GATE Chemistry 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. What is the fundamental structural building block of all silicates, and how do they link?

   The basic unit is the $\ce{SiO4^{4-}}$ tetrahedron (Si at center, 4 O at corners). Tetrahedra link by sharing corner (bridging) oxygen atoms to form chains, rings, sheets, or 3-D frameworks.

2. Classify silicates by the number of shared corners: orthosilicate, pyrosilicate, cyclic/chain, sheet, and 3-D.

   Orthosilicate $\ce{SiO4^{4-}}$ (0 shared O); Pyrosilicate $\ce{Si2O7^{6-}}$ (1 shared); Cyclic/single-chain $\ce{(SiO3^{2-})_n}$ (2 shared); Sheet/phyllosilicate $\ce{(Si2O5^{2-})_n}$ (3 shared); 3-D framework $\ce{SiO2}$ (4 shared, e.g. quartz, zeolites).

3. Boron nitride exists in two main forms analogous to carbon. Name them and their structures.

   Hexagonal BN ('white graphite') — layered sheets of alternating B and N, soft lubricant, analogous to graphite (but layers eclipsed with B over N). Cubic BN (borazon) — sphalerite/diamond-like structure, extremely hard abrasive.

4. Why is hexagonal boron nitride an electrical insulator while graphite conducts?

   In graphite the delocalized $\pi$ electrons are free to move, giving conduction. In h-BN the $\pi$ electrons are localized on the more electronegative nitrogen atoms (the B–N bond is polar, large band gap), so there are no mobile electrons — it is an insulator.

5. What is borazine, its formula, and why is it called 'inorganic benzene'?

   Borazine is $\ce{B3N3H6}$, a six-membered planar ring of alternating B and N atoms each bearing one H, isoelectronic and isostructural with benzene ($\ce{C6H6}$). It has delocalized $\pi$ character, hence the nickname.

6. How is borazine synthesized, and how does its reactivity differ from benzene?

   $\ce{3 B2H6 + 6 NH3 -> 2 B3N3H6 + 12 H2}$ (via the diammoniate/aminoborane intermediate). Unlike benzene, borazine readily undergoes addition reactions (e.g. with HCl across B–N bonds) because the ring is polar and less aromatic.

7. What are phosphazenes, give the general formula of the cyclic trimer, and how is it made?

   Phosphazenes contain the $\ce{-N=P-}$ repeat unit. The cyclic trimer is hexachlorocyclotriphosphazene $\ce{(NPCl2)3}$, made by $\ce{n PCl5 + n NH4Cl -> (NPCl2)_n + 4n HCl}$.

8. What useful materials are obtained from cyclic chlorophosphazenes, and how?

   Heating $\ce{(NPCl2)3}$ gives linear high polymer poly(dichlorophosphazene) $\ce{[NPCl2]_n}$ ('inorganic rubber'). Replacing the reactive Cl with $\ce{-OR}$ or $\ce{-OC6H5}$/$\ce{-OCH2CF3}$ groups yields stable polyphosphazene elastomers and fire-resistant materials.

9. Name the principal crystalline allotropes of carbon and the hybridization of carbon in each.

   Diamond — $sp^3$, 3-D tetrahedral network (hardest, insulator). Graphite — $sp^2$, layered sheets (soft, conductor). Fullerenes (e.g. $\ce{C60}$) — $sp^2$ closed cages. Carbon nanotubes and graphene — $sp^2$ cylindrical/single-sheet forms.

10. Describe the structure of $\ce{C60}$ (buckminsterfullerene).

    $\ce{C60}$ is a truncated icosahedron (soccer-ball) of 60 $sp^2$ carbons forming 20 hexagonal and 12 pentagonal faces. Each C is bonded to 3 others; it has 90 edges and the icosahedral $I_h$ symmetry.

11. Compare the structures of white, red and black phosphorus.

    White phosphorus: discrete tetrahedral $\ce{P4}$ molecules (strained $60^\circ$ angles, very reactive, toxic). Red phosphorus: polymeric chains of linked $\ce{P4}$ units (amorphous, stable). Black phosphorus: layered, graphite-like puckered sheets (most thermodynamically stable, semiconductor).

12. Name the common allotropes of sulfur and the molecular unit of rhombic sulfur.

    Rhombic ($\alpha$) and monoclinic ($\beta$) sulfur both consist of crown-shaped $\ce{S8}$ rings; rhombic is stable below $95.5\,^\circ\text{C}$. Plastic (amorphous) sulfur is made of long $\ce{S_n}$ chains formed by quenching molten sulfur.

13. Write the contact process for the industrial manufacture of sulfuric acid, including the key catalyzed step.

    $\ce{S + O2 -> SO2}$; then the catalyzed step $\ce{2SO2 + O2 ->[V2O5] 2SO3}$; $\ce{SO3}$ is absorbed in $\ce{H2SO4}$ to give oleum $\ce{H2S2O7}$, then diluted: $\ce{H2S2O7 + H2O -> 2H2SO4}$.

