The d-and f-Block Elements — NCERT Solutions

Given: Silver (Ag, Z = 47) has ground state electronic configuration [Kr] $4d^{10}$ $5s^1$.

Concept: A transition element is defined as one which has an incompletely filled $d$ orbital in its ground state OR in any of its commonly occurring oxidation states.

Working:
Although Ag has completely filled $4d^{10}$ orbitals in its ground state, it can exhibit a +2 oxidation state (Ag²⁺). In the +2 state, the electronic configuration becomes [Kr] $4d^9$, which has an incompletely filled $d$ orbital.

Conclusion: Since Ag can exist in the +2 oxidation state with an incompletely filled $d$ subshell, it qualifies as a transition element.

Given: Zn has the lowest enthalpy of atomisation (126 kJ mol⁻¹) in the 3d series.

Concept: Enthalpy of atomisation depends on the strength of metallic bonding, which in turn depends on the number of electrons available for metallic bonding (including $d$ electrons).

Explanation:
Zinc has the electronic configuration [Ar] $3d^{10}$ $4s^2$. Its $3d$ orbitals are completely filled. Therefore, no electrons from the $3d$ orbitals are available for metallic bonding — only the two $4s$ electrons participate.

In all other metals of the 3d series (Sc to Cu), electrons from the $d$ orbitals are also involved in metallic bonding, resulting in stronger interatomic interactions and higher enthalpies of atomisation.

Conclusion: Since Zn uses only $4s$ electrons for metallic bonding (no $3d$ electron involvement), its interatomic interaction is the weakest, giving it the lowest enthalpy of atomisation.

Answer: Manganese (Mn, Z = 25) exhibits the largest number of oxidation states.

Electronic configuration of Mn: [Ar] $3d^5$ $4s^2$

Reason:
Manganese has the maximum number of unpaired electrons in its atom (five unpaired $3d$ electrons + two $4s$ electrons = 7 electrons available for bonding). It can therefore lose electrons ranging from 1 to 7, giving oxidation states from +1 to +7 (i.e., +2, +3, +4, +5, +6, +7 are all known).

The elements at the beginning of the series (Sc, Ti) have too few $d$ electrons to show many oxidation states, while elements at the end (Cu, Zn) have too many filled $d$ electrons. Mn, being near the middle with a half-filled $3d^5$ configuration, can exhibit the widest range of oxidation states.

Given: $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}$ (positive value).

Concept: The standard electrode potential $E^\circ$ for $\text{M}^{2+}/\text{M}$ depends on:

1. Enthalpy of atomisation ($\Delta_a H^\circ$) — energy required to convert solid metal to gaseous atoms.
2. Sum of first and second ionisation enthalpies ($\Delta_i H_1 + \Delta_i H_2$) — energy to form $\text{M}^{2+}$.
3. Hydration enthalpy ($\Delta_{\text{hyd}} H^\circ$) — energy released when $\text{M}^{2+}$ is hydrated.

Explanation:
Copper has a high enthalpy of atomisation and relatively high sum of ionisation enthalpies. The energy required to transform Cu(s) to Cu²⁺(aq) is very large. This large energy input is NOT sufficiently compensated by the hydration enthalpy of Cu²⁺, which is comparatively less negative (low $\Delta_{\text{hyd}}H^\circ$).

As a result, the overall process $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2e^-$ is energetically unfavourable, making $E^\circ$ positive.

Consequence: A positive $E^\circ$ means Cu cannot liberate $\text{H}_2$ from dilute acids; only oxidising acids (like $\text{HNO}_3$ or hot conc. $\text{H}_2\text{SO}_4$) react with Cu.

Concept: Ionisation enthalpy depends on nuclear charge, atomic radius, and the stability of the electronic configuration.

Explanation of irregular variation:

1. General trend: As we move from Sc to Zn, nuclear charge increases, so ionisation enthalpies generally increase. However, the increase is not regular.

