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JEE Main Inorganic Chemistry Syllabus

The Inorganic Chemistry syllabus for JEE Main in full: 4 chapters, 10 topics and 18 sub-topics, every one listed below, plus 50 Inorganic Chemistry flashcards with answers.

4Chapters
10Topics
18Sub-topics
50Flashcards

Inorganic Chemistry syllabus: all 4 chapters

Expand any chapter to see its topics and sub-topics.

  1. Classification of Elements and Periodicity in Properties

     3 topics
    • Modern periodic law and present form of the periodic table
    • s, p, d and f block elements
    • Periodic trends in properties of elements
      • Atomic and ionic radii
      • Ionization enthalpy
      • Electron gain enthalpy
      • Valence
      • Oxidation states and chemical reactivity

  2. P-Block Elements

     1 topic
    • Group -13 to Group 18 Elements
      • Electronic configuration and general trends
      • Unique behaviour of the first element in each group

  3. d - and f- Block Elements

     2 topics
    • Transition Elements
      • General introduction, electronic configuration, occurrence and characteristics
      • General trends in properties of the first-row transition elements
      • Physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloy formation
      • Preparation, properties, and uses of K2Cr2O7, and KMnO4
    • Inner Transition Elements
      • Lanthanoids - Electronic configuration, oxidation states, and lanthanoid contraction
      • Actinoids - Electronic configuration and oxidation states

  4. Co-ordination Compounds

     4 topics
    • Introduction to coordination compounds
      • Werner's theory
      • Ligands, coordination number, denticity, chelation
    • IUPAC nomenclature of mononuclear co-ordination compounds, isomerism
    • Bonding-Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties
    • Importance of co-ordination compounds
      • In qualitative analysis
      • Extraction of metals
      • In biological systems

JEE Main Inorganic Chemistry flashcards — 50 questions with answers

  1. State the modern periodic law.

    The physical and chemical properties of elements are a periodic function of their atomic numbers (number of protons), i.e. properties repeat at regular intervals when elements are arranged in order of increasing atomic number $Z$.

  2. How is the long form (modern) periodic table organized in terms of periods and groups?

    It has 7 horizontal periods and 18 vertical groups. The period number equals the principal quantum number $n$ of the outermost shell being filled; elements in a group have similar valence-shell electronic configurations and hence similar properties.

  3. On what basis are the s, p, d and f blocks defined?

    They are named after the subshell into which the differentiating (last) electron enters: s-block (last electron in $ns$), p-block ($np$), d-block ($(n-1)d$), and f-block ($(n-2)f$).

  4. Which groups make up the s-block, and what is its general valence-shell configuration?

    Groups 1 (alkali metals) and 2 (alkaline earth metals). General configuration $ns^{1\text{--}2}$.

  5. Which groups make up the p-block and what is its general valence configuration?

    Groups 13 to 18. General valence-shell configuration $ns^{2}np^{1\text{--}6}$ (for He it is $1s^{2}$).

  6. What is the general valence configuration of d-block (transition) elements?

    $(n-1)d^{1\text{--}10}\,ns^{0\text{--}2}$, where $n$ is the outermost principal shell. They lie in groups 3 to 12.

  7. What is the general valence configuration of f-block elements and what are the two series called?

    $(n-2)f^{1\text{--}14}(n-1)d^{0\text{--}1}ns^{2}$. The two series are the lanthanoids ($\ce{^{4f}}$, Ce to Lu) and actinoids ($\ce{^{5f}}$, Th to Lr); they are the inner transition elements.

  8. Define atomic radius and state how it varies across a period and down a group.

    Atomic radius is roughly the distance from the nucleus to the outermost electron shell. It decreases across a period (increasing effective nuclear charge with same shell) and increases down a group (new shells added).

  9. Why is the ionic radius of a cation smaller and an anion larger than the parent atom?

    A cation loses electrons, increasing effective nuclear charge per electron and often removing a shell, so it shrinks. An anion gains electrons, increasing electron-electron repulsion and effective size, so it grows larger than the neutral atom.

  10. For an isoelectronic series such as $\ce{O^{2-}, F^-, Na^+, Mg^{2+}, Al^{3+}}$, how does ionic radius vary?

    All have the same number of electrons (10). Ionic radius decreases as nuclear charge $Z$ increases, so $\ce{O^{2-}} > \ce{F^-} > \ce{Na^+} > \ce{Mg^{2+}} > \ce{Al^{3+}}$.

