Ammonia (NH3) is present in Nessler's reagent (ammoniacal K2HgI4) and in Tollen's reagent (ammoniacal Ag+ complex) and is also used frequently in reagents for group III analysis (e.g., as NH4OH). Group IV basic radical analysis uses strong bases like NaOH rather than NH3, so NH3 is not used in option (b).
Nitrogen (2s2 2p3) has available p orbitals and can form multiple bonds with itself through p–p π overlap (as in N≡N). Options (a),(b),(c) are incorrect: nitrogen is not the least electronegative, its first ionisation enthalpy is higher than oxygen's, and it has no d orbitals (period 2 element).
Group 15, 3rd period element is phosphorus (P). Its configuration is 1s2 2s2 2p6 3s2 3p3 (option d).
White phosphorus (P4) reacts with NaOH and water to give phosphine (PH3), a foul–smelling gas that can ignite spontaneously producing smoky rings: P4 + 3 NaOH + 3 H2O → PH3 + 3 NaH2PO2. Hence A = P4(white), B = PH3 (option b).
PCl3 hydrolyses to phosphorous acid: PCl3 + 3 H2O → H3PO3 + 3 HCl. So product is H3PO3 (option a).
P4O6 is the anhydride of phosphorous acid: P4O6 + 6 H2O → 4 H3PO3. So it gives H3PO3 (option a).
Basicity = number of ionisable (–OH) hydrogens. In H4P2O5 (pyrophosphorous acid) only two hydrogens are acidic (–OH), the P–H hydrogens are non‑acidic. Hence basicity = 2 (option b).
H3PO4 is triprotic (basicity = 3). Normality = Molarity × basicity = 2 × 3 = 6 N (option a).
The assertion is false: F2 has a lower bond dissociation energy than Cl2 due to strong lone‑pair repulsion between small F atoms. The reason is also false because fluorine, not chlorine, shows greater electronic repulsion. Therefore both assertion and reason are false (option d).
Fluorine (F2) is the strongest oxidizing agent among halogens because it has the highest (most positive) reduction potential and highest electronegativity (option b).
Thermal stability increases with increasing H–X bond strength. HF has the strongest H–X bond, then HCl, HBr and HI. So HF > HCl > HBr > HI (option b).
Xenon forms many compounds (XeF2, XeOF4, XeO3). Neon is chemically inert and does not form NeF2 under normal conditions; thus NeF2 is not formed (option d).
Liquefaction ease increases with stronger intermolecular (van der Waals) forces i.e., larger atomic size. Among the given noble gases Kr (largest) is most easily liquefied (option d).
Complete hydrolysis of XeF6 yields xenon trioxide: XeF6 + 3 H2O → XeO3 + 6 HF. So product is XeO3 (option c).
Among hydrogen halides, acid strength increases down the group as H–X bond strength decreases. HI has the weakest H–I bond and is the strongest acid (option a).
Due to small‑atom lone pair repulsion F2 has unusually low BDE; actual order of bond dissociation enthalpies is Cl2 > Br2 > F2 > I2 (option d).
Acidity of oxyacids of the same central atom increases with oxidation state (and number of P=O or Cl=O bonds). Thus HClO < HClO2 < HClO3 < HClO4 (option d).
Concentrated HNO3 oxidizes Cu to Cu2+ producing nitrogen dioxide (brown fumes): Cu + 4 HNO3(conc) → Cu(NO3)2 + 2 NO2 + 2 H2O. So products include Cu(NO3)2 and NO2 (option c).
Concise explanation: In heavier p‑block elements (down a group) the ns2 electrons are increasingly reluctant to participate in bonding due to poor shielding by inner d/f electrons, relativistic stabilization and increased effective nuclear attraction; hence lower oxidation states (e.g., Pb(II) rather than Pb(IV)) are stabilized. The effect increases down the group.