2. Shielding by 3d electrons: As electrons are added to the inner $3d$ orbitals, they shield the outer $4s$ electrons from the increasing nuclear charge more effectively than outer electrons shield each other. This reduces the rate of increase of ionisation enthalpy.

3. Stability of specific configurations: Certain $d$-configurations are exceptionally stable:

• $d^0$ (empty), $d^5$ (half-filled), $d^{10}$ (completely filled) configurations have extra stability.
• For example, Mn has $3d^5$ $4s^2$ configuration. Its first ionisation enthalpy is higher than expected because removing an electron disturbs the stable half-filled $3d^5$ arrangement.
• Similarly, Zn ($3d^{10}$ $4s^2$) has a higher ionisation enthalpy due to the stability of completely filled $d$ orbitals.

4. Second ionisation enthalpy: The second ionisation enthalpy shows a break at Mn²⁺ (d⁵) and Fe³⁺ (d⁵) because these ions have extra stable half-filled $d^5$ configurations, making removal of the next electron more difficult.

Conclusion: The irregular variation is mainly due to the varying degree of stability of different $3d$ configurations ($d^0$, $d^3$, $d^5$, $d^{10}$ are exceptionally stable), which affects the ease of electron removal.

Concept: The ability of a non-metal to oxidise the transition metal to its highest oxidation state depends on the electronegativity and size of the non-metal.

Explanation:
Oxygen and fluorine are the two most electronegative elements with very small atomic sizes. Due to these properties:

1. High electronegativity: Both O and F can attract electrons strongly from the metal, thereby oxidising it to its highest possible oxidation state.

2. Small size: Their small size allows them to accommodate around the metal ion and form stable compounds even with metals in very high oxidation states.

3. Multiple bonding (for oxygen): Oxygen can form multiple bonds (double bonds) with metals (e.g., in $\text{Mn}_2\text{O}_7$, $\text{CrO}_3$), which further stabilises the high oxidation state. This ability exceeds that of fluorine.

Example: The highest Mn fluoride is $\text{MnF}_4$ (Mn in +4 state), whereas the highest oxide is $\text{Mn}_2\text{O}_7$ (Mn in +7 state).

Conclusion: Because of their small size and high electronegativity, oxygen and fluorine can oxidise metals to their highest oxidation states, which other non-metals cannot achieve.

Answer: $\text{Cr}^{2+}$ is a stronger reducing agent than $\text{Fe}^{2+}$.

Electronic configurations:

• $\text{Cr}^{2+}$: [Ar] $3d^4$ — oxidised to $\text{Cr}^{3+}$: [Ar] $3d^3$
• $\text{Fe}^{2+}$: [Ar] $3d^6$ — oxidised to $\text{Fe}^{3+}$: [Ar] $3d^5$

Reason:
When $\text{Cr}^{2+}$ is oxidised to $\text{Cr}^{3+}$, the configuration changes from $d^4$ to $d^3$. The $d^3$ configuration has a half-filled $t_{2g}$ level (in octahedral field), which is particularly stable. This extra stability of the product ($\text{Cr}^{3+}$) drives the oxidation of $\text{Cr}^{2+}$ readily, making it a strong reducing agent.

In contrast, when $\text{Fe}^{2+}$ ($d^6$) is oxidised to $\text{Fe}^{3+}$ ($d^5$), the product has a half-filled $d^5$ configuration which is stable, but the driving force is less compared to $\text{Cr}^{2+} \rightarrow \text{Cr}^{3+}$ because $d^3$ (half-filled $t_{2g}$) provides greater crystal field stabilisation energy.

Conclusion: $\text{Cr}^{2+}$ is a stronger reducing agent because its oxidation to $\text{Cr}^{3+}$ ($d^3$) is more favourable than the oxidation of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ ($d^5$).

Given: $Z = 27$ is Cobalt (Co). The ion is $\text{Co}^{2+}$.