  11. Define first ionization enthalpy ($\Delta_i H_1$).

    The energy required to remove the most loosely bound electron from one mole of isolated gaseous atoms in their ground state: $\ce{X(g) -> X^+(g) + e^-}$. It is always positive (endothermic).

  12. How does ionization enthalpy generally vary across a period and down a group?

    It increases across a period (rising $Z_{eff}$, smaller size) and decreases down a group (larger size and increased shielding make outer electron easier to remove).

  13. Why is the first ionization enthalpy of nitrogen higher than that of oxygen?

    $\ce{N}$ has a stable half-filled $2p^{3}$ configuration which is hard to disturb, whereas removing an electron from $\ce{O}$ ($2p^{4}$) relieves electron-electron repulsion in a paired orbital, so O loses its electron more readily.

  14. Why is the first ionization enthalpy of boron lower than that of beryllium?

    In $\ce{Be}$ ($2s^{2}$) the electron is removed from a stable fully filled 2s orbital, whereas in $\ce{B}$ ($2s^{2}2p^{1}$) the 2p electron is higher in energy and better shielded, so it is removed more easily.

  15. Define electron gain enthalpy ($\Delta_{eg}H$) and give its sign convention.

    The enthalpy change when an electron is added to a neutral gaseous atom: $\ce{X(g) + e^- -> X^-(g)}$. It is negative if energy is released (favourable) and positive if energy must be supplied.

  16. Why is the electron gain enthalpy of chlorine more negative than that of fluorine?

    Fluorine's small 2p subshell causes strong electron-electron repulsion on adding an electron, reducing the energy released. Chlorine's larger 3p orbital accommodates the extra electron with less repulsion, giving a more negative $\Delta_{eg}H$.

  17. Why do noble gases have large positive electron gain enthalpies?

    They have completely filled stable octet (or duplet) configurations; the incoming electron must enter a higher-energy new shell, so energy is absorbed rather than released.

  18. Define valence in terms of electronic configuration and state how it varies for s- and p-block elements across a period.

    Valence is the combining capacity, related to the number of valence electrons or electrons needed to complete the octet. For representative elements it rises from 1 to 4 and then falls to 0 across a period (e.g. Na=1, ..., C=4, ..., Ne=0).

  19. What is the general trend in metallic and non-metallic character across a period and down a group?

    Metallic (electropositive) character decreases across a period and increases down a group; non-metallic character increases across a period and decreases down a group.

  20. Why do elements show variable oxidation states, and how does this relate to chemical reactivity of metals vs non-metals?

    Variable oxidation states arise when an element can lose/share different numbers of electrons (common in d-block due to close $(n-1)d$ and $ns$ energies). Metallic reactivity (tendency to lose electrons) increases down a group; non-metal reactivity (tendency to gain electrons) increases up a group toward fluorine.

Show the other 30 Inorganic Chemistry flashcards

  1. Name the Group 13 to Group 18 element families.

    Group 13: Boron family; Group 14: Carbon family; Group 15: Pnictogens (nitrogen family); Group 16: Chalcogens (oxygen family); Group 17: Halogens; Group 18: Noble gases.

  2. What is the inert pair effect and where is it most prominent?

    The reluctance of the outer $ns^{2}$ electrons to participate in bonding due to poor shielding by intervening d and f electrons. It increases down groups 13-16, stabilizing lower oxidation states (e.g. $\ce{Pb^{2+}} > \ce{Pb^{4+}}$, $\ce{Tl^+} > \ce{Tl^{3+}}$).

  3. State the general valence configurations and characteristic stable oxidation states for Group 13 and Group 17.

    Group 13: $ns^{2}np^{1}$, common oxidation state $+3$ (also $+1$ lower down). Group 17 (halogens): $ns^{2}np^{5}$, characteristic oxidation state $-1$ (and positive states up to $+7$ for the heavier ones except F).

  4. Why does the first element of each group (e.g. Li, Be, B, C, N, O, F) differ markedly from the rest of its group?

    Because of its small size, high charge/radius ratio, high electronegativity, and absence of vacant d-orbitals in its valence shell. This causes anomalous behaviour and a diagonal relationship with the element below-right (e.g. Li-Mg, Be-Al, B-Si).

  5. State two examples of the anomalous behaviour of the first element of a group.

    (1) Nitrogen forms strong $p\pi$-$p\pi$ triple bonds ($\ce{N#N}$) and cannot expand its octet, unlike P which forms $\ce{PCl5}$. (2) Fluorine shows only $-1$ state and forms abnormally weak F-F bonds due to lone-pair repulsion in the small molecule.

