CUET UG Chemistry Inorganic Chemistry p-Block Elements Group 15 16 17 18 Oxoacids Anomalous Properties

Must-Know (15–20 detailed bullets)

• Nitrogen exhibits anomalous behavior due to absence of d-orbitals, small size, high electronegativity, and high ionization enthalpy; for example, it forms pπ-pπ multiple bonds (N≡N), while P forms P₄ tetrahedral structure.
• Nitrogen shows maximum covalency of 4 (e.g., NH₄⁺), whereas other Group 15 elements can expand covalency beyond 4 due to d-orbital availability (e.g., PF₆⁻).
• HNO₃ has a planar structure with nitrogen in +5 oxidation state; it acts as a strong oxidizing agent and decomposes on exposure to light: 4HNO₃ → 4NO₂ + O₂ + 2H₂O.
• Phosphorus forms two common oxoacids: H₃PO₃ (phosphorous acid, diprotic, +3 oxidation state) and H₃PO₄ (orthophosphoric acid, triprotic, +5 oxidation state).
• H₃PO₃ is diprotic because only two –OH groups are ionizable; the H directly bonded to P is not acidic.
• In Group 16, oxygen differs from others due to absence of d-orbitals, high electronegativity (3.5), and ability to form hydrogen bonds; e.g., H₂O is liquid, H₂S is gas at room temperature.
• Ozone (O₃) is an allotrope of oxygen, bent structure (116.8°), resonance hybrid, and acts as a powerful oxidizing agent; it absorbs UV radiation in stratosphere.
• SO₂ is a bent molecule (O–S–O angle ≈ 119°), acidic oxide, reducing agent; turns acidified K₂Cr₂O₇ paper green.
• Sulphuric acid (H₂SO₄) is manufactured via Contact Process: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅ catalyst, 2 bar, 720 K); SO₃ absorbed in H₂SO₄ to form oleum.
• H₂SO₄ is dibasic, highly corrosive, and acts as dehydrating agent (e.g., converts HCOOH to CO); concentrated H₂SO₄ has low volatility due to hydrogen bonding.
• In Group 17, fluorine shows anomalous behavior: highest electronegativity (4.0), low electron gain enthalpy (due to small size, interelectronic repulsion), and absence of d-orbitals (no higher oxidation states).
• HF is a weak acid (pKa = 3.2) due to strong H–F bond and extensive hydrogen bonding; other HX are strong acids.
• Interhalogen compounds like ClF₃ (T-shaped, sp³d hybridization) and IF₇ (pentagonal bipyramidal, sp³d³d²) are more reactive than halogens except F₂.
• Bleaching action of Cl₂ is due to oxidation: Cl₂ + H₂O → HCl + HOCl; HOCl → HCl + [O]; colored substance + [O] → colorless.
• Chlorine oxoacids: HClO (hypochlorous acid, +1), HClO₂ (chlorous acid, +3), HClO₃ (chloric acid, +5), HClO₄ (perchloric acid, +7); acidity increases with oxidation state.
• HClO₄ is the strongest oxoacid of chlorine due to maximum oxidation state (+7) and greater delocalization of negative charge in conjugate base.
• Noble gases have very high ionization enthalpies; He has highest (2372 kJ/mol), making it chemically inert under normal conditions.
• Xenon forms compounds with highly electronegative elements: XeF₂ (linear, sp³d), XeF₄ (square planar, sp³d²), XeF₆ (distorted octahedral, sp³d³).
• XeF₂ hydrolyzes: XeF₂ + 2H₂O → 2HF + Xe + O₂ (partial hydrolysis); complete hydrolysis gives XeO₃, which is explosive.
• verify from NCERT: The order of thermal stability of hydrides of Group 16: H₂O > H₂S > H₂Se > H₂Te > H₂Po.

Difficulty Level

Intermediate — requires understanding of periodic trends, molecular geometry, and acid strength trends, but most facts are directly from NCERT.

Common CUET Traps

• Trap: Assuming H₃PO₃ is triprotic like H₃PO₄.
  Avoid: Remember only –OH groups are acidic; H₃PO₃ has two –OH and one P–H bond → diprotic.

