CBSE Class 12

by Rezyo Staff WriterJanuary 16, 2026 CBSE Class 12, Chemistry0 comments

Class XII Chemistry Unit 4: d and f Block Elements | Trends, Compounds & Exam Q&A (2025-26)

Transition and Inner Transition elements often present a challenge due to their variable oxidation states and irregular atomic trends. This guide targets the core concepts of Unit 4 for the 2025-26 Class XII curriculum, breaking down the 3d series, Lanthanoids, and Actinoids.

We examine the specific anomalies in melting points and ionization enthalpies, visualize the geometry of chromate and permanganate ions, and outline the reaction pathways for major industrial compounds. Use the interactive modules to track electronic configurations and predict magnetic properties for board exam preparation.

CLASS XII-2025-26 Unit 4 – d and f Block Elements
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Unit 4: d- and f-Block Elements

A definitive guide to Transition and Inner Transition elements. Updated for the 2025-26 academic curriculum. Explore electronic configurations, magnetic properties, and the nuanced chemistry of Lanthanoids.

The 3d Transition Series

Select an element to inspect its electronic configuration.

Definition Check

A transition element has an incompletely filled d-subshell in its ground state or any common oxidation state.

Zn, Cd, Hg are typically excluded from “true” transition metals because they have full d10 configurations.

Sc
Scandium
Atomic Number: 21
Configuration [Ar] 3d1 4s2
Oxidation States +3
Magnetic Nature Paramagnetic

Melting Point Trends

Melting points indicate the strength of metallic bonding. Stronger bonding occurs when more electrons (from both ns and (n-1)d) participate.

The Mn/Tc Anomaly

Manganese (Mn) and Technetium (Tc) show a sharp dip in melting point. This is because the stable half-filled d5 configuration holds electrons more tightly, making them less available for metallic delocalization.

Oxidation State Landscape

Transition elements exhibit variable oxidation states because both ns and (n-1)d electrons participate in bonding. The number of states increases to a maximum at Manganese, then decreases.

Sc Ti V Cr Mn Fe Co Ni Cu Zn
+7
+6
+6
+6
+5
+5
+5
+4
+4
+4
+4
+4
+4
+4
+3
+3
+3
+3
+3
+3
+3
+3
+2
+2
+2
+2
+2
+2
+2
+2
+2
+1
Most Common State
Strong Oxidizing Agent

Industrial Preparation Pathways

Preparation of Potassium Dichromate

From Chromite Ore
1

Fusion

Chromite ore + Na₂CO₃ + Air

Na₂CrO₄

(Sodium Chromate – Yellow)

2

Acidification

Filter & treat with H₂SO₄

Na₂Cr₂O₇

(Sodium Dichromate – Orange)

3

Crystallization

Treat with KCl

K₂Cr₂O₇

(Crystals separate out)

Preparation of Potassium Permanganate

From Pyrolusite
1

Oxidative Fusion

MnO₂ + KOH + O₂/KNO₃

K₂MnO₄

(Potassium Manganate – Green)

2

Electrolytic Oxidation

In Alkaline solution

KMnO₄

(Permanganate Ion – Purple)

Note: In the laboratory, Mn²⁺ salt is oxidized by Peroxodisulphate to Permanganate.

Reaction Profiles

Dichromate Reactions

CrO₄²⁻ ⇌ Cr₂O₇²⁻

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Permanganate Reactions

Acidic pH Mn²⁺ (Colorless)
Neutral/Alkaline pH MnO₂ (Brown Solid)

Volumetric Analysis: Redox Titrations

High Yield

Potassium Permanganate acts as a self-indicator in acidic medium. Here are the specific half-reactions you must know for titrations.

Oxidation of Ferrous Salts Mohr’s Salt Titration
5Fe²⁺ → 5Fe³⁺ + 5e⁻

Green Fe²⁺ turns to Yellow Fe³⁺.

Oxidation of Oxalates Needs heating to 60°C
C₂O₄²⁻ → 2CO₂ + 2e⁻

Slow reaction initially; autocatalyzed by Mn²⁺.

Oxidation of Iodide
2I⁻ → I₂ + 2e⁻

Forms Iodine (Purple/Brown in solution).

Oxidation of Sulphite
5SO₃²⁻ + H₂O → 5SO₄²⁻ + 2H⁺ + 2e⁻

Sulphite oxidizes to Sulphate.

Master Equation (Acidic): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Why do Transition Metals form Complexes?

Transition metals are unique in their ability to form a vast number of complex compounds (like [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺). This property drives biological systems (Hemoglobin) and industrial catalysis.

  • Small Size & High Charge: High ionic charge density attracts ligands strongly.
  • Availability of d-Orbitals: Empty (n-1)d orbitals of appropriate energy can accept lone pairs from ligands.
Mⁿ⁺

The Copper Anomaly (E° Values)

Why is E°(Cu²⁺/Cu) Positive (+0.34V)?

Unlike other 3d metals which have negative E° values (meaning they release H₂ from acids), Copper has a positive E°. This means Copper is less reactive than Hydrogen.

The stability of an ion in solution depends on the energy balance:
Energy Required (Sublimation + Ionization) vs Energy Released (Hydration).

For Copper, the high energy required to transform Cu(s) to Cu²⁺(g) is not fully compensated by its Hydration Enthalpy.

High Cost

Sublimation + IE₁ + IE₂

Lower Gain

Hydration Enthalpy

Visual representation of energy imbalance for Cu

Formation of Colored Ions

Why are some ions colored while others are white?

Concept Module

The d-d Transition Mechanism

In the presence of ligands (like water), the degenerate d-orbitals split into two energy levels (t2g and eg).

When white light falls on the ion, electrons absorb specific wavelengths to jump from the lower to higher d-orbital. The transmitted light is the complementary color we see.

Condition for Color:

Presence of at least one unpaired d-electron.

Ions with d0 or d10 are colorless (e.g., Sc3+, Zn2+).

