Theoretical Foundation: Electronic Configuration and Stability

The stability of any electronic configuration in transition metal ions is primarily determined by the exchange energy, the crystal field stabilization energy (CFSE), and the effective nuclear charge. For a $d^1$ configuration, we examine the behavior of transition metal ions (e.g., $Ti^{3+}$, $V^{4+}$) in aqueous or coordination environments.

Step 1: Crystal Field Splitting Analysis

In an octahedral coordination environment, the five degenerate $d$-orbitals split into two distinct energy levels due to the electrostatic field of the ligands: the $t_{2g}$ set (lower energy) and the $e_g$ set (higher energy) [Per Crystal Field Theory].

Step 2: Analytical Comparison of Configurations

The stability of the $d^1$ configuration is assessed by its tendency to attain a more stable state through oxidation or reduction. The $d^1$ state represents a single electron outside of a stable closed-shell or half-filled configuration.

Step 3: Justification of $d^1$ Instability

• High Ionization Potential: Removing the sole $d$-electron allows the ion to achieve the stable, lower-energy noble gas core configuration ($d^0$), releasing significant energy.
• Oxidation Potential: Because the transition from $d^1 \rightarrow d^0$ results in a substantial decrease in electronic potential energy, ions with a $d^1$ configuration act as strong reducing agents [Per thermodynamic feasibility of oxidation].
• Exchange Energy Deficit: With only one electron, there are zero pairs of electrons with parallel spins that can undergo exchange, meaning the configuration lacks the "Exchange Energy" stabilization found in $d^3$ or $d^5$ states [Per Quantum Mechanical Exchange Theorem].

Final Solution: The $d^1$ configuration is unstable because the ion can achieve a more stable noble gas configuration ($d^0$) by losing its single electron. This process is energetically favorable as it eliminates the high-energy state of a single electron in an otherwise empty $d$-subshell, providing a strong driving force for oxidation.

1. Theoretical Foundation of Oxidising Action

Potassium dichromate ($K_2Cr_2O_7$) acts as a powerful oxidising agent in an acidic medium. The oxidising action is primarily due to the reduction of the dichromate ion ($Cr_2O_7^{2-}$) to the chromic ion ($Cr^{3+}$). [Per the half-reaction approach in redox chemistry].

The fundamental half-reaction for the reduction of dichromate in an acidic medium is expressed as:

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ [Standard Electrode Potential $E^\circ = 1.33 \, V$]

2. Reaction with Iron(II) Solution

When potassium dichromate reacts with iron(II) sulfate (or any $Fe^{2+}$ source) in the presence of an acid (typically $H_2SO_4$), the iron(II) ions are oxidised to iron(III) ions, while the dichromate ions are reduced to chromium(III) ions.

3. Derivation of the Ionic Equation

Step 1: Write the Oxidation Half-Reaction

The iron(II) ion loses an electron to become an iron(III) ion:

$Fe^{2+} \rightarrow Fe^{3+} + e^-$ [Eq. 1]

Step 2: Write the Reduction Half-Reaction

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ [Eq. 2]

Step 3: Balance the Electrons

To cancel the electrons, we must multiply the oxidation half-reaction (Eq. 1) by a factor of 6:

$6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$ [Eq. 3]

Step 4: Combine the Half-Reactions

Adding [Eq. 2] and [Eq. 3] results in the final balanced ionic equation:

$Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$

4. Graphical Representation of Electron Transfer

5. Summary of Oxidation States

Final Solution: The balanced ionic equation for the reaction of potassium dichromate with iron(II) solution is: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.

Step 1: Establishing the Neutral Atomic State

To determine the electronic configuration of the ion $Ce^{4+}$, we must first establish the configuration of the neutral Cerium (Ce) atom. Cerium is a lanthanide element with an atomic number ($Z$) of $58$.

The electronic configuration of a neutral atom is governed by the Aufbau Principle, which dictates that orbitals are filled in order of increasing energy ($1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$).

The noble gas core preceding Cerium is Xenon ($Xe$, $Z=54$).

Neutral $Ce$ ($Z=58$): $[Xe] \, 4f^1 \, 5d^1 \, 6s^2$ [Per the Aufbau Principle, accounting for the specific orbital occupancy of the lanthanide series].

Step 2: Ionization Process of $Ce^{4+}$

The formation of a cation involves the removal of electrons from the highest energy orbitals (outermost shells) first. Specifically, electrons are removed from the $6s$ orbital before the $5d$ or $4f$ orbitals due to higher principal quantum numbers ($n$) and shielding effects.

The configuration of neutral Cerium is $[Xe] \, 4f^1 \, 5d^1 \, 6s^2$.

To obtain $Ce^{4+}$, we must remove $4$ electrons in the following sequence:

• First two electrons: Removed from the $6s$ orbital. Configuration becomes $[Xe] \, 4f^1 \, 5d^1$.
• Third electron: Removed from the $5d$ orbital. Configuration becomes $[Xe] \, 4f^1$.
• Fourth electron: Removed from the $4f$ orbital. Configuration becomes $[Xe]$.

Step 3: Verification of the Final State

Upon the removal of all four electrons, the resulting electronic configuration corresponds to the stable noble gas configuration of Xenon ($Z=54$).

Summary of Electronic Configuration

Final Solution: The electronic configuration of $Ce^{4+}$ is $[Xe]$.