  6. Give the general definition of a transition element (per IUPAC).

    An element whose atom has a partially filled d-subshell, or which can give rise to cations with a partially filled d-subshell. This excludes Zn, Cd, Hg (group 12) which have $d^{10}$ configuration in atoms and common ions.

  7. Write the electronic configurations of $\ce{Cr}$ and $\ce{Cu}$ and explain the anomaly.

    $\ce{Cr}: [Ar]3d^{5}4s^{1}$ and $\ce{Cu}: [Ar]3d^{10}4s^{1}$. The extra stability of exactly half-filled ($d^{5}$) and fully filled ($d^{10}$) subshells causes one 4s electron to shift into 3d.

  8. Name the first-row (3d) transition series elements in order.

    Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), Zinc (Zn).

  9. Why do transition metals show high melting points and high enthalpies of atomization?

    They have strong metallic bonding involving both $ns$ and $(n-1)d$ electrons. The greater the number of unpaired d-electrons available for bonding, the stronger the bonding; values peak around the middle of the series (Cr, ~the maximum).

  10. What is the trend in atomic radii across the first transition series, and why is it small?

    Atomic radii decrease slightly from Sc to about Mn/Fe, then stay nearly constant, rising a little at Cu/Zn. The poor shielding by d-electrons roughly balances the increasing nuclear charge, so the variation is small.

  11. Why do transition metals exhibit variable oxidation states?

    The $(n-1)d$ and $ns$ orbitals are very close in energy, so a variable number of electrons (from both subshells) can be lost. This gives a range of oxidation states differing by one, e.g. $\ce{Mn}$ shows $+2$ to $+7$.

  12. What is the most common oxidation state of the 3d series, and which element shows the maximum oxidation state of $+7$?

    $+2$ is the most common (loss of two 4s electrons). Manganese shows the maximum oxidation state of $+7$ (e.g. in $\ce{KMnO4}$).

  13. Why are most transition metal ions coloured?

    Their partially filled d-orbitals split into different energy levels in a ligand field; absorption of visible light promotes electrons via $d$-$d$ transitions, and the complementary colour is transmitted. Ions with $d^{0}$ or $d^{10}$ (e.g. $\ce{Sc^{3+}}$, $\ce{Zn^{2+}}$) are colourless.

  14. How is the magnetic moment of a transition metal ion calculated from unpaired electrons?

    Using the spin-only formula $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM), where $n$ is the number of unpaired electrons. Species with unpaired electrons are paramagnetic; those with none are diamagnetic.

  15. Calculate the spin-only magnetic moment of $\ce{Fe^{2+}}$ ($3d^{6}$).

    $\ce{Fe^{2+}}$ has 4 unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM.

  16. Why do transition metals and their compounds act as good catalysts?

    Because of their variable oxidation states (they can readily gain/lose electrons to form intermediates) and their ability to provide a large surface that adsorbs reactants, weakening their bonds. Examples: Fe in Haber process, $\ce{V2O5}$ in contact process.

  17. What are interstitial compounds? Give two characteristic properties.

    Compounds formed when small atoms (H, C, N, B) occupy interstitial sites in the metal lattice. They are very hard, have high melting points, retain metallic conductivity, and are chemically rather inert. Example: steel/cementite ($\ce{Fe3C}$), $\ce{TiC}$.

  18. Why do transition metals readily form alloys?

    Their atomic radii are very similar, so atoms of one metal can readily replace atoms of another in the crystal lattice, forming substitutional solid solutions (alloys) such as brass, bronze, and steel.

  19. Why do transition metals readily form complex (coordination) compounds?

    Because of their small size, high charge/ionic charge density, and availability of vacant d-orbitals of suitable energy that can accept lone pairs of electrons from ligands, e.g. $\ce{[Fe(CN)6]^{4-}}$, $\ce{[Cu(NH3)4]^{2+}}$.

  20. How is potassium dichromate ($\ce{K2Cr2O7}$) prepared from chromite ore (industrial steps)?

    Chromite $\ce{FeCr2O4}$ is fused with $\ce{Na2CO3}$ in air to give sodium chromate; this is acidified to sodium dichromate, then treated with KCl: $\ce{Na2Cr2O7 + 2KCl -> K2Cr2O7 + 2NaCl}$, and $\ce{K2Cr2O7}$ crystallizes out.