Inert pair effect is the tendency of the two electrons in the outermost ns2 subshell of heavy p‑block elements to remain non‑ionized or non‑shared, causing lower than expected oxidation states.
Reason: Elements of group 16 have outer electronic configuration of the form ns2 np4, i.e., their highest energy electrons are in p orbitals. By definition, elements whose valence electrons lie in p subshell belong to the p‑block; therefore chalcogens are p‑block elements.
Chalcogens (group 16 elements) belong to the p‑block because their valence electrons occupy the ns2 np4 configuration; the outermost electrons are in p orbitals.
Fluorine is the most electronegative element (highest ability to attract electrons) and has seven valence electrons (2s2 2p5). It always gains one electron to attain the noble gas configuration and so has oxidation state −1. Also F has no vacant d-orbitals and a very high ionization energy, so it cannot be oxidized (i.e. cannot exhibit positive oxidation states) in compounds.
-1
Use algebraic sum of oxidation numbers = 0. a) OF2: x + 2(−1) = 0 ⇒ x(O)=+2 so F = −1. b) O2F2 (F–O–O–F): each F = −1. c) Cl2O3: 2(Cl) + 3(−2) = 0 ⇒ 2Cl = +6 ⇒ Cl = +3. d) I2O4: 2I + 4(−2)=0 ⇒ 2I = +8 ⇒ I = +4.
Interhalogen compounds are binary compounds of two different halogens. General formulas: XY, XY3, XY5, XY7 (X = less electronegative halogen, Y = more electronegative). Examples: ClF, BrCl (XY); IF3, ClF3 (XY3); BrF5, SbF5? (XY5); IF7 (XY7). They are molecular, often more reactive than constituent halogens and are strong halogenating agents.
Compounds formed between different halogens, e.g. XY, XY3, XY5, XY7 (ClF, BrCl, ClF3, BrF5, IF7).
Factors: (1) Highest electronegativity and high electron affinity — strong tendency to gain an electron. (2) Very small atomic size → strong attraction for incoming electron. (3) The F–F bond is unusually weak (large lone-pair repulsion between close fluorine atoms), so bond breaking is easier than for other X–X bonds. These combine to make fluorine the most reactive halogen.
Because of highest electronegativity, small size and weak F–F bond facilitating bond cleavage.
Common uses of helium include:
- Filling balloons and airships (non-flammable, lighter-than-air gas).
- Breathing gas mixtures for deep-sea diving to avoid nitrogen narcosis.
- Liquid helium for cryogenics and cooling superconducting magnets (e.g., MRI machines).
- Inert shielding gas in welding and in semiconductor production.
- Leak detection and as an inert atmosphere for manufacture of reactive metals.
Iodine in IF7 has seven bonding pairs (no lone pairs) → seven electron domains → sp3d3 hybridisation. Molecular shape is pentagonal bipyramidal: five equatorial F atoms form a planar pentagon around I and two F atoms occupy axial positions above and below the plane.
Hybridisation sp3d3; geometry: pentagonal bipyramidal (five equatorial F in a plane forming a pentagon, two axial F).
Chlorine disproportionates in alkaline medium. Cold NaOH (dilute, 0–10 °C) yields hypochlorite: Cl2 + 2 NaOH → NaCl + NaOCl + H2O. Hot/concentrated NaOH (heating) gives chlorate: 3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O.
Cold: Cl2 + 2 NaOH → NaCl + NaOCl + H2O Hot: 3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O
Common lab method: add concentrated HCl to manganese(IV) oxide (pyrolusite): MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O. Evolved Cl2 (greenish-yellow, denser than air) is collected by downward displacement of air or over water (if dry Cl2 required, dry over CaCl2; avoid P2O5 which oxidizes).
React MnO2 with concentrated HCl: MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O (collect by downward displacement).
Major uses: (1) Manufacture of phosphatic fertilizers (superphosphate). (2) Production of chemicals (HCl, nitric acid, dyes, detergents). (3) Petroleum refining (acid catalysts). (4) As a dehydrating and oxidizing agent in industry (e.g. dehydration of sugars). (5) Electrolyte in lead–acid batteries. (6) Pickling and processing of metals.
