Q.How would you account for the following:

The key idea is stability of oxidation states in transition metals, governed by the electronic configuration and the tendency to achieve half-filled or fully-filled $d$ subshells.

(i) $Cr^{2+}$ has a $d^4$ configuration. It is strongly reducing because losing one electron gives $Cr^{3+}$ ($d^3$), which has a stable half-filled $t_{2g}$ level. Conversely, $Mn^{3+}$ ($d^4$) is strongly oxidising because gaining one electron gives $Mn^{2+}$ ($d^5$), a stable half-filled $d$ subshell.

(ii) $Co^{2+}$ ($d^7$) is stable in water because the hydration energy is sufficient. However, in the presence of strong-field ligands (like $NH_3$), the pairing energy is overcome, and $Co^{2+}$ is easily oxidised to $Co^{3+}$ ($d^6$), which gains extra crystal field stabilisation energy (CFSE) in a low-spin octahedral complex. …

The Core Idea: Stability and the $d$-Subshell

The key to understanding these observations lies in the stability associated with half-filled and fully-filled $d$-orbitals. A $d^5$ configuration (half-filled) has all five electrons unpaired, one in each orbital — this gives maximum exchange energy and extra stability. A $d^{10}$ configuration (fully-filled) is also exceptionally stable. Ions that are one electron away from these configurations tend to be either strongly reducing (if they can lose an electron to reach $d^5$ or $d^{10}$) or strongly oxidising (if they can gain an electron to reach $d^5$ or $d^{10}$).

Let’s apply this to each case.

(i) $Cr^{2+}$ is strongly reducing; $Mn^{3+}$ is strongly oxidising

Both $Cr^{2+}$ and $Mn^{3+}$ have a $d^4$ configuration. But their chemical behaviour is opposite. Why?

Step 1: Identify the electronic configurations.

• $Cr$ (atomic number 24): $[Ar] 3d^5 4s^1$. $Cr^{2+}$ loses the $4s^1$ and one $3d$ electron → $[Ar] 3d^4$.
• $Mn$ (atomic number 25): $[Ar] 3d^5 4s^2$. $Mn^{3+}$ loses both $4s$ electrons and one $3d$ electron → $[Ar] 3d^4$.

So both are $d^4$ ions. But look at what they want to become.

Step 2: The driving force — reaching $d^3$ or $d^5$.

• $Cr^{2+}$ ($d^4$) readily loses one electron to become $Cr^{3+}$ ($d^3$). In an octahedral field, $d^3$ means the $t_{2g}$ set is exactly half-filled ($t_{2g}^3$), which carries extra stability from exchange energy. This matches NCERT's own reasoning (Example 4.4): $Cr^{2+}$ is reducing because its configuration changes from $d^4$ to the extra-stable half-filled-$t_{2g}$ $d^3$.

A negative potential confirms $Cr^{2+}$ is a good reducing agent — it wants to give away an electron to reach that stable $d^3$ state.

• $Mn^{3+}$ ($d^4$) readily gains one electron to become $Mn^{2+}$ ($d^5$). Here $d^5$ is the half-filled whole $d$-subshell (not just $t_{2g}$) — the single most stable $d^n$ configuration there is. So $Mn^{3+}$ readily accepts an electron, acting as a strong oxidising agent.

Step 3: The numbers confirm it.

A large positive potential means $Mn^{3+}$ is a strong oxidising agent — it pulls electrons from others.

(ii) Cobalt(II) is stable in water but easily oxidised in presence of complexing reagents

Step 1: The aqueous ion.

$Co^{2+}$ has a $d^7$ configuration. In water, it forms the hexaaqua complex $[Co(H_2O)_6]^{2+}$. Water is a weak field ligand, so the electrons occupy orbitals according to Hund’s rule — high spin configuration: $t_{2g}^5 e_g^2$. This is reasonably stable.

Step 2: Why is it stable in water?

The standard reduction potential for $Co^{3+}/Co^{2+}$ in water is:

This is highly positive, meaning $Co^{3+}$ is a very strong oxidising agent in water — it would oxidise water itself. So $Co^{2+}$ is the stable form in aqueous solution because $Co^{3+}$ is too reactive.

Step 3: What changes with complexing reagents?

