Q.Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

The key idea is the increasing effective nuclear charge across the series, which binds the $3d$ electrons progressively more tightly — so removing a third electron becomes harder and harder, and the +2 state (loss of the two $4s$ electrons only) becomes relatively more stable.

Reasoning:

1. Across the first transition series (Sc to Zn), each added proton increases $Z_{\text{eff}}$ for the $3d$ electrons, while the $4s$ electrons are less shielded and feel this pull more strongly.
2. The $3d$ orbital energy drops relative to $4s$. For early elements (e.g., Sc, Ti), the $3d$ electrons are relatively high in energy and easily lost, so higher oxidation states (like +3, +4) are common.
3. As $Z_{\text{eff}}$ rises, the $3d$ electrons become more tightly bound. By the middle of the series (Mn, Fe, Co, Ni), it becomes energetically less favourable to remove a $3d$ electron, so the +2 state (which involves losing only the two $4s$ electrons) becomes more stable relative to higher states.

The question asks about a trend that many students find counterintuitive at first. You might think: if we're removing electrons, shouldn't it get harder to form +2 as we add more protons? Actually, the opposite happens — and the reason lies in the subtle dance between nuclear charge, shielding, and the energies of the 4s and 3d orbitals.

Let's walk through this step by step.

1. What does "stability of +2 state" mean here?

   In aqueous chemistry, a +2 oxidation state is considered "stable" if the ion doesn't easily get oxidised further (to +3) or reduced (to +0). For the first-row transition metals (Sc through Zn), the +2 state's stability is measured by how readily the metal loses its two 4s electrons. Two quantities matter: the sum of the first two ionisation energies ($IE_1 + IE_2$), which sets the cost of forming M²⁺, and the third ionisation energy ($IE_3$), which sets the cost of oxidising it further — the +2 state is "stable" when $IE_3$ is prohibitively high.

2. The electronic structure trap

   Many students memorise that the 4s orbital fills before 3d, and then assume 4s electrons are "outer" and easy to remove. But here's the critical insight: once you start removing electrons, the 4s and 3d energy levels shift. In the neutral atom, 4s is lower in energy than 3d (for K and Ca). But from Sc onward, the 3d orbital dips below 4s in energy. When you ionise, you always remove the highest-energy electrons first — which are the 4s electrons for the neutral atom, but after the first removal, the remaining electron configuration rearranges.

3. The real driver: increasing nuclear charge

   As you move from Sc ($Z = 21$) to Mn ($Z = 25$), the nuclear charge increases by 4 units. The 3d electrons are poor at shielding each other (they're in the same shell), so each added proton pulls the 4s electrons tighter. This means:

   • $IE_1$ (removing the first 4s electron) generally increases.
   • $IE_2$ (removing the second 4s electron) also increases, but more sharply.
   • $IE_3$ (removing a 3d electron to get M³⁺) increases even faster.

   The +2 state becomes more stable because the third ionisation energy rises so steeply that it becomes much harder to go to +3. In other words, the +2 state is "protected" by the rising cost of further oxidation.

4. Let's look at the numbers (all in kJ/mol):

Notice that $IE_1 + IE_2$ generally increases (Sc: 1866 → Mn: 2226), meaning it's actually harder to form M²⁺ as we go right. But the +2 state becomes more stable relative to +3 because $IE_3$ rises even faster (Sc: 2393 → Mn: 3260). The gap between $IE_3$ and $IE_2$ widens dramatically.

5. The special case of chromium and manganese

   Cr²⁺ is 3d⁴, and losing one electron gives Cr³⁺ (3d³) — a half-filled $t_{2g}$ level in an octahedral field, itself an extra-stable arrangement — which is why Cr²⁺ is readily oxidised to Cr³⁺. Mn²⁺, however, has the half-filled 3d⁵ configuration, giving it maximum exchange-energy stabilisation — that's why Mn²⁺ is so common in nature and why the +2 state's stability peaks at Mn.