Ti³⁺(aq)

Purple Solution

Analysis Unpaired e⁻: 1

d-d Transition possible

Absorbs in visible region

Magnetic Moment (Spin Only):

1.73 B.M.

Structures of Oxoanions

Cr

Chromate Ion (CrO₄²⁻)

Tetrahedral Geometry

Yellow Color

Dichromate Ion (Cr₂O₇²⁻)

Two tetrahedra sharing one oxygen at 126° angle.

Orange Color

Mn

Permanganate Ion (MnO₄⁻)

Tetrahedral Geometry (π-bonding involved).

Intense Purple

Why are they Catalysts?

Reason 1: Variable Oxidation States

Transition metals can form unstable intermediates by changing oxidation states, lowering activation energy.

Ex: V₂O₅ in Contact Process (V⁵⁺ ↔ V⁴⁺)

Reason 2: Large Surface Area

Provide surface for reactants to adsorb and weaken bonds.

Ex: Finely divided Iron in Haber Process

Alloy Formation

Transition metals have very similar atomic radii (within 15%). Atoms of one metal can easily replace atoms of another in the crystal lattice.

Brass Cu + Zn
Bronze Cu + Sn
Stainless Steel Fe + Cr + Ni

The Lanthanoid Contraction

When moving from the 4d series (Yttrium to Cadmium) to the 5d series (Lanthanum to Mercury), we expect a size increase due to the new shell. This does not happen.

The filling of 14 electrons in the inner 4f orbitals occurs before the 5d series. 4f electrons have a diffuse shape and shield the nucleus poorly. This results in a higher effective nuclear charge pulling the outer electrons inward.

Result: The atomic radii of 4d and 5d metals are nearly identical (e.g., Zr ≈ 160 pm, Hf ≈ 159 pm). This makes separation extremely difficult.

Titanium (Group 4, 3d) 132 pm
Zirconium (Group 4, 4d) 160 pm
Hafnium (Group 4, 5d) 159 pm

Similarity in size between Zr and Hf is due to Lanthanoid Contraction.

The Inner Transition Matrix

Comparison between 4f (Lanthanoids) and 5f (Actinoids) series. This differentiation is critical for board exams.

Oxidation States
+3 is dominant. Occasionally +2 (Eu, Yb) or +4 (Ce, Tb).
Show greater range (+3 to +7). 5f, 6d, and 7s levels have comparable energies.
Chemical Reactivity
Less reactive. Resemble Aluminum in behavior.
Highly reactive, especially in finely divided states.
Magnetic Properties
Simple magnetic behavior.
Complex magnetic behavior due to stronger spin-orbit coupling.
Contraction Effect
Lanthanoid Contraction (poor shielding by 4f).
Actinoid Contraction is greater from element to element due to even poorer shielding by 5f.

High-Yield Q&A

Click to reveal answers

Transition elements are defined as having incompletely filled d-orbitals in their ground state or common oxidation states.

Zn, Cd, and Hg have fully filled d-orbitals (n-1)d10 in both their ground state and stable ions (e.g., Zn2+ is 3d10). Thus, they do not fit the definition.

Scandium (Z=21) has the config [Ar] 3d1 4s2. Sc3+ loses all 3 valence electrons, resulting in [Ar] 3d0.

Since there are zero unpaired electrons, the magnetic moment is 0 B.M. It is diamagnetic.

Cr2+ to Cr3+: Cr2+ (d4) wants to lose an electron to become Cr3+ (d3). The d3 configuration is highly stable in aqueous solutions because it has a half-filled t2g level. Thus, it reduces others and oxidizes itself.

Mn3+ to Mn2+: Mn3+ (d4) wants to gain an electron to become Mn2+ (d5). The d5 configuration is a stable half-filled subshell. Thus, it oxidizes others and reduces itself.

Conceptual Deep Dive

Addressing the most common points of confusion in exams.

Why is the E° value for Mn³⁺/Mn²⁺ couple much more positive than for Cr³⁺/Cr²⁺?

A large positive E° value indicates a strong tendency to get reduced. Mn³⁺ (d⁴) is desperate to become Mn²⁺ (d⁵) because d⁵ is a half-filled, stable configuration. In contrast, Cr³⁺ (d³) is already very stable in aqueous medium (t₂g³), so it has less tendency to accept an electron compared to Mn³⁺.

Why is Cu⁺ unstable in aqueous solution?

Although Cu⁺ has a stable d¹⁰ configuration, it is unstable in water and disproportionates:
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s).
This happens because the Hydration Enthalpy of Cu²⁺ is much higher (more negative) than Cu⁺ due to its smaller size and higher charge. This energy release compensates for the 2nd ionization enthalpy of Copper.

What are Interstitial Compounds?

Transition metals have crystal lattices with empty spaces (voids). Small atoms like H, C, or N get trapped here.

Key Properties:

  • High melting points (higher than pure metal).
  • Very hard (some approach diamond hardness).
  • Retain metallic conductivity.
  • Chemically inert.

Why do transition metals exhibit variable oxidation states?

This is due to the very small energy gap between the (n-1)d and ns orbitals. Electrons from both shells can participate in bonding.

However, oxidation states differ by unity (e.g., Fe²⁺, Fe³⁺) unlike non-transition elements which differ by 2 units (e.g., Pb²⁺, Pb⁴⁺) due to the inert pair effect.

Unit Summary & Outcomes

You have navigated the complexities of d and f block elements. The key takeaways for your board exams are:

Trends Matter

Master the irregularities in Ionization Enthalpy and Atomic Radii, especially around Zn and Mn.

Color & Magnetism

Always check for unpaired electrons. No unpaired e⁻ = Colorless & Diamagnetic.

Redox Kings

KMnO₄ and K₂Cr₂O₄ are your go-to oxidizers. Remember their pH-dependent products.

