(a) Account for the following : (i) Manganese shows maximum number of oxidation states in 3d series. (ii) $E^{\circ}$ value for $Mn^{3+}/Mn^{2+}$ couple is much more positive than that for $Cr^{3+}/Cr^{2+}$. (iii) $Ti^{4+}$ is colourless whereas $V^{4+}$ is coloured in an aqueous

Question

(a) Account for the following : (i) Manganese shows maximum number of oxidation states in 3d series. (ii) $E^{\circ}$ value for $Mn^{3+}/Mn^{2+}$ couple is much more positive than that for $Cr^{3+}/Cr^{2+}$. (iii) $Ti^{4+}$ is colourless whereas $V^{4+}$ is coloured in an aqueous solution. (b) Write the chemical equations for the preparation of $KMnO_4$ from $MnO_2$. Why does purple colour of acidified permanganate solution decolourise when it oxidises $Fe^{2+}$ to $Fe^{3+}$ ? OR (a) Write one difference between transition elements and p-block elements with reference to variability of oxidation states. (b) Why do transition metals exhibit higher enthalpies of atomization ? (c) Name an element of lanthanoid series which is well known to shown +4 oxidation state. Is it a strong oxidising agent or reducing agent ? (d) What is lanthanoid contraction ? Write its one consequence. (e) Write the ionic equation showing the oxidation of Fe(II) salt by acidified dichromate solution.

Accepted Answer

> [!TLDR]
> **Part (a):** Mn ($3d^5 4s^2$) shows $+2$ to $+7$; $E^\circ(Mn^{3+}/Mn^{2+})$ is large and positive because $Mn^{2+}$ is a stable $d^5$ ion; $Ti^{4+}(d^0)$ is colourless while $V^{4+}(d^1)$ is coloured; $KMnO_4$ is made $MnO_2	o K_2MnO_4	o KMnO_4$ and its purple colour fades as $MnO_4^-	o Mn^{2+}$. **Part (b):** transition metals vary oxidation state by 1 (p-block by 2), have high atomization enthalpies (strong metallic bonding), Ce$^{4+}$ is a strong oxidiser, lanthanoid contraction makes Zr≈Hf, and dichromate oxidises $Fe^{2+}$ to $Fe^{3+}$.

## Part (a)

### (i) Maximum oxidation states of manganese
Manganese, $[Ar]3d^5 4s^2$, has seven valence electrons ($5d+2s$). It can lose them **progressively** without a prohibitive energy jump, giving oxidation states from $+2$ (loss of $4s^2$) up to $+7$ (as in $MnO_4^-$). No other 3d metal has as many accessible states.

### (ii) $E^\circ(Mn^{3+}/Mn^{2+})$ vs $E^\circ(Cr^{3+}/Cr^{2+})$
A more positive reduction potential means the higher state is a better oxidiser. $Mn^{3+}(d^4)+e^-	o Mn^{2+}(d^5)$ forms the **exceptionally stable half-filled $d^5$** ion, so the reduction is very favourable ($E^\circ\approx+1.5$ V). By contrast $Cr^{3+}(d^3, t_{2g}^3)$ is already stable, so $Cr^{3+}	o Cr^{2+}(d^4)$ is unfavourable ($E^\circ\approx-0.4$ V); indeed $Cr^{2+}$ is a reducing agent.

### (iii) Colour of $Ti^{4+}$ and $V^{4+}$
Colour in these ions comes from **d–d transitions**. $Ti^{4+}$ is $3d^0$ — no d-electron, no transition, hence **colourless**. $V^{4+}$ (as $VO^{2+}$) is $3d^1$; the single d-electron can be excited between split d-orbitals, absorbing visible light, so it is **coloured** (blue).

### (b) $KMnO_4$ from $MnO_2$ and the decolourisation
Fusion with alkali and an oxidant gives green manganate:
$$2MnO_2+4KOH+O_2\xrightarrow{	ext{fuse}}2K_2MnO_4+2H_2O.$$
Manganate is then converted to permanganate by disproportionation (or electrolytic oxidation):
$$3MnO_4^{2-}+4H^+\longrightarrow 2MnO_4^-+MnO_2

Element	Common oxidation states
Sc	+3
Ti	+3, +4
V	+2, +3, +4, +5
Cr	+2, +3, +6
Mn	+2, +3, +4, +6, +7
Fe	+2, +3
Co	+2, +3
Ni	+2
Cu	+1, +2
Zn	+2

Q.(a) Account for the following :

Part (a)Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

The Core Intuition

The Precise Statement

Examples That Make It Concrete

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

The Core Idea: Energy Minimisation

Key Principle 1: Inert Pair Effect (for p-block elements)

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Key Principle 3: Lattice Energy / Hydration Energy Compensation

Key Principle 4: Disproportionation

Part (b)Concept understanding — Variable Oxidation States

Variable Oxidation States – The Intuition

The Precise Statement

Why This Happens – The Energy Picture

The Pattern Across the Series

Stability and the Environment

Part (a)

Part (a)

(i) Maximum oxidation states of manganese

(ii) $E^\circ(Mn^{3+}/Mn^{2+})$ vs $E^\circ(Cr^{3+}/Cr^{2+})$

(iii) Colour of $Ti^{4+}$ and $V^{4+}$

(b) $KMnO_4$ from $MnO_2$ and the decolourisation

Why the medium matters

Step-by-step reasoning

Why transition metals show variable oxidation states

Analyzing each element

Q.(a) Account for the following :

Part (a)

(i) Maximum oxidation states of manganese

(ii) E∘(Mn3+/Mn2+)E^\circ(Mn^{3+}/Mn^{2+})E∘(Mn3+/Mn2+) vs E∘(Cr3+/Cr2+)E^\circ(Cr^{3+}/Cr^{2+})E∘(Cr3+/Cr2+)

(iii) Colour of Ti4+Ti^{4+}Ti4+ and V4+V^{4+}V4+

(b) KMnO4KMnO_4KMnO4​ from MnO2MnO_2MnO2​ and the decolourisation

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(ii) $E^\circ(Mn^{3+}/Mn^{2+})$ vs $E^\circ(Cr^{3+}/Cr^{2+})$

(iii) Colour of $Ti^{4+}$ and $V^{4+}$

(b) $KMnO_4$ from $MnO_2$ and the decolourisation

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