Lowest Atomisation Enthalpy in 3d Metals

❓ Concept Question

In 3d transition metals, how do we identify the element with lowest enthalpy of atomisation and determine its valence electrons?

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🖼 Concept Image

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✍️ Short Concept

Atomisation enthalpy depends on:

👉 Metallic bond strength
👉 Number of unpaired d-electrons

More unpaired electrons ⇒ stronger bonding ⇒ higher enthalpy

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🔷 Step 1 — What is Enthalpy of Atomisation? 💯

Energy required to convert:

$$\text{Metal} \rightarrow \text{gaseous atoms}$$

👉 Higher value = stronger metallic bonding

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🔷 Step 2 — d-Block Golden Rule

$$\text{Bond strength} \propto \text{unpaired d-electrons}$$

👉 More unpaired electrons

⇒ More exchange energy

⇒ Stronger bonding

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🔷 Step 3 — Trend in 3d Series

From:

Cr → Fe → Co → Ni

👉 Initially unpaired electrons increase
👉 Then pairing starts

Pairing causes:

$$\text{Bond strength} \downarrow$$

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🔷 Step 4 — Lowest Atomisation Enthalpy

Occurs when:

👉 Maximum pairing
👉 Minimum unpaired electrons

👉 Weak metallic bonding

In given set → Ni

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🔷 Step 5 — Valence Electron Concept

For transition metals:

$$\text{Valence electrons} = ns + (n-1)d$$

For Ni:

$$3d^8 4s^2$$
$$\text{Total} = 10$$

⚠️ JEE trap: Don’t count only s-electrons.

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✅ Final Takeaway

• Weakest bonding → Ni
• Valence electrons:

$$\boxed{10}$$

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⭐ Golden JEE Insight

In d-block:

👉 More unpaired electrons = stronger bonding

👉 More pairing = weaker bonding