© 2026 Rezyo.in | Class XII Education Series

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by Rezyo Staff WriterJanuary 16, 2026 CBSE, CBSE Class 12, Chemistry0 comments

Class 12 Chemistry Unit 3: Chemical Kinetics – Practice Questions Quiz (2025-26)

Thermodynamics predicts if a reaction occurs, but Chemical Kinetics determines the speed. For Class XII students tackling the 2025-26 syllabus, mastering the “time” variable in chemistry is required for solving numerical problems and interpreting rate graphs correctly.

This guide breaks down the mathematics of reaction rates, comparing Average and Instantaneous speeds while distinguishing experimental Order from theoretical Molecularity. From deriving Integrated Rate Equations for Zero and First Order reactions to calculating Activation Energy using the Arrhenius formula, these sections provide the specific tools needed for board exams.

Explore how concentration, temperature, and catalysts dictate reaction velocity through interactive visualizers designed to clarify the mechanics behind the math.

Unit 3 Chemical Kinetics Class 12 (2025-26) | Rezyo.in
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Class XII Unit 3: Chemical Kinetics

Chemical Kinetics: The Dimension of Time

Thermodynamics predicts if a reaction is feasible, but Kinetics defines the speed. This guide covers the complete 2025-26 syllabus for Unit 3, identifying how concentration, temperature, and catalysts dictate the pace of chemical change.

Knowledge Check

Pop-out quiz with timer to test your speed.

Speed & Accuracy Challenge

Ready to test your mastery of Chemical Kinetics? You have 20 questions. The clock starts when you click below.

Rate Concepts & Stoichiometry

Average Rate

Measured over a time interval. Ignores rate fluctuations.

\( r_{av} = \frac{\Delta [P]}{\Delta t} \)

Instantaneous Rate

Rate at a specific moment (\(\Delta t \to 0\)). The slope of the tangent.

\( r_{inst} = \frac{d[P]}{dt} \)

Stoichiometric Normalization

For \( aA + bB \to cC \), the unique rate of reaction is found by dividing the rate of change of any species by its stoichiometric coefficient:

Rate = \( -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} \)

Concept: Pseudo First Order Reaction

A reaction that is theoretically higher order but behaves as first order because one reactant is present in large excess.
Example: Acid hydrolysis of ethyl acetate. Water is in excess, so its concentration remains effectively constant.
\( Rate = k'[Ester][H_2O] \approx k[Ester] \) where \( k = k'[H_2O] \).

Decay Curves

Visualizing Concentration [R] vs Time (t)

Comparison Logic

Zero Order

Rate is constant. Concentration drops linearly. Slope = \(-k\).

\( [R] = -kt + [R]_0 \)

First Order

Rate decreases as conc. drops. Exponential decay. Constant Half-Life.

\( \ln[R] = -kt + \ln[R]_0 \)

Half-Life Dynamics

Zero Order: Half-life decreases as concentration drops.
First Order: Half-life remains constant regardless of concentration.

Select a reaction order above.

Arrhenius Equation & Temperature

The Arrhenius equation relates rate constant (\(k\)) to temperature (\(T\)).

\( \ln k = -\frac{E_a}{R}\frac{1}{T} + \ln A \)

Plot of \(\ln k\) vs \(1/T\) (Linear)

Key Parameters

  • Slope (\(m\)): Represents \( -E_a/R \). A steeper slope means higher Activation Energy (\(E_a\)).
  • Intercept (\(c\)): Represents \( \ln A \). \(A\) is the Frequency Factor.

Temperature Coefficient (\(\mu\))

For most reactions, rate constant nearly doubles for a 10° rise in temperature.
\( \mu = \frac{k_{T+10}}{k_T} \approx 2 \) to \( 3 \).

Collision Theory & Catalysis

Reactions occur when molecules collide. However, not all collisions lead to products. Two barriers must be overcome:

  1. Energy Barrier: Kinetic energy \(\ge\) Activation Energy (\(E_a\)).
  2. Orientation Barrier: Molecules must align correctly (Steric factor \(P\)).

Mathematical Form

Rate = \( P \cdot Z_{AB} \cdot e^{-E_a/RT} \)
  • \(Z_{AB}\): Collision Frequency
  • \(P\): Probability/Steric Factor
  • \(e^{-E_a/RT}\): Fraction of energetic collisions

Effect of Catalyst

A catalyst provides an alternate pathway with lower \(E_a\). It does not change the Enthalpy (\(\Delta H\)) or Gibbs Energy (\(\Delta G\)) of the reaction. It simply speeds up the attainment of equilibrium.

Numerical Cheat Sheet

First Order Rate
\( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)

Unit of k: \( s^{-1} \)

Half Life (1st Order)
\( t_{1/2} = \frac{0.693}{k} \)

Independent of concentration

Arrhenius (2 Temps)
\( \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} [\frac{T_2-T_1}{T_1 T_2}] \)

T in Kelvin, R = 8.314

Zero Order Rate
\( [R] = -kt + [R]_0 \)

Unit of k: \( mol L^{-1} s^{-1} \)

Conceptual Doubts (FAQ)

Q: Why does a 10° rise in temperature nearly double the rate?

Collision frequency (\(Z\)) only increases by ~3%. The primary reason is the exponential increase in the fraction of effective collisions. A small temperature rise pushes significantly more molecules over the Activation Energy (\(E_a\)) barrier.

Q: Can Molecularity be zero or fractional?

No. Molecularity refers to the count of reacting species colliding simultaneously in an elementary step. You cannot have zero molecules colliding, nor half a molecule. It must be a positive integer.

Q: Do Zero Order reactions go to completion?

Yes. Unlike First Order reactions (which theoretically take infinite time), Zero Order reactions consume reactants at a constant rate and reach completion in finite time: \( t_{completion} = [R]_0 / k \).

Q: Does a catalyst change the Equilibrium Constant (Kc)?

No. A catalyst lowers the activation energy for both forward and reverse reactions equally. It speeds up the attainment of equilibrium but does not shift the position of equilibrium or change \(\Delta G\).