Theoretical Definition of Disproportionation

Disproportionation is a specific type of redox reaction in which a chemical species, usually an element, is simultaneously oxidized and reduced. In such reactions, the reactant species exists in an intermediate oxidation state and transforms into at least two different products, one having a higher oxidation state (oxidation) and one having a lower oxidation state (reduction). [Per Redox Chemistry Principles].

General Representation

The general form of a disproportionation reaction can be represented as:

$2A^{n} \rightarrow A^{(n+x)} + A^{(n-y)}$

Where $A^{n}$ represents the intermediate oxidation state of element $A$.

Example 1: Disproportionation of Copper(I) ions in Aqueous Solution

In aqueous solution, the $Cu^{+}$ ion is unstable and undergoes disproportionation to form solid copper metal ($Cu^0$) and copper(II) ions ($Cu^{2+}$).

Reaction Equation:

$2Cu^{+}(aq) \rightarrow Cu(s) + Cu^{2+}(aq)$

Oxidation State Analysis:

• $Cu^{+} \rightarrow Cu^{0}$: Decrease in oxidation state (Reduction)
• $Cu^{+} \rightarrow Cu^{2+}$: Increase in oxidation state (Oxidation)

Example 2: Disproportionation of Manganate(VI) ions in Acidic Medium

Manganate(VI) ions ($MnO_{4}^{2-}$) are stable in alkaline solutions but undergo disproportionation in acidic or neutral media to produce Permanganate ($MnO_{4}^{-}$) and Manganese dioxide ($MnO_{2}$).

Reaction Equation:

$3MnO_{4}^{2-}(aq) + 4H^{+}(aq) \rightarrow 2MnO_{4}^{-}(aq) + MnO_{2}(s) + 2H_{2}O(l)$

Oxidation State Analysis:

Visual Representation of Disproportionation (Example: $Cu^+$)

Final Solution: Disproportionation is a redox reaction where an element in an intermediate oxidation state simultaneously undergoes oxidation and reduction. Key examples include the decomposition of $Cu^{+}$ into $Cu$ and $Cu^{2+}$, and the reaction of $MnO_{4}^{2-}$ into $MnO_{4}^{-}$ and $MnO_{2}$ in acidic conditions.

Step 1: Electronic Configuration of Mn²⁺ and Fe²⁺

To determine the relative stability of these ions towards oxidation, we must first analyze their ground-state electronic configurations based on the Aufbau principle and Hund’s rule of maximum multiplicity.

• Manganese ($Z=25$): The neutral atom configuration is $[Ar] 3d^5 4s^2$. Upon forming the $Mn^{2+}$ ion, the two electrons are removed from the $4s$ orbital.
  Configuration of $Mn^{2+}$: $[Ar] 3d^5$
• Iron ($Z=26$): The neutral atom configuration is $[Ar] 3d^6 4s^2$. Upon forming the $Fe^{2+}$ ion, the two electrons are removed from the $4s$ orbital.
  Configuration of $Fe^{2+}$: $[Ar] 3d^6$

Step 2: Stability Analysis via Exchange Energy

The stability of $d$-orbital configurations is governed by the symmetry and filling of the $d$-subshell. [Per Hund’s Rule and the principles of Exchange Energy]:

• The $Mn^{2+}$ ion possesses a $3d^5$ configuration. This is a half-filled $d$-subshell. According to symmetry principles, half-filled and fully-filled shells possess extra stability due to the symmetrical distribution of electron density and maximum exchange energy.
• The $Fe^{2+}$ ion possesses a $3d^6$ configuration. Removal of one electron ($Fe^{2+} \rightarrow Fe^{3+} + e^-$) results in a $3d^5$ configuration. Since $3d^5$ is highly stable, there is a strong thermodynamic driving force for $Fe^{2+}$ to lose an electron and achieve the $3d^5$ state.

Step 3: Evaluation of Oxidation Potentials

The oxidation process involves the transformation of the divalent ion to the trivalent state:

The conversion of $Mn^{2+} (3d^5)$ to $Mn^{3+} (3d^4)$ involves the disruption of the stable half-filled configuration, which requires a significant amount of energy. Conversely, the conversion of $Fe^{2+} (3d^6)$ to $Fe^{3+} (3d^5)$ is energetically favorable because it results in the formation of a stable, half-filled $d$-subshell.

Step 4: Conclusion

Because $Mn^{2+}$ has a stable half-filled $d$-orbital configuration, it resists oxidation as any change would decrease its stability. $Fe^{2+}$, having an extra electron beyond the half-filled state, readily loses it to attain the stable half-filled configuration. Therefore, $Mn^{2+}$ is more stable towards oxidation than $Fe^{2+}$.

Final Solution: $Mn^{2+}$ compounds are more stable because the $Mn^{2+}$ ion has a $3d^5$ electronic configuration, which is a half-filled subshell providing extra stability. Oxidation to $Mn^{3+}$ would disrupt this configuration, whereas oxidation of $Fe^{2+}$ ($3d^6$) to $Fe^{3+}$ ($3d^5$) allows the ion to achieve this stable half-filled state.

