JEE Main Inorganic Chemistry Flashcards

24 sample cards from the Inorganic Chemistry deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. 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.

2. 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).

3. 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).

4. 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.

5. 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 $+7$.

6. 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 $\ce{KMnO4}$).

7. 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.

8. 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.

9. 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.

10. 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.

11. 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 ($\ce{Fe3C}$), $\ce{TiC}$.

12. 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.

13. 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. $\ce{[Fe(CN)6]^{4-}}$, $\ce{[Cu(NH3)4]^{2+}}$.

14. 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.

15. 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.

16. 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 $+3$.

17. 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 $\ce{KMnO4}$: $\ce{3MnO4^{2-} + 4H+ -> 2MnO4^- + MnO2 + 2H2O}$.

18. 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).

19. 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).

20. 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.

21. 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.

22. 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.

23. 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.

24. 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.

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