Q: What is the unit of k for an \(n^{th}\) order reaction?

The general formula is \( (mol \ L^{-1})^{1-n} \ s^{-1} \).
For \(n=0\): \(mol \ L^{-1} \ s^{-1}\).
For \(n=1\): \(s^{-1}\).
For \(n=2\): \(mol^{-1} \ L \ s^{-1}\).

Chapter Summary

We have explored the dynamics of chemical change. From defining rates to analyzing the temperature dependence via Arrhenius, you now possess the tools to calculate when a reaction completes. Remember: Order is experimental, Molecularity is theoretical, and Temperature increases rate by increasing effective collisions, not just total collisions.

Slope = -Ea/R t1/2 = 0.693/k Rate = k[A]^x

© 2026 Rezyo.in. Aligned with CBSE Class XII Unit 3.

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by Rezyo Staff WriterJanuary 16, 2026 CBSE, CBSE Class 12, Chemistry0 comments

Class 12 Electrochemistry Practice Exam: Nernst Calculator, Logic Lab & Board Simulator

Electrochemistry links spontaneous chemical reactions to electrical output. This Unit 2 guide moves beyond static notes, offering a hands-on learning dashboard. Students can gauge their readiness with a timed board-exam simulator, solve numericals instantly using dedicated calculators, and predict electrolysis products through logic trees. Whether reviewing the Nernst equation or comparing commercial batteries, these tools support practical understanding and exam performance.

Class 12 Electrochemistry (Unit 2) – Master Exam Suite | Rezyo.in
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Class 12 • Electrochemistry

Board Exam Simulator

Ready to test your
Electrochemistry concepts?

Time Limit

25 Minutes

Questions

25 MCQs

Target

Board Level

Battery Technology Hub

Dry Cell

Primary
  • Anode: Zinc Cup
  • Cathode: Graphite Rod
  • Electrolyte: NH₄Cl + ZnCl₂ paste
  • Volt: ~1.5 V

Mercury Cell

Primary
  • Anode: Zn-Hg Amalgam
  • Cathode: HgO + Carbon
  • Volt: 1.35 V (Constant)
  • Use: Watches, Hearing aids

Lead Storage

Secondary
  • Anode: Spongy Lead (Pb)
  • Cathode: PbO₂ Grid
  • Electrolyte: 38% H₂SO₄
  • Recharge: PbSO₄ dissolves back

H₂-O₂ Fuel Cell

Efficiency
  • Anode: H₂ fed (Oxidation)
  • Cathode: O₂ fed (Reduction)
  • Byproduct: Pure Water
  • Efficiency: η = ΔG/ΔH (~70%)

Conductivity Measurement

To find conductivity (κ), we first measure Resistance (R) using a Wheatstone Bridge with AC current (to prevent electrolysis).

Step 1: Find Resistance (R) R = R_known × (l1 / l2)
Step 2: Cell Constant (G*) G* = l / A
Step 3: Conductivity (κ) κ = G* / R
Conductivity Cell
(Rx)
AC Source

Electrolysis Logic Lab

Predict the anode product using step-by-step reasoning.

Λm vs √C Trends

Nernst Calc

— V

Kohlrausch Calc (Λ°m)

— S cm² mol⁻¹

Can We Store It?

Check if a solution can be kept in a metal container.

—

Faraday’s 2nd Law: Series Connection

AgNO₃
CuSO₄
AlCl₃

Charge (Q) is same for all.

Mass ∝ Equivalent Weight

m₁/m₂ = E₁/E₂

Corrosion Prevention Strategies

1

Sacrificial Protection

Coating with a more reactive metal (e.g., Zinc on Iron). Zinc oxidizes (E° = -0.76V) instead of Iron (-0.44V). This is Galvanization.

2

Cathodic Protection

Connecting iron pipes to Mg or Zn blocks. The pipe becomes the cathode (protected) and the block becomes the anode (sacrificed).

Quick Revision Formulae

Nernst Equation
E = E° – (0.059/n)logQ
Gibbs Energy
ΔG° = -nFE°cell
Kohlrausch Law
Λ°m = ν₊λ°₊ + ν₋λ°₋
Faraday’s 1st Law
m = ZIt

Conceptual FAQs

Why does conductivity decrease with dilution?
Conductivity is conductance per unit volume. As volume increases, number of ions per cm³ decreases.
Can we store CuSO₄ in Zinc pot?
No. Zn is more reactive than Cu (lower E°) and will displace it, destroying the pot.

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by Rezyo Staff WriterJanuary 16, 2026 CBSE, CBSE Class 12, Chemistry0 comments

Class 12 Electrochemistry Notes & Calculator | Unit 2 Nernst Equation, Conductance & Batteries Guide

Electrochemistry connects spontaneous chemical reactions to electrical energy production. This Unit 2 syllabus guide simplifies electrolytic conductance, the Nernst equation, and the working principles of commercial batteries.

Try our Practice Exams

Students can compute cell potentials directly with the built-in calculator, track ion migration through visual animations, and test their grasp of Faraday’s Laws with the integrated quiz. From the standard hydrogen electrode to the chemistry of rusting, these notes prioritize exam-focused concepts and numerical problem-solving.

Class 12 Unit 2: Electrochemistry – Rezyo.in
Class 12 • Unit 2 • 2025-26 Syllabus

Electrochemistry:
From Ions to Energy

A comprehensive guide to understanding how chemical energy powers our world. Covers conduction, Nernst equation, commercial batteries, and corrosion.

Filter View:

01 Electrolytic Conductance

Unlike metallic wires where electrons flow through a lattice, electrolytic conduction relies on the physical movement of ions. This creates a unique relationship between concentration and resistance.

Conductivity (κ):
κ = G × G*
Where G is Conductance and G* is the Cell Constant (l/A)
Molar Conductivity (Λm):
Λm = (κ × 1000) / C
Units: S cm2 mol-1

Visualizing Dilution Effect

Animation: As water is added (Dilution), ions move apart.
κ decreases (fewer ions/vol), but Λm increases (higher mobility).

