Unpaired Electrons in Coordination Complexes

 

❓ Question
The number of unpaired electrons responsible for the paramagnetic nature of the following complex ions are, respectively:
$[{\text{Fe(CN)}}_6{]}^{3-<br>}$$[ \text{FeF}_6 ]^{3-}$
$[ \text{CoF}_6 ]^{3-}$
$[ \text{Mn(CN)}_6 ]^{3-}$

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Crystal Field Theory (CFT)
👉 Strong field vs weak field ligands
👉 Unpaired electrons.

Main idea:

• $CN^-$ → strong field ligand → pairing occurs
• $F^-$ → weak field ligand → high spin complex

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🔷 Step 1 — $[Fe(CN)_6]^{3-}$ 💯

Oxidation state of Fe:

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

So:

$$Fe^{3+}: 3d^5$$

Ligand:

$$CN^-$$

is strong field.

Thus low-spin octahedral:

$$t_{2g}^5e_g^0$$

Unpaired electrons:

$$\boxed{1}$$

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🔷 Step 2 — $[FeF_6]^{3-}$

Again:

$$Fe^{3+}=3d^5$$

But:

$$F^-$$

is weak field ligand.

Thus high-spin complex:

$$t_{2g}^3e_g^2$$

Unpaired electrons:

$$\boxed{5}$$

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🔷 Step 3 — $[CoF_6]^{3-}$

Oxidation state:

$$Co^{3+}$$

Electronic configuration:

$$Co^{3+}:3d^6$$

$F^-$ is weak field → high spin.

Configuration:

$$t_{2g}^4e_g^2$$

Unpaired electrons:

$$\boxed{4}$$

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🔷 Step 4 — $[Mn(CN)_6]^{3-}$

Oxidation state:

$$Mn^{3+}$$

Electronic configuration:

$$Mn^{3+}:3d^4$$

$CN^-$ is strong field → low spin.

Configuration:

$$t_{2g}^4e_g^0$$

Unpaired electrons:

$$\boxed{2}$$

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🔷 Step 5 — Final Sequence

$$\boxed{ 1,\ 5,\ 4,\ 2 }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $CN^-$ko weak field maan lena

❌ Oxidation state calculation mistake

❌ High-spin and low-spin confuse kar dena

Remember:

Strong field ligand

$$\boxed{ CN^- \rightarrow \text{pairing} }$$

Weak field ligand

$$\boxed{ F^- \rightarrow \text{high spin} }$$

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✅ Final Answer

$$\boxed{ 1,\ 5,\ 4,\ 2 }$$

(Option 2)

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