When you add strong field ligands (like $NH_3$, $CN^-$, $en$), the crystal field splitting $\Delta_o$ increases. For $Co^{3+}$ ($d^6$), a strong field forces a low spin configuration: $t_{2g}^6$ — all electrons paired in the lower set. This gives huge extra stabilisation (CFSE). For $Co^{2+}$ ($d^7$), even with strong field, you get $t_{2g}^6 e_g^1$ — still one electron in the higher $e_g$ level, less stable.

So the complex of $Co^{3+}$ becomes much more stable than that of $Co^{2+}$ under strong field ligands. The equilibrium shifts:

Step 4: Real example. …

Here is a clear, concept-first solution for each part, using the Standard Electrode Potential (E°) and Electronic Configuration method.

Core Concept: Stability & E° Values

The stability of an oxidation state in aqueous solution is determined by its standard reduction potential ($E^\circ$).

• A high positive $E^\circ$ (e.g., $M^{3+} \rightarrow M^{2+}$) means the higher state is strongly oxidising (it wants to gain electrons).
• A high negative $E^\circ$ means the lower state is strongly reducing (it wants to lose electrons).
• Stability is also linked to achieving half-filled ($d^5$) or fully-filled ($d^{10}$) configurations.

(i) $Cr^{2+}$ (reducing) vs $Mn^{3+}$ (oxidising)

Method: Electronic Configuration & E° Comparison

Steps:

1. Write the configurations:

   • $Cr^{2+}$: $[Ar]\,3d^4$
   • $Mn^{3+}$: $[Ar]\,3d^4$
   • Both are $d^4$ species.

2. Identify the stable target:

   • $Cr^{2+}$ wants to lose one electron to become $Cr^{3+}$ ($d^3$). This is not a half-filled shell, but it is a stable configuration.
   • $Mn^{3+}$ wants to gain one electron to become $Mn^{2+}$ ($d^5$). $d^5$ is a half-filled, highly stable configuration.

3. Apply the E° logic:

   • $Cr^{2+}$ is reducing: The reaction $Cr^{3+} + e^- \rightarrow Cr^{2+}$ has $E^\circ = -0.41\,V$. The negative value means the reverse reaction ($Cr^{2+} \rightarrow Cr^{3+} + e^-$) is spontaneous. Thus, $Cr^{2+}$ loses electrons easily (acts as a reducing agent).
   • $Mn^{3+}$ is oxidising: The reaction $Mn^{3+} + e^- \rightarrow Mn^{2+}$ has $E^\circ = +1.57\,V$. The large positive value means $Mn^{3+}$ gains electrons very easily (acts as a strong oxidising agent) to achieve the stable $d^5$ state.

Result: $Cr^{2+}$ is reducing because it tends to lose an electron. $Mn^{3+}$ is strongly oxidising because gaining an electron gives it the stable half-filled $d^5$ configuration.

(ii) Cobalt(II) stability with complexing reagents

Method: Crystal Field Theory & Ligand Field Stabilisation Energy (LFSE)

Steps:

1. Identify the species:

   • $Co^{2+}$: $d^7$ configuration.
   • $Co^{3+}$: $d^6$ configuration.

2. Analyse the aqueous state (weak field ligand - $H_2O$):

   • In water, $Co^{2+}$ exists as $[Co(H_2O)_6]^{2+}$ (high spin, $t_{2g}^5 e_g^2$).
   • Water is a comparatively weak field ligand: it does not stabilise the +3 state enough to offset cobalt's high third ionisation enthalpy.
   • The $E^\circ$ for $Co^{3+}/Co^{2+}$ is $+1.82\,V$ (literature value; NCERT's Table 4.2 prints $+1.97\,V$), meaning aqueous $Co^{3+}$ is strongly oxidising and readily reduces to $Co^{2+}$. Hence, $Co^{2+}$ is stable in water.

3. Analyse with strong field ligands (e.g., $NH_3$, $CN^-$):

   • Strong ligands cause large crystal field splitting ($\Delta_0$).
   • $Co^{3+}$ ($d^6$): Pairs up completely to become low spin ($t_{2g}^6 e_g^0$). This gives a huge LFSE (very stable).
   • $Co^{2+}$ ($d^7$): Becomes low spin ($t_{2g}^6 e_g^1$), but the LFSE gain is less than for $Co^{3+}$.

