p-Block Elements (Halogen & Noble Gases) Exercises

p-Block Elements (Halogen & Noble Gases)  Marked questions are recommended for Revision. PART - I : SUBJECTIVE QUESTIONS Section (AI + AII) : Elements: Occurence & Isolation AI-1. Name the compound which is used to obtain fluorine gas on electrolysis. At which electrode does F 2 appears? AI-2. Name the valuable halogen impurities present in chile salt petre (NaNO 3). AII-3. What idea lead to the discovery of Xenon fluorides? Section (BI) : Based on Periodic Trends BI-1. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F2 and Cl2. BI-2. Write all the common oxidation states of halogens. BI-3. Write and explain the order of X–X bond energy for halogen down the group. Section (CI + CII) : Based on Chemical Bonding C-1. Draw the Lewis dot structures of the following compounds : HClO3, XeOF4, XeO2F2, ICl2–, ICl4–. CI-2. Draw the Lewis dot structures of the following multicentred compounds: I2O5, Cl2O7. CI-3. Give the formula and describe the structure of a noble gas species which is isostructural with: (i) Cl4– (ii) Br2– (iii) BrO3– CII-4. Arrange the XeF2, XeF4 in decreasing order of Xe–F bond length, give reason also. Section (DI + DII) : Properties of elements DII-1. Answer the following with relevant reason. (i) The boiling points of noble gases increase with increase in atomic number. (ii) Why helium and neon do not form clathrate compounds with quinol ? DI-2. Why are halogens coloured ? DI-3. Write the reactions of F2 and Cl2 with water. DI-4. State what happens when halogens react with a cold dilute solution of NaOH ? DI-5. State what happens when halogens (X = Cl / Br / I) react with hot and conc. solution of NaOH ? DI-6. Which halogen is oxidised by conc. HNO3? Give reaction. DI-7. Explain the following with proper reason : (i) Fluorine cannot be prepared from fluorides by chemical oxidation. (ii) Fluorine does not form F3– (polyhalide) ion. Section (EI) : Oxides, Hydroxides & Oxyacids EI-1. Write chemical reactions involved in preparation of HClO 3 & HClO4 by displacement from their salts. EI-2. When a blue litmus is dipped into a solution of hypochlorous acid, it turns red and then gets decolourised. Explain. EI-3. Explain why fluorine forms only one oxyacid, HOF. p-Block Elements (Halogen & Noble Gases) EI-4. Predict the products when the following reactions are carried out : (i) In acidic medium when SO2 is passed through NaClO3. (ii) HCl + KIO3 + KI ⎯→ EI-5. What happen when ClO2 dissolves in NaOH? Section (FI) : Hydracids FI-1. Arrange the following in the order of : (i) Acidic strength (ii) Reducing behaviour (HI, HBr, HCl & HF) FI-2. How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only. FI-3. HI can not be prepared by heating NaI with concentrated H2SO4. Give the method which is preferred for the preparation of HI. FI-4. What is aqua regia? Write its reaction product with gold and platinum. FI-5. Explain the following with proper reason : (i) Anhydrous HCl is a bad conductor of electricity while aqueous HCl is a good conductor. (ii) HF is not stored in glass bottles but kept in wax lined bottles. (iii) HF has a greater electronegativity difference and more ionic character than HCl, HBr and HI but it is the weakest acid. FI-6. Fill in the blanks : (i) Among halogen acids (hydrogen halides) .............. is the strongest reducing agent. (ii) H2SO4 + H ⎯→ ________ + ___________ + ________ FI-7. Predict the products when the following reactions are carried out : (i) Red lead is boiled with concentrated HCl. (ii) SiO2 + HF ⎯→ Section (GII) : Halides & Oxyhalides GII-1. Write the method of preparation of XeF2, XeF4 & XeF6. GII-2. How is XeOF4 prepared ? GII-3. Does the hydrolysis of XeF4 lead to a redox reaction ? GII-4. Write the complete and the partial hydrolysis product of XeF6. GII-5. Complete the following reactions : (i) XeF2 + H2 → (ii) XeF6 + SiO2 → (iii) XeF6 + SbF5 → Section (HI) : Other compounds (Inter halogens, Bleaching powder, Pseudo halides & Poly halides) HI-1. Complete the following reactions : 473 K ⎯ → (a) Cl2 + F2 ⎯⎯⎯ (b) I2 + 3Cl2 ⎯→ (Equal volume) (c) Br2 + 3F2 ⎯→ (Excess) (d) Br2 + 5F2 ⎯→ (Excess) HI-2. Why ICl is more reactive than I2. HI-3. (a) Name two interhalogens of AB3 type. (b) Write the hydrolysis product of ICl ? HI-4. Explain the following with proper reason : (i) Bleaching of flowers by chlorine is permanent while after bleaching with SO 2, the colour returns. (ii) Iodine dissolves more in K solution than in water. HI-5. What happens when ? (Give balanced equations) (i) Sodium iodate is treated with sodium bisulphite solution. (ii) Chlorine is passed over slaked lime. p-Block Elements (Halogen & Noble Gases) PART - II : ONLY ONE OPTION CORRECT TYPE Section (AI + AII) : Elements : Occurence and Isolation AII-1. AII-2. AII-3. AI-4. AI-5. Which of the following gaseous molecules is monoatomic ? (A) chlorine (B) helium (C) oxygen (D) nitrogen Which one of the following noble gases is not found in atmosphere ? (A) Rn (B) Kr (C) Ne (D) Ar The inert gas abundantly found in atmosphere is : (A) Ar (B) Kr (C) He (D) Xe Electrolysis of aqueous solution of Brine (NaCl) will give: (A) Cl2 (B) H2 (C) NaOH (D) all of these The catalyst used in the Deacon’s process for the manufacture of chlorine is : (A) Cu (B) An alloy of copper (C) CuCl2 (D) CuS Section (BI + BII) : Based on Periodic Trends BII-1. Which one of the following configuration represents a noble gas ? (A) 1s2 2s2 p6, 3s2 (B) 1s2 2s2 p6, 3s1 (C) 1s2 2s2 p6 (D) 1s2 2s2 p6, 3s2 p6, 4s2 BI-2. Astatine is the element below iodine in the group VIIA of the periodic table. Which of the following statements is not true for astatine ? (A) It is less electronegative than iodine. (B) It will exhibit only –1 oxidation state. (C) Intermolecular forces between the astatine molecules will be larger than that between iodine molecules. (D) None