Factors Affecting Electrolytic Conductance

1. Nature of Electrolyte

Strong electrolytes dissociate completely (high conductance). Weak electrolytes dissociate partially (low conductance).

2. Size of Ions & Solvation

Larger ions (due to hydration) move slower. Example: Li+ is heavily hydrated and moves slower than Cs+.

3. Nature of Solvent & Viscosity

Polar solvents favor ionization. Higher viscosity resists ion flow, lowering conductance.

4. Temperature

Unlike metals, electrolytic conductance increases with temperature as kinetic energy increases and interionic attractions weaken.

Strong vs. Weak Electrolytes

Feature Strong Electrolyte (e.g., KCl) Weak Electrolyte (e.g., CH3COOH)
Dissociation Complete at all concentrations. Partial. Increases with dilution.
Conductivity Rise Gradual linear increase. Steep, exponential increase near zero conc.
Reason Reduced interionic drag. Increase in number of ions (Ostwald Law).
Limiting Value Found by extrapolation of graph. Found using Kohlrausch Law.

02 Conductivity Trends

Note for Competitive Exams: The linear graph for strong electrolytes follows the Debye-Hückel-Onsager equation: Λm = Λ°m – A√C. The slope is ‘-A’.

03 Kohlrausch’s Law of Independent Migration

At infinite dilution, every ion makes a definite contribution towards the molar conductivity of the electrolyte, regardless of the nature of the other ion it is associated with.

Λ°m = ν+λ°+ + ν–λ°–

Application: This law is critical for calculating the limiting molar conductivity (Λ°m) for weak electrolytes, which cannot be determined graphically.

04 Nernst Equation & Gibbs Energy

The Nernst equation links the potential of a cell to the concentration of species involved. It bridges thermodynamics and electrochemistry.

Nernst Equation (at 298K):
Ecell = E°cell – (0.0591 / n) log Q
Gibbs Energy Relation:
ΔG° = -nFE°cell
n = moles of electrons, F = 96500 C

Equilibrium Constant & Cell Potential

When an electrochemical cell reaches equilibrium, the reaction stops flowing. At this stage, Ecell = 0 and the reaction quotient Q becomes Kc (Equilibrium Constant).

E°cell = (0.0591 / n) log Kc
at 298 K

05 Nernst Equation Calculator

Calculate the cell potential ($E_{cell}$) for a reaction at 298K.

Cell Potential ($E_{cell}$)

— V

06 Electrochemical Series Explorer

Values at 298K relative to SHE (0.00V)

The Reference: Standard Hydrogen Electrode (SHE)

H₂ Gas (1 bar)
H⁺ (1M)

Since individual half-cell potentials cannot be measured, the SHE is the universal reference.

  • Setup: Platinum foil coated with platinum black dipped in 1M H⁺ solution.
  • Convention: Its potential is arbitrarily fixed at 0.00 V at all temperatures.
  • Role: Acts as Anode or Cathode depending on the other half-cell.

07 Faraday’s Laws Calculator

First Law: The mass of substance deposited is directly proportional to the charge passed. (m = ZIt).

Mass Deposited

0.00 g

Formula Used: m = (M × I × t) / (n × 96500)

08 Deep Dive: Electrolysis Products

Predicting products of electrolysis requires analyzing Standard Potentials and considering Overpotential.

Case A: Molten NaCl

Only two ions exist: Na+ and Cl–.

  • Cathode: Na+ + e– → Na(s)
  • Anode: Cl– → ½Cl2 + e–

Case B: Aqueous NaCl (Brine)

Competition with Water molecules.

  • At Cathode: H+ discharges (E°=0.0V) instead of Na+ (E°=-2.71V).
    Product: H2 gas.
  • At Anode: Cl– oxidizes to Cl2. Oxygen is expected (lower E°), but Overpotential makes Cl2 easier to form.
    Product: Cl2 gas.

Comparison: The Two Cell Types

Parameter Electrochemical (Galvanic) Cell Electrolytic Cell
Energy Conversion Chemical Energy → Electrical Energy Electrical Energy → Chemical Energy
Spontaneity Spontaneous (ΔG < 0) Non-Spontaneous (ΔG > 0)
Anode Charge Negative (-) Positive (+)
Cathode Charge Positive (+) Negative (-)

09 Commercial Batteries

Dry Cell (Leclanché)

Primary
  • • Anode: Zinc Container
  • • Cathode: Carbon (Graphite) rod
  • • Voltage: ~1.5 V (Non-constant)

Mercury Cell

Primary
  • • Anode: Zn-Hg Amalgam
  • • Cathode: Paste of HgO + C
  • • Voltage: 1.35 V (Constant)
  • • Key Fact: No ions in overall reaction.

Lead Storage Battery

Secondary
  • • Anode: Pb (Lead)
  • • Cathode: PbO2
  • • Electrolyte: 38% H2SO4

Nickel-Cadmium (NiCd)

Secondary
  • • Anode: Cadmium (Cd)
  • • Cathode: Ni(OH)3
  • • Voltage: ~1.4 V
  • • Life: Longer life than Lead storage.

Fuel Cell Efficiency Focus

Fuel cells, like the H2-O2 cell used in the Apollo space program, convert energy of combustion directly into electricity.

Thermal Efficiency Formula:
η = (ΔG / ΔH) × 100
Since ΔG (useful work) is less than ΔH (total heat), efficiency is typically around 70-80%, far higher than thermal plants (~40%).

10 Corrosion: Electrochemical Theory

Rusting of iron is essentially setting up of a tiny electrochemical cell on the surface of the metal.

Anode (Oxidation): 2Fe(s) → 2Fe2+ + 4e– (E° = -0.44V)
Cathode (Reduction): O2(g) + 4H+ + 4e– → 2H2O(l) (E° = 1.23V)

Prevention: Barrier protection (painting), Galvanization (coating with Zinc), or Cathodic Protection (connecting to Mg or Zn).