• Trap: Thinking bleaching action of SO₂ and Cl₂ are the same.
  Avoid: SO₂ bleaches by reduction (temporary), Cl₂ by oxidation (permanent); SO₂ decolorizes acidified KMnO₄, Cl₂ does not.

• Trap: Believing noble gases cannot form compounds.
  Avoid: Xenon forms stable compounds with F and O (e.g., XeF₄, XeO₃); He, Ne, Ar do not form compounds under normal conditions.

Practice MCQs

1. Which of the following oxoacids of phosphorus is diprotic?
   
   A. H₃PO₄
   
   B. H₄P₂O₇
   
   C. H₃PO₃
   
   D. H₃PO₂
   Answer: C
   Explanation: H₃PO₃ has two ionizable –OH groups and one P–H bond, making it diprotic.
   Why others fail: H₃PO₄ is triprotic, a common confusion due to similar naming.

2. The bleaching action of chlorine is permanent because:
   
   A. It reduces the colored substance
   
   B. It forms a colorless complex
   
   C. It oxidizes the colored substance
   
   D. It absorbs light
   Answer: C
   Explanation: Cl₂ produces nascent oxygen via HOCl, which oxidizes the colored material.
   Why others fail: SO₂ bleaches by reduction (temporary), leading to confusion with Cl₂.

3. Which interhalogen compound has a T-shaped geometry?
   
   A. ClF
   
   B. ClF₃
   
   C. BrF₅
   
   D. IF₇
   Answer: B
   Explanation: ClF₃ has sp³d hybridization and T-shaped geometry due to two lone pairs on Cl.
   Why others fail: BrF₅ is square pyramidal; students often confuse geometries of interhalogens.

4. Among the following, which acid is the strongest?
   
   A. HClO
   
   B. HClO₂
   
   C. HClO₃
   
   D. HClO₄
   Answer: D
   Explanation: Acid strength increases with oxidation state of Cl; HClO₄ (+7) is strongest.
   Why others fail: HClO₂ is less acidic than HClO₃ and HClO₄, but students misremember the order.

5. Which noble gas compound is isostructural with ICl₃⁻?
   
   A. XeF₂
   
   B. XeF₄
   
   C. XeO₃
   
   D. XeF₆
   Answer: A
   Explanation: ICl₃⁻ has linear geometry (sp³d, two lone pairs), same as XeF₂.
   Why others fail: XeF₄ is square planar; students often misapply hybridization rules.

Last‑Minute Revision

• ⚠️ H₃PO₃ is diprotic; H₃PO₂ is monoprotic — count –OH groups only.
• ⚠️ Ozone layer absorbs UV-B (280–320 nm) in stratosphere.
• ⚠️ H₂S has bent shape, ~92° bond angle — no hydrogen bonding, weak dipole.
• ⚠️ Nitrogen cannot form pentahalides (no d-orbitals); PCl₅ exists.
• ⚠️ HF stored in wax bottles — attacks glass: SiO₂ + 4HF → SiF₄ + 2H₂O.
• ⚠️ XeF₄ has two lone pairs, square planar (sp³d²).
• ⚠️ H₂SO₄: dibasic, dehydrating, oxidizing agent when hot/concentrated.
• ⚠️ O₂ molecule is paramagnetic — two unpaired electrons in π* orbitals.
• ⚠️ Cl₂ + 2NaOH (cold, dil.) → NaCl + NaClO + H₂O.
• ⚠️ Cl₂ + 6NaOH (hot, conc.) → 5NaCl + NaClO₃ + 3H₂O.
• ⚠️ Bleaching powder: Ca(OCl)Cl — active ingredient OCl⁻.
• ⚠️ H₂O has higher boiling point than H₂S due to H-bonding.
• ⚠️ NH₃ has higher boiling point than PH₃ — hydrogen bonding in NH₃.
• ⚠️ OF₂ is oxygen difluoride; oxygen has +2 oxidation state.
• ⚠️ XeO₃ is explosive, trigonal pyramidal, sp³ hybridized.
• ⚠️ Mnemonic: "I See 6 F's" — XeF₆ has 6 F atoms, distorted octahedral.