Q.Compare the stability of +2 oxidation state for the elements of the first transition series.

The key idea is that the stability of the +2 oxidation state in the first transition series depends on the electronic configuration of the $\text{M}^{2+}$ ion — specifically, whether it achieves a half-filled ($d^5$) or fully-filled ($d^{10}$) d-subshell, which confers extra stability.

Reasoning:

1. All first-row transition metals (Sc to Zn) can lose their two 4s electrons to form $\text{M}^{2+}$ ions. The stability of this state is highest when the resulting $d^n$ configuration is especially stable.
2. $\text{Mn}^{2+}$ ($d^5$, half-filled) and $\text{Zn}^{2+}$ ($d^{10}$, fully-filled) have symmetrical, exchange-stabilised configurations, making them very stable.
3. $\text{Fe}^{2+}$ ($d^6$) is also relatively stable because it is one electron away from half-filled $d^5$, but it is easily oxidised to $\text{Fe}^{3+}$ ($d^5$). In contrast, $\text{Cr}^{2+}$ ($d^4$) is readily oxidised to $\text{Cr}^{3+}$ ($d^3$) due to the high stability of the latter. …

The +2 oxidation state is the most common oxidation state for first-row transition metals, but its stability varies significantly across the series. To understand why, we need to think about what "stability" means in this context — it refers to the tendency of the element to form and retain the +2 ion in aqueous solution or in compounds. The key factors are the energy required to remove two electrons (first and second ionization energies) and the energy released when the resulting ion interacts with water molecules or other ligands (hydration enthalpy).

Let’s walk through the series element by element, from scandium to zinc.

1. Scandium (Sc): Sc has electronic configuration $[Ar] 3d^1 4s^2$. Removing two electrons gives $Sc^{2+}$ with configuration $[Ar] 3d^1$. However, Sc strongly prefers the +3 state because removing the third electron (from 3d) is relatively easy due to the stability of the noble gas configuration $[Ar]$ in $Sc^{3+}$. The +2 state is rarely observed and is highly unstable — it readily oxidizes to +3. So, stability of +2 is very low.

2. Titanium (Ti): Ti is $[Ar] 3d^2 4s^2$. $Ti^{2+}$ has $[Ar] 3d^2$, but it is a strong reducing agent and easily oxidizes to $Ti^{3+}$ or $Ti^{4+}$. The +2 state exists in some compounds (like TiCl₂) but is not stable in water — it reduces water to hydrogen. Stability is low.

3. Vanadium (V): V is $[Ar] 3d^3 4s^2$. $V^{2+}$ has $[Ar] 3d^3$ — not a half-filled d-subshell, but a configuration with appreciable exchange-energy stabilisation (a half-filled $t_{2g}^3$ set in an octahedral field). $V^{2+}$ is a strong reducing agent and is easily oxidized to higher states. Stability is moderate but still low compared to later elements.

4. Chromium (Cr): Cr is $[Ar] 3d^5 4s^1$ (anomalous due to half-filled stability). $Cr^{2+}$ has $[Ar] 3d^4$, which is not particularly stable. However, $Cr^{2+}$ is a strong reducing agent and is easily oxidized to $Cr^{3+}$ (which has $d^3$ — a stable half-filled $t_{2g}^3$ level in water, with high exchange energy). In practice, $Cr^{2+}$ is unstable in air and water, though it can exist in acidic solutions. Stability is moderate.

5. Manganese (Mn): Mn is $[Ar] 3d^5 4s^2$. $Mn^{2+}$ has $[Ar] 3d^5$ — a half-filled d-subshell, which is exceptionally stable due to maximum exchange energy and symmetrical distribution. This is a turning point. $Mn^{2+}$ is quite stable in acidic solutions and resists oxidation to $Mn^{3+}$ (which would lose the half-filled stability). Stability is high.

6. Iron (Fe): Fe is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ has $[Ar] 3d^6$, which is one electron beyond half-filled. It is reasonably stable, though it can be oxidized to $Fe^{3+}$ (d⁵, half-filled) under oxidizing conditions. In water, $Fe^{2+}$ is stable in the absence of oxygen. Stability is high.

7. Cobalt (Co): Co is $[Ar] 3d^7 4s^2$. $Co^{2+}$ has $[Ar] 3d^7$, which is stable. $Co^{3+}$ is a strong oxidizing agent and tends to reduce back to $Co^{2+}$. So $Co^{2+}$ is very stable in aqueous solution. Stability is high.

8. Nickel (Ni): Ni is $[Ar] 3d^8 4s^2$. $Ni^{2+}$ has $[Ar] 3d^8$, which is stable. $Ni^{3+}$ is rare and unstable. $Ni^{2+}$ is the dominant oxidation state for nickel. Stability is high.

9. Copper (Cu): Cu is $[Ar] 3d^{10} 4s^1$ (anomalous). $Cu^{2+}$ has $[Ar] 3d^9$, which is stable in water. However, $Cu^+$ (d¹⁰) also exists but disproportionates in water. $Cu^{2+}$ is the common state. Stability is moderate to high, but note that $Cu^{2+}$ can be reduced to $Cu^+$ or Cu metal under some conditions.