4. Conclusion: …

Here are the common mistakes students make on this topic, broken down by each sub-part, along with the correct conceptual approach to avoid them.

General Mistake: Memorizing without understanding the why

Students often try to memorize that "Cr²⁺ is reducing" and "Mn³⁺ is oxidizing" without linking it to the electronic configuration and the stability of the half-filled or fully-filled $d$-subshell. This leads to confusion when a similar question is asked for a different element.

How to avoid: Always connect stability to the exchange energy and symmetry of the $d$-orbital configuration. The most stable configurations are $d^0$, $d^5$ (half-filled), and $d^{10}$ (fully-filled).

(i) $Cr^{2+}$ is reducing, $Mn^{3+}$ is oxidizing

Common Mistake 1: Confusing the direction of electron transfer.

Students often say "Cr²⁺ is reducing because it wants to lose electrons" but fail to specify what it reduces (it reduces other species by getting oxidized itself). Similarly, they say "Mn³⁺ is oxidizing because it wants to gain electrons" without linking it to the stable product.

How to avoid: Be precise. A reducing agent gets oxidized (loses electrons). A oxidizing agent gets reduced (gains electrons).

• Cr²⁺ ($d^4$) loses one electron to become Cr³⁺ ($d^3$). The $d^3$ configuration has a half-filled $t_{2g}$ level in an octahedral field, which is extra stable due to high exchange energy. So Cr²⁺ readily gives up an electron.
• Mn³⁺ ($d^4$) gains one electron to become Mn²⁺ ($d^5$). The $d^5$ configuration is a half-filled $d$-subshell, which is exceptionally stable. So Mn³⁺ readily accepts an electron.

Common Mistake 2: Forgetting the standard electrode potentials.

Students might not realize that the tendency is quantified by $E^\circ$ values.

• $E^\circ (Cr^{3+}/Cr^{2+}) = -0.41 \text{ V}$ (negative, so Cr²⁺ is a good reducing agent).
• $E^\circ (Mn^{3+}/Mn^{2+}) = +1.57 \text{ V}$ (highly positive, so Mn³⁺ is a strong oxidizing agent).

How to avoid: Memorize the trend: For $d^4$ ions, the $M^{3+}/M^{2+}$ potential is very negative for Cr (easy to oxidize) and very positive for Mn (easy to reduce).

(ii) Cobalt(II) stability in water vs. easy oxidation with ligands

Common Mistake 1: Thinking "stable in water" means it never oxidizes.

Students often assume Co²⁺ is inert. The key is comparison: Co²⁺ is stable in water relative to Co³⁺, but in the presence of strong ligands, the stability order reverses.

How to avoid: Understand the role of crystal field stabilization energy (CFSE) and ligand field strength.

• In water (weak field ligand):
  • Co²⁺ ($d^7$): High spin, CFSE is moderate.
  • Co³⁺ ($d^6$): High spin, CFSE is lower than expected. The standard potential $E^\circ (Co^{3+}/Co^{2+})$ is about $+1.82 \text{ V}$ in water, meaning Co³⁺ is a very strong oxidizing agent and readily gets reduced to Co²⁺. So Co²⁺ is stable.
• With strong ligands (e.g., NH₃, CN⁻):
  • Co³⁺ ($d^6$): Low spin ($t_{2g}^6$). This gives a huge CFSE (pairing energy is overcome by the large splitting). The $t_{2g}^6$ configuration is very stable.
  • Co²⁺ ($d^7$): Low spin ($t_{2g}^6 e_g^1$). The extra electron in the $e_g$ level reduces stability.
  • Result: Co²⁺ is now easily oxidized to the very stable Co³⁺ complex.

Common Mistake 2: Ignoring the spin-state change.

Students forget that the same ion can have different spin states depending on the ligand. They assume Co³⁺ is always unstable.

How to avoid: Always check the ligand. Strong field ligands (like CN⁻, NH₃, en) cause low spin configurations, which can dramatically stabilize a higher oxidation state.

(iii) The $d^1$ configuration is very unstable in ions

Common Mistake 1: Thinking "unstable" means it doesn't exist. …