11 Quick Check Quiz

1. For a spontaneous reaction, ΔG must be:

Positive
Negative

2. In an Electrolytic cell, the Anode is:

Positive
Negative

3. The unit of Molar Conductivity is:

S cm-1
S cm2 mol-1

12 Exam Focus Q&A & Doubts

Q1: Why does conductivity decrease but molar conductivity increase with dilution?

+

Think about ions per unit volume vs total volume.

Conductivity (κ) decreases because the number of ions per unit volume decreases on dilution.
Molar Conductivity (Λm) increases because the volume containing one mole of electrolyte increases significantly, outweighing the decrease in κ.

What is the exact function of a Salt Bridge?

1. Completes the electrical circuit.
2. Maintains electrical neutrality by allowing flow of ions.

Difference between EMF and Potential Difference?
  • EMF: Potential diff when no current flows (Max voltage).
  • Potential Diff: Potential diff when current is flowing (Less than EMF).

⚡ Key Formula Cheat Sheet

Resistance & Conductance
R = ρ (l/A)
G = 1/R = κ (A/l)
Cell Constant (G*) = l/A
Molar Conductivity
Λm = (κ × 1000) / C
α = Λm / Λ°m
Nernst Equation
Ecell = E° – (0.059/n) log Q
ΔG° = -nFE°cell
Faraday’s First Law
m = ZIt = (M × I × t) / (nF)

Glossary of Terms

Anode: Electrode where oxidation (loss of electrons) occurs.
Cathode: Electrode where reduction (gain of electrons) occurs.
Electrolyte: Substance that conducts electricity in molten or aqueous state via ions.
Overpotential: Extra potential required to drive a reaction (especially gas evolution).
SHE: Standard Hydrogen Electrode (Reference, 0.00V).
Fuel Cell: Cell that converts combustion energy directly to electricity.

Conclusion

Electrochemistry connects chemical reactions with electricity. This master guide has covered everything from fundamental conductance to complex commercial batteries. Remember to practice the numericals using the provided calculators and review the cheat sheet before exams.

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by Rezyo Staff WriterJanuary 15, 2026 CBSE, CBSE Class 12, Chemistry0 comments

Class 12 Chemistry Unit 1: Solutions – Practice Questions Quiz (2025-26)

Physical Chemistry often feels abstract until you see how the variables interact. Unit 1: Solutions establishes the mathematical framework for the Class 12 syllabus, linking concentration terms to observable changes in vapor pressure and boiling points. This page moves beyond static textbook definitions to offer a hands-on learning experience.

Here, you will find a dynamic dashboard to test your grasp of Molarity, Molality, and Henry’s Law in real-time. We break down the thermodynamics behind Ideal and Non-Ideal solutions using comparison matrices and deviation logic trees, allowing you to predict positive or negative deviations without rote memorization. From visualizing surface barriers in vapor pressure to calculating the Van’t Hoff factor for complex ions, this resource aligns strictly with the 2025-26 academic standards to prepare you for board exams and competitive tests.

Class XII 2025-26 Unit 1: Solutions | Rezyo.in
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Class XII Chemistry Unit 1 • 2025-26

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Thermodynamics of Solution Mixing

Property
Ideal
Positive Dev
Negative Dev
Interaction
A-B ≈ A-A
A-B < A-A
A-B > A-A
Enthalpy ($\Delta H_{mix}$)
0
+ve (Endo)
-ve (Exo)
Volume ($\Delta V_{mix}$)
0
+ve (Expands)
-ve (Shrinks)
Entropy ($\Delta S_{mix}$)
+ve (Always Increases for Spontaneous Mixing)
Gibbs Energy ($\Delta G_{mix}$)
-ve (Always Decreases for Spontaneous Mixing)

Solubility Dynamics

S

Solid in Liquid

  • Effect of Pressure Negligible
  • Temp (Endothermic $\Delta H > 0$) Increases
  • Temp (Exothermic $\Delta H < 0$) Decreases
G

Gas in Liquid

  • Effect of Pressure Increases (Henry’s Law)
  • Effect of Temp Decreases
  • Henry’s Constant ($K_H$) Higher $K_H$ = Lower Solubility

Azeotrope Classification Hub

Minimum Boiling Azeotrope

Deviation: Large Positive

Example: Ethanol (95.5%) + Water (4.5%)

BP: 351.15 K (Lower than water 373K)

Maximum Boiling Azeotrope

Deviation: Large Negative

Example: Nitric Acid (68%) + Water (32%)

BP: 393.5 K (Higher than water 373K)

Visual Concept: Surface Barrier Effect

Why does adding a non-volatile solute lower vapour pressure?

Gray particles occupy surface area, preventing Blue particles from escaping.

Raoult’s Law Deviations

Deviation Logic Tree

Compare Interactions
Solute (A) & Solvent (B)
A-B < A-A, B-B
↓
Positive Dev ΔH > 0
A-B ≈ A-A
↓
Ideal ΔH = 0
A-B > A-A
↓
Negative Dev ΔH < 0

Conceptual Doubts (FAQ)

Q: Why does adding salt to ice melt it (Depression in Freezing Point)?

Adding a non-volatile solute (salt) lowers the freezing point of the solvent (water). If the ambient temperature is higher than this new, lowered freezing point (e.g., -5°C), the ice cannot exist as a solid and melts into liquid water.

Q: Why are aquatic species more comfortable in cold water?

This relates to Henry’s Law. The value of Henry’s constant ($K_H$) increases with temperature. Since solubility is inversely proportional to $K_H$, the solubility of oxygen decreases as water gets warmer.

Q: What is the medical condition “Edema”?

Excess salt intake increases ion concentration in tissue fluids. Water moves out of cells via osmosis to balance this, causing water retention and puffiness known as Edema.

