Skip to content
Exercises · 4.21

Q.How would you account for the following:

(i) Of the d4d^4 species, Cr2+Cr^{2+} is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d1d^1 configuration is very unstable in ions.
CBSENCERTSubjective· 3mImportance★★★★★
31% · 41/132 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The stability of oxidation states in transition metals is governed by the electronic configuration of the ion and its tendency to achieve a half-filled (d5d^5) or fully-filled (d10d^{10}) dd-subshell. Cr2+Cr^{2+} (d4d^4) is strongly reducing because losing an electron gives Cr3+Cr^{3+} (d3d^3, a half-filled t2gt_{2g} level); Mn3+Mn^{3+} (d4d^4) is strongly oxidising because gaining an electron gives Mn2+Mn^{2+} (d5d^5); Co2+Co^{2+} (d7d^7) is stable in water but oxidises to d6d^6 in complexes; and d1d^1 ions are unstable because they easily lose their single dd-electron to reach the stable d0d^0 state.

The Core Idea: Stability and the dd-Subshell

The key to understanding these observations lies in the stability associated with half-filled and fully-filled dd-orbitals. A d5d^5 configuration (half-filled) has all five electrons unpaired, one in each orbital — this gives maximum exchange energy and extra stability. A d10d^{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 d5d^5 or d10d^{10}) or strongly oxidising (if they can gain an electron to reach d5d^5 or d10d^{10}).

Let’s apply this to each case.

(i) Cr2+Cr^{2+} is strongly reducing; Mn3+Mn^{3+} is strongly oxidising

Both Cr2+Cr^{2+} and Mn3+Mn^{3+} have a d4d^4 configuration. But their chemical behaviour is opposite. Why?

Step 1: Identify the electronic configurations.

  • CrCr (atomic number 24): [Ar]3d54s1[Ar] 3d^5 4s^1. Cr2+Cr^{2+} loses the 4s14s^1 and one 3d3d electron → [Ar]3d4[Ar] 3d^4.
  • MnMn (atomic number 25): [Ar]3d54s2[Ar] 3d^5 4s^2. Mn3+Mn^{3+} loses both 4s4s electrons and one 3d3d electron → [Ar]3d4[Ar] 3d^4.

So both are d4d^4 ions. But look at what they want to become.

Step 2: The driving force — reaching d3d^3 or d5d^5.

  • Cr2+Cr^{2+} (d4d^4) readily loses one electron to become Cr3+Cr^{3+} (d3d^3). In an octahedral field, d3d^3 means the t2gt_{2g} set is exactly half-filled (t2g3t_{2g}^3), which carries extra stability from exchange energy. This matches NCERT's own reasoning (Example 4.4): Cr2+Cr^{2+} is reducing because its configuration changes from d4d^4 to the extra-stable half-filled-t2gt_{2g} d3d^3.

E∘(Cr3+/Cr2+)=−0.41 VE^\circ (Cr^{3+}/Cr^{2+}) = -0.41\ \text{V}

A negative potential confirms Cr2+Cr^{2+} is a good reducing agent — it wants to give away an electron to reach that stable d3d^3 state.

  • Mn3+Mn^{3+} (d4d^4) readily gains one electron to become Mn2+Mn^{2+} (d5d^5). Here d5d^5 is the half-filled whole dd-subshell (not just t2gt_{2g}) — the single most stable dnd^n configuration there is. So Mn3+Mn^{3+} readily accepts an electron, acting as a strong oxidising agent.
Watch out

A common mistake is to think both d4d^4 ions behave the same. The difference is which neighbouring configuration is more stable: Cr2+Cr^{2+} oxidises to the half-filled-t2gt_{2g} d3d^3, while Mn3+Mn^{3+} reduces to the fully half-filled d5d^5. Both moves are driven by reaching a more stable configuration, just in opposite directions.

Step 3: The numbers confirm it.

E∘(Mn3+/Mn2+)=+1.57 VE^\circ (Mn^{3+}/Mn^{2+}) = +1.57\ \text{V}

A large positive potential means Mn3+Mn^{3+} is a strong oxidising agent — it pulls electrons from others.

Tip

The d4d^4 configuration is inherently unstable because it is one electron short of d5d^5 (half-filled) or one electron away from d3d^3 (which is also relatively stable in some cases). The actual behaviour depends on which neighbour (d3d^3 or d5d^5) is more stable in that element’s context.

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

Step 1: The aqueous ion.

Co2+Co^{2+} has a d7d^7 configuration. In water, it forms the hexaaqua complex [Co(H2O)6]2+[Co(H_2O)_6]^{2+}. Water is a weak field ligand, so the electrons occupy orbitals according to Hund’s rule — high spin configuration: t2g5eg2t_{2g}^5 e_g^2. This is reasonably stable.

Step 2: Why is it stable in water?

The standard reduction potential for Co3+/Co2+Co^{3+}/Co^{2+} in water is:

E∘(Co3+/Co2+)=+1.82 V    (literature value; NCERT’s own Table 4.2 prints +1.97 V for this couple)E^\circ (Co^{3+}/Co^{2+}) = +1.82\ \text{V} \;\; \text{(literature value; NCERT's own Table 4.2 prints } +1.97\ \text{V for this couple)}

This is highly positive, meaning Co3+Co^{3+} is a very strong oxidising agent in water — it would oxidise water itself. So Co2+Co^{2+} is the stable form in aqueous solution because Co3+Co^{3+} is too reactive.

Step 3: What changes with complexing reagents?

When you add strong field ligands (like NH3NH_3, CN−CN^-, enen), the crystal field splitting Δo\Delta_o increases. For Co3+Co^{3+} (d6d^6), a strong field forces a low spin configuration: t2g6t_{2g}^6 — all electrons paired in the lower set. This gives huge extra stabilisation (CFSE). For Co2+Co^{2+} (d7d^7), even with strong field, you get t2g6eg1t_{2g}^6 e_g^1 — still one electron in the higher ege_g level, less stable.

So the complex of Co3+Co^{3+} becomes much more stable than that of Co2+Co^{2+} under strong field ligands. The equilibrium shifts:

[Co(H2O)6]2+→ligand[CoL6]3+ (easily oxidised)[Co(H_2O)_6]^{2+} \xrightarrow{\text{ligand}} [CoL_6]^{3+} \text{ (easily oxidised)}

The CFSE for d6d^6 low spin (octahedral) is −2.4Δo+2P-2.4\Delta_o + 2P (where PP is pairing energy), while for d7d^7 high spin it is −0.8Δo-0.8\Delta_o. The difference favours d6d^6 when Δo\Delta_o is large.

Step 4: Real example. …

Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question4.20 — Compare the chemistry of actinoids with that of the lanthanoids with s…→Next question4.22 — What is meant by 'disproportionation'? Give two examples of disproport…