Step 1: Electronic configuration of Co (Z = 27):
$$\text{Co}: [\text{Ar}]\ 3d^7\ 4s^2$$

Step 2: Electronic configuration of Co²⁺:
Remove 2 electrons from $4s$ first:
$$\text{Co}^{2+}: [\text{Ar}]\ 3d^7$$

Step 3: Find number of unpaired electrons in $3d^7$:
Filling $3d$ orbitals using Hund's rule:
$$\underbrace{\uparrow\downarrow}_{} \underbrace{\uparrow\downarrow}_{} \underbrace{\uparrow}_{} \underbrace{\uparrow}_{} \underbrace{\uparrow}_{}$$
Number of unpaired electrons, $n = 3$

Step 4: Apply spin-only formula:
$$\mu = \sqrt{n(n+2)}\ \text{BM}$$
$$\mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15}$$
$$\boxed{\mu \approx 3.87\ \text{BM}}$$

Concept: Lanthanoid contraction and actinoid contraction both arise due to imperfect shielding of one electron by another in the same subshell ($4f$ and $5f$ respectively).

Explanation:

1. In lanthanoids, $4f$ electrons are added successively. The $4f$ orbitals are deeply buried inside the atom and are well shielded by the outer electrons. However, $4f$ electrons themselves provide poor shielding to each other from the nuclear charge.

2. In actinoids, $5f$ electrons are added. The $5f$ orbitals are more diffuse (larger and more extended in space) compared to $4f$ orbitals. As a result, $5f$ electrons are even less effective at shielding each other from the nuclear charge than $4f$ electrons.

3. Because of poorer shielding by $5f$ electrons, the effective nuclear charge experienced by the outer electrons increases more sharply from one actinoid to the next compared to lanthanoids.

Conclusion: Since $5f$ electrons provide poorer shielding than $4f$ electrons, the increase in effective nuclear charge per element is greater in actinoids, leading to a greater contraction in atomic/ionic size from element to element in the actinoid series compared to the lanthanoid series.

Method: Write the ground state configuration of the neutral atom, then remove electrons starting from the outermost shell ($ns$ before $(n-1)d$; $6s$ before $4f$ for lanthanoids/actinoids).

(i) $\text{Cr}^{3+}$ (Z = 24):
Neutral Cr: [Ar] $3d^5$ $4s^1$
Remove 3 electrons (1 from $4s$, 2 from $3d$):
$$\text{Cr}^{3+}: [\text{Ar}]\ 3d^3$$

(ii) $\text{Pm}^{3+}$ (Z = 61):
Neutral Pm: [Xe] $4f^5$ $5d^0$ $6s^2$
Remove 3 electrons (2 from $6s$, 1 from $4f$):
$$\text{Pm}^{3+}: [\text{Xe}]\ 4f^4$$

(iii) $\text{Cu}^+$ (Z = 29):
Neutral Cu: [Ar] $3d^{10}$ $4s^1$
Remove 1 electron (from $4s$):
$$\text{Cu}^+: [\text{Ar}]\ 3d^{10}$$

(iv) $\text{Ce}^{4+}$ (Z = 58):
Neutral Ce: [Xe] $4f^1$ $5d^1$ $6s^2$
Remove 4 electrons (2 from $6s$, 1 from $5d$, 1 from $4f$):
$$\text{Ce}^{4+}: [\text{Xe}]$$
(i.e., same as Xenon core — no $4f$ electrons)

(v) $\text{Co}^{2+}$ (Z = 27):
Neutral Co: [Ar] $3d^7$ $4s^2$
Remove 2 electrons (from $4s$):
$$\text{Co}^{2+}: [\text{Ar}]\ 3d^7$$

(vi) $\text{Lu}^{2+}$ (Z = 71):
Neutral Lu: [Xe] $4f^{14}$ $5d^1$ $6s^2$
Remove 2 electrons (from $6s$):
$$\text{Lu}^{2+}: [\text{Xe}]\ 4f^{14}\ 5d^1$$

(vii) $\text{Mn}^{2+}$ (Z = 25):
Neutral Mn: [Ar] $3d^5$ $4s^2$
Remove 2 electrons (from $4s$):
$$\text{Mn}^{2+}: [\text{Ar}]\ 3d^5$$

