🇮🇳 JEE Advanced · flashcards
JEE Advanced Chemistry Flashcards
50 question-and-answer cards covering Chemistry as it is examined in JEE Advanced. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Chemistry deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Define oxidation and reduction in terms of electron transfer.
Oxidation is the loss of electrons; reduction is the gain of electrons. Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
Define oxidising agent and reducing agent.
An oxidising agent accepts electrons (and is itself reduced). A reducing agent donates electrons (and is itself oxidised).
Define oxidation and reduction in terms of oxidation number.
Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.
What is a redox (oxidation-reduction) reaction?
A reaction in which oxidation and reduction occur simultaneously, involving transfer of electrons and changes in oxidation numbers. Example: $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$.
State the main rules for assigning oxidation numbers.
Free element $= 0$; monatomic ion $=$ its charge; $\ce{O} = -2$ (except peroxides $-1$, $\ce{OF2}$ $+2$); $\ce{H} = +1$ (except metal hydrides $-1$); sum in a neutral molecule $= 0$; sum in a polyatomic ion $=$ its charge.
What is a disproportionation reaction? Give an example.
A redox reaction in which the same element is simultaneously oxidised and reduced. Example: $$\ce{Cl2 + 2OH- -> Cl- + ClO- + H2O}$$ where $\ce{Cl}$ goes from $0$ to both $-1$ and $+1$.
Define a neutralisation reaction and write its net ionic equation.
A reaction between an acid and a base producing salt and water. Net ionic equation for strong acid + strong base: $$\ce{H+ + OH- -> H2O}$$
What is the enthalpy of neutralisation for a strong acid and strong base, and why is it constant?
It is approximately $-57.1\ \text{kJ mol}^{-1}$. It is constant because the only reaction occurring is $\ce{H+ + OH- -> H2O}$, regardless of the specific strong acid or base used.
Define a displacement reaction and give the two main types.
A reaction in which a more reactive element displaces a less reactive one from its compound. Types: single displacement (e.g. $\ce{Fe + CuSO4 -> FeSO4 + Cu}$) and double displacement (e.g. $\ce{AgNO3 + NaCl -> AgCl + NaNO3}$).
Write an example of a metal displacing hydrogen from an acid.
$$\ce{Zn + 2HCl -> ZnCl2 + H2 ^}$$ Zinc, being more reactive than hydrogen, displaces it.
What does the reactivity series predict about displacement reactions?
A metal higher in the reactivity series can displace a metal lower in the series from its salt solution. For example, $\ce{Zn}$ displaces $\ce{Cu}$ but $\ce{Cu}$ cannot displace $\ce{Zn}$.
State Boyle's law with its mathematical expression.
At constant temperature and amount of gas, the volume of a gas is inversely proportional to its pressure: $$P \propto \frac{1}{V} \quad \Rightarrow \quad PV = \text{constant}, \quad P_1V_1 = P_2V_2$$
State Charles's law with its mathematical expression.
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature: $$V \propto T \quad \Rightarrow \quad \frac{V}{T} = \text{constant}, \quad \frac{V_1}{T_1} = \frac{V_2}{T_2}$$
State Gay-Lussac's (pressure-temperature) law.
At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature: $$\frac{P}{T} = \text{constant}, \quad \frac{P_1}{T_1} = \frac{P_2}{T_2}$$
State Avogadro's law.
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Hence $V \propto n$ at constant $T$ and $P$.
Write the ideal gas equation and name each term.
$$PV = nRT$$ where $P$ = pressure, $V$ = volume, $n$ = moles, $R$ = universal gas constant, $T$ = absolute temperature.
Give the value of the universal gas constant $R$ in common units.
$R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1} = 2\ \text{cal K}^{-1}\text{mol}^{-1}$.
Express the ideal gas equation in terms of density and molar mass.
Since $n = \frac{m}{M}$, $$PM = \frac{m}{V}RT = dRT \quad \Rightarrow \quad d = \frac{PM}{RT}$$
State the combined gas law.
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ combining Boyle's, Charles's, and Gay-Lussac's laws for a fixed amount of gas.
What is the absolute (Kelvin) scale of temperature and absolute zero?
The absolute scale measures temperature from absolute zero ($0\ \text{K} = -273.15\,^{\circ}\text{C}$), the temperature at which the volume/pressure of an ideal gas extrapolates to zero. Conversion: $T(\text{K}) = t(^{\circ}\text{C}) + 273.15$.
What is the compressibility factor $Z$ and its value for an ideal gas?
$$Z = \frac{PV}{nRT}$$ For an ideal gas $Z = 1$ at all conditions. Deviations ($Z \neq 1$) indicate real (non-ideal) gas behaviour.
What do $Z > 1$ and $Z < 1$ indicate about a real gas?
$Z < 1$: gas is more compressible than ideal (attractive forces dominate, common at moderate pressures). $Z > 1$: gas is less compressible than ideal (repulsive forces/molecular volume dominate, common at high pressures).
Under what conditions do real gases behave most ideally?
At low pressure and high temperature, where molecules are far apart, intermolecular forces are negligible, and molecular volume is small compared with the container volume.
Write the van der Waals equation for $n$ moles of a real gas and explain the correction terms.
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$ The term $\frac{an^2}{V^2}$ corrects pressure for intermolecular attraction; $nb$ corrects volume for the finite size of molecules. $a$ and $b$ are van der Waals constants.
What this deck covers
The Chemistry deck follows the JEE Advanced Chemistry syllabus — 37 chapters and 201 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 1.4 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 165 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Chemistry flashcards FAQ
How many Chemistry flashcards are in this JEE Advanced deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these JEE Advanced flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Chemistry cards cover?
They follow the JEE Advanced Chemistry syllabus — 37 chapters and 201 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.