Start here

Welcome to Class 12 Chemistry

Class 11 taught you what matter is made of and why reactions happen. Class 12 asks how much is dissolved, how much electrical energy a reaction can give, how fast it goes, and how real molecules (metals, complexes, organic compounds and the molecules of life) behave.

1. Where Class 11 left you

In Class 11 you built the foundations: the mole and stoichiometry, atomic structure and the periodic table, chemical bonding, thermodynamics ($\Delta H$, $\Delta S$, $\Delta G$), equilibrium and pH, redox reactions, and the first ideas of organic chemistry: IUPAC names, electronic effects and hydrocarbons. Those topics answer what substances are and whether a reaction can go.

Class 12 puts these foundations to work in three blocks. Physical chemistry (Chapters 1–3) makes three quantities precise: concentration, the electrical work a reaction can do, and the rate at which it happens. Inorganic chemistry (Chapters 4–5) studies the transition and inner-transition metals and the coordination compounds they form, which give gemstones, paints and haemoglobin their colours. Organic chemistry (Chapters 6–10) follows functional groups step by step, from haloalkanes through alcohols, carbonyl compounds and amines to the biomolecules in your own cells. Throughout, you will learn to predict products by understanding reaction mechanisms, not by memorising them.

2. A story that needs three chapters at once

A Daniell cell is a zinc rod in zinc sulphate solution, joined by a salt bridge to a copper rod in copper sulphate solution. It lights a small bulb. While it runs, zinc atoms become $\mathrm{Zn^{2+}}$ ions and $\mathrm{Cu^{2+}}$ ions become copper metal: $$\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}.$$ The blue colour of the copper solution slowly fades. How long will the cell last, and does its voltage stay steady?

Answering this takes concentrations (Chapter 1: how many moles of $\mathrm{Cu^{2+}}$ are left in the beaker), electrochemistry (Chapter 2: the Nernst equation $E=E^{\circ}_{cell}-\tfrac{0.059}{2}\log\tfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}$ gives the voltage from those concentrations) and rates (Chapter 3: the current fixes how fast the ions are used up, since $2F$ coulombs of charge reduce one mole of $\mathrm{Cu^{2+}}$). Run the lab: the voltage hardly moves for most of the cell's life, then falls away more and more steeply near the end.

Interactive 1 — A Daniell Cell Running Down

Each half-cell holds $1.00\ \mathrm{L}$ of solution, starting at $1.00\ \mathrm{M}$ $\mathrm{Cu^{2+}}$ and $1.00\ \mathrm{M}$ $\mathrm{Zn^{2+}}$. A constant current $I$ is drawn. After time $t$, $\xi=\dfrac{It}{2F}$ mol of $\mathrm{Cu^{2+}}$ has been reduced, so $[\mathrm{Cu^{2+}}]=1-\xi$ and $[\mathrm{Zn^{2+}}]=1+\xi$. The voltage follows the Nernst equation with $E^{\circ}_{cell}=1.10\ \mathrm{V}$ at $298\ \mathrm{K}$ ($F=96\,487\ \mathrm{C\,mol^{-1}}$). Watch the blue fade and the voltage curve.

2.0
time $t$ (h)—
$[\mathrm{Cu^{2+}}]$ (M)—
$[\mathrm{Zn^{2+}}]$ (M)—
cell voltage $E$ (V)—
✏️ Example 1 — How long, and at what voltage?
The cell above delivers a steady $2.00\ \mathrm{A}$. (a) How long does it take for $[\mathrm{Cu^{2+}}]$ to fall to $0.100\ \mathrm{M}$? (b) What is the voltage then?
Moles used (Ch. 1) $[\mathrm{Cu^{2+}}]$ falls from $1.00$ to $0.100\ \mathrm{M}$ in $1.00\ \mathrm{L}$, so $\xi=0.900\ \mathrm{mol}$ of $\mathrm{Cu^{2+}}$ has been reduced, and $[\mathrm{Zn^{2+}}]=1.90\ \mathrm{M}$.
(a) Charge and time (Ch. 2–3) Each $\mathrm{Cu^{2+}}$ needs 2 electrons: $Q=2F\xi=2\times96\,487\times0.900=1.737\times10^{5}\ \mathrm{C}$. Then $t=\dfrac{Q}{I}=\dfrac{1.737\times10^{5}}{2.00}=8.68\times10^{4}\ \mathrm{s}\approx24.1\ \mathrm{h}$.
(b) Nernst equation $$E=1.10-\frac{0.059}{2}\log\frac{1.90}{0.100}=1.10-0.0295\times1.279=1.06\ \mathrm{V}.$$
Meaning After 90% of the copper ions are used up, the voltage has dropped by less than 4%. That is why the voltage of a cell is a poor guide to how much charge it has left, and why it drops steeply only at the very end, when $[\mathrm{Cu^{2+}}]$ approaches zero and the log term grows without limit.

3. The ten chapters at a glance

The course follows the current NCERT (rationalised) Class 12 syllabus, chapter by chapter, in the same order.

Physical: 1–3→Inorganic: 4–5→Organic: 6–9→Biomolecules: 10
Chapter 1

Solutions

Concentration units, Henry's and Raoult's laws, non-ideal solutions and colligative properties.

$\Delta T_{f}=i\,K_{f}\,m$
Open chapter → Chapter 2

Electrochemistry

Galvanic cells, electrode potentials, the Nernst equation, conductance, electrolysis, batteries and corrosion.