(viii) $\text{Th}^{4+}$ (Z = 90):
Neutral Th: [Rn] $6d^2$ $7s^2$
Remove 4 electrons (2 from $7s$, 2 from $6d$):
$$\text{Th}^{4+}: [\text{Rn}]$$
(same as Radon core — no $5f$ or $6d$ electrons)

Electronic configurations:

• $\text{Mn}^{2+}$: [Ar] $3d^5$ (half-filled $d$ subshell)
• $\text{Fe}^{2+}$: [Ar] $3d^6$

Explanation:
$\text{Mn}^{2+}$ has a $3d^5$ configuration, which is a half-filled $d$ subshell. Half-filled subshells have extra stability due to:

1. Symmetrical distribution of electrons
2. Maximum exchange energy

To oxidise $\text{Mn}^{2+}$ to $\text{Mn}^{3+}$, one electron must be removed from this stable half-filled $3d^5$ configuration, giving $3d^4$. This requires a large amount of energy, making the oxidation difficult.

In contrast, $\text{Fe}^{2+}$ has $3d^6$ configuration. Oxidation to $\text{Fe}^{3+}$ gives $3d^5$ (half-filled, extra stable). The product $\text{Fe}^{3+}$ is more stable than the reactant $\text{Fe}^{2+}$, so the oxidation is thermodynamically favourable.

Conclusion: $\text{Mn}^{2+}$ (with stable $d^5$) resists oxidation, while $\text{Fe}^{2+}$ is readily oxidised to $\text{Fe}^{3+}$ (stable $d^5$). Hence $\text{Mn}^{2+}$ compounds are more stable than $\text{Fe}^{2+}$ towards oxidation.

Elements considered: Sc (Z=21) to Mn (Z=25) — first half of 3d series.

Electronic configurations of M²⁺ ions:

Explanation:
As we move from Sc to Mn, the number of $d$ electrons in the M²⁺ ion increases from $d^1$ to $d^5$. The $d^5$ configuration (half-filled) of Mn²⁺ is exceptionally stable due to symmetrical electron distribution and maximum exchange energy.

The successive M²⁺ ions become progressively more stable because:

1. Nuclear charge increases, holding the $d$ electrons more firmly.
2. The $d$ electron count increases towards the stable half-filled $d^5$ configuration.
3. The third ionisation enthalpy (required to go from +2 to +3 state) increases across the series, making it harder to oxidise M²⁺ further.

Conclusion: The +2 state becomes increasingly stable from Sc to Mn because the $3d^n$ configuration of M²⁺ becomes progressively more stable, culminating in the extra-stable half-filled $3d^5$ of Mn²⁺.

Concept: The stability of an oxidation state is strongly influenced by the electronic configuration of the resulting ion, particularly the stability of $d^0$, $d^5$, and $d^{10}$ configurations.

Key configurations and their stability:

1. $d^0$ configuration (empty $d$ subshell): Very stable. Example: $\text{Sc}^{3+}$ ([Ar] $3d^0$) — Sc almost exclusively shows +3 state.

2. $d^5$ configuration (half-filled $d$ subshell): Extra stable due to maximum exchange energy and symmetrical distribution. Examples:

• $\text{Mn}^{2+}$ ([Ar] $3d^5$) — very stable; $\text{Mn}^{2+}$ compounds resist oxidation to $\text{Mn}^{3+}$.
• $\text{Fe}^{3+}$ ([Ar] $3d^5$) — stable; $\text{Fe}^{2+}$ is readily oxidised to $\text{Fe}^{3+}$.

3. $d^{10}$ configuration (completely filled $d$ subshell): Extra stable. Example: $\text{Cu}^+$ ([Ar] $3d^{10}$) — Cu⁺ is stable in solid state (e.g., $\text{Cu}_2\text{O}$, $\text{CuCl}$); $\text{Zn}^{2+}$ ([Ar] $3d^{10}$) — Zn exclusively shows +2 state.