Chapter Summary & Outcomes

You have navigated the complexities of Solution Chemistry. From mastering concentration units to applying the Van’t Hoff factor for ionic compounds, these concepts form the bedrock of Physical Chemistry.

Mass % is T-Independent VP ∝ Mole Fraction i > 1 for Dissociation

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by Rezyo Staff WriterJanuary 15, 2026 CBSE Class 12, Chemistry0 comments

Class 12 Chemistry Unit 7: Alcohols, Phenols & Ethers | Notes, Mechanisms & Reaction Maps (2025-26)

Organic compounds containing Carbon-Oxygen bonds form a massive part of industrial and biological chemistry. Unit 7 focuses on three specific classes: Alcohols, Phenols, and Ethers.

This guide breaks down their structural differences, nomenclature rules, and preparation methods, such as Hydroboration-Oxidation and the Cumene process. You will study specific reaction mechanisms, including the acid-catalyzed hydration of ethene, and learn to distinguish between primary, secondary, and tertiary alcohols using the Lucas Test.

Try Practice Questions and Answers with Hints

From understanding the acidity of phenols to mastering the cleavage of ethers with hydrogen iodide, these resources cover the essential concepts required for board exams and competitive entrance tests.

Class XII 2025-26: Alcohols, Phenols, and Ethers | Rezyo.in
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Unit 7: Alcohols, Phenols, and Ethers

Class XII Chemistry (2025-26 Session). Focusing on Competency-Based Assessment, Molecular Classification, and Reaction Mechanisms.

2025-26 Curricular Updates

The current session emphasizes competency-based education. Students must apply concepts rather than rely on rote memorization.

CBSE Focus

  • Green Chemistry context.
  • Mechanism identification.
  • Material science applications.

ISC Focus

  • Dow’s process specifics.
  • Aryl ether substitution.
  • Detailed mechanism coverage.

WBCHSE Focus

  • Semester-based integration.
  • Acidic nature of phenol.
  • Traditional functional tests.

Nomenclature Pitfalls

Confusion often arises between Common and IUPAC names. Below are high-frequency exam compounds.

Structure Common Name IUPAC Name
HO-CH2-CH2-OH Ethylene Glycol Ethane-1,2-diol
CH2(OH)-CH(OH)-CH2(OH) Glycerol Propane-1,2,3-triol
OH-C6H4-CH3 (ortho) o-Cresol 2-Methylphenol
C6H5-O-CH3 Anisole Methoxybenzene
C6H5-O-C2H5 Phenetole Ethoxybenzene

Structural Geometry & Hybridization

Molecule Angle Reason
Methanol108.9°Lone pair repulsion.
Phenol109.0°sp2 carbon effect.
Ether111.7°Steric repulsion.

Physical Properties

Boiling Point Trends

Alcohols have significantly higher boiling points than hydrocarbons and ethers of comparable molecular mass due to Intermolecular Hydrogen Bonding.

Solubility Dynamics

Lower Alcohols

Highly soluble in water due to H-bonds with water molecules.

Effect of Mass

Solubility decreases as the hydrophobic alkyl group size increases.

Separation: o-Nitrophenol vs p-Nitrophenol

Ortho-Nitrophenol

Volatile (Steam Distillable)

Due to Intramolecular H-Bonding (within the same molecule), it does not associate strongly with other molecules, lowering its boiling point.

Para-Nitrophenol

Non-Volatile (Remains in flask)

Due to Intermolecular H-Bonding (between different molecules), it forms associated aggregates, significantly raising the boiling point.

Mechanism Explorer

Interactive

Topic: Acid Catalysed Hydration of Ethene to Ethanol.

Step 1: Protonation

Hydronium ion transfers a proton to the double bond.

CH2=CH2 + H+  →  CH3-CH2+

Step 2: Nucleophilic Attack

Water attacks the carbocation.

R+ + H2O  →  R-OH2+

Step 3: Deprotonation

Loss of proton regenerates catalyst.

R-OH2+  →  R-OH + H+

Preparation Methods

1. From Cumene (Commercial)

High Yield

Most phenols produced via this method; Acetone is a valuable by-product.

Benzene + Propene
Cumene
Phenol + Acetone

2. Hydroboration-Oxidation

Alkene + (BH3)2 → Alcohol

Rule: Follows Anti-Markovnikov addition.

Commercial & Green Chemistry

Methanol

CO + 2H2 → CH3OH. Highly toxic (causes blindness).

Ethanol

Fermentation of molasses. Denatured with CuSO4/Pyridine.

Green Context

Ethanol is used as a bio-solvent replacing toxic chlorides, and as a fuel blend (Gasohol) to reduce emissions.

Reaction Mechanisms

Grignard Synthesis

ReactantResult
HCHOPrimary Alcohol
RCHOSecondary Alcohol
RCORTertiary Alcohol

Ether Cleavage Decision Tree

Reacting R-O-R’ with HI. Where does the Iodide go?

Reaction: R-O-R’ + HI
Is one R group Tertiary (3°)?
YES
Mechanism: SN1
Iodide attacks Tertiary Carbon.
NO
Mechanism: SN2
Iodide attacks Smaller/Less hindered group.

The Temperature Trap: 413K vs 443K

The dehydration of alcohols with concentrated H2SO4 yields completely different products based on the temperature.

At 443 K (170°C)

C2H5OH → CH2=CH2 + H2O

Mechanism: Elimination.
Product: Alkene (Ethene).
Thermodynamically favored at higher temperatures (entropy increases).

At 413 K (140°C)

2 C2H5OH → C2H5-O-C2H5

Mechanism: Nucleophilic Substitution (SN2).
Product: Ether (Diethyl Ether).
Requires excess alcohol.

Synthesis of Aspirin (Acetylsalicylic Acid)

Produced by the acetylation of Salicylic acid using Acetic Anhydride in the presence of acid catalyst.

