Correct Option: (d)

Explanation: A transition element is defined as one which has incompletely filled (partially filled) $d$-subshell in its ground state or in any of its common oxidation states. Zinc ($\text{Zn}$) has the electronic configuration:

Because the $3d$ subshell remains completely filled ($d^{10}$) in both its elemental form and its only stable ionic state, zinc (along with cadmium, $\text{Cd}$, and mercury, $\text{Hg}$) is not classified as a transition element.

Correct Option: (b)

Explanation: Manganese ($\text{Mn}$, $Z = 25$) has the outer electronic configuration of $3d^5 4s^2$. Because it contains the maximum number of unpaired electrons ($5$) in its $3d$ orbitals in addition to the $2$ electrons in the $4s$ orbital, it can share up to $7$ electrons for bonding. It shows a complete range of variable oxidation states from $+2$ to $+7$ (e.g., in $\text{MnO}_4^{-}$).

Reason: The variability of oxidation states in transition elements is due to the participation of both **$(n-1)d$** and **$ns$** subshell electrons in chemical bonding.

This is possible because the energy difference between $(n-1)d$ and $ns$ orbitals is extremely small. Initially, only $ns$ electrons are lost (showing $+1$ or $+2$ states), but as the reaction conditions change, $(n-1)d$ electrons also take part in bonding step-by-step, resulting in a wide array of stable oxidation states.

(i) Chromium Ion ($\text{Cr}^{3+}$):

The ground state configuration of neutral chromium (anomalous half-filled stability) is:

To form $\text{Cr}^{3+}$, we remove 1 electron from the $4s$ orbital and 2 electrons from the $3d$ orbital:

(ii) Copper (I) Ion ($\text{Cu}^{+}$):

The ground state configuration of neutral copper (anomalous fully filled stability) is:

To form $\text{Cu}^{+}$, we remove 1 electron from the outermost $4s$ orbital:

Correct Option: (a)

Explanation: The catalytic properties of transition metals are highly pronounced because of their ability to show **variable oxidation states** and **provide a large surface area** for adsorption. They have vacant $d$-orbitals which allow them to bond with reactant molecules to form highly unstable active intermediates, lowering the overall activation energy ($E_a$) of the reaction.

Correct Option: (a)

Explanation:

• $\text{Cr}^{2+}$ ($3d^4$) acts as a **reducing agent** because it easily loses one electron to form $\text{Cr}^{3+}$ ($3d^3$). In an aqueous medium, $d^3$ is exceptionally stable because it represents a half-filled $t_{2g}$ sub-level ($t_{2g}^3$) under crystal field splitting theory.
• $\text{Mn}^{3+}$ ($3d^4$) acts as an **oxidizing agent** because it easily gains one electron to form $\text{Mn}^{2+}$ ($3d^5$). The $d^5$ state is highly stable because it has a perfectly symmetrical, half-filled $d$-shell.

Hence, both statements are correct, and the reason perfectly explains the assertion.

Reason: The high melting and boiling points of transition metals are due to **strong interatomic metallic bonding**.

Unlike alkali and alkaline earth metals, transition elements can utilize a greater number of electrons from both the **$(n-1)d$** subshell (unpaired electrons) and the outermost **$ns$** subshell for metallic bonding. This results in highly covalent, dense, and tough metallic lattices which require enormous thermal energy to break.

Reason: The standard electrode potential ($E^{\circ}$) of a metal is determined by the net sum of three thermodynamic steps:

1. Enthalpy of Atomization ($\Delta_{\text{a}}H$) to turn solid metal into gas (always endothermic).
2. Ionization Enthalpy ($\Delta_{\text{i}}H$) to remove electrons (highly endothermic).
3. Hydration Enthalpy ($\Delta_{\text{hyd}}H$) when gaseous ions dissolve in water (highly exothermic).