6. The trend in the first half

   From Sc to Mn, the +2 state becomes progressively more stable because:

   • Nuclear charge increases, pulling all electrons tighter.
   • The 3d electrons are less effective at shielding each other, so the effective nuclear charge felt by the 4s electrons rises.
   • The third ionisation energy (to remove a 3d electron) increases faster than the second, making M³⁺ harder to achieve.
   • By Mn, the +2 state (3d⁵) is so stable that Mn³⁺ is a strong oxidising agent.

Method: Electronic Configuration & Stability of Half-Filled d-Subshell

Why the +2 State Becomes More Stable Across the First Half of 3d Series

The stability of the +2 oxidation state in the first-row transition elements (Sc to Mn) increases because the third ionisation enthalpy rises steeply across this stretch (removing a 3d electron gets harder and harder), while the +2 ion's configuration moves toward the extra-stable half-filled d⁵ reached at Mn²⁺.

Step-by-Step Explanation

Step 1: Write the electronic configurations of the atoms and their +2 ions

Step 2: Identify the key stability factor — half-filled d-subshell

• A half-filled d⁵ configuration has extra stability due to:
  • Symmetrical distribution of electrons
  • Maximum exchange energy (Hund's rule)
• Mn²⁺ achieves this exact configuration: 3d⁵
• Cr²⁺ (3d⁴) is one electron short of half-filled, so it is less stable than Mn²⁺

Step 3: Track the trend in ionization energies (Table 4.2)

• The sum of the first two ionization enthalpies (IE₁ + IE₂) actually rises from Sc to Mn (1866 → 1965 → 2064 → 2245 → 2226 kJ/mol) — forming M²⁺ does not get cheaper.
• What changes decisively is the third ionization enthalpy (IE₃): it climbs even faster (2393 → 2657 → 2833 → 2990 → 3260 kJ/mol), so removing a 3d electron to reach +3 becomes disproportionately expensive.
• Result: the +2 state becomes progressively more stable relative to the +3 state

Step 4: Observe the experimental evidence

• Sc²⁺ is rare (Sc prefers +3)
• Ti²⁺ is known but easily oxidised
• V²⁺ is moderately stable
• Cr²⁺ (3d⁴) is a strong reducing agent — it is readily oxidised to Cr³⁺ (3d³, a stable half-filled t₂g level in water)
• Mn²⁺ is the most stable — it resists further oxidation due to the half-filled d⁵ stability

Summary of the Trend

Across Sc → Mn:

• IE₃ rises much faster than IE₁ + IE₂ → oxidation beyond +2 becomes increasingly difficult
• the +2 ion's d-configuration approaches half-filled (d⁵) → maximum stability at Mn²⁺
• Mn²⁺ is the most stable +2 ion in the first half

Key Takeaway for Exams

The +2 state stability increases because the d-configuration of M²⁺ moves toward the highly stable half-filled d⁵, while the third ionisation enthalpy rises so steeply that further oxidation to +3 becomes progressively harder.

Here is a breakdown of the common mistakes students make regarding the stability of the +2 state in the first half of the 3d series, along with how to avoid each.

The Core Concept (The "Why")

The stability of the +2 state increases from Sc to Mn because of increasing Ionization Enthalpy and Exchange Energy.

• Ionization Enthalpy: The sum of the first two ionization enthalpies ($IE_1 + IE_2$) rises across Sc → Mn (with only a small dip from Cr to Mn), but the third ionization enthalpy rises even faster — so losing a third electron becomes disproportionately harder.
• Exchange Energy: In the $d^5$ configuration (Mn$^{2+}$), the exchange energy is maximum, making the +2 state exceptionally stable.

Common Mistake #1: Confusing "Stability" with "Oxidizing Power"

The Mistake: Students often think that if the +2 state is "more stable," it means it is harder to oxidize (i.e., it is a weaker reducing agent). They then incorrectly state that Mn$^{2+}$ is the least stable because it is easily oxidized to Mn$^{3+}$.

The Truth: In this context, "stability of the +2 state" means the tendency of the element to form and retain the +2 oxidation state relative to other oxidation states.