4. $d^3$ configuration: Relatively stable (half-filled $t_{2g}$ in octahedral field). Example: $\text{Cr}^{3+}$ ([Ar] $3d^3$) — very stable; Cr³⁺ is the most common and stable state of chromium.

Limitations: Electronic configuration alone does not fully determine stability. Other factors like hydration enthalpy, lattice energy, and ionisation enthalpy also play important roles. For example, $\text{Cu}^{2+}$ is more stable in aqueous solution than $\text{Cu}^+$ despite $\text{Cu}^+$ having the stable $d^{10}$ configuration, because the much higher hydration enthalpy of $\text{Cu}^{2+}$ compensates for the higher ionisation energy.

Conclusion: Electronic configurations play a major but not exclusive role in deciding the stability of oxidation states.

Method: The ground state configuration of the atom includes $ns^2$ electrons. The stable oxidation state corresponds to the ion with a stable $d$ configuration.

(i) Atom with $3d^3$ configuration:
Ground state of atom: [Ar] $3d^3$ $4s^2$ → This is Vanadium (V, Z=23).
Losing 2 electrons ($4s^2$) gives V²⁺: $3d^3$ — but $3d^3$ is relatively stable.
Losing 3 electrons gives V³⁺: $3d^2$.
Losing 5 electrons gives V⁵⁺: $3d^0$ — also stable.
Most stable oxidation state: +3 (V³⁺ with $3d^2$) or +5 (common for V).
Note: The element is V and its most characteristic stable state is +3.

(ii) Atom with $3d^5$ configuration:
Ground state: [Ar] $3d^5$ $4s^2$ → This is Manganese (Mn, Z=25).
Losing 2 electrons gives Mn²⁺: $3d^5$ — half-filled, extra stable.
Most stable oxidation state: +2 (Mn²⁺ with $3d^5$).

(iii) Atom with $3d^8$ configuration:
Ground state: [Ar] $3d^8$ $4s^2$ → This is Nickel (Ni, Z=28).
Losing 2 electrons gives Ni²⁺: $3d^8$.
Most stable oxidation state: +2 (Ni²⁺ with $3d^8$).

(iv) Atom with $3d^9$ configuration:
Ground state: [Ar] $3d^9$ $4s^2$ → This is Copper (Cu, Z=29) — but Cu actually has $3d^{10}$ $4s^1$. If we take the configuration as given ($3d^9$ $4s^2$), losing 2 electrons gives M²⁺: $3d^9$.
Most stable oxidation state: +2 (Cu²⁺ with $3d^9$).

Summary:

• $3d^3$: +3
• $3d^5$: +2
• $3d^8$: +2
• $3d^9$: +2

Concept: The oxidation state of the metal in the oxoanion equals its group number.

Identification:

Verification:

• In $\text{VO}_4^{3-}$: Let V = $x$; $x + 4(-2) = -3 \Rightarrow x = +5$ ✓ (Group 5)
• In $\text{CrO}_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$ ✓ (Group 6)
• In $\text{MnO}_4^-$: $x + 4(-2) = -1 \Rightarrow x = +7$ ✓ (Group 7)

Answer: The oxometal anions are vanadate ($\text{VO}_4^{3-}$), chromate/dichromate ($\text{CrO}_4^{2-}$/$\text{Cr}_2\text{O}_7^{2-}$), and permanganate ($\text{MnO}_4^-$).

Definition of Lanthanoid Contraction:
The gradual decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number (from La to Lu) is called lanthanoid contraction.

Cause:
As we move from La (Z=57) to Lu (Z=71), electrons are successively added to the inner $4f$ orbitals. The $4f$ electrons have poor shielding effect on each other (due to the shape of $f$ orbitals). Therefore, the effective nuclear charge experienced by the outer electrons increases progressively, causing a gradual contraction in size.

Consequences of Lanthanoid Contraction:

1. Similarity in properties of 4d and 5d transition metals:

Due to lanthanoid contraction, the elements of the second (4d) and third (5d) transition series in the same group have nearly identical atomic radii and ionic radii. For example:

• Zr (4d) and Hf (5d) have almost the same radius (~160 pm).
• This makes their separation very difficult.