  • Reactant: Salicylic Acid (2-Hydroxybenzoic acid)
  • Reagent: Acetic Anhydride + Conc H2SO4
  • Product: Aspirin (Analgesic & Antipyretic)
  • By-product: Acetic Acid (Vinegar smell)
Salicylic Acid + (CH3CO)2O Acetylsalicylic Acid + CH3COOH

Resonance in Phenol & Phenoxide

Why is Phenol acidic and Ortho/Para directing?

1. Acidic Nature

The Phenoxide ion formed after losing H+ is resonance stabilized. The negative charge is delocalized over the benzene ring.

Negative charge moves to:
Ortho → Para → Ortho positions.

Comparison: Alkoxide ion (R-O–) has no resonance, making alcohols weaker acids than phenol.

2. Ortho/Para Directing

The lone pair on Oxygen is pushed into the ring (+R Effect), increasing electron density.

Electron density is highest at:
Ortho (2,6) and Para (4) positions.

Electrophiles (E+) attack these electron-rich positions.

Key Conversion Pathways

Convert Ethanol to Acetone (Propanone)

Ethanol →[Cu/573K] Ethanal →[1. CH3MgBr 2. H3O+] Propan-2-ol →[Cu/573K] Acetone

Convert Phenol to Aspirin

Phenol →[NaOH] Sod. Phenoxide →[CO2 / H+] Salicylic Acid →[(CH3CO)2O / H+] Aspirin

The Haloform (Iodoform) Test

Key Identification Test

Reagents: NaOH + I2 (or NaOI).
Positive Result: Formation of Yellow Precipitate (CHI3).

Condition: The molecule must contain the CH3-CH(OH)- group (methyl carbinol) or CH3-C=O group.

Positive Ethanol, Propan-2-ol
Negative Methanol, Propan-1-ol

Oxidation Pathways

TypeMild (PCC)Strong (KMnO4)Cu / 573K
1°AldehydeAcidAldehyde
2°KetoneKetoneKetone
3°NoneMix of AcidsAlkene

Electrophilic Substitution

Phenol (OH is Activator)

  • Br2 / CS2: Mono-substituted (o/p-Bromophenol).
  • Br2 Water: Poly-substituted (2,4,6-Tribromophenol – White PPT).
  • Conc. HNO3: Picric Acid (2,4,6-Trinitrophenol).

Anisole (OR is Activator)

  • Friedel-Crafts: CH3Cl/AlCl3 gives o/p-Methoxy toluene.
  • Nitration: Gives mixture of o/p-Nitroanisole.

Named Reactions

Reimer-Tiemann

Phenol + CHCl3 + NaOH → Salicylaldehyde.

Intermediate: Dichlorocarbene (:CCl2).

Kolbe’s Reaction

Sod. Phenoxide + CO2 → Salicylic Acid.

Phenol Conversion Roadmap

Phenol
Zn Dust
Benzene
Na2Cr2O7
Benzoquinone
Br2(aq)
2,4,6-Tribromo…
Conc HNO3
Picric Acid

Interactive Lab Bench

Select a test and a sample to see the virtual result.

1. Select Reagent
2. Select Sample

Ready

Reference Material

Distinction Table

TestAlcoholPhenol
LitmusNeutralBlue → Red
FeCl3No RxnViolet Color
Br2 WaterNo RxnWhite PPT

Reagent Cheat Sheet

PCC
1° Alc → Aldehyde
LiAlH4
Acid → Alcohol
Zn Dust
Phenol → Benzene
H2SO4 443K
Alc → Alkene
H2SO4 413K
Alc → Ether
Na2Cr2O7
Phenol → Quinone

Flashcards

Tap to flip
Lucas Reagent
Conc. HCl + ZnCl2
Picric Acid
2,4,6-Trinitrophenol
Wood Spirit
Methanol (CH3OH)
Esterification Order
1° > 2° > 3°
Anisole Name
Methoxybenzene
Denaturation
Add CuSO4 + Pyridine

Quiz

1. Reagent to convert Phenol to Benzene?

Zn Dust
NaBH4

2. Electrophile in Reimer-Tiemann?

:CCl2
CO2

3. Product of 3° Alcohol + Cu/573K?

Ketone
Alkene

Frequently Asked Questions

Q: Why is Phenol more acidic than Ethanol?

A: The phenoxide ion formed after losing a proton is resonance stabilized (negative charge delocalized over the benzene ring). The ethoxide ion is not stabilized and is actually destabilized by the +I effect of the ethyl group.

Q: Why do boiling points of alcohols decrease with branching?

A: Branching makes the molecule more spherical, reducing its surface area. This decreases the Van der Waals forces between molecules, lowering the boiling point.

Q: Why can’t we prepare Anisole from Bromobenzene and Sodium Methoxide?

A: Bromobenzene (Aryl Halide) has a partial double bond character in the C-Br bond due to resonance, making it extremely unreactive towards nucleophilic substitution. The correct method uses Sodium Phenoxide and Methyl Bromide.

Q: Why is sulphuric acid used in esterification?

A: It acts as a dehydrating agent to remove the water formed, shifting the equilibrium to the right (forward direction) to yield more ester, and also acts as a catalyst.

Q: How to distinguish between 1°, 2°, and 3° alcohols?

A: Use the Lucas Test (ZnCl2 + HCl). 3° alcohols react immediately (turbidity), 2° react in ~5 mins, and 1° do not react at room temperature.

Summary & Conclusion

This unit has covered the critical aspects of Alcohols, Phenols, and Ethers, moving from basic nomenclature to complex reaction mechanisms.

Key Takeaways:

  • Structure Dictates Property: Hydrogen bonding elevates boiling points and solubility.
  • Resonance Rules: The acidity of phenol and its electrophilic substitution patterns are driven by resonance.
  • Reagent Specificity: The outcome of reactions (like Oxidation or Dehydration) depends heavily on the specific reagent and conditions (Temperature, Strength).

Mastering these concepts requires understanding the “Why” behind the reactions—steric hindrance, electronic effects, and stability of intermediates—rather than just memorizing equations.

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