅ Counts

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4️⃣ $[Co(C_2O_4)_3]^{3-}$

Co³⁺:

$$3d^6$$

Oxalate causes pairing.

Hybridization:

$$d^2sp^3$$

All electrons paired.

❌ Diamagnetic

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5️⃣ $[MnCl_6]^{3-}$

$Cl^-$ weak ligand.

Outer orbital:

$$sp^3d^2$$

❌ Not counted

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6️⃣ $[CoF_6]^{3-}$

$F^-$ weak ligand.

Outer orbital:

$$sp^3d^2$$

❌ Not counted

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🔷 Step 3 — Count Correct Complexes

Paramagnetic + $d^2sp^3$:

✅ $[Fe(CN{)}_6{]}^{3-}$

✅ $[Mn(CN{)}_6{]}^{3-}$

Total:

$$\boxed{2}$$

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✅ Final Answer

$$\boxed{2}$$

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⭐ Golden JEE Insight

For octahedral complexes:

Strong ligand:

$$d^2sp^3$$

Weak ligand:

$$sp^3d^2$$

Always check BOTH:

👉 Hybridization
👉 Paramagnetic nature

before counting.

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🔗 Realted Concepts

• Van't Hoff factor  
• First Order Growth
• Paramagnetic Complexes
• Hybridization in Coordination Compounds