of these. BI-3. Which is wrong statement ? (A) Basic nature of X– is in order F– > Cl– > Br– > I– (B) Electron gain enthalpy in order F > Cl > Br > I. (C) The ionic character of M—X bond decreases in the order M—F > M—Cl > M—Br > M—I (D) Among F–, Cl–, Br– and I–, F– has the highest enthalpy of hydration. Section (CI + CII) : Based on Chemical Bonding CI-1. Which of the following structure is correct for BrF3 ? (A) CII-2. CII-3. (B) (C) Among the following molecules, (i) XeO3 (ii) XeOF4 (iii) XeF6 those having same number of lone pairs on Xe are : (A) (i) and (ii) only (B) (i) and (iii) only (C) (ii) and (iii) only Select the incorrect match (A) XeF2 : linear (C) XeF6 : distorted octahedral (D) none of these (D) (i) , (ii) and (iii) (B) XeF4 : square planar (D) XeO3 : trigonal planar Section (DI) : Properties of elements DI-1. DII-2. DI-3. Correct order of boiling point of halogens is (A) F2 < Cl2 < Br2 < I2 (B) Cl2 > F2 > Br2 > I2 (C) F2 > Cl2 > Br2 > I2 Which of the following statement is correct ? (A) Helium has abnormal behaviour on liquefication (B) Iodine is readily soluble in CS2 and the solution is purple in colour (C) Helium do not form any clathrate (D) All of these Chlorine gas is dried over : (A) CaO (B) NaOH (C) conc. H2SO4 (D) Cl2 < F2 < Br2 < I2 (D) dil. H2SO4 p-Block Elements (Halogen & Noble Gases) DI-4. F2 reacts with H2O as follows : F2 + H2O ⎯→ H+ + F– + O2 Which of the following halogens shows same reaction but in opposite direction ? (A) Br2 (B) Cl2 (C) I2 (D) All DI-5. Chlorine acts as a bleaching agent only in presence of : (A) dry air (B) moisture (C) sunlight (D) pure oxygen Section (EI) : Oxides, Hydroxides & Oxyacids EI-1. Select the incorrect statement : (A) Perchloric acid is a stronger acid than sulphuric acid (B) Only one oxyacid [HOF] is formed by fluorine (C) The most stable oxy-acid of chlorine is perchloric acid (D) None of these EI-2. Cl2O6 reacts with water and alkali to give : (A) Only sodium chlorate (C) Both sodium chlorate and sodium perchlorate (B) Only sodium perchlorate (D) None of these On heating KClO3 we get : (A) KClO2 + O2 (B) KCl + O2 (D) KCl + O2 + O3 EI-3. (C) KCl + O3 EI-4. The following acids have been arranged in order of decreasing acid strength. Identify the correct order. ClOH (), BrOH (), IOH () (A)  >  >  (B)  >  >  (C)  >  >  (D)  >  >  EI-5. ClO3– ion leads with I2 to form (A) ClO4– (B) IO3– and Cl2 (C) lCl and O2 (D) ICl and O3 The strongest acid amongst the following is : (A) HClO4 (B) HClO3 (C) HClO2 (D) HClO EI-6. Section (FI) : Hydracids FI-1. Which of the following halogen hydrides will have the weakest conjugate base ? (A) HF (B) HCl (C) HBr (D) HI FI-2. Concentrated H2SO4 cannot be used to prepare HBr from NaBr, because it : (A) reduces HBr (B) oxidises HBr (C) disproportionates HBr (D) reacts slowly with NaBr FI-3. Hydrogen bromide is dried by passing the gas through : (A) quick line (B) anhydrous calcium chloride (C) potassium hydroxide pellet (D) con. H2SO4 FI-4. Which one of the hydracid does not form any precipitate with AgNO 3 ? (A) HF (B) HCl (C) HBr (D) HI Which can do glass etching ? (A) HIO4 (B) HF (D) SiF4 FI-5 FI-6. FI-7. FI-8. FI-9. FI-10. (C) HNO3 Identify A and B in following reaction, H2SO4 + HBr ⎯→ A + B + H2O (A) Br2 , SO3 (B) Br2 , S (C) BrO3– , SO3 (D) Br2 , SO2 Which of the following is obtained when gold is treated with aquaregia. (A) AuCl4 (B) AuCl3 (C) [AuCl4]– (D) [AuCl4]+ Which of the following hydrogen halide is most volatile. (A) HCl (B) HF (C) HI Which of the following has maximum bond strength : (A) HI (B) HCl (C) HF Which of the following is the strongest acid ? (A) HBr (B) HF (C) H2S (D) HBr (D) HBr (D) PH3 p-Block Elements (Halogen & Noble Gases) Section (GII) : Halides and oxyhalides GII-1. GII-2. GII-3. The number of lone pairs on central atom in XeF2 , XeF4 and XeF6 are : (A) 1,2,3 (B) 3,2,1 (C) 2,2,1 (D) 1,3,2 Of the following species, one which is non-existent : (A) XeF6 (B) XeF5 (C) XeF4 (D) XeF2 XeF2 on complete hydrolysis gives : (A) Xe (B) XeO2 (D) XeO4 (C) XeO2F2 GII-4. Hydrolysis of XeF4 and CaCN2 gives respectively : (A) XeO3 and CaCO3 (B) XeO2 and Ca(OH)2 (C) XeOF3 and Ca(OH)2 (D) XeOF2 and CaCO3 GII-5. Xenon hexafluoride undergoes hydrolysis in strong alkaline medium: 2 XeF6 + 16 OH– ⎯→ 8H2O + 12F– + A + B + C then reaction product may be: (A) perxenate ion (B) O2 (C) Xenon (D) All of the above Section (HI) : Other compounds (Inter halogens, Bleaching powder, Pseudo halides & Poly halides) HI-1. In the inter halogen compounds of AB3/AB5 form which is correct : (A) A is large size halogen (B) B is large size halogen (C) B is small size halogen (D) Both (A) & (C) HI-2. Which one of the following is not a pseudohalide ? (A) CNO– (B) RCOO– (C) OCN– (D) NNN– Which of the following behaves like pseudohalogen compound: (A) NCCN (B) CN– (C) N3– (D) I3– HI-3. HI-4. Which of the following statement is correct. (A) All interhalogen compounds are gas at room temperature. (B) Interhalogen are either gas or liquid at room temperature. (C) Interhalogens can be solid or liquid or gas at room temperature. (D) All interhalogen compounds are liquid at room temperature. PART - III : MATCH THE COLUMN 1. Match the reactions listed in column-I with the product(s) listed in column-II. Column-I Column-II (X = Halogen) (A) (p) N2 NH3 + F2 ⎯→ (B) (q) HX NH3 (excess.) + Cl2 ⎯→ (C) (r) NH4X NH3 + Br2 (excess.) ⎯→ (D) (s) Explosive NH3 (aq.) + I2 ⎯→ 2. Match the reactions listed in column-I with the product(s) listed in column-II. Column - I Column - II (A) (p) ClO2 Cl2O6 + H2O → (B) (q) HClO3 NaClO4(s) + HCl(conc.) → (C) (r) Cl2O KClO3 + (COOH)2 → 573K (D) (s) HClO4 → HgO + Cl2 ⎯⎯⎯ 3. Match the compounds listed in column-I with characteristic(s) / type of reaction(s) listed in column-II. (A) (B) (C) (D) Column–I XeF2 XeF4 XeF6 XeO3 (p) (q) (r) (s) (t) Column–II Undergoes hydrolysis with water. Acts as oxidising agent. Undergoes addition reaction. Has lone pair(s) of electrons. Gives disproportionation reaction with H2O or OH–. p-Block Elements (Halogen & Noble Gases)  Marked questions are recommended for Revision. PART - I : ONLY ONE OPTION CORRECT TYPE 1. 