14. Describe the Haber–Bosch process for ammonia, including conditions and catalyst.

    $\ce{N2 + 3H2 <=> 2NH3}$, $\Delta H = -92\ \text{kJ mol}^{-1}$. Conditions: $\sim 200\ \text{atm}$, $\sim 400\text{–}450\,^\circ\text{C}$, finely divided iron catalyst (promoted with $\ce{K2O}$/$\ce{Al2O3}$). A compromise temperature balances rate and yield.

15. Outline the Ostwald process for nitric acid manufacture.

    Catalytic oxidation of ammonia: $\ce{4NH3 + 5O2 ->[Pt/Rh] 4NO + 6H2O}$; then $\ce{2NO + O2 -> 2NO2}$; then $\ce{3NO2 + H2O -> 2HNO3 + NO}$ (the NO is recycled).

16. How is chlorine and sodium hydroxide produced industrially by the chlor-alkali (membrane) process?

    Electrolysis of brine ($\ce{NaCl}$ solution): anode $\ce{2Cl^- -> Cl2 + 2e^-}$; cathode $\ce{2H2O + 2e^- -> H2 + 2OH^-}$. Net: $\ce{2NaCl + 2H2O -> Cl2 + H2 + 2NaOH}$; a cation-exchange membrane keeps products separated.

17. Describe the industrial preparation of phosphoric acid by the wet process and the thermal process.

    Wet process: $\ce{Ca3(PO4)2 + 3H2SO4 -> 2H3PO4 + 3CaSO4}$. Thermal (furnace) process: burn elemental $\ce{P4}$ to $\ce{P4O10}$, then hydrate: $\ce{P4O10 + 6H2O -> 4H3PO4}$ (gives high-purity acid).

18. What was the first noble-gas compound prepared, by whom, and its formula?

    $\ce{Xe+[PtF6]^-}$ (xenon hexafluoroplatinate), prepared by Neil Bartlett in 1962, after he noted that $\ce{O2}$ and $\ce{Xe}$ have similar first ionization energies and $\ce{PtF6}$ oxidized $\ce{O2}$.

19. Give the preparation conditions and products for $\ce{XeF2}$, $\ce{XeF4}$ and $\ce{XeF6}$.

    Direct combination of $\ce{Xe}$ and $\ce{F2}$ at different ratios/conditions: low $\ce{F2}$ ratio → $\ce{XeF2}$ ($\ce{Xe + F2 -> XeF2}$); excess $\ce{F2}$, higher T/P → $\ce{XeF4}$ and $\ce{XeF6}$. Higher fluorine and pressure favor higher fluorides.

20. What are the products of complete and partial hydrolysis of $\ce{XeF6}$?

    Partial hydrolysis: $\ce{XeF6 + H2O -> XeOF4 + 2HF}$; further $\ce{XeOF4 + H2O -> XeO2F2 ...}$; complete hydrolysis gives explosive xenon trioxide: $\ce{XeF6 + 3H2O -> XeO3 + 6HF}$.

21. Why do the noble gases form so few compounds, and why is Xe the most reactive of the stable ones?

    They have stable, fully filled valence shells (very high ionization energies, near-zero electron affinity). Reactivity increases down the group as IE falls; Xe (and Kr to a lesser extent) has a low enough IE to be oxidized by very strong oxidizers like $\ce{F2}$ and $\ce{PtF6}$.

22. Compare the basicity/reducing nature trend of Group 15 hydrides $\ce{NH3}$ to $\ce{BiH3}$.

    Basicity (lone-pair donor ability) decreases down the group: $\ce{NH3} > \ce{PH3} > \ce{AsH3} > \ce{SbH3} > \ce{BiH3}$, while reducing character and thermal instability increase down the group ($\ce{BiH3}$ is the strongest reductant, least stable).

23. What is the hybridization and shape of the silicate metasilicate chain anion, and how does asbestos relate to silicate structure?

    Each Si is $sp^3$ in a $\ce{SiO4}$ tetrahedron sharing two corners to give single-chain pyroxenes $\ce{(SiO3^{2-})_n}$. Amphibole asbestos is a double-chain silicate $\ce{(Si4O11^{6-})_n}$; the fibrous chain structure gives its characteristic fibres.

24. State the structure and bonding type of $\ce{P4O10}$ and $\ce{P4O6}$.

    Both derive from the $\ce{P4}$ tetrahedron. $\ce{P4O6}$ has an O atom bridging each of the 6 P–P edges (each P pyramidal). $\ce{P4O10}$ adds 4 terminal $\ce{P=O}$ oxygens (one per P), so each P is tetrahedrally bonded to 4 O. They are the anhydrides of $\ce{H3PO3}$ and $\ce{H3PO4}$ respectively.

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