For copper, the exceptionally high energy required to transform solid copper into gaseous ions (high $\Delta_{\text{a}}H$ and high sum of first and second $\Delta_{\text{i}}H$) is **not compensated** by its relatively weak, less negative hydration enthalpy ($\Delta_{\text{hyd}}H$). Consequently, the overall potential remains positive, indicating that copper is relatively stable and cannot easily release hydrogen gas from dilute acids.

Explanation: In an aqueous solution, the $\text{Cu}^{+}$ ion is highly unstable and undergoes spontaneous **disproportionation** (self-redox reaction) to form $\text{Cu}^{2+}$ and solid metallic copper ($\text{Cu}$).

Thermodynamic Reason: Although removing an electron from $\text{Cu}^{+}$ requires a high second ionization enthalpy, the resulting $\text{Cu}^{2+}$ ion has a much smaller ionic radius and higher charge density. This allows $\text{Cu}^{2+}$ to form extremely strong electrostatic bonds with water molecules, releasing a highly negative, exothermic **hydration energy ($\Delta_{\text{hyd}}H$)** which more than compensates for the second ionization cost.

General Concept of Color: Transition metal complexes are colored due to **$d-d$ electronic transitions**.

When ligands surround a transition metal ion, the degenerate $d$-orbitals split into groups of different energy levels (like $t_{2g}$ and $e_g$). Unpaired electrons in lower-energy $d$-orbitals absorb specific wavelengths of visible light to jump to empty higher-energy $d$-orbitals. The remaining transmitted light appears as the complementary color.

Comparison:

• $\text{Sc}^{3+}$: Configuration is $[\text{Ar}] 3d^0$. Since there are **no electrons** in its $3d$ shell ($d^0$), no $d-d$ transitions can occur, rendering the solution completely **colorless**.
• $\text{Cr}^{3+}$: Configuration is $[\text{Ar}] 3d^3$. It contains **3 unpaired electrons** in its $3d$ shell. These electrons readily undergo $d-d$ electronic transitions, giving the solution a rich violet/green color.

Reason: Zinc ($\text{Zn}$), cadmium ($\text{Cd}$), and mercury ($\text{Hg}$) have **completely filled $(n-1)d$ and $ns$ subshells** ($d^{10} s^2$ configuration) in their elemental states.

Because there are **no unpaired electrons** in their $d$-subshell, they cannot participate in covalent interatomic sharing or form localized d-orbital networks. Consequently, the interatomic metallic bonding in these lattices is extremely weak (relying mostly on basic $s$-orbital electron-sea attraction), making them soft metals with low melting points.

Step 1: Find the electronic configuration of $\text{M}^{2+}$ where $Z = 27$ (Cobalt).

Neutral atom configuration:

For the divalent cation ($\text{M}^{2+}$), remove the two $4s$ electrons:

Step 2: Determine the number of unpaired electrons ($n$).

According to Hund's rule, distribute 7 electrons across the 5 orbitals of the $3d$ subshell:

Thus, there are $n = 3$ unpaired electrons.

Step 3: Apply the spin-only magnetic moment formula.

Answer: The spin-only magnetic moment is $3.87\text{ B.M.}$

Step 1: Write the half-reactions.

• Reduction half-reaction: Permanganate ion ($\text{MnO}_4^{-}$) is reduced to $\text{Mn}^{2+}$ in acidic medium:
  $$ \text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \cdots\text{ (Equation 1)} $$
• Oxidation half-reaction: Ferrous ion ($\text{Fe}^{2+}$) is oxidized to ferric ion ($\text{Fe}^{3+}$):
  $$ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-} \quad \cdots\text{ (Equation 2)} $$

Step 2: Equalize electrons and add half-reactions.

Multiply Equation 2 by $5$ to balance the electron transfer:

Add the two balanced half-reactions together:

Step 1: Write the half-reactions.

• Reduction half-reaction:
  $$ \text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \cdots\text{ (Equation 1)} $$
• Oxidation half-reaction: Oxalate ion ($\text{C}_2\text{O}_4^{2-}$) is oxidized to carbon dioxide gas ($\text{CO}_2$):
  $$ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^{-} \quad \cdots\text{ (Equation 2)} $$

Step 2: Equalize electrons (LCM of 5 and 2 is 10) and combine.