2. 3. 4. The X – X bond dissociation energy is minimum in : (A) F2 (B) Cl2 (C) Br2 (D) I2 Iodine is liberated from KI solution when treated with : (A) ZnSO4 (B) CuSO4 (C) NiSO4 (D) FeSO4 Which of the following is not oxidised by MnO2 ? (A) F– (B) Cl– (C) Br– (D) I– F2 + dil NaOH ⎯⎯→ A + NaF + H2O F2 + conc. NaOH ⎯⎯→ B + NaF + H2O A and B respectively are : (A) OF2 and O2 (B) O2 and OF2 (D) Both OF2 (C) Both O2 5. When thiosulphate ion is oxidised by iodine, the new product formed is : (A) SO32– (B) SO42– (C) S4O62– (D) S2O62– 6. NH3(excess) + Cl2 ⎯⎯→ NH4Cl + A(gas) NH3 + Cl2(excess) ⎯⎯→ B + HCl Incorrect statement regarding A and B. (A) A is highly reactive gas at room temperature. (B) Bond order of gas ‘A’ is same as C22–. (C) Compound ‘B’ is explosive. (D) Bond angle of compound B is greater than bond angle of NF3. 7. Which amongst the following reactions cannot be used for the respective preparation ? (A) 2KBr + H2SO4 (conc.) ⎯⎯→ K2SO4 + 2HBr (B) NaCl + H2SO4 (conc.) ⎯⎯→ NaHSO4 + HCl (C) NaHSO4 + NaCl ⎯⎯→ Na2SO4 + HCl (D) CaF2 + H2SO4(conc.) ⎯⎯→ CaSO4 + 2HF 8. Hvaporisation (KJ/mol) are given for the hydrides of halogens in the following graph. The hydride HF will correspond to (A) P (B) Q (C) R (D) S 9. Alkali metal hydrogen fluorides have a formula M[HF 2]. They are found to contain a linear symmetrical anion having an overall F–H–F distance of 2.26 Å which may be compared with the H–F bond length of 0.92 Å in the moment. Which of the following is false for the anion ? (A) Hydrogen bonding plays a significant role in the existence of the anion (B) Average bond length (H–F) is 1.13 Å in the anion HF2– (C) The stretching of the H–F bond in the anion is 0.21 Å (D) H-atom is bonded to two F-atoms through two T bonds in the anion. A certain hypohalite on treating with hot and conc. NaOH forms anions P and Q. More stable anion among P and Q can be obtained by neutralizing its conjugate acid X. Upon heating X to very high temperature, a compound Y is formed, which is used in estimation of a toxic gas which have 300 times stronger affinity for haemoglobin than dioxygen. Then which of the following statements are true. (A) X is HI. (B) Y is Cl2O5 (C) Final product Y, has total no. of bonds 5. (D) Y on reacting with toxic gas produces a gas which is used in fire extinguisher. 10. p-Block Elements (Halogen & Noble Gases) 11. When F2 is passed into a solution of mineral acid X, a greenish yellow gas Y is formed. Which on treating with slaked lime forms “Z”. When Red litmus is kept in contact with Z, it changes into (A) Blue colour (B) No change in colour (C) White in colour (D) None of these 12. Which statement regarding iodine trichloride is incorrect. (A) It forms dimer in gaseous state. (B) In dimer I2Cl6, the bridge bonds are longer than the terminal bonds. (C) In solid state, it exist as planar molecule. (D) On hydrolysis, it gives a mixture of iodous acid and hydrochloric acid. 13. The order of solubility of noble gases in water is (A) He > Ar > Kr > Ne > Xe (B) He > Ne > Ar > Kr > Xe (C) Xe > Ar > Kr > He > Ne (D) Xe > Kr > Ar > Ne > He 14. Consider following properties of the noble gases. I : They readily form compounds which are colourless. II : They generally do not form ionic compounds. III : Xenon has variable oxidation states in its compounds. IV : the smaller He and Ne do not form clathrate compounds. Select correct properties. (A) I, II, III (B) II, III, IV (C) I, III, IV O2+ (D) All [PtF6]– 15. The formation of is the basis for the formation of xenon fluorides. This is because : (A) O2 and Xe have comparable sizes. (B) both O2 and Xe are gases. (C) O2 and Xe have comparable ionisation energies. (D) O2 and Xe have comparable electronegativities. 16. [HXeO4]– + OH– ⎯→ [X] + [Y] + O2 + H2O The products [X] and [Y] in unbalanced reaction are : (A) [XeO6]4– & Xe (B) [XeO6]4– & XeO3 (C) XeO3 & Xe (D) H2XeO4 & Xe PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE 1. The total number of electrons present in 4th shell of Astatatine (85At) are : 2. How many of the following properties of halogen increases with increase in atomic number. (a) Number of valence electron (b) Metallic nature (c) Boiling points (d) Atomic radii (e) Density (f) Inoisation enthalpies (g) Electronegativities (h) Reactivity (i) Oxidising nature. (j) egH (magnitude wise) 3. A gas P is obtained at anode during the electrolysis of brine. The gas P when treated with excess of NH3 released a diatomic gas Q. Find the value of (x – y) where x & y are the molar mass of P and Q. 4. How many orders are correct : (a) H–F < H–Cl < H–Br < H–I (Bond length) (b) H–F < H–I < H–Br < H–I (Acidic strength) (c) H–I < H–Br < H–Cl < H–F (Bond strength) (d) H–F > H–Cl > H–Br > H–I (Thermodynamic stability) (e) H–F < H–Cl < H–Br < H–I (Reducing power) (f) H–F > H–I > H–Br > H–Cl (Boiling point) HCl + HNO3 ⎯⎯ → 5. (Conc.) (Conc.) In this reaction change in oxidation number of N is ______ 6. How many of the following reactions would have HCl as one of the products ? (a) CH4 + Cl2 ⎯→ (b) FeSO4 + H2SO4 + Cl2 ⎯→ (c) I2 + Cl2 + H2O ⎯→ (d) Cl2 + H2O ⎯→ (e) H2O + SO2 + Cl2 ⎯→ (f) SO3 + Cl2 ⎯→ Electrolysis ⎯ → (g) NaCl(aq) ⎯⎯⎯⎯ (h) Cl2O7 + H2O ⎯→ (i) Cl2 + NaOH(conc.) ⎯→ 7. The number of mixed anhydride among the following are Cl2O ; ClO2 ; Cl2O6 ; Cl2O7 ; N2O5 ; NO2 ; N2O p-Block Elements (Halogen & Noble Gases) 8. Hydrolysis → B+C Br2 + F2 (excess) ⎯→ A ⎯⎯⎯⎯ The summation of atomicities of compound A, B and C is : 9. The number of lone pairs of electrons present in central atom of ClF3 is : 10. Which of the following interhalogens exist at room temperature and have central atom hybridization sp3d2 (a) ClF3 (b) ClF5 (c) BrCl3 (d) IF3 (e) IF5 (f) BrF5 (g) IF7 (h) ICl5 (i) IBr5 (j) BrF3 (k) ClBr5 11. How many of the following properties of noble gases would increase from Helium to Radon ? Boiling point, First Ionisation enthalpy, Atomic volume, Abundance in atmosphere, Density at STP, Valence electrons, Critical temperature. 