2. Difficulty in separation of lanthanoids:

The lanthanoid elements themselves have very similar sizes and chemical properties, making their separation from each other extremely difficult. Techniques like ion-exchange chromatography are required.

3. Basicity of lanthanoid hydroxides:

As the ionic radius decreases from La to Lu, the basic character of the hydroxides $\text{Ln(OH)}_3$ decreases. $\text{La(OH)}_3$ is the most basic and $\text{Lu(OH)}_3$ is the least basic among lanthanoid hydroxides.

4. Effect on post-lanthanoid elements:

The elements following the lanthanoids (Hf, Ta, W, etc.) have smaller radii than expected due to lanthanoid contraction, which affects their chemical properties.

Why called Transition Elements:
Transition elements are those elements which have incompletely filled $d$ orbitals in their ground state or in any of their commonly occurring oxidation states. They are called 'transition' elements because they occupy a transitional position between the highly electropositive $s$-block metals and the less electropositive $p$-block metals in the periodic table.

Characteristics of Transition Elements:

1. Variable oxidation states: They exhibit multiple oxidation states due to the availability of both $ns$ and $(n-1)d$ electrons for bonding.

2. Formation of coloured ions: Most transition metal ions are coloured in solution due to $d$–$d$ electronic transitions.

3. Paramagnetic behaviour: Most transition metals and their compounds are paramagnetic due to the presence of unpaired $d$ electrons.

4. High melting and boiling points: Due to strong metallic bonding involving $d$ electrons.

5. Formation of complex compounds: Transition metals have small ionic size, high charge, and available empty $d$ orbitals, enabling complex formation.

6. Catalytic properties: Many transition metals and their compounds act as catalysts (e.g., Fe in Haber process, V₂O₅ in Contact process).

7. Formation of interstitial compounds: Small atoms like H, C, N can occupy interstitial sites in the metallic lattice.

8. Alloy formation: Transition metals readily form alloys with each other due to similar atomic sizes.

9. High enthalpies of atomisation: Due to strong interatomic bonding.

$d$-Block Elements NOT Regarded as Transition Elements:
Zinc (Zn, $3d^{10}$ $4s^2$), Cadmium (Cd, $4d^{10}$ $5s^2$), and Mercury (Hg, $5d^{10}$ $6s^2$) are NOT transition elements because they have completely filled $d$ orbitals in both their ground state and in their common oxidation state (+2), giving $d^{10}$ configuration in all cases. They do not satisfy the definition of transition elements.

Electronic Configuration of Transition Elements:
Transition elements (d-block) have the general electronic configuration:
$$[\text{Noble gas}]\ (n-1)d^{1-10}\ ns^{0-2}$$
The distinguishing feature is the presence of partially filled $(n-1)d$ orbitals (in ground state or in common oxidation states).

Electronic Configuration of Non-Transition Elements:

• $s$-block elements: $[\text{Noble gas}]\ ns^{1-2}$ (outermost $s$ orbital being filled)
• $p$-block elements: $[\text{Noble gas}]\ ns^2\ np^{1-6}$ (outermost $p$ orbital being filled)

Key Differences:

Example:

• Fe (transition): [Ar] $3d^6$ $4s^2$ — inner $3d$ being filled
• Ca (non-transition, $s$-block): [Ar] $4s^2$ — outermost $4s$ being filled
• Cl (non-transition, $p$-block): [Ne] $3s^2$ $3p^5$ — outermost $3p$ being filled

Principal Oxidation State:
The most common and characteristic oxidation state of all lanthanoids is +3.

This is because the sum of the first three ionisation enthalpies is low enough to be compensated by the lattice energy or hydration energy, making the +3 state stable for all lanthanoids.

Other Oxidation States:

• +4 oxidation state: Exhibited by Ce (Z=58), Pr (Z=59), Nd (Z=60), and Tb (Z=65).
• Ce⁴⁺ is the most stable +4 ion (configuration becomes [Xe] — noble gas configuration).