Multiply Equation 1 by $2$ and Equation 2 by $5$:

Adding these gives the net balanced ionic equation:

Step 1: Write the half-reactions.

• Reduction half-reaction: Dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) is reduced to $\text{Cr}^{3+}$ in acidic medium:
  $$ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \quad \cdots\text{ (Equation 1)} $$
• Oxidation half-reaction: Iodide ion ($\text{I}^{-}$) is oxidized to elemental iodine ($\text{I}_2$):
  $$ 2\text{I}^{-} \rightarrow \text{I}_2 + 2e^{-} \quad \cdots\text{ (Equation 2)} $$

Step 2: Equalize electrons and combine (multiply Equation 2 by $3$):

Add both half-reactions to obtain the net equation:

Observation: Thermal decomposition of ammonium dichromate is highly exothermic and represents a **green volcanic eruption** simulation. It decomposes to release nitrogen gas, water vapor, and a fluffy green powder of chromium (III) oxide.

Where $\text{Cr}_2\text{O}_3$ is the green solid residue.

Explanation: In an aqueous solution, yellow chromate ($\text{CrO}_4^{2-}$) and orange dichromate ($\text{Cr}_2\text{O}_7^{2-}$) exist in a dynamic chemical equilibrium that is highly sensitive to the pH of the medium.

• Effect of Acid (Low pH / $\text{pH} \lt 7$): On adding an acid, the high concentration of $\text{H}^{+}$ shifts the equilibrium in the **forward direction** (Le Chatelier's principle), converting yellow chromate into **orange dichromate**.
• Effect of Base (High pH / $\text{pH} \gt 7$): On adding a base, $\text{OH}^{-}$ ions neutralize the free $\text{H}^{+}$ ions. This shifts the equilibrium in the **backward direction**, converting orange dichromate back into **yellow chromate**:
  $$ \text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^{-} \rightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O} $$

Definition: Lanthanoid contraction refers to the steady, progressive decrease in the atomic and ionic radii of the lanthanoid elements with an increase in atomic number from Lanthanum ($\text{La}^{3+}$, $Z = 57$) to Lutetium ($\text{Lu}^{3+}$, $Z = 71$).

Primary Cause: The contraction is caused by the **poor shielding effect of $4f$ electrons**.

As we move across the series, the nuclear charge increases by $+1$ at each step, and the incoming electron enters the inner $4f$ subshell. Because $f$-orbitals have highly diffused shapes, they shield outer electrons very poorly from the positive nucleus. Consequently, the effective nuclear charge pulling the valence shell inward increases continuously, resulting in a gradual decrease in size.

The two main consequences of lanthanoid contraction are:

1. Identical Sizes of $4d$ and $5d$ Elements: Under normal trends, atomic size increases down a group. However, due to the lanthanoid contraction spanning before the $5d$ series, the expected size increase is completely offset. As a result, pairs of elements like Zirconium ($\text{Zr}$, $4d$ series, $r \approx 160\text{ pm}$) and Hafnium ($\text{Hf}$, $5d$ series, $r \approx 159\text{ pm}$) have virtually identical physical dimensions. This makes their chemical separation extremely difficult.
2. Decrease in Basicity of Hydroxides: As the size of the lanthanoid ion decreases from $\text{La}^{3+}$ to $\text{Lu}^{3+}$, the covalent character of the $\text{M-OH}$ bond increases (according to Fajan's Rules). Consequently, their tendency to release $\text{OH}^{-}$ ions decreases, making $\text{La(OH)}_3$ the **most basic** and $\text{Lu(OH)}_3$ the **least basic** (most covalent).

(A) One Major Similarity:

• Both lanthanoids and actinoids exhibit a dominant and highly stable **oxidation state of $+3$**.
• Both series undergo a progressive contraction in atomic and ionic size (Lanthanoid contraction and Actinoid contraction, respectively) due to the poor shielding of inner $f$ electrons.

(B) One Major Difference: