Highest Enthalpy of Atomization in Transition Series

 

❓ Concept Question

A transition metal:

• Has highest enthalpy of atomisation in 1st transition series
• Forms green aqueous ion $M^{2+}$
  

Identify:

👉 Metal $M$
👉 Oxide formed from $M^{2+}$
👉 Nature of the oxide

---

🖼 Concept Image

---

✍️ Short Concept

This question combines:

👉 Metallic bonding
👉 Colour of transition metal ions
👉 Nature of oxides

---

🔷 Step 1 — Highest Enthalpy of Atomisation 💯

In first transition series:

👉 Maximum enthalpy of atomisation:

$$\boxed{Cr}$$

Reason:

• Maximum unpaired electrons
• Strong metallic bonding

---

🔷 Step 2 — Green Aquated Ion

Given:

$$M^{2+}(aq)$$

is green.

Chromium ions:

$$Cr^{2+} \rightarrow \text{Blue/Green}$$
$$Cr^{3+} \rightarrow \text{Green}$$

So metal:

$$\boxed{M = Cr}$$

---

🔷 Step 3 — Oxide from $M^{2+}$

If:

$$M^{2+} = Cr^{2+}$$

Corresponding oxide:

$$\boxed{CrO}$$

---

🔷 Step 4 — Nature of $CrO$

Trend in transition metal oxides:

👉 Lower oxidation state → more basic

👉 Higher oxidation state → more acidic

Since chromium is in:

$$+2$$

state in CrO:

$$\boxed{CrO \text{ is basic}}$$

---

🔷 Step 5 — JEE Trap Alert 🚨

❌ Green ion dekh ke $Fe^{2+}$ assume kar lena

❌ $Cr_2O_3$ mark kar dena (amphoteric yaad karke)

Remember:

👉 Oxide asked is from:

$$Cr^{2+}$$

So:

$$CrO$$

---

✅ Final Takeaway

$$\boxed{M = Cr}$$
$$\boxed{\text{Oxide} = CrO}$$
$$\boxed{\text{Nature} = \text{Basic}}$$

---

⭐ Golden JEE Insight

Transition metal oxide trend:

$$\text{Lower oxidation state} \rightarrow \text{Basic}$$
$$\text{Higher oxidation state} \rightarrow \text{Acidic}$$