• +2 oxidation state: Exhibited by Sm (Z=62), Eu (Z=63), and Yb (Z=70).
• Eu²⁺ is relatively stable (configuration: [Xe] $4f^7$ — half-filled).
• Yb²⁺ is stable (configuration: [Xe] $4f^{14}$ — completely filled).

Summary:
$$\text{Lanthanoids: Oxidation states} = +2,\ +3\ (\text{most common}),\ +4$$

The +4 and +2 states are exhibited only by a few lanthanoids where the resulting $4f$ configuration is particularly stable ($4f^0$, $4f^7$, or $4f^{14}$).

(i) Paramagnetic behaviour:

Reason: Paramagnetism arises due to the presence of unpaired electrons. Transition metals have incompletely filled $d$ orbitals containing one or more unpaired electrons. When placed in a magnetic field, these unpaired electrons (with their spin magnetic moments) are attracted towards the field, causing paramagnetic behaviour.

The magnetic moment is given by: $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons.

Example: Fe²⁺ ($3d^6$) has 4 unpaired electrons and is paramagnetic.

(ii) High enthalpies of atomisation:

Reason: Enthalpy of atomisation is a measure of the strength of metallic bonding. In transition metals, both the outer $ns$ electrons AND the inner $(n-1)d$ electrons participate in metallic bonding. The large number of unpaired $d$ electrons leads to stronger interatomic interactions (stronger metallic bonds).

The maxima in enthalpies of atomisation occur near the middle of each transition series (around $d^5$), where the number of unpaired electrons is maximum, confirming that unpaired $d$ electrons contribute significantly to metallic bonding.

(iii) Formation of coloured compounds:

Reason: Transition metal ions have incompletely filled $d$ orbitals. When ligands (like water) surround the metal ion, the $d$ orbitals split into two sets of different energies (crystal field splitting). An electron from the lower energy $d$ orbital can be excited to the higher energy $d$ orbital by absorbing visible light ($d$–$d$ transition). The colour observed is the complementary colour of the absorbed light.

Example: $\text{Cu}^{2+}$ ($3d^9$) absorbs red light and appears blue.

Note: Ions with $d^0$ (Sc³⁺) or $d^{10}$ (Zn²⁺) configurations are colourless as no $d$–$d$ transition is possible.

(iv) Catalytic properties:

Reason: Transition metals act as good catalysts due to:

1. Variable oxidation states: They can form intermediate compounds with reactants by changing their oxidation state, providing an alternative reaction pathway with lower activation energy.
2. Ability to adsorb reactants: Transition metals have the ability to adsorb reactant molecules on their surface (due to partially filled $d$ orbitals), bringing them close together and facilitating the reaction (heterogeneous catalysis).

Examples: Fe in Haber process, $\text{V}_2\text{O}_5$ in Contact process, Ni in hydrogenation of oils.

Definition of Interstitial Compounds:
Interstitial compounds are those formed when small atoms such as hydrogen (H), carbon (C), nitrogen (N), or boron (B) are trapped inside the voids (interstitial sites) of the metallic crystal lattice of transition metals without displacing any metal atoms.

Examples: TiC, Mn₄N, Fe₃H, TiH₂, VH₀.₅₆

Properties of Interstitial Compounds:

1. They are chemically inert compared to the parent metal.
2. They have high melting points, higher than the pure metal.
3. They are very hard (e.g., steel is harder than iron due to interstitial carbon).
4. They retain metallic conductivity.
5. They are non-stoichiometric in nature.

Why well known for transition metals:
Transition metals form interstitial compounds readily because:

1. Suitable atomic radii: Transition metals have large atomic radii with interstitial voids of appropriate size to accommodate small atoms like H, C, N, and B.
2. Partially filled $d$ orbitals: The empty or partially filled $d$ orbitals of transition metals can interact with the electrons of the small atoms (H, C, N), providing some bonding character that stabilises the interstitial compound.
3. Metallic crystal structure: Transition metals have close-packed metallic structures (bcc, hcp, ccp) with well-defined interstitial sites.

