Coordination Chemistry PYQ: Unpaired Electrons in Fe–Co–Mn Complexes! ⚗️

 

❓ Question

The number of unpaired electrons responsible for the paramagnetic nature of the following complex ions are, respectively:

1. $[{\text{Fe(CN)}}_6{]}^{3-}$

2. $[ \text{FeF}_6 ]^{3-}$
   

3. $[ \text{CoF}_6 ]^{3-}$
   

4. $[ \text{Mn(CN)}_6 ]^{3-}$
   

---

🖼️ Question Image

---

✍️ Short Solution

We analyze each complex using:

• Oxidation state

• Electronic configuration

• Nature of ligand (strong field or weak field)

• High-spin vs low-spin

• Count unpaired electrons

---

✅ (1) $[ \text{Fe(CN)}_6 ]^{3-}$

Step-1: CN⁻ is a strong-field ligand → large Δ₀ → low-spin.

Oxidation state:

$$\text{Fe} + 6(-1) = -3 \Rightarrow \text{Fe}^{3+}$$
$$\text{Fe}^{3+} : 3d^5$$

In low-spin d⁵, electrons pair as much as possible:
Configuration: $t_{2g}^5 e_g^0$

Unpaired electrons = 1

---

✅ (2) $[ \text{FeF}_6 ]^{3-}$

F⁻ is a weak-field ligand → small Δ₀ → high-spin.

Oxidation state of Fe is same: Fe³⁺ → $3d^5$.

In high-spin d⁵:
Configuration: $t_{2g}^3 e_g^2$

Unpaired electrons = 5

---

✅ (3) $[ \text{CoF}_6 ]^{3-}$

Ligand: F⁻ → weak-field → high-spin.

Oxidation state:

$$\text{Co} + 6(-1) = -3 \Rightarrow \text{Co}^{3+}$$
$$\text{Co}^{3+} : 3d^6$$

High-spin d⁶ distribution:
$t_{2g}^4 e_g^2$

Unpaired electrons = 4

---

✅ (4) $[ \text{Mn(CN)}_6 ]^{3-}$

CN⁻ → strong-field, low-spin.

Oxidation state:

$$\text{Mn} + 6(-1) = -3 \Rightarrow \text{Mn}^{3+}$$
$$\text{Mn}^{3+} : 3d^4$$

Low-spin d⁴ → electrons first fill t₂g:
$t_{2g}^4 e_g^0$

Unpaired electrons = 2

---

🧮 Image Solution

---

✅ Final Answer

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

---

📘 Concept Recap

• CN⁻ = strong field, causes pairing → low-spin.

• F⁻ = weak field, no pairing → high-spin.

• Paramagnetism depends on number of unpaired electrons.

• More unpaired electrons → stronger paramagnetism.