12. The number of compounds/elements oxidised by XeF2 among following is: HF, HBr, HCl, HI, NH3, CrF2, Pt, S8 13. The oxidation state of xenon in perxenate ion is +n. Give the value of ‘n’. 14. The summation of total no. of lone pairs and  bonds in in species (A, B and C) is. 15. How many of the given compounds can produce XeO 3. XeF6, XeF4, XeO2F2, XeOF4 PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE 1. Which of the following salts will evolve halogen on treatment with conc. H 2SO4 ? (A) NaCl (B) KI (C) NaBr (D) none of these 2. Which of the following reactions are correct ? (A) NaIO3 + 5NaI + 6H2SO4 ⎯→ 6NaHSO4 3H2O + I2 (B) 2KBr + MnO2 + 3H2SO4 ⎯→ 2KHSO4 + MnSO4 + 2H2O + Br2  (C) K2Cr2O7 + 14HCl ⎯→ 2 KCl + 2CrCl3 + 7H2O + 3Cl2 (D) 2K2MnF6 + 4SbF5 ⎯→ 4KSbF6 + MnF3 + F2 3. Which of the following will not displace the halogen from the solution of the halide ? (A) Br2 added to NaI (B) Br2 added to NaCl (C) F2 added to KCl (D) Cl2 added to NaF 4. Iodine reacts with hypo to give : (A) NaI (B) Na2SO3 (C) Na2S4O6 (D) Na2SO4 5. Cl2 + NH3(excess) ⎯→ A + B (A) One of the product is also obtained by decomposition of (NH4)2Cr2O7. (B) Bond order in one of the product is 3 (C) Both products contain chlorine. (D) If Br2 is used instead of Cl2, one of product remain same 6. In which of the following case disproportionation reaction take place. (A) F2 + Hot water ⎯→ products (B) Cl2 + Cold and dilute NaOH ⎯→ products (C) Cl2 + Hot and conc. NaOH ⎯→ products (D) Cl2 + NH3(excess) ⎯→ products 7. In which following statement are correct. (a) Anhydrous hydrogen fluroide is a liquid at ordianry temperature while other halogen acid are gases. (b) Lead acetate does not form any precipitate with HF acid but other halogen acid from precipitates. (c) HF is heated with a mixture of MnO2 and H2SO4, no gas is evloved but in case of HCl, HBr and HI acids evolved gases. (d) HF is not oxidised by strong oxidising agent but other halogen acid are oxidised. (A) a (B) b (C) c (D) d p-Block Elements (Halogen & Noble Gases) 8. A pungent smelling gas X after being dried by concentrated H 2SO4 was dissolved in water to give strongly acidic solution. The gas also gives dense white fumes with NH3. X is also a constituent of aqua-regia. Which of the following is/are true for X ? (A) X is HCl (B) X is Cl2 (C) X is the most volatile among the hydrides of halogens (D) Solution of X in water can liberate CO2 from the solution of sodium hydrogencarbonates. 9. Among the following which reactions are possible (A) F2 + H2O ⎯→ HF + O2 (B) Cl2 + H2O ⎯→ HCl + HClO (C) Br2 + H2O ⎯→ HBr + HBrO (D) I2 + H2O ⎯→ HI + HIO 10. Which of the following statements are true : (A) Strength of oxyacids : HClO4 > HClO3 > HClO2 > HClO (B) Strength of oxyacids : HClO4 > HBrO4 > HIO4 (C) Number of p-p bonds : HClO4 > HClO3 > HClO2 > HClO (D) Percentage s-character of central atom : HClO4 > HClO3 > HClO2 > HClO. 11. Which of the following interhalogen compounds is/are possible : (A) ClF3 (B) IF5 (C) FCl3 (D) BrF5 12. Which of the following statement are true regareding interhalogens : (A) Thermal stability order IF > BrF > ClF (B) Hydrolysis of IF7 produces H5IO6 and HF as products. (C) Interhalogen compounds are diamagnetic in nature. (D) IF7 have pentagonal bipyramidal structure. 13. Which of the following inert gas(es) form(s) clathrate compoud(s) with quinol ? (A) Helium (B) Xenon (C) Krypton (D) Neon 14. Consider the following reactions Ni →A Xe + F2 ⎯⎯ 2 : 1 Xe + F2 ⎯⎯→ B 1 : 20 Xe + F2 ⎯⎯→ C 1 : 5 Select incorrect statements : (A) A, B and C all are non-polar and planner molecule (C) The order of Xe-F bond length is A > C > B 15. (B) B has no lone pair of electrons (D) A, B and C are act as lewis base. Which of the following statements(s) is /are true for XeF 6 ? (A) Its partial hydrolysis gives XeOF4. (B) Its reaction with silica gives XeOF4 (C) It is prepared by the reaction of XeF4 and O2F2 (D) Its reaction with XeO3 gives XeOF4. PART - IV : COMPREHENSION Read the following passage carefully and answer the questions. Comprehension # 1 A red liquid (A) when treated with sodium carbonate gives a mixture of two salts (B) and (C) in the solution in which (C) contains oxygen. The mixture then on acidification with sulphuric acid and distillation produces the red liquid (A) again. 1. Select the correct statement for the liquid (A). (A) It acts as an oxidising agent, (B) It is sparingly soluble in water (C) It converts the yellow-dye stuff fluorescein (I) into red colour compound (D) All of these 2. Which of the following statement is false for salt (B) ? (A) Its solution in water gives pale yellow precipitate with silver nitrate solution (B) Its solution in water gives white precipitate with lead nitrate solution (C) Its acidified solution (with conc. H2SO4) liberates a coloured gas which produces orange red spots on starch paper (D) None p-Block Elements (Halogen & Noble Gases) 3. Which of the following statement is correct ? (A) Liquid (A) undergoes disproportionation reaction in aqueous solution of sodium carbonate (B) The anion of compound (C) has sp3 hybridisation and is trigonal pyramidal in shape (C) (A) and (B) both (D) None of these Comprehension # 2 Pseudo halides are anions having resemblance with halide ions. Group I metals can form salts with pseudo halides. Pseudo halogens can acts as ligands and form coordinate complexes. Their hydrides are weakly acidic and can be prepared in analogous way as halogen hydrides are prepared. Azides, cyanides, selonocyanides are example of pseudo halides. 