Step 1: Data Extraction and Thermodynamic Framework

To evaluate the ease of oxidation for the metals Cr, Mn, and Fe from the neutral state ($M^0$) to the dipositive state ($M^{2+}$), we must analyze the standard electrode potentials ($E^o$) provided for the $M^{2+}/M$ redox couple. The oxidation half-reaction is represented as:

$M(s) \rightarrow M^{2+}(aq) + 2e^-$

The standard potential for this oxidation is given by $E^o_{ox} = -E^o_{red}$. [Per the principles of Electrochemistry, a more negative reduction potential $E^o_{M^{2+}/M}$ signifies a greater tendency to undergo oxidation.]

Step 2: Comparative Analysis of $E^o$ Values

Based on the provided data, we extract the reduction potentials ($E^o_{M^{2+}/M}$):

Step 3: Calculating Oxidation Potentials

The oxidation potential ($E^o_{ox}$) represents the ease of losing electrons. The higher the positive value of $E^o_{ox}$, the easier the metal is to oxidize.

• For $Cr$: $E^o_{ox} = -(-0.90\text{ V}) = +0.90\text{ V}$
• For $Mn$: $E^o_{ox} = -(-1.18\text{ V}) = +1.18\text{ V}$
• For $Fe$: $E^o_{ox} = -(-0.44\text{ V}) = +0.44\text{ V}$

Step 4: Interpretation of Results

Comparing the oxidation potentials: $E^o_{Mn} (+1.18\text{ V}) > E^o_{Cr} (+0.90\text{ V}) > E^o_{Fe} (+0.44\text{ V})$.

The magnitude of the oxidation potential is directly proportional to the thermodynamic feasibility of the oxidation process. Since the oxidation potential of Iron ($+0.44\text{ V}$) is significantly lower than that of both Manganese ($+1.18\text{ V}$) and Chromium ($+0.90\text{ V}$), it follows that Iron is the most resistant to oxidation among the three.

Final Solution: Iron ($Fe$) is less easily oxidized than Chromium ($Cr$) or Manganese ($Mn$) because its standard oxidation potential ($+0.44\text{ V}$) is lower than the oxidation potentials of both Chromium ($+0.90\text{ V}$) and Manganese ($+1.18\text{ V}$), indicating a lesser thermodynamic tendency to lose electrons compared to the other two metals.

Step 1: Identifying the Atomic Number and Neutral Configuration

The element Cobalt ($Co$) possesses an atomic number ($Z$) of $27$. [Per the Aufbau Principle, electrons are filled into orbitals in order of increasing energy]. The electronic configuration for a neutral Cobalt atom in its ground state is:

$Co (Z=27): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7$

Alternatively, using the nearest noble gas (Argon, $Ar$, $Z=18$) core notation:

$Co: [Ar] 3d^7 4s^2$

Step 2: Determining the Configuration of the Ion

To derive the electronic configuration of the cation $Co^{2+}$, we must remove $2$ electrons from the neutral atom. [Per the shell ionization rule, electrons are removed from the orbital with the highest principal quantum number ($n$) first].

• In $Co$, the $4s$ orbital ($n=4$) is higher in energy/further from the nucleus than the $3d$ orbital ($n=3$).
• Therefore, electrons are removed from the $4s^2$ subshell before the $3d$ subshell.

Step 3: Calculating Final Electronic Configuration

Subtracting $2$ electrons from the $4s$ subshell:

$Co^{2+} = [Ar] 3d^7 4s^{2-2}$

$Co^{2+} = [Ar] 3d^7$

Visual Representation of Orbital Occupancy

Final Solution: The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$ or $1s^2 2s^2 2p^6 3s^2 3p^6 3d^7$.

Definition of Disproportionation

A disproportionation reaction is a specific type of redox reaction in which a single substance acts as both the oxidizing agent and the reducing agent. In this process, an element in one oxidation state is simultaneously oxidized to a higher oxidation state and reduced to a lower oxidation state. This phenomenon occurs frequently with transition metal ions in aqueous solutions where multiple stable oxidation states exist.

Example 1: Disproportionation of Copper(I) Ion

The copper(I) ion ($Cu^+$) is unstable in an aqueous medium. It undergoes disproportionation to form copper(II) ($Cu^{2+}$) and metallic copper ($Cu^0$).

Reaction Equation:

$2Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s)$

• Oxidation: $Cu^+ \rightarrow Cu^{2+} + e^-$ (Increase in oxidation state from $+1$ to $+2$)
• Reduction: $Cu^+ + e^- \rightarrow Cu^0$ (Decrease in oxidation state from $+1$ to $0$)

[Justification: The hydration energy of $Cu^{2+}(aq)$ is significantly higher than that of $Cu^+(aq)$, making the $Cu^{2+}$ ion more stable in aqueous solution.]

Example 2: Disproportionation of Manganate(VI) Ion

The manganate ion ($MnO_4^{2-}$) is stable only in highly alkaline solutions. In neutral or acidic media, it undergoes disproportionation into permanganate ($MnO_4^-$) and manganese dioxide ($MnO_2$).

Reaction Equation:

$3MnO_4^{2-}(aq) + 4H^+(aq) \rightarrow 2MnO_4^-(aq) + MnO_2(s) + 2H_2O(l)$

• Oxidation: The oxidation state of Mn changes from $+6$ in $MnO_4^{2-}$ to $+7$ in $MnO_4^-$.
• Reduction: The oxidation state of Mn changes from $+6$ in $MnO_4^{2-}$ to $+4$ in $MnO_2$.