Non-transition metals generally do not form such compounds because their atomic sizes and electronic structures are not suitable.

Variability in Oxidation States — Comparison:

Transition Metals:

1. Transition metals show a wide range of oxidation states because both $ns$ and $(n-1)d$ electrons are available for bonding.
2. The oxidation states differ by unity (i.e., by 1).
3. Example: Vanadium shows +2, +3, +4, +5 oxidation states (differ by 1).
4. Example: Manganese shows +2, +3, +4, +5, +6, +7 oxidation states.
5. The variable oxidation states arise from the involvement of $d$ electrons, which have similar energies to $s$ electrons.

Non-Transition Metals:

1. Non-transition metals (mainly $p$-block) also show variable oxidation states, but the states differ by two (due to the inert pair effect).
2. Example: Tin (Sn) shows +2 and +4 oxidation states (differ by 2).
3. Example: Lead (Pb) shows +2 and +4 oxidation states.
4. Example: Sulphur shows +2, +4, +6 oxidation states (differ by 2).
5. The variability arises from the involvement of $ns^2$ pair (inert pair effect) in heavier $p$-block elements.

Key Difference:
$$\text{Transition metals: oxidation states differ by 1 (e.g., V: +2, +3, +4, +5)}$$
$$\text{Non-transition metals: oxidation states differ by 2 (e.g., Sn: +2, +4)}$$

Preparation of Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$):

Step 1: Fusion of chromite ore with alkali in air
Chromite ore ($\text{FeCr}_2\text{O}_4$) is fused with sodium carbonate ($\text{Na}_2\text{CO}_3$) in the presence of air (oxygen) at high temperature:
$$4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \xrightarrow{\Delta} 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$$
Yellow sodium chromate ($\text{Na}_2\text{CrO}_4$) is formed.

Step 2: Filtration and acidification
The yellow solution of sodium chromate is filtered to remove $\text{Fe}_2\text{O}_3$ and then acidified with dilute sulphuric acid:
$$2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}$$
Orange sodium dichromate ($\text{Na}_2\text{Cr}_2\text{O}_7 \cdot 2\text{H}_2\text{O}$) crystallises out.

Step 3: Conversion to potassium dichromate
Sodium dichromate is treated with potassium chloride:
$$\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}$$
Potassium dichromate (less soluble) crystallises out as orange crystals.

Effect of Increasing pH on Potassium Dichromate Solution:

In solution, dichromate ($\text{Cr}_2\text{O}_7^{2-}$, orange) and chromate ($\text{CrO}_4^{2-}$, yellow) ions exist in equilibrium:
$$\text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \rightleftharpoons 2\text{CrO}_4^{2-} + 2\text{H}^+$$

On increasing pH (adding alkali, i.e., decreasing $[\text{H}^+]$):

• The equilibrium shifts to the right (towards chromate).
• The orange colour of dichromate changes to yellow colour of chromate.

On decreasing pH (adding acid):

• The equilibrium shifts to the left (towards dichromate).
• The yellow colour changes back to orange.

Oxidising Action of Potassium Dichromate:

In acidic medium, the dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) acts as a strong oxidising agent. Chromium is reduced from +6 to +3 state:
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

(i) Reaction with iodide ions ($\text{I}^-$):

Iodide is oxidised to iodine ($\text{I}^-$ → $\text{I}_2$, change: $-1$ to $0$, loses 1e⁻ per I):
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{I}_2$$

(ii) Reaction with iron(II) solution ($\text{Fe}^{2+}$):

Fe²⁺ is oxidised to Fe³⁺ (loses 1e⁻ per Fe):
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$$

(iii) Reaction with $\text{H}_2\text{S}$:

$\text{H}_2\text{S}$ is oxidised to sulphur (S is oxidised from $-2$ to $0$, loses 2e⁻ per S):
$$\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S}$$