10. Zinc (Zn): Zn is $[Ar] 3d^{10} 4s^2$. $Zn^{2+}$ has $[Ar] 3d^{10}$ — a completely filled d-subshell, which is very stable. $Zn^{2+}$ is the only common oxidation state for zinc, and it is highly stable in water. However, note that Zn has high second ionization energy, but the hydration enthalpy compensates. Stability is high.

Now, let's quantify this using the standard reduction potentials ($E^\circ$ for $M^{2+}/M$). A more negative value means the reduction $M^{2+} + 2e^- \rightarrow M$ is less favourable — the metal is easily oxidised to $M^{2+}$ and the $M^{2+}$ ion resists being reduced back to the metal. A positive value (Cu) means the ion is readily reduced. Note that this couple measures the stability of $M^{2+}$ with respect to the metal; stability against further oxidation to +3 is governed by the $M^{3+}/M^{2+}$ couple instead.

Method: Electronic Configuration & Hydration Enthalpy Analysis

This method explains the stability of the +2 oxidation state by linking electronic configuration of the ion with hydration enthalpy trends.

Steps

1. Write the general electronic configuration For first transition series (Sc to Zn):

In the +2 oxidation state, the two 4s electrons are lost first:

2. Identify the most stable configurations

   The $M^{2+}$ ions with half-filled ($3d^5$) or fully-filled ($3d^{10}$) d-subshells have extra stability due to exchange energy and symmetry.

   • Mn²⁺ ($3d^5$) — half-filled
   • Zn²⁺ ($3d^{10}$) — fully filled
   • Fe²⁺ ($3d^6$) — close to half-filled, also stable

3. Consider hydration enthalpy

   For ions in aqueous solution, higher hydration enthalpy (more negative $\Delta H_{\text{hyd}}$) stabilizes the +2 state.

   • Smaller ionic radius → higher charge density → more negative hydration enthalpy
   • Example: Ni²⁺ has high hydration enthalpy, making Ni²⁺ stable in water

4. Compare across the series

   Element	$M^{2+}$ config	Stability of +2 state
   Sc	$3d^1$	Least stable (easily oxidised to Sc³⁺)
   Ti	$3d^2$	Moderate
   V	$3d^3$	Moderate
   Cr	$3d^4$	Unstable (Cr³⁺ more stable)
   Mn	$3d^5$	Very stable (half-filled)
   Fe	$3d^6$	Stable
   Co	$3d^7$	Stable
   Ni	$3d^8$	Very stable (high hydration enthalpy)

Common Mistakes: Stability of +2 Oxidation State in First Transition Series

Students often struggle with this concept because it requires understanding electronic configurations, hydration energy, and lattice energy together. Here are the most frequent errors and how to avoid them.

✗ Mistake 1: Assuming All +2 States Are Equally Stable

Why it's wrong:

The stability of $M^{2+}$ varies significantly across the series due to changes in $d$-orbital occupancy and hydration enthalpy.

How to avoid:

Remember the trend:

• Sc²⁺ barely exists (scandium goes straight to Sc$^{3+}$), while Zn$^{2+}$ is very stable ($d^{10}$) — the two ends of the series behave completely differently.
• Mn$^{2+}$ and Zn$^{2+}$ are exceptionally stable because they have half-filled ($d^5$) and fully filled ($d^{10}$) configurations respectively.
• Fe$^{2+}$ is stable but oxidises easily to Fe$^{3+}$ ($d^5$).
• Co$^{2+}$ and Ni$^{2+}$ are moderately stable.
• Cu$^{2+}$ is more stable than Cu$^+$ due to higher hydration energy.

Key rule: Stability increases with higher hydration enthalpy and favourable electronic configuration.

✗ Mistake 2: Ignoring Hydration Energy Trends

Why it's wrong:

Hydration energy ($\Delta H_{hyd}$) decreases across the series as ionic radius decreases, but the magnitude of change affects stability.

How to avoid:

• Hydration energy is highest for Cu$^{2+}$ (small size, high charge density).
• This makes Cu$^{2+}$ more stable than Cu$^+$ in aqueous solution.
• For Mn$^{2+}$, hydration energy is lower, but the $d^5$ configuration compensates.

Formula to remember:

✗ Mistake 3: Forgetting the Role of Crystal Field Stabilisation Energy (CFSE)

Why it's wrong:

CFSE varies with $d$-electron count and affects the relative stability of +2 states.

How to avoid:

• $d^4$ (Cr$^{2+}$) and $d^9$ (Cu$^{2+}$) ions gain a little extra stabilisation from Jahn–Teller distortion — this is separate from CFSE, which depends on the $t_{2g}/e_g$ occupancy.
• For $d^5$ (Mn$^{2+}$), CFSE is zero (high-spin), but the half-filled stability dominates.
• For $d^{10}$ (Zn$^{2+}$), CFSE is zero, but full-filled stability dominates.

Quick reference:

✗ Mistake 4: Confusing Stability in Solid vs. Aqueous State

Why it's wrong:

Stability in solid compounds depends on lattice energy, while in solution it depends on hydration energy.

How to avoid:

• In solid state: Lattice energy dominates → smaller ions form more stable compounds.
• In aqueous solution: Hydration energy dominates → Cu$^{2+}$ is more stable than Cu$^+$ because Cu$^{2+}$ has much higher hydration energy.
• Exam tip: Always check whether the question asks about aqueous or solid state.

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