• Correct Logic: As we move from Sc to Mn, the +2 state becomes more stable. This means it becomes harder to remove a third electron.
• Example: Sc easily loses three electrons (Sc$^{3+}$ is most stable). Mn easily loses two electrons (Mn$^{2+}$ is most stable). Mn$^{2+}$ is more stable than Sc$^{2+}$.

How to Avoid: Always ask: "Stable relative to what?" Here, it is stable relative to the +3 state. A more stable +2 state means a higher third ionization enthalpy ($IE_3$).

Common Mistake #2: Ignoring the Role of Exchange Energy

The Mistake: Students only consider ionization enthalpy trends and forget the crucial stabilizing effect of exchange energy for half-filled and fully-filled subshells.

The Truth: The exchange energy is a quantum mechanical stabilization that occurs when electrons of the same spin exchange positions. It is proportional to the number of pairs of electrons with parallel spins.

• Why it matters: Mn$^{2+}$ has a $d^5$ configuration (five unpaired electrons). This gives it the maximum exchange energy among the first half of the series. This extra stabilization makes Mn$^{2+}$ exceptionally stable.
• The Trap: Without this, you might think Fe$^{2+}$ (which comes later) is more stable, but the question specifically asks about the first half (Sc to Mn).

How to Avoid: Memorize the exchange energy trend: $d^5$ > $d^4$ > $d^3$ > $d^2$ > $d^1$. Always mention this when explaining the peak stability at Mn$^{2+}$.

Common Mistake #3: Forgetting the "First Half" Limitation

The Mistake: Students extend the smooth Sc → Mn logic across the entire 3d series and miss that the pattern breaks at iron.

The Truth: The smooth rise breaks at Fe. Fe$^{2+}$ is less stable relative to +3 than Mn$^{2+}$ is, because Fe$^{3+}$ has the extra-stable half-filled $d^5$ configuration — so Fe$^{2+}$ is fairly easily oxidised. Beyond Fe, however, the +2 state grows more dominant again: the third ionisation enthalpies of Co, Ni, Cu and Zn are so high that the +3 state becomes increasingly rare and strongly oxidising, until +2 is effectively the only stable state at Zn.

• Correct Trend (First Half): Sc < Ti < V < Cr < Mn (Stability increases).
• Correct Trend (Second Half): a dip at Fe (stable Fe$^{3+}$, $d^5$), then the +2 state increasingly dominant from Co → Zn (very high $IE_3$ makes +3 harder and harder to reach).

How to Avoid: Read the question carefully. If it says "first half," stop at Mn. Do not discuss Fe, Co, Ni, etc., unless asked.

Common Mistake #4: Misinterpreting the "Stability" of Cr$^{2+}$

The Mistake: Students think Cr$^{2+}$ is unstable because it is a strong reducing agent (easily oxidized to Cr$^{3+}$).

The Truth: Cr$^{2+}$ is indeed a strong reducing agent, but this is because Cr$^{3+}$ (with a $d^3$ configuration) is even more stable due to its half-filled $t_{2g}$ level in an octahedral field. The +2 state of Cr is still more stable than the +2 state of Sc or Ti.

How to Avoid: Distinguish between thermodynamic stability (the +2 state itself is stable relative to the metal) and kinetic/redox stability (it easily changes to another state). The question asks about the former.

Summary Table: How to Avoid Each Mistake

Final Tip for Exams

When writing your answer, use this structure:

1. State the trend: Stability of +2 state increases from Sc to Mn.
2. Give the reason (Ionization): $IE_3$ rises much more sharply than $IE_1 + IE_2$ across Sc → Mn, making it harder and harder to remove a third electron.
3. Give the reason (Exchange Energy): The $d^5$ configuration of Mn$^{2+}$ provides maximum exchange energy, giving it exceptional stability.
4. Conclude: Therefore, Sc$^{2+}$ is least stable (easily oxidizes to Sc$^{3+}$), while Mn$^{2+}$ is most stable.