$\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$
Open chapter → Chapter 3

Chemical Kinetics

Rates and rate laws, integrated rate equations and half-lives, the Arrhenius equation and collision theory.

$k=Ae^{-E_{a}/RT}$
Open chapter → Chapter 4

The d- and f-Block Elements

Transition-metal trends, colour and magnetism, $\mathrm{K_{2}Cr_{2}O_{7}}$ and $\mathrm{KMnO_{4}}$, lanthanoids and actinoids.

$\mu=\sqrt{n(n+2)}\ \mathrm{BM}$
Open chapter → Chapter 5

Coordination Compounds

Werner's theory, IUPAC names, isomerism, valence bond and crystal field theories.

$\Delta_{t}=\tfrac49\Delta_{o}$
Open chapter → Chapter 6

Haloalkanes and Haloarenes

Preparation, SN1 and SN2 mechanisms, stereochemistry, elimination and polyhalogen compounds.

$\mathrm{R{-}X+Nu^{-}\rightarrow R{-}Nu+X^{-}}$
Open chapter → Chapter 7

Alcohols, Phenols and Ethers

Preparation, acidity, reactions of the O–H and C–O bonds, and Williamson ether synthesis.

$\mathrm{C_{6}H_{5}OH}$
Open chapter → Chapter 8

Aldehydes, Ketones and Carboxylic Acids

Nucleophilic addition, named reactions, tests for carbonyl groups, and the acidity of carboxylic acids.

$\mathrm{R{-}CHO\rightarrow R{-}COOH}$
Open chapter → Chapter 9

Amines

Preparation, basicity, tests that tell 1°, 2° and 3° amines apart, and diazonium-salt chemistry.

$\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}}$
Open chapter → Chapter 10

Biomolecules

Carbohydrates, proteins and enzymes, vitamins, nucleic acids and hormones.

$\mathrm{C_{6}H_{12}O_{6}}$
Open chapter →
🗺️ How the chapters connect
  1. Solutions (Ch. 1) supply the concentrations that appear in the Nernst equation, in conductivity (Ch. 2) and in every rate law (Ch. 3).
  2. Electrochemistry (Ch. 2) links thermodynamics to measurable voltages: $\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$ and $E^{\circ}_{cell}=\tfrac{0.059}{n}\log K$ at $298\ \mathrm{K}$.
  3. Kinetics (Ch. 3) adds the dimension of time. Thermodynamics tells you whether a reaction can go; kinetics tells you whether it goes fast enough to matter.
  4. d- and f-block and coordination chemistry (Ch. 4–5) use electrode potentials from Ch. 2 to explain which oxidation states are stable, and use d-orbital splitting to explain colour and magnetism.
  5. Organic chapters (Ch. 6–9) are one long story of functional-group conversions. Haloalkanes make alcohols; alcohols make carbonyl compounds and ethers; carbonyl compounds make acids; nitro compounds and amides make amines; amines make diazonium salts. Mechanisms (SN1, SN2, nucleophilic addition, electrophilic substitution) tie it together.
  6. Biomolecules (Ch. 10) show the same functional groups at work in sugars (hemiacetals), proteins (amides) and nucleic acids (phosphate esters).

Threads to watch for: numbers always carry units and sensible significant figures; every reaction is balanced for atoms and charge; and in organic chemistry, always ask where the electrons come from and where they go.

✏️ Example 2 — Three exams, one idea
The same concept is asked in different styles. Answer all three. (a) Board: State Henry's law. (b) NEET: At the same partial pressure, which is more soluble in water, a gas with $K_{H}=1.67\ \mathrm{kbar}$ or one with $K_{H}=76.5\ \mathrm{kbar}$? (c) JEE: By what factor does the dissolved amount of the second gas change when its partial pressure rises from $0.78$ to $3.12\ \mathrm{bar}$ at constant temperature?
(a) At constant temperature, the mole fraction of a gas dissolved in a liquid is proportional to the partial pressure of the gas above it: $p=K_{H}\,x$.
(b) $x=p/K_{H}$, so the gas with the smaller $K_{H}$ ($1.67\ \mathrm{kbar}$, which is $\mathrm{CO_{2}}$) is about $76.5/1.67\approx46$ times more soluble.
(c) $x\propto p$, so the factor is $3.12/0.78=4.0$. This is the nitrogen in a diver's blood at about $30\ \mathrm{m}$ depth; see Section 1.1.
🧭 How to use this course
  • Read with a pen. Try every worked example before you open its solution, then compare your working step by step. Physical chemistry is learned by calculating; organic chemistry by drawing structures and mechanisms.
  • Play with the labs. Each section has two interactives. Change the settings until you can predict what will happen before you press ▶ Play.
  • Do the exercises in order. They run from routine to challenging, and every part has a full solution.
  • Use the 🎯 Exam corner. Each section ends with problems in the styles of the Board, JEE Main, JEE Advanced and NEET papers, each tagged and fully solved. Board answers need clear definitions and stated units. JEE and NEET reward speed and spotting traps.
  • Use the quiz as a gate. Five questions end every section. If you score below 4, reread before moving on.
  • Key terms. Dotted words show their meaning when you hover over or tap them. The Glossary (with flashcards) and the Formula Sheet collect everything in one place for revision.
← Class 12 Chemistry — ContentsChapter 1 — Solutions →