[Justification: Per the principles of chemical thermodynamics, the disproportionation is driven by the relative stability of the Mn species in acidic/neutral media compared to the highly basic state.]

Visual Representation of Redox Electron Transfer

Comparative Summary of States

Final Solution: Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced. Examples include $2Cu^+ \rightarrow Cu^{2+} + Cu$ and $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$.

Preparation of Potassium Permanganate ($KMnO_4$)

The industrial production of potassium permanganate involves a two-step oxidative fusion process starting from the mineral Pyrolusite ($MnO_2$).

Step 1: Conversion of $MnO_2$ to Potassium Manganate ($K_2MnO_4$)
Pyrolusite is fused with potassium hydroxide ($KOH$) in the presence of atmospheric oxygen or an oxidizing agent like potassium nitrate ($KNO_3$).
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ [Oxidation of $Mn(IV)$ to $Mn(VI)$]

Step 2: Oxidation of Potassium Manganate to Potassium Permanganate
The green potassium manganate is extracted with water and electrolytically oxidized or treated with ozone/chlorine to produce the purple $KMnO_4$. In acidic or neutral media, the disproportionation reaction occurs:
$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$

Reactions of Acidified Permanganate ($MnO_4^-$)

In acidic solution ($H^+$), the permanganate ion acts as a strong oxidizing agent, typically undergoing the reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. [Standard Electrode Potential $E^\circ = +1.51V$].

Analysis of Red-Ox Processes

• Iron(II) to Iron(III): The $Fe^{2+}$ ion is oxidized to $Fe^{3+}$ while $Mn$ is reduced from $+7$ to $+2$.
• Sulfur Dioxide: $SO_2$ acts as a reducing agent where $S^{IV}$ is oxidized to $S^{VI}$ (sulfate ion).
• Oxalic Acid: The oxalate ion ($C_2O_4^{2-}$) is oxidized to carbon dioxide ($CO_2$). This reaction is auto-catalyzed by the $Mn^{2+}$ ions produced during the reaction.

Final Solution: The preparation of $KMnO_4$ involves the fusion of $MnO_2$ with $KOH$ to form $K_2MnO_4$, followed by oxidation. Acidified $KMnO_4$ acts as a powerful oxidizer, reducing to $Mn^{2+}$ while oxidizing $Fe^{2+}$ to $Fe^{3+}$, $SO_2$ to $SO_4^{2-}$, and oxalic acid to $CO_2$.

Step 1: Analyzing $3d^3$ Configuration

The ground state configuration of $3d^3$ corresponds to the element Vanadium ($Z=23$), which has the configuration $[Ar] 3d^3 4s^2$.

• The loss of two $4s$ electrons and one $3d$ electron yields a stable half-filled $t_{2g}$ subshell ($d^3$ is a stable configuration due to crystal field stabilization energy in octahedral complexes).
• Common and stable oxidation states are $+2, +3, +4,$ and $+5$.
• The $+5$ state is particularly stable as it corresponds to the noble gas configuration ($[Ar] 3d^0 4s^0$) following the removal of all valence electrons.

Step 2: Analyzing $3d^5$ Configuration

The $3d^5$ configuration can correspond to Chromium ($Z=24$, $[Ar] 3d^5 4s^1$) or Manganese ($Z=25$, $[Ar] 3d^5 4s^2$).

• Due to the stability of the half-filled $d$-subshell ($d^5$), this configuration exhibits high stability.
• For Chromium, $+3$ and $+6$ are prominent. The $+3$ state is stable due to the $t_{2g}^3$ configuration [Per Crystal Field Theory].
• For Manganese, $+2$ is exceptionally stable because it retains the $d^5$ configuration after losing both $4s$ electrons. The $+7$ state is also stable in oxyanions like $MnO_4^-$.

Step 3: Analyzing $3d^8$ Configuration

The $3d^8$ configuration corresponds to Nickel ($Z=28$, $[Ar] 3d^8 4s^2$).

• Nickel primarily exhibits the $+2$ oxidation state.
• The loss of two $4s$ electrons results in the $3d^8$ ion ($Ni^{2+}$), which is the most stable state for nickel in aqueous chemistry.
• While higher states are theoretically possible, they are extremely rare or unstable under standard laboratory conditions.

Step 4: Analyzing $3d^4$ Configuration

The $3d^4$ configuration (excluding $4s$ electrons) often refers to $Cr^{2+}$ or $Mn^{3+}$.

• $3d^4$ is inherently unstable as it is only one electron away from the highly stable half-filled $d^5$ configuration.
• Consequently, species with $3d^4$ act as strong reducing agents (e.g., $Cr^{2+}$) to achieve the $d^3$ or $d^5$ state.
• The stable oxidation states for elements possessing this configuration are typically $+2$ and $+3$.

Summary of Stability

Final Solution: The stable oxidation states are: $3d^3 \rightarrow +3, +5$; $3d^5 \rightarrow +2, +7$; $3d^8 \rightarrow +2$; $3d^4 \rightarrow +2, +3$.

Step 1: Theoretical Foundation of Electronic Configurations

The electronic configuration of lanthanoids and actinoids is governed by the progressive filling of the $4f$ and $5f$ orbitals, respectively. Both series arise from the necessity to accommodate electrons in inner-shell orbitals after the $6s$ or $7s$ subshells have been partially filled.