4. Cyanide, CN– is a pseudo halide. When cyanogen is heated with alkali solution, the products are : (A) HCN, H2O (B) NH3, (NH4)2CO3 (C) NaCN, NaOCN (D) HCOONa, NH3 5. When NaCN reacts with H2SO4, the products are : (A) HCN and Na2SO4 (B) HCN and NaHSO4 (C) (CN)2 and Na2SO4.H2O (D) None of these 6. When sodium pseudo halides are dissolved in water, it resembles with : (A) NaCl (B) NaBr (C) NaF (D) NaI 7. (CN)2 when react with Na metal, the product is : (A) NaCN (B) Na2C2 and N2 (C) NaN3 and C black (D) Na(CN)2– Comprehension # 3 The ionisation energy of dioxygen (O 2) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and F2 are mixed and reacted at different temperatures to give XeF 2, XeF4 and XeF6. Xe also forms an unstable gaseous XeO4 and solid XeO3 which is a very powerful explosive at higher temperatures. Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compoudns with transition metals. Bigger members of rare gases do not form such compounds because of their large size. 8. Xenon forms the larges number of compounds only with oxygen and fluorine because : (i) oxygen and fluorine have very high electronegativity. (ii) ionisation energy of Xe is the largest among rare gases. (iii) ionisation energy of Xe is low compared to those of other rare gases. (iv) low dissociation energy of fluorine molecule compared to those of Cl 2 and Br2 (A) (i), (ii), (iii) (B) (i), (iii), (iv) (C) (iii), (iv) (D) (i), (iv) 9. XeF6 cannot be prepared by the method : 475 −532 K (A) Xe + 3F2 ⎯⎯⎯⎯⎯ → XeF6 500 K (B) XeF2 + 2F2 ⎯⎯⎯⎯ → XeF6 475 K (C) XeF4 + F2 ⎯⎯⎯⎯ → XeF6 → XeF6 + 3H2O (D) XeO3 + 6HF ⎯⎯⎯⎯ 50 atm 10. 475 K  He and Ne do not form any clathrates because : (A) He and Ne are very large in size. (B) being neutral they cannot form any polar bonds with the host molecules. (C) being too small, they cannot be entrapped in the cages of water, phenol or quinol. (D) clathrated with He and Ne are highly explosive. Comprehension # 4 Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in the three columns of the following table. Observe the three columns in which column-1 represents Oxy acid, column-2 represents Oxidation State of Cl while column-3 represents facts. Column-1 Column-2 Column-3 (Oxy acid) (Oxidation State of Cl) (I) HClO2 (P) +1 (i) Chlorine has highest electron affinity (II) HClO3 (Q) +3 (ii) Fluorine never exhibit oxidation state > 0 (III) HClO4 (R) +5 (iii) HI is strong hydra acid The decreasing order of thermal stability is (IV) HClO (S) +7 (iv) IF > BrF > ClF p-Block Elements (Halogen & Noble Gases) 11. 12. 13. Which of the following set of combination is correct? (A) I–(s)–(i) (B) (II)–R–(ii) (C) (III)–Q–(iii) (D) (IV)–R–(iv) Which of the following set of combination is incorrect? (A) (I)–Q–(iv) (B) (II)–R–(iii) (C) (III)–P–(ii) (D) (IV)–P–(i) Which of the following set of combination is correct? (A) (IV)–S–(i) (B) (III)–S–(iii) (C) (II)–Q–(ii) (D) (I)–Q–(iv) Comprehension # 5 Answer Q.14, Q.15 and Q.16 by appropriately matching the information given in the three columns of the following table. Observe the three columns in which column-1 represents Compounds, column-2 represents Hybridisation while column-3 represents facts. Column-1 Column-2 Column-3 (Compounds) (Hybridisation) 14. 15. 16. (I) XeF4 (P) sp3d3 (i) Neon is used in fluorescent bulbs. (II) XeF2 (Q) sp3d2 (ii) Helium do not form clatharate compounds (III) XeF6 (R) sp3d (iii) XeOF4 has square pyramidal structure (IV) XeO3 (S) sp3 (iv) Reacts with H2 produces Xe & HF. Which of the following set of combination is correct? (A) (I)–(S)–(iv) (B) (II)–(R)–(iii) (C) (III)–(Q)–(ii) (D) (IV)–(P)–(i) Which of the following set of combination is incorrect? (A) (IV)– (S)–(i) (B) (III)–(P)–(ii) (C) (II)–(R)–(iii) (D) (I)–(S)–(iv) Which of the following set of combination is correct? (A) (II)–(Q)–(ii) (B) (III)–(R)–(iii) (C) (IV)–(S)–(iv) (D) (I)–(P)–(i) * Marked Questions may have more than one correct option. PART - I : (ADVANCED) / PROBLEMS (PREVIOUS YEARS) Section (A) : Group 17th 1. Give an example of oxidation of one halide by another halogen. Explain the feasibility of reaction. 2. The set with correct order of acidity is : (A) HClO < HClO2 < HClO3 < HClO4 (C) HClO < HClO4 < HClO3 < HClO2 (B) HClO4 < HClO3 < HClO2 < HClO (D) HClO4 < HClO2 < HClO3 < HClO 3. The reaction, 3ClO– (aq) → ClO3– (aq) + 2Cl– (aq) is an example of : (A) oxidation reaction (B) reduction reaction (C) disproportionation reaction (D) decomposition reaction 4. A gas ‘X’ is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas ‘Y’. Identify ‘X’ and ‘Y’. (A) X = CO2, Y = Cl2 (B) X = Cl2, Y = CO2 (C) X = Cl2, Y = H2 (D) X = H2, Y = Cl2 p-Block Elements (Halogen & Noble Gases) Section (B) : Group 18th Paragraph for Question Nos. 5 to 7 The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell. 5. 6. 7. 8. Argon is used in arc welding because of its : (A) low reactivity with metal (C) flammability (B) ability to lower the melting point of metal (D) high calorific value The structure of XeO3 is : (A) linear (B) planar (C) pyramidal (D) T-shaped XeF4 and XeF6 are expected to be : (A) oxidizing (B) reducing (C) unreactive (D) strongly basic All the compounds listed in Column I react with the appropriate options listed in Column II. Column I (A) (CH3)2SiCl2 (B) XeF4 (C) Cl2 (D) VCl5 water. Match the result of the respective reactions with Column II (p) Hydrogen halide formation (q) Redox reaction (r) Reacts with glass (s) Polymerization (t) O2 formation Paragraph for Questions 9 to 10 The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts to two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts with white phosphours to give a compound S. On hydrolysis, S gives an oxoacid of phosphours T. 9. P and Q, respectively, are the sodium salts of : (A) hypochlorus and chloric acids (B) hypochlorus and chlorus acids (C) chloric and perchloric acids (D) chloric and hypochlorus acids 10. R, S and T, respectively, are : (A) SO2Cl2, PCl5 and H3PO4 (C) SOCl2, PCl3 and H3PO2 11. (B) SO2Cl2, PCl3 and H3PO3 (D) SOCl2, PCl5 and H3PO4 The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists : List-I List-II ? P. PbO2 + H2SO4 ⎯⎯ → PbSO4 + O2 + other product 1. NO ? Q. Na2S2O3 + H2O ⎯⎯ → NaHSO4 + other product 2. I2 ? R. N2H4 ⎯⎯ → N2 + other product ? S. XeF2 ⎯⎯ → Codes : P (A) 4 (C) 1 3. Warm Xe + other product Q 2 4 R 3 2 S 1 3 4. Cl2 (B) (D) P 3 3 Q 2 4 R 1 2 S 4 1 p-Block Elements (Halogen & Noble Gases) 12. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is : (A) 0 (B) 1 (C) 2 (D) 3 PART - II : (MAIN) / PROBLEMS (PREVIOUS YEARS) (MAIN) OFFLINE PROBLEMS 1. In case of nitrogen, NCl3 is possible but no NCl5 while in case of phosphorus, PCl3 as well as PCl5 are possible. It is due to : (1) availability of vacant d-orbital in P but not in N. (2) lower electronegativity of P then N. (3) lower tendency of H bond formation in P than N. (4) occurrence of P in solid while N in gaseous state at room temperature. 2. Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. This is due to : (1) strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke. (2) strong affinity for water, conc. HCl pulls moisture of air towards self. The moisture forms droplets of water and hence the cloud. (3) conc. HCl emits strongly smelling HCl gas all the time. (4) oxygen in air reacts with emitted HCl gas to form a cloud of chlorine gas. 3. The substance used in holmes singnals of the ship is a mixture of : (1) CaC2 + Ca3P2 (2) Ca3(PO4)2 + Pb3O4 (3) H3PO4 + CaCl2 (4) NH3 + HOCl 4. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid ? (1) Cr3+ and Cr2O72– are formed (2) Cr2O72– and H2O are formed 2– (3) CrO4 is reduced to + 3 state of Cr (4) CrO42– is oxidized to + 7 state of Cr 5. Which one of the following statements regarding helium is incorrect ? (1) It is used to produce and sustain powerful superconducting magnets (2) It is used as a cryogenic agent for carrying out experiments at low temperatures (3) It is used to fill gas balloons instead of hydrogen because it is lighter and non-inflammable (4) It is used in gas-cooled nuclear reactors 6. Which among the following factors is the most important in making fluorine the strongest oxidizing halogen? (1) Hydration enthalpy (2) Ionization enthalpy (3) Electron affinity (4) Bond dissociation energy 7. The correct order of the thermal stability of hydrogen halides (H–X) is : (1) HI > HBr > HCl > HF (2) HF > HCl > HBr > HI (3) HCl < HF < HBr < HI (4) HI > HCl < HF < HBr 8. Which of the following statements is true? (1) H3PO3 is a stronger acid than H2SO3 (3) HCIO4 is a weaker acid than HClO3 9. (2) In aqueous medium HF is a stronger acid than HCl (4) HNO3 is a stronger acid than HNO2 What products are expected from the disproportionation reaction of hypochlorous acid? (1) HClO3 and Cl2O (2) HClO2 and HClO4 (3) HCl and Cl2O (4) HCl and HClO3 p-Block Elements (Halogen & Noble Gases) 10. Identify the incorrect statement among the following. (1) Cl2 reacts with excess of NH3 to give N2 and HCl. (2) Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO 4 and H2O. (3) Ozone reacts with SO2 to given SO3. (4) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O. 11 Which one of the following reactions of Xenon compounds is not feasible ? (1) 3XeF4 + 6H2O → 2Xe + XeO3 + 12HF + 1.5O2 (2) 2XeF2 + 2H2O → 2Xe + 4HF + O2 (3) XeF6 + RbF → Rb[XeF7] (4) XeO3 + 6HF → XeF6 + 3H2O 12. Which among the following is the most reactive ? (1) Cl2 (2) Br2 (3) 2 13. (4) Cl The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are : (1) ClO2– and ClO3– (2) Cl– and ClO– (3) Cl– and ClO2– (4) ClO– and ClO3– (MAIN) ONLINE PROBLEMS 1. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated? (1) IF7 : pentagonal bipyramid (3) BrF3 : planar T-shaped 2. Which of the following xenon-OXO compounds may not be obtained by hydrolysis of xenon fluorides ? (1) XeO2F2 3. 4. 5. (2) XeOF4 The least number of oxyacids are formed by: (1) Nitrogen (2) Fluorine (3) XeO3 (4) XeO4 (3) Chlorine (4) Sulphur Chlorine water on standing loses its colour and forms: (1) HCl only (2) HCl and HClO2 (3) HCl and HOCl (4) HOCl and HOCl2 The non-metal that does not exhibit positive oxidation state is : (1) Fluorine 6. (2) BrF5 : trigonal bipyramid (4) ICl3 : planar dimeric (2) Oxygen (3) Chlorine (4) Iodine The following statements concern elements in the periodic table. Which of the following is true? (1) The Group 13 elements are all metals. (2) All the elements in Group 17 are gases. (3) Elements of Group 16 have lower ionization enthalpy values compared to those of Group 15 in the corresponding periods. (4) For Group 15 elements, the stability of +5 oxidation state increases down the group. 7. XeF6 on partial hydrolysis with water produces a compound 'X'. The same compound 'X' is formed when XeF6 reacts with silica. The compound 'X' is : 17), 4/120] (1) XeO3 (2) XeF4 (3) XeF2 (4) XeOF4 8. Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y'. Compounds 'X' and 'Y' and the oxidation state of Xe are respectively : (1) XeOF4 (+6) and XeO3 (+6) (2) XeO2 (+4) and XeO3 (+6) (3) XeOF4 (+6) and XeO2F2 (+6) (4) XeO2F2 (+6) and XeO2 (+4) 9. Among the following reactions of hydrogen with halogens, the one that requires a catalyst is: (1) H2 + F2 → 2HF 10. (2) H2 + Cl2 → 2HCl (3) H2 + I2 → 2HI (4) H2 + Br2 → 2HBr Chlorine on reaction with hot and concentrated sodium hydroxide gives: (1) ClO3– and ClO2– (2) Cl– and ClO– (3) Cl– and ClO3– (4) Cl– and ClO2– p-Block Elements (Halogen & Noble Gases) EXERCISE - 1 PART - I AI-1. A solution of potassium hydrogen fluoride (KHF2) in anhydrous HF. electrolysis KF + HF ⎯→ K [HF2] ⎯⎯⎯⎯⎯ → H2 + F 2 On Electrolysis : Anode : HF2– ⎯→ F2 + H+ + 2e– 2HF + H+ + 2e – ⎯→ H2 + HF2– Cathode : AI-2. It contains trace amounts of NaIO3 and NaIO4. An important property of chile salt petre is that it is soluble in water while its impurities are not. AII-3. Neil Bartlett obtained an orange yellow solid from the reaction of xenon with PtF 6to have Xe+[PtF6]–. He noticed earlier that O2 with PtF6 reacts to give O2+[PtF6]– and since ionisation enthalpies of O2 and Xe were close to each other (Xe = 1170, O2 = 1175 kJ mol–1), he could succeed to prepare Xe+[PtF6]– from Xe and PtF6. BI-1. The electrode potential of F2 (+2.87 V) is much higher than that of Cl2 (+1.36 V). 