Fertiliser production, chemical manufacture, petroleum refining, dehydrating agent, lead–acid batteries, metal processing.
Example: Concentrated H2SO4 removes water from sucrose, producing carbon and steam: C12H22O11 (sucrose) + H2SO4 → 12 C (black) + 11 H2O (g) + H2SO4 (regenerated). Also it dehydrates ethanol to give ethene: C2H5OH → C2H4 + H2O (H2SO4 catalyst). These show its powerful dehydrating property.
Conc. H2SO4 removes water from organic compounds (e.g. sugar → carbon + water), demonstrating strong dehydrating action.
Nitrogen differs from other group-15 elements because of its small atomic radius and high electronegativity. It lacks accessible d-orbitals, so it forms strong pπ–pπ multiple bonds (e.g. N≡N), exhibits limited catenation compared with heavier pnictogens, shows different oxidation states and bonding (e.g. forms stable N2), and often behaves non-metallic while heavier congeners are more metallic.
Small size, high electronegativity, absence of d-orbitals, strong tendency to form pπ–pπ multiple bonds and N≡N triple bond.
a) Nitric acid: HNO3. Structural description: HO–N(=O)–O with resonance among oxygens. b) Dinitrogen pentoxide: N2O5. Structure: two NO3 units joined by a bridging O: O2N–O–NO2 (each N in +5 oxidation state). c) Phosphoric acid: H3PO4. Structural formula: P(=O)(OH)3 (tetrahedral P). d) Phosphine: PH3. Molecular formula PH3; shape trigonal pyramidal with a lone pair on P.
Uses of argon include:
- Inert shielding gas in arc welding (prevents oxidation of the weld).
- Filling incandescent and fluorescent lamps to prevent filament oxidation and prolong life.
- Providing an inert atmosphere in the production of reactive metals and in semiconductor manufacturing.
- Used in gas-discharge tubes and as an inert blanket in laboratory and industrial processes.
Group-15 (pnictogens) valence configuration is ns2 np3. Examples: N → 2s2 2p3; P → 3s2 3p3; As → 4s2 4p3; Sb → 5s2 5p3; Bi → 6s2 6p3.
ns2 np3 (e.g. N: 2s2 2p3; P: 3s2 3p3; As: 4s2 4p3; Sb: 5s2 5p3; Bi: 6s2 6p3).
Two illustrative reactions: (1) Basicity/protonation: PH3 + HCl → PH4Cl (phosphonium chloride). (2) Oxidation (reducing property): 4 PH3 + 8 O2 → P4O10 + 6 H2O (balanced). These show PH3 can be protonated and is a reducing agent.
PH3 + HCl → PH4Cl (forms phosphonium salt) 4 PH3 + 8 O2 → P4O10 + 6 H2O (PH3 is oxidised; reducing behaviour)
Example: Reaction of basic oxide copper(II) oxide with nitric acid: CuO + 2 HNO3 → Cu(NO3)2 + H2O. This illustrates acid + basic oxide → salt + water.
CuO + 2 HNO3 → Cu(NO3)2 + H2O (or Na2O + 2 HNO3 → 2 NaNO3 + H2O)
On heating, phosphorus pentachloride undergoes thermal decomposition (endothermic, reversible): PCl5 → PCl3 + Cl2. In solid state PCl5 exists partly as ionic [PCl4+][PCl6−], but heating shifts equilibrium to molecular PCl3 and Cl2.
PCl5 decomposes to PCl3 and Cl2: PCl5 ⇌ PCl3 + Cl2 (upon heating, decomposition to PCl3 + Cl2).