The general valence electronic configurations are defined as:

• Lanthanoids: $[Xe] 4f^{0-14} 5d^{0-1} 6s^2$
• Actinoids: $[Rn] 5f^{0-14} 6d^{0-1} 7s^2$

Step 2: Comparative Analysis of Orbital Filling

The primary difference in the electronic configuration arises from the energy gap between the orbitals being filled:

Step 3: Visualizing Energy Distribution

Step 4: Logical Justification of Observed Irregularities

The electronic configuration of actinoids is less regular than that of lanthanoids due to the following reasons:

1. Energy Equivalence: In the actinoid series, the $5f$, $6d$, and $7s$ subshells possess comparable energies [Per the Aufbau Principle under relativistic effects]. This allows for the promotion of electrons into various subshells.
2. Relativistic Effects: As the nuclear charge ($Z$) increases in the actinoid series ($Z=89-103$), relativistic effects become significant, causing the $7s$ orbital to contract and the $5f$ orbitals to expand.
3. Shielding Efficiency: The $5f$ electrons provide poor shielding compared to $4f$ electrons, leading to the "Actinoid Contraction," which influences the electronic arrangement of the subsequent elements in the period.

Final Solution: While both series involve the filling of $(n-2)f$ orbitals, lanthanoids demonstrate a more consistent $4f^n 6s^2$ configuration due to the localized nature of $4f$ orbitals. Conversely, actinoids exhibit varied electronic configurations because the $5f$, $6d$, and $7s$ subshells possess nearly identical energy levels, facilitating greater participation of $5f$ electrons in chemical bonding.

Step 1: Conversion of Chromite Ore to Sodium Chromate

The process begins with the fusion of powdered chromite ore ($FeCr_{2}O_{4}$) with sodium carbonate ($Na_{2}CO_{3}$) in the presence of excess air. This step facilitates the oxidation of the Chromium(III) ion to Chromium(VI).

Chemical Reaction:

$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$

[Justification: The iron (II) is oxidized to iron (III) oxide, and the chromium (III) is oxidized to the chromate ion ($CrO_{4}^{2-}$), which is soluble in water.]

Step 2: Acidification of Sodium Chromate to Sodium Dichromate

The resulting yellow solution of sodium chromate ($Na_{2}CrO_{4}$) is filtered and acidified with concentrated sulfuric acid ($H_{2}SO_{4}$). This shifts the equilibrium to form the orange-colored sodium dichromate ($Na_{2}Cr_{2}O_{7}$).

Chemical Reaction:

$2Na_{2}CrO_{4} + 2H_{2}SO_{4} \rightarrow Na_{2}Cr_{2}O_{7} + 2Na_{2}SO_{4} + H_{2}O$

[Justification: Per Le Chatelier's Principle, increasing the concentration of $H^{+}$ ions (acidification) promotes the dimerization of chromate ions to dichromate ions.]

Step 3: Conversion to Potassium Dichromate

Sodium dichromate is highly soluble and difficult to crystallize. To obtain potassium dichromate ($K_{2}Cr_{2}O_{7}$), the solution is treated with potassium chloride ($KCl$). Potassium dichromate is significantly less soluble than sodium dichromate at low temperatures and precipitates out upon cooling.

Chemical Reaction:

$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$

[Justification: Based on the difference in solubility products ($K_{sp}$), $K_{2}Cr_{2}O_{7}$ crystallizes preferentially from the solution as it cools, leaving $NaCl$ in the mother liquor.]

Visual Representation of the Structural Interconversion

Final Solution: The preparation of $K_{2}Cr_{2}O_{7}$ involves the fusion of chromite ore with $Na_{2}CO_{3}$ to form sodium chromate, subsequent acidification to produce sodium dichromate, and final double decomposition with $KCl$ to precipitate $K_{2}Cr_{2}O_{7}$.

Step 1: Establishing the Neutral Electronic Configuration of Lutetium ($Lu$)

Lutetium ($Lu$) is the final element of the lanthanoid series with an atomic number ($Z$) of $71$. To determine its configuration, we apply the Aufbau principle, filling orbitals in order of increasing energy ($1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p$).

The ground state configuration of neutral Lutetium ($Lu, Z=71$) is:

$[Xe] 4f^{14} 5d^1 6s^2$

[Per the Aufbau Principle and the stability of the fully filled $4f$ subshell].

Step 2: Ionization Process for $Lu^{2+}$

To form the $Lu^{2+}$ cation, we must remove two electrons from the neutral atom. According to the rules of ionization for transition and inner-transition metals, electrons are removed from the orbital with the highest principal quantum number ($n$) first.

• The outermost shell for Lutetium is $n=6$.
• There are two electrons in the $6s$ orbital.
• Removing these two electrons ($6s^2 \rightarrow 6s^0$) yields the $Lu^{2+}$ ion.

Step 3: Determination of the Resultant Configuration

Subtracting the two electrons from the $6s$ subshell:

Neutral $Lu$: $[Xe] 4f^{14} 5d^1 6s^2$

Subtract $2e^-$ from $6s$: $[Xe] 4f^{14} 5d^1$

Step 4: Summary Table of Shell Occupation

Final Solution: The electronic configuration of $Lu^{2+}$ is $[Xe] 4f^{14} 5d^1$.