1 H = Ex-x – |E.A| – |H°Hy| 2 1 – H – x (aq.) x2(g) + e (aq.) 2 1 H°Hy Ex-x 2 (X(g) E.A. x – (g) F2 Cl2 158.8 242.6 F2 < Cl2 Magnitude of E.A/kJmol–1 333 349 F2 < Cl2 Magnitude of H°Hyd./kJmol–1 515 381 F2 > Cl2 Bond dissociation enthalpy/kJmol–1 Order Oxidizing power F2 > Cl2 Therefore F2 is much stronger oxidising agent than Cl2. BI-2. F : –1, 0 Cl : –1, 0, +3, +5, +7 Br : –1, 0, +3, +5, +7 I : –1, 0, +3, +5, +7 BI-3. Enthalpy of dissociation (x-x)/kJ mol–1 Cl–Cl > Br–Br > F–F > I–I, Smaller enthalpy of dissociation of F2 is due to relatively large electronselectrons repulsion among the lone pairs in F2 molecule. C-1. ; ; ; ; p-Block Elements (Halogen & Noble Gases) CI-2. CI-3. Cl ; Cl (i) Structure of Cl4– – No. of electrons in the valence shell of the central I atom = 7. Cl Cl No. of electrons provided by four Cl atoms = 4×1 = 4 Charge on the central atom = 1 I  Total no. of electrons around the central atom = 7 + 4 + 1 = 12 Total no. of electron pairs around the central atom = 12 /2 = 6 But the no. of bond pairs = 4 ( there are four I-Cl bonds) Cl Cl  No. of lone pairs = 6 – 4 = 2 Thus, I in Cl4– has 4 bond pairs and 2 lone pairs. Therefore, according to Square planar VSEPR theory, it should be square planar. Now a noble gas compound having 12 electrons in the valence shell of the central atom is XeF 4 (8 + 1 × 4 = 12). Like Cl4– , it also has 4 bond pairs and 2 lone pairs. Therefore, like, XeF 4 is also square planar. (ii) Structure of Br2– . No. of electrons in the valence shell of the central I atom = 7 No. of electrons provided by two Br atoms = 2 × 1 = 2 Charge on the central I atom is = 1  Total no. of electrons around the central I atom = 7 + 2 + 1 = 1 But the no. of bond pairs = 2 ( there are two I – Br bonds)  No. of lone pairs = 5 – 2 = 3 Br – I Br LINEAR Thus, I in Br2– has two bond pairs and three lone pairs, Therefore, according to VSEPR theory, it should be linear. Now a noble gas compound having 10 electrons in the valence shell of the central atom is XeF 2 (8 + 1 × 2 =10). Like IBr–, it also has 2 bond pairs and 3 lone pairs. (iii) Structure of BrO3– . In BrO3– , since O is more electronegative than Br, therefore, –ve charge stays on the O atom. Br O O - O PYRAMIDAL – 3 Therefore, in BrO , there are two Br = O bonds and one bond Br – O– bond. Now according to VSEPR theory, double bonds do not contribute any electron while single bonds contribute one electron towards the total number of the central atom. However, both double and single bonds contribute one bond pair. Thus, total number of electrons is the valence shell of the central Br atom = 7 + 2 × 0 + 1 × 1 = 8  No. of electron pairs around Br atom = 8/2 = 4 But total number of bond pairs = 2 × 1 (Br = 0) + 1 × 1 (Br – O–) = 3 and lone pairs = 4 – 3 = 1. Thus, BrO3– has 3 bond pairs and one lone pair. Therefore, according to VSEPR theory, it should be pyramidal. Now a noble gas compound having 8 electrons in the valence shell of the central atom is XeO 3 (8 × 1 + 3 × 0 = 8). Like BrO3– , it also has 3 bond pairs and one lone pair. Therefore, like BrO3– , XeO3 is also pyramidal. p-Block Elements (Halogen & Noble Gases) CII-4. XeF2 > XeF4 2.00 Å 1.94 Å (Order of Xe–F bond length) XeF2 : sp3d, axial bonds are long & weak XeF4 : sp3d2 square planar : all bonds identical. DII-1. (i) Van der Waal’s forces of attraction among molecules increase with the increase of atomic masses. Thus, higher energy is required to separate these molecules as to get gaseous state. Hence, the boiling points increase with increase of atomic masses. (ii) The size of cavities formed during crystallisation of quinol is more than the size of helium and neon atoms. DI-2. All the halogens are coloured. The colour is due to the allowed  The colour of halogen is actually the colour of transmitted light. →  molecular orbital transitions. E DI-3. F2 being a strong oxidizing agent H2O to O2 or O3. 2F2 (g) + 2H2O(l) ⎯→ 4H+ (aq.) + 4F–(aq.) + O2(g) 3F2(g) + 3H2O(l) ⎯→ 6 H+(aq.) + 6F–(aq.) + O3(g) Cl2, on the other hand, reacts with H2O to form hydrochloric acid and hypochlorous acid to a small extent. Cl2(g) + H2O(l) HCl(aq.) + HOCl (aq.) Hydrochloric acid DI-4. Hydrochlorous acid 2F2 + 2NaOH ⎯→ OF2 + 2NaF + H2O Cold Other halogens (X = Cl / Br / I) disproportionate : X2 + 2NaOH ⎯⎯⎯ → NaX + NaOX + H2O DI-5. Hot X2 + 2NaOH ⎯⎯⎯ → NaX + NaXO3 + H2O DI-6. I2 , I2 + HNO3 (conc.) ⎯→ HIO3 + NO2 + H2O DI-7. (i) The standard reduction potential of fluorine is maximum. + e– ⎯→ Thus, it cannot be oxidised by any other reagent. F – ion is very stable due to small size and high electronegativity of fluorine atom. (ii) F-does not have empty d-orbital liked other halogen. The formation of X 3– ions involves sp3dhybridizaion. EI-1. Ba (ClO3)2 + H2SO4 ⎯→ BaSO4 + 2HClO3 KCIO4 + H2SO4 ⎯→ KHSO4 + HCIO4 EI-2. HOCl acts as an acid, HOCl ⎯→ H+ + OCl–, and thus turns blue litmus to red. It also work as bleaching agent, HOCl ⎯→ HCl + [O], and the red litmus is decolorized by nascent oxygen. EI-3. Chlorine, bromine and iodine form four series of oxyacids of the general formula HOX , HOXO, HOXO 2 and HOXO3 in which the oxidation states of halogen (X = Cl, Br or I) is + 1, + 3 , and + 5 and + 7 respectively. However, due to high electronegativity, small size and absence of d-orbitals, F cannot act as central atom in higher oxacids such as HOFO. HOFO 2 and HOFO3 In which the oxidation state of F would be + 3, + 5 and + 7. It just forms one oxoacid. i.e., HOF in which the oxidation state of F is –1 and oxygen is in +1. EI-4. (i) 2NaClO3 + SO2 + H2SO4 ⎯→ 2NaHSO4 + 2ClO2 (ii) 5– + IO3– + 6H+ ⎯→ 3I2 + 3 H2O EI-5. ClO2 (s) is a mixed anhydride of HClO 2 and HClO3 because on dissolving in water it gives a mixture of these two acids. 