  21. Write the equilibrium between chromate and dichromate ions and state the effect of pH.

    $\ce{2CrO4^{2-} + 2H+ <=> Cr2O7^{2-} + H2O}$. Orange $\ce{Cr2O7^{2-}}$ predominates in acidic solution; yellow $\ce{CrO4^{2-}}$ predominates in alkaline solution.

  22. Write the half-reaction showing the oxidizing action of $\ce{K2Cr2O7}$ in acidic medium.

    $\ce{Cr2O7^{2-} + 14H+ + 6e^- -> 2Cr^{3+} + 7H2O}$, with $E^\circ = +1.33\,\text{V}$. Chromium changes from $+6$ to $+3$.

  23. How is potassium permanganate ($\ce{KMnO4}$) prepared from pyrolusite?

    $\ce{MnO2}$ is fused with KOH and an oxidant ($\ce{KNO3}$ or air) to give green $\ce{K2MnO4}$: $\ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O}$. The manganate is then oxidized (electrolytically or by $\ce{Cl2}$) to purple $\ce{KMnO4}$: $\ce{3MnO4^{2-} + 4H+ -> 2MnO4^- + MnO2 + 2H2O}$.

  24. Write the half-reaction and oxidation-state change for $\ce{KMnO4}$ acting as an oxidant in acidic medium.

    $\ce{MnO4^- + 8H+ + 5e^- -> Mn^{2+} + 4H2O}$, $E^\circ = +1.51\,\text{V}$. Mn goes from $+7$ to $+2$ (5-electron change).

  25. What are the products when $\ce{KMnO4}$ acts as an oxidant in neutral/faintly alkaline medium?

    It is reduced to brown $\ce{MnO2}$ (Mn $+7 \to +4$): $\ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-}$ (a 3-electron change).

  26. Name the inner transition elements and state which series corresponds to filling of 4f and 5f orbitals.

    They are the lanthanoids (lanthanides) and actinoids (actinides). Lanthanoids (Ce-Lu) involve filling of the $4f$ orbitals; actinoids (Th-Lr) involve filling of the $5f$ orbitals. They form the f-block.

  27. What is the general electronic configuration of the lanthanoids and their most common oxidation state?

    $[\text{Xe}]\,4f^{1\text{--}14}\,5d^{0\text{--}1}\,6s^{2}$. The characteristic and most stable oxidation state is $+3$; some also show $+2$ or $+4$ (e.g. $\ce{Eu^{2+}}$, $\ce{Ce^{4+}}$) when it gives a stable $f^{0}$, $f^{7}$ or $f^{14}$ configuration.

  28. Define lanthanoid contraction and state its cause.

    The steady decrease in atomic and ionic ($\ce{Ln^{3+}}$) radii from La to Lu across the lanthanoid series. It is caused by the poor shielding of one 4f electron by another, so the increasing nuclear charge progressively contracts the electron cloud.

  29. State two important consequences of the lanthanoid contraction.

    (1) The second- and third-row transition elements of a group have nearly identical radii (e.g. Zr~Hf, Nb~Ta), making them hard to separate. (2) It causes the very similar properties of the lanthanoids themselves, making their separation difficult.

  30. Why is $\ce{Ce^{4+}}$ a good oxidizing agent and $\ce{Eu^{2+}}$ a good reducing agent?

    $\ce{Ce^{4+}}$ readily gains an electron to revert to the very stable $+3$ state ($4f^{1} \to$ favourable), so it oxidizes others. $\ce{Eu^{2+}}$ ($4f^{7}$, half-filled) readily loses an electron to form $\ce{Eu^{3+}}$, so it acts as a reducing agent; both tend toward the stable $+3$ state.

JEE Main Inorganic Chemistry FAQ

What is in the JEE Main Inorganic Chemistry syllabus?

JEE Main Inorganic Chemistry is split into 4 chapters: Classification of Elements and Periodicity in Properties, P-Block Elements, d - and f- Block Elements and Co-ordination Compounds. Together they contain 10 topics and 18 sub-topics.

How many chapters are in JEE Main Inorganic Chemistry?

4 chapters, containing 10 topics and 18 sub-topics. The largest chapter is Co-ordination Compounds, with 4 topics.

Are there flashcards for JEE Main Inorganic Chemistry?

Yes — a 50-card Inorganic Chemistry deck, printed in full in the flashcards section of this page and available in the Examius app with spaced repetition. The free plan covers one exam of your choice, including its flashcards; Examius Pro, a one-time purchase, unlocks every other exam and deck.