1. Identification of Lanthanoids with Non-Standard Oxidation States

While the most stable and characteristic oxidation state of lanthanoids ($Ln$) is $+3$, certain elements exhibit stability in the $+4$ or $+2$ states due to the attainment of stable electronic configurations (empty, half-filled, or fully-filled $f$-orbitals).

2. Correlation with Electronic Configurations

The stability of these oxidation states is governed by the exchange energy associated with $f$-orbital occupancy [Per Hund’s Rule and Orbital Stability Theory].

• Cerium ($Ce$, $Z=58$): Electronic configuration $[Xe] 4f^1 5d^1 6s^2$. By losing four electrons, it achieves the stable noble gas configuration of Xenon ($4f^0$), which is highly stable [Ref: Energetic feasibility of noble gas core].
• Europium ($Eu$, $Z=63$): Electronic configuration $[Xe] 4f^7 6s^2$. By losing two electrons, it attains the $4f^7$ configuration. This half-filled $f$-subshell provides extra stability [Per Exchange Energy Maximization].
• Ytterbium ($Yb$, $Z=70$): Electronic configuration $[Xe] 4f^{14} 6s^2$. By losing two electrons, it achieves the $4f^{14}$ configuration. This completely filled $f$-subshell offers significant stability [Ref: Symmetry-based stability].
• Terbium ($Tb$, $Z=65$): Electronic configuration $[Xe] 4f^9 6s^2$. By losing four electrons, it reaches the $4f^7$ configuration, again benefiting from the stability of the half-filled subshell.

3. Visualizing Stability via Orbital Energy Levels

4. Summary of Analytical Findings

The deviation from the common $+3$ oxidation state is not random. It is systematically linked to the energy required to remove electrons versus the energy gained by reaching a stable $4f^0$, $4f^7$, or $4f^{14}$ configuration. Elements like $Ce^{4+}$ act as strong oxidizing agents to return to the $+3$ state, whereas $Eu^{2+}$ and $Yb^{2+}$ act as strong reducing agents to reach the stable $+3$ state.

Final Solution: The lanthanoids exhibiting $+4$ states include $Ce$, $Pr$, $Nd$, $Tb$, and $Dy$, primarily driven by the stability of $4f^0$ and $4f^7$ configurations. Those exhibiting $+2$ states include $Nd$, $Sm$, $Eu$, $Tm$, and $Yb$, driven by the stability of $4f^7$ and $4f^{14}$ configurations.

Step 1: Analyzing the First Transition Series

The first transition series (3d series) consists of elements from Scandium ($Sc, Z=21$) to Zinc ($Zn, Z=30$). The electronic configurations follow the general pattern $[Ar] 3d^{n} 4s^{2}$, with exceptions for Chromium ($3d^{5} 4s^{1}$) and Copper ($3d^{10} 4s^{1}$).

Step 2: Identifying the Metal

Among the elements of the first transition series, Copper ($Cu$) is the only element that exhibits a stable $+1$ oxidation state frequently ($Cu^{+}$).

Step 3: Electronic Configuration and Stability Analysis

The ground state electronic configuration of Copper is $[Ar] 3d^{10} 4s^{1}$.

When Copper loses one electron to form the $Cu^{+}$ ion, its configuration becomes:

$Cu^{+} = [Ar] 3d^{10}$

[Per the Aufbau Principle and Hund’s Rule, a completely filled d-subshell confers exceptional electronic stability to the ion.]

Step 4: Energetic Justification

The stability of an oxidation state in aqueous solution is determined by the balance between the Ionization Enthalpy required to remove electrons and the Hydration Enthalpy released when the ion is solvated.

• While $Cu^{+}$ has a stable $3d^{10}$ configuration, it is often unstable in aqueous solutions because it undergoes disproportionation: $2Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)$
• However, $Cu^{+}$ is frequently observed in solid-state compounds and in specific complexes where the high hydration energy of $Cu^{2+}$ (which usually favors $Cu^{2+}$ in water) is offset by lattice energy or specific ligand-field stabilization effects.

Step 5: Visual Representation of Electron Distribution

Summary Table of Oxidation States

Final Solution: Copper ($Cu$) is the metal that exhibits the $+1$ oxidation state most frequently because the loss of the single $4s$ electron results in a stable, completely filled $3d^{10}$ electronic configuration.

Step 1: Conversion of Chromite Ore to Sodium Chromate

The primary source of chromium is the mineral chromite ore ($FeCr_2O_4$). The ore is fused with sodium carbonate ($Na_2CO_3$) in the presence of free access to air (excess $O_2$). This process results in the formation of yellow sodium chromate ($Na_2CrO_4$).

Chemical Equation: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$ [Per the principle of oxidative roasting]

Step 2: Conversion of Sodium Chromate to Sodium Dichromate

The yellow solution of sodium chromate is filtered and acidified with concentrated sulphuric acid ($H_2SO_4$). This shifts the chemical equilibrium to produce orange-colored sodium dichromate ($Na_2Cr_2O_7$).

Chemical Equation: $2Na_2Cr_2O_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$ [By the principle of acid-base equilibrium]

Sodium sulphate ($Na_2SO_4$) crystallizes out due to its lower solubility, allowing the separation of $Na_2Cr_2O_7$.

Step 3: Conversion to Potassium Dichromate

Sodium dichromate is highly soluble and hygroscopic, making it unsuitable for certain laboratory applications. It is treated with potassium chloride ($KCl$) to precipitate potassium dichromate ($K_2Cr_2O_7$) via double displacement and fractional crystallization.