Reasons: (1) H–F bond enthalpy is very high → difficult to break and dissociate. (2) HF forms strong hydrogen-bonded networks (associates) in aqueous solution which reduces free H+ concentration. (3) F− is small and strongly solvated, but the dominant effect is the strong H–F bond and hydrogen bonding, leading to partial ionization (weak acid) compared with HCl, HBr, HI which dissociate fully.
HF has a very strong H–F bond and forms strong hydrogen-bonded associates in water; both make dissociation (H+ release) difficult, so HF is weak compared to other H–X acids.
Assign oxidation numbers: H = +1, F = −1 (always). For HOF: (+1) + x + (−1) = 0 ⇒ x = 0. Thus oxygen has oxidation state 0 in HOF.
0
a) BrF5: bromine has 6 electron domains (5 bonds + 1 lone pair) → sp3d2 hybridisation; geometry derived from octahedral → square pyramidal. b) BrF3: bromine has 5 electron domains (3 bonds + 2 lone pairs) → sp3d hybridisation; geometry derived from trigonal bipyramid with lone pairs equatorial → T-shaped molecule.
1. 4 NaCl + MnO2 + 4 H2SO4 → MnCl2 + 4 NaHSO4 + 2 H2O + Cl2 (in practice HCl is generated from NaCl + H2SO4, then MnO2 + 4HCl → MnCl2 + 2H2O + Cl2)
2. NaNO3 + HCl → HNO3 + NaCl (acidification of nitrate gives nitric acid)
3. P4 + 3 NaOH + 3 H2O → PH3 + 3 NaH2PO2 (white phosphorus in alkaline medium yields phosphine and hypophosphite)
4. 3 AgNO3 + PH3 → Ag3P + 3 HNO3 (phosphine reacts with Ag+ to give silver phosphide; equation shown stoichiometrically)
5. (a) With dilute HNO3: Mg + 2 HNO3 (dilute) → Mg(NO3)2 + H2 (b) With concentrated HNO3 (oxidising): Mg + 4 HNO3 (conc) → Mg(NO3)2 + 2 NO2 + 2 H2O
6. 2 KClO3 Δ → 2 KCl + 3 O2 (thermal decomposition of potassium chlorate)
7. Cu + 2 H2SO4 (hot, conc) → CuSO4 + SO2 + 2 H2O (hot conc. sulphuric acid oxidises copper to CuSO4 producing SO2)
8. 2 Sb + 3 Cl2 → 2 SbCl3 (antimony forms the trichloride under chlorine)
9. 2 HBr + H2SO4 (conc) → Br2 + SO2 + 2 H2O (conc. H2SO4 oxidises bromide to bromine and itself is reduced to SO2)
10. XeF6 + 3 H2O → XeO3 + 6 HF (hydrolysis of xenon hexafluoride gives xenon trioxide and HF; some texts write formation of xenic acid H2XeO4)
11. (OCR of the question was ambiguous.) A typical redox behaviour: XeO3 is a strong oxidant. As an example reaction with Mn2+ in acidic medium (qualitative representation): 5 XeO3 + 12 Mn^{2+} + 9 H2O → 5 Xe (reduced) + 12 MnO4^- + 18 H^+ (This is a stoichiometric example showing oxidation of Mn^{2+} to MnO4^- by XeO3; exact balanced equation depends on reaction conditions.)
12. XeOF4 + SiO2 → XeO2F2 + SiF4 (xenon oxyfluorides react with silica to give silicon tetrafluoride and higher xenon oxyfluoride species; stoichiometry may vary with conditions and starting xenon fluoride)
13. Xe + 2 F2 --Ni, 200 atm, 400 °C--> XeF4 (direct fluorination of xenon under these conditions yields xenon tetrafluoride)
Note: Items 11 and 12 in the provided OCR were ambiguous; for 11 a representative redox transformation is given (actual classroom/ textbook answer may show a specific balanced redox equation), and for 12 a plausible product (SiF4 and a xenon oxyfluoride) is shown. For uncertain items consult the textbook reaction statements.
See balanced equations below (with brief notes).