NCERT Solutions for Class 12 Chemistry Chapter 4 d- and f- Block Elements
NCERT Solutions for Class 12 Chemistry Chapter 4 d- and f- Block Elements provides complete coverage of electronic configurations, general properties of transition elements, catalytic activity, interstitial compounds, potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), potassium permanganate ($\text{KMnO}_4$), Lanthanoids, Actinoids, and Lanthanoid Contraction.
Overview of Topics & Subtopics (Click to Jump)
1. General Properties & Trends of Transition Elements (d-Block)
Short Note:
• Electronic Configuration: General valence shell electronic configuration: $(n-1)d^{1-10} ns^{1-2}$. Transition metals are defined as elements having partially filled $d$-orbitals in their ground state or in any common oxidation state (Note: $\text{Zn, Cd, Hg}$ are not regarded as transition elements due to completely filled $d$-orbitals).
• Variable Oxidation States: Due to small energy difference between $(n-1)d$ and $ns$ orbitals, electrons from both subshells participate in bond formation.
• Magnetic Properties: Paramagnetism arises due to unpaired electrons. Magnetic moment ($\mu$) is calculated using the spin-only formula:
$\mu = \sqrt{n(n+2)}\text{ BM}$ (where $n$ is number of unpaired electrons and BM is Bohr Magneton).
• Formation of Coloured Ions: Caused by $d$-$d$ electronic transitions of unpaired electrons upon absorption of visible light energy.
• Catalytic Activity: Ability to show variable oxidation states and form reaction intermediates (e.g., $\text{V}_2\text{O}_5$ in Contact process, Fine $\text{Fe}$ in Haber process).
2. Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) & Potassium Permanganate ($\text{KMnO}_4$)
Short Note:
• Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$):
- Prepared from chromite ore ($\text{FeCr}_2\text{O}_4$).
- pH Dependent Equilibrium: Chromate ($\text{CrO}_4^{2-}$, yellow) and dichromate ($\text{Cr}_2\text{O}_7^{2-}$, orange) interconvert depending on pH:
$2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$
- Powerful oxidizing agent in acidic medium: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
• Potassium Permanganate ($\text{KMnO}_4$):
- Prepared from pyrolusite ore ($\text{MnO}_2$).
- Oxidizing Action in Acidic Medium ($E^\circ = +1.51\text{ V}$):
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
3. Inner Transition Elements (f-Block): Lanthanoids & Actinoids
Short Note:
• Lanthanoids ($4f$-series, $\text{Ce}_{58}$ to $\text{Lu}_{71}$):
- Common oxidation state is $+3$. $+2$ and $+4$ states exist if they yield extra stable $4f^0, 4f^7, 4f^{14}$ configurations (e.g., $\text{Ce}^{4+}, \text{Eu}^{2+}$).
- Lanthanoid Contraction: Steady decrease in atomic and ionic radii ($\text{M}^{3+}$) from $\text{La}$ to $\text{Lu}$ due to poor shielding effect of $4f$-electrons.
- Consequences: Similar chemical properties, difficult separation, equal atomic radii of $4d$ and $5d$ series elements (e.g., $\text{Zr} \approx \text{Hf}$).
• Actinoids ($5f$-series, $\text{Th}_{90}$ to $\text{Lr}_{103}$):
- All elements are radioactive. Show greater variety of oxidation states ($+3, +4, +5, +6, +7$) due to comparable energies of $5f, 6d$, and $7s$ subshells.
Intext & Conceptual Questions
• Definition: The steady and continuous decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number from Lanthanum ($\text{La}$) to Lutetium ($\text{Lu}$) is known as Lanthanoid Contraction. It occurs because $4f$ electrons have poor nuclear shielding efficacy.
• Consequences:
1. Similarity in Radii of $4d$ and $5d$ Transition Series: Elements belonging to the 2nd and 3rd transition series of the same group exhibit almost identical atomic radii (e.g., $\text{Zr} = 160\text{ pm}$ and $\text{Hf} = 159\text{ pm}$), making their chemical properties virtually identical.
2. Difficulty in Separation: Due to nearly identical ionic radii and chemical properties, separating individual lanthanoid elements in pure form is extremely difficult.
3. Trend in Basic Strength: Basic strength of hydroxides decreases from $\text{La(OH)}_3$ (most basic) to $\text{Lu(OH)}_3$ (least basic) as covalent character increases.
| Property | Scientific Reason |
|---|---|
| Variable Oxidation States | Energy levels of $(n-1)d$ and $ns$ subshells are very close. Therefore, electrons from both $ns$ and $(n-1)d$ subshells can participate in chemical bonding. |
| Complex Formation | Driven by: (i) Small ionic size of metal cations, (ii) High ionic charge density, and (iii) Availability of vacant $d$-orbitals to accept electron pairs donated by ligands. |
| Catalytic Activity | Ability to adopt multiple oxidation states, form variable intermediates, and provide large active surface areas for reactant adsorption. |
Textbook Chapter End Exercise Questions
Given:
• Atomic Number ($Z$) = 25 (Manganese, $\text{Mn}$)
• Divalent ion = $\text{M}^{2+}$
Step 1: Write Electronic Configuration
• Neutral atom ($Z=25$): $[\text{Ar}] 3d^5 4s^2$
• Divalent cation ($\text{M}^{2+}$): $[\text{Ar}] 3d^5 4s^0$
Step 2: Determine Number of Unpaired Electrons ($n$)
• The $3d^5$ subshell contains 5 unpaired electrons ($n = 5$).
Step 3: Calculate Spin-Only Magnetic Moment ($\mu$)
$\mu = \sqrt{n(n+2)}\text{ BM}$
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx \mathbf{5.92\text{ BM}}$
• The magnetic moment of the divalent ion is 5.92 BM (Bohr Magnetons).
(i) Reaction of Dichromate ion with Ferrous ion in Acidic Medium:
• Dichromate oxidizes $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ while getting reduced to $\text{Cr}^{3+}$.
$$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 6\text{Fe}^{2+}\text{(aq)} + 14\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 6\text{Fe}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$$
(ii) Reaction of Permanganate ion with Oxalate ion in Acidic Medium:
• Permanganate oxidizes oxalate ion ($\text{C}_2\text{O}_4^{2-}$) to carbon dioxide ($\text{CO}_2$) while getting reduced to $\text{Mn}^{2+}$.
$$2\text{MnO}_4^-\text{(aq)} + 5\text{C}_2\text{O}_4^{2-}\text{(aq)} + 16\text{H}^+\text{(aq)} \rightarrow 2\text{Mn}^{2+}\text{(aq)} + 10\text{CO}_2\text{(g)} + 8\text{H}_2\text{O(l)}$$
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