Chemical Equation: $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$ [By the principle of solubility product constant ($K_{sp}$) values]

Step 4: Effect of pH on Potassium Dichromate

In aqueous solution, there exists a pH-dependent equilibrium between the dichromate ion ($Cr_2O_7^{2-}$, orange) and the chromate ion ($CrO_4^{2-}$, yellow). The equilibrium is represented as follows:

$Cr_2O_7^{2-} (orange) + H_2O \rightleftharpoons 2CrO_4^{2-} (yellow) + 2H^+$

• Increasing pH (Addition of Base): Adding a base (e.g., $OH^-$ ions) removes $H^+$ ions from the solution. According to Le Chatelier’s Principle, the equilibrium shifts to the right to compensate for the loss of $H^+$, causing the solution to turn from orange to yellow as $Cr_2O_7^{2-}$ converts to $CrO_4^{2-}$.
• Decreasing pH (Addition of Acid): Adding an acid increases the concentration of $H^+$, shifting the equilibrium to the left, favoring the formation of the orange $Cr_2O_7^{2-}$ ion.

Final Solution: Potassium dichromate is produced from chromite ore via oxidative fusion to sodium chromate, acidification to sodium dichromate, and treatment with $KCl$. Increasing the pH of a $K_2Cr_2O_7$ solution shifts the equilibrium toward the formation of the yellow chromate ion ($CrO_4^{2-}$).

1. Electronic Configuration and Atomic Size

The first transition series (3d-series: $Sc$ to $Zn$) exhibits significant differences from the 4d and 5d series due to the lack of filled $f$-orbitals. In the 3d series, the $d$-electrons are more exposed to the periphery of the atom, whereas in the 4d and 5d series, the intervening $f$-orbitals provide poor shielding [Per Slater's Rules regarding Screening Effect].

Consequently, the atomic radii of the 4d and 5d series elements are nearly identical (the Lanthanoid Contraction), whereas the 3d series shows a distinct trend of decrease followed by an increase. This disparity impacts the metallic bonding strength and melting points.

2. Comparative Analysis of Stability and Enthalpies

The enthalpy of atomization for the 3d series is lower compared to the 4d and 5d series. This is attributed to the fact that $d-d$ overlap is much more effective in the 4d and 5d series than in the 3d series [Per Molecular Orbital Theory applied to Metallic Bonding].

3. Chemical Reactivity and Coordination Chemistry

Step 1: Oxidation States: In the 3d series, lower oxidation states are more stable, whereas, in the 4d and 5d series, higher oxidation states are more stable (e.g., $W(VI)$ is more stable than $Cr(VI)$). This is because the bond energies for transition metal-ligand bonds are higher for the 4d and 5d series due to better spatial extension of $4d/5d$ orbitals.

Step 2: Magnetic and Spectral Properties: The 3d series exhibits more pronounced crystal field splitting energies ($\Delta_o$). However, the spin-orbit coupling is significantly higher in the 4d and 5d series, leading to distinct magnetic behaviors not observed in the 3d series [Per Russell-Saunders Coupling].

4. Visual Representation of d-Orbital Extension

5. Conclusion

The fundamental differences arise from:

• The absence of $f$-orbitals in the 3d series, eliminating lanthanoid contraction effects.
• The higher effective nuclear charge and poorer shielding of $4f$ electrons in heavier series.
• Increased metal-metal bonding in the 4d/5d series due to higher spatial extension of $d$-orbitals, leading to higher melting points and higher oxidation state stability.

Final Solution: The elements of the first transition series differ from heavier series due to the lack of inner $f$-orbital occupancy, smaller atomic radii, lower enthalpy of atomization, and lower stability of higher oxidation states compared to the 4d and 5d series.

1. Theoretical Basis for Variability in Oxidation States

The variability of oxidation states in transition elements ($d$-block) versus non-transition elements ($p$-block) is dictated by the energy levels of the available valence orbitals and the mode of electron participation in bonding.

2. Transition Metals: The $(n-1)d$ and $ns$ Interplay

In transition metals, the energy gap between the $(n-1)d$ subshell and the $ns$ subshell is small. Consequently, electrons from both subshells are available for chemical bonding [Per the Aufbau Principle and Hund's Rule].

• Gradual Change: Oxidation states change in steps of 1 because of the sequential removal or participation of single electrons from the $d$-orbitals.
• Example: Manganese ($Mn$, atomic number $25$) exhibits states from $+2$ to $+7$ ($[Ar] 3d^5 4s^2$). It utilizes all $5d$ and $2s$ electrons to reach the $+7$ state in $KMnO_4$.

3. Non-Transition Metals: The Inert Pair Effect

In heavier $p$-block elements, the variability is governed by the Inert Pair Effect, where the $ns^2$ electron pair becomes reluctant to participate in bonding due to poor shielding by $d$ and $f$ orbitals [Per Slater’s Rules].

• Jump of Two: Oxidation states typically differ by units of 2. This corresponds to the presence or absence of the $ns^2$ electron pair participation.
• Example: Lead ($Pb$, group 14) exhibits states $+2$ and $+4$. In $PbCl_2$, the $6s^2$ pair is inert, whereas in $PbCl_4$, it participates in bonding.

4. Visualizing Energy Trends

5. Comparative Conclusion

The distinction lies fundamentally in the electronic structure. Transition metals show a continuity of oxidation states because of the energy degeneracy of $d$-electrons, allowing for a "sliding scale" of participation. In contrast, non-transition metals show "discrete jumps" due to the stability of the $ns^2$ configuration in heavier elements, which resists ionization or hybridization [Per Quantum Mechanical Shell Theory].

Final Solution: Transition metals display variable oxidation states differing by unity due to the small energy gap between $(n-1)d$ and $ns$ orbitals, while non-transition metals show states differing by two units, primarily attributed to the inert pair effect in higher periods.

Step 1: Data Synthesis and Theoretical Foundation

To analyze the stability and redox behavior of transition metals, we evaluate the standard electrode potential ($E^{\circ}$) values provided. A higher positive value for the $M^{3+}/M^{2+}$ couple indicates a greater tendency to reduce to the $M^{2+}$ state, whereas a more negative $E^{\circ}$ for the $M^{2+}/M$ couple indicates a greater tendency for the metal to undergo oxidation.

Step 2: Stability Analysis of $M^{3+}$ ions

The stability of the $M^{3+}$ oxidation state in an aqueous acidic solution is inversely proportional to the $E^{\circ} (M^{3+}/M^{2+})$ value. A more positive value signifies that the $M^{3+}$ ion is a strong oxidizing agent and is easily reduced to $M^{2+}$ [Per the Nernst Equation and reduction potential trends].

• For $Mn^{3+}$ ($E^{\circ} = +1.57$ V): This value is very high, indicating $Mn^{3+}$ is extremely unstable and is easily reduced to $Mn^{2+}$. This is due to the extra stability of the $d^5$ configuration of $Mn^{2+}$.
• For $Fe^{3+}$ ($E^{\circ} = +0.77$ V): This value is positive but lower than that of $Mn^{3+}$. Thus, $Fe^{3+}$ is more stable than $Mn^{3+}$ but acts as a moderate oxidizing agent.
• For $Cr^{3+}$ ($E^{\circ} = -0.41$ V): The negative value indicates that $Cr^{3+}$ is the most stable among the three, as it does not easily reduce to $Cr^{2+}$. In fact, $Cr^{2+}$ is a strong reducing agent and tends to oxidize to $Cr^{3+}$ ($d^3$ configuration, $t_{2g}^3$ stability).

Step 3: Ease of Oxidation of Metal to $M^{2+}$

The ease of oxidation of the metal ($M \rightarrow M^{2+} + 2e^-$) is determined by the $E^{\circ} (M^{2+}/M)$ value. A more negative (or less positive) value indicates a greater thermodynamic tendency for the metal to be oxidized [By the Electrochemical Series convention].

• Manganese ($E^{\circ} = -1.18$ V): Being the most negative, $Mn$ is the easiest to oxidize among the three.
• Chromium ($E^{\circ} = -0.91$ V): Also highly negative, indicating $Cr$ is very easily oxidized.
• Iron ($E^{\circ} = -0.44$ V): This value is significantly less negative than $Mn$ or $Cr$. Therefore, iron is the most difficult to oxidize among the three metals considered.

Step 4: Summary Comparison

Final Solution:
(i) The stability of $M^{3+}$ ions follows the order $Cr^{3+} > Fe^{3+} > Mn^{3+}$. $Mn^{3+}$ is the least stable due to the high $E^{\circ}$ value favoring reduction to the $d^5$ stable $Mn^{2+}$, while $Cr^{3+}$ is highly stable.
(ii) The ease of oxidation of the metals follows the order $Mn > Cr > Fe$. Consequently, iron is the least susceptible to oxidation among the three metals.

Definition of Alloys

An alloy is a homogeneous mixture or a solid solution composed of two or more metallic elements, or a metallic element combined with non-metallic elements. The properties of the resulting alloy, such as hardness, corrosion resistance, or ductility, are often superior to those of the pure base metal [Per the Hume-Rothery rules for solid solution formation]. Alloys are formed by melting the components together and allowing the mixture to cool and solidify, creating a uniform distribution of atoms within the metallic lattice.

Identification of Mischmetal

An industrially significant alloy containing lanthanoid metals is known as Mischmetal. Its composition is as follows:

Technical Schematic of Mischmetal Application

Mischmetal is primarily valued for its pyrophoric property, which is utilized in the manufacturing of flint for lighters and ignition devices.

Uses of Mischmetal

• Pyrophoric Alloys: Because of the high reactivity of lanthanoids, Mischmetal produces sparks when struck, making it indispensable for cigarette lighters and gas torches [Justified by the high heat of combustion of Lanthanoid oxides].
• Metallurgical Applications: It is added to steel to produce specialized alloys. Specifically, it is used to improve the workability and heat resistance of iron-based alloys.
• Desulfurization and Deoxidation: The lanthanoids in Mischmetal act as scavengers for oxygen and sulfur in molten metals, effectively cleaning the metal melt during production.
• Bullet Manufacturing: It is used in the production of bullets, shells, and lighter flints to improve mechanical performance under high-impact conditions.

Final Solution: An alloy is a solid solution of metals. The most significant lanthanoid-containing alloy is Mischmetal, which consists of approximately 95% lanthanoid metals and 5% iron. It is primarily used in the production of pyrophoric materials like lighter flints and as a metallurgical additive to improve the quality of steel and other alloys.