2ClO2 + 2NaOH ⎯→ NaClO2 + NaClO3 + H2O FI-1. (i) HI > HBr > HCl > HF ; Due to decrease in bond energy (ii) HI is the strongest reducing agent. p-Block Elements (Halogen & Noble Gases) FI-2. Conc. HCl can be oxidized to Cl2 by heating with a number of oxidizing agents such as MnO2, KMnO4, K2Cr2O7, etc. MnO2 + 4 HCl ⎯→ MnCl2 + Cl2 + 2 H2O Cl2 can reduced to HCl by its reaction with H2 in presence of diffused sunlight. Diffused sunlight → 2 HCl H2 + Cl2 ⎯⎯⎯⎯⎯⎯ FI-3. HI is a strong reducing agent and, therefore, reduces H2SO4 to SO2 and itself get oxidised to 2.  3Na + conc. H3PO4 ⎯⎯ → K3PO4 + H FI-4. Aqua regia is three parts of concentrated HCl and one part of concentrated HNO 3. 2Au + 11HCl + 3HNO3 ⎯→ 2H[AuCl4] + 3NOCl + 6H2O 8HCl + 2HNO3 + Pt ⎯→ H2[PtCl6] + 2NOCl + 4H2O FI-5. (i) In anhydrous state, HCl is a covalent molecule. In aqueous solution, HCl combines with water molecule to form H3O+ and Cl– ions. HCl (g) + H2O( ) ⎯→ H3O+ (aq) + Cl– (aq) (ii) HF attacks glass bottles. The sodium and potassium silicates are converted into flurosilicates Na2 SiO3 + 6HF ⎯→ Na2 SiF6 + 3H2O (iii) The weakest acidic nature of HF is due to : High bond dissociation enthalpy of HF bond. HF has conjugate base F –. Small size and concentrated charge make F– a Lewis base. Hence, it can accept back H+.  (ii) 2 + SO2 + H2O FI-6. (i) H (hydrogen iodide) FI-7. (i) Pb3O4 + 8HCl ⎯→ 3PbCl2 + 4H2O + Cl2 (ii) 3 [SiO2 + 4HF ⎯→ SiF4 + 2H2O] 3SiF4 + 3H2O ⎯→ 2H2SiF6 + H2SiO3 _____________________________________ Overall : 3SiO2 + 12HF ⎯→ 2H2SiF6 + H2SiO3 +3H2O White waxy deposit GII-1. 673 K, 1 bar Xe(g) + F2(g) ⎯⎯⎯⎯⎯ → XeF2(s) (Xenon in excess) 873 K, 7 bar Xe(g) + 2F2(g) ⎯⎯⎯⎯⎯ → XeF4(s) (1 : 5 ratio) 573 K, 60−70 bar Xe(g) + 3F2(g) ⎯⎯⎯⎯⎯⎯⎯ → XeF6(s) (1 : 20 ratio) GII-2. Partial hydrolysis of XeF6 gives XeOF4. XeF6 + H2O ⎯→ XeOF4 + 2 HF GII-3. Yes, hydrolysis of XeF4 leads to a redox reaction(disproportination) 6XeF4 + 12 H2O ⎯→ 4 Xe + 2XeO3 + 24 HF + 3O2 GII-4. XeF6 + 3 H2O ⎯→ XeO3 + 6 HF XeF6 + H2O ⎯→ XeOF4 + 2 HF XeF6 + 2H2O ⎯→ XeO2F2 + 4 HF GII-5. (i) (ii) (iii) HI-1. XeF2 + H2 → Xe + 2HF 2XeF6 + SiO2 → 2XeOF4 + SiF4 ; 2XeOF4 + SiO2 → 2XeO2F2 + SiF4 2XeO2F2 + SiO2 → 2XeO3 (Explosive) + SiF4 XeF6 + SbF5 → [XeF5]+ [SbF6]– [Addition ionic product is formed] (a) Cl2 + F2 473 K ⎯⎯⎯ ⎯ → 2ClF (Equal volume) (c) Br2 + 3F2 ⎯⎯ → 2BrF3 (b) I2 + 3Cl2 ⎯⎯→ 2ICl3 (Excess) (d) Br2 + 5F2 (Excess) ⎯⎯ → 2BrF5 p-Block Elements (Halogen & Noble Gases) HI-2. ICl is more reactive than I2 because I-Cl bond is weaker than I-I bond. Consequently, ICl breaks easily to form halogen atoms which readily bring about the reactions. HI-3. (a) Chlorine trifluoride (ClF3); iodine trichloride (Cl3 or 2Cl6) (b) ICl + H2O ⎯→ HCl + HOI HI-4. (i) Chlorine bleaching action is due to oxidation while that of sulphur dioxide is due to reduction. Hence, the substance bleached by SO2 is reoxidised by the oxygen of the air to its original state. Cl2 bleaches coloured material by oxidation and thus bleaching is permanent. Cl2 + H2O ⎯→ 2HCl + [O] Coloured material + [O] ⎯→ Colourless On the other hand beaching by SO2 is by reduction and thus temporary because colourless articles are further oxidised by air. SO2 + 2H2O ⎯→ 2H2SO4 + 2[H] [O] Coloured material + [H] ⎯→ Colourless ⎯⎯→ coloured (ii) 2 is a covalent molecule. Thus, its solubility is less in polar solvent, i.e., water. Potassium iodide combines with iodine and forms a polyhalide which is an ionic compound. Being ionic, K3 is more soluble. K + 2 ⎯⎯→ K3 (K+ 3–) brown solution HI-5. (i) (ii) 2NaO3 + 5NaHSO3 ⎯→ 3 NaHSO4 + 2Na2SO4 + H2O + 2 Ca(OH)2 + Cl2 ⎯→ Ca(OCl)Cl + H2O PART - II AII-1. (B) AII-2. (A) AII-3. (A) AI-4. (D) AI-5. (C) BII-1. (C) BI-2. (B) BI-3. (B) CI-1. (C) CII-2. (D) CII-3. (D) DI-1. (A) DII-2. (D) DI-3. (C) DI-4. (C) DI-5. (B) EI-1. (D) EI-2. (C) EI-3. (B) EI-4. (A) EI-5. (B) EI-6. (A) FI-1. (D) FI-2. (B) FI-3. (B) FI-4. (A) FI-5 (B) FI-6. (D) FI-7. (C) FI-8. (A) FI-9. (C) FI-10. (A) GII-1. (B) GII-2. (B) GII-3. (A) GII-4. (A) GII-5. (D) HI-1. (D) HI-2. (B) HI-3. (A) HI-4. (C) PART - III 1. (A - p,q) ; (B - p,r) ; (C - q,s) ; (D - p,q,r,s) 2. 3. (A - p,q,r,s) ; (B - p,q,r,s,t); (C - p,q,r,s) ; (D - q,s,t) (A - q,s) ; (B - s) ; (C - p) ; (D - r) EXERCISE - 2 PART - I 1. (D) 2. (B) 3. (A) 4. (A) 5. (C) 6. (A) 7. (A) 8. (A) 9. (D) 10. (D) 11. (C) 12. (A) 13. (D) 14. (B) 15. (C) 16. (A) PART - II 1. 32 2. 4 3. 43 4. 6 5. 2 6. 11. 5 4 7. 12. 3 7 8. 13. 13 8 9. 14. 2 20 10. 15. 3 4 4. (AC) 5. (ABD) PART - III 1. (BC) 2. (ABCD) 3. (BD) p-Block Elements (Halogen & Noble Gases) 6. (BC) 7. (ACD) 8. (ACD) 9. (ABC) 10. (ABC) 11. (ABD) 12. (ABCD) 13. (BC) 14. (AB) 15. (ABCD) PART - IV 1. (D) 2. (D) 3. (C) 4. (C) 5. (A) 6. (C) 7. (A) 8. (B) 9. (D) 10. (C) 11. (B) 12. (C) 13. (B) 14. (B) 15. (D) 16. (C) EXERCISE – 3 PART - I 1. 2K(aq.) + Cl2 ⎯→ 2KCl(aq.) + 2 In the reaction Cl2 oxidises iodide ion (–1 oxidation state) to 2 (0 oxidation state). Cl2 has higher oxidation potential than 2 and thus oxidises iodide to iodine getting itself reduced to chloride ion. Similarly, 2I– (aq.) ⎯→ I2 (s) + 2e– Cl2 (g) + 2e– ⎯→ 2Cl– (aq) _______________________________ 2I– (aq) + Cl2 (g) ⎯→ I2 (s) + 2Cl– (aq). 2. (A) 3. (C) 7. (A) 8. (A - p, s) ; (B - p, q, r, t); (C - p, q) ; (D - p) 10. (A) 11. (D) 4. 12. (C) 5. (A) 6. (C) 9. (A) (C) PART – II (MAIN) OFFLINE PROBLEMS 1. (1) 2. (4) 3. (1) 4. (2) 5. (3) 6. (1) 7. (2) 8. (4) 9. (4) 10. (2) 11 (4) 12. (4) 13. (2) (MAIN) ONLINE PROBLEMS 1. (2) 2. (4) 3. (2) 4. (3) 5. (1) 6. (3) 7. (4) 8. (3) 9. (3) 10. (3)