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Cu(OH)2 × CuSO 4 + 8NH3 ¾®2[Cu(NH3 )4 ]SO 4 + 2OH– + SO 2–, 4, Tetraammine copper (II) (Deep blue solution), , Q. 3 In a cyclotrimetaphosphoric acid, , molecule, how many single and double, , bonds are present?, (a) 3 double bonds; 9 single bonds, (b) 6 double bonds; 6 single bonds, (c) 3 double bonds; 12 single bonds, (d) Zero double bond; 12 single bonds, , Ans. (c), , Cyclotrimetaphosphoric acid contains three double bonds and 9 single bonds as, shown below, O a, , 1 O, , c, , 2, , P, H, , 10, , O, , P, , 9 8, , 3, , 4, O, , O, 7, , P, , b, , O, , O, , 12, , H, , 5, , 6, , O, , O, , 11, , H, , Cyclotrimetaphosphoric acid (HPO 3 ) 3, a, b, c are three p bonds and numerics 1 to 12 are sigma (s) bonds., , Q. 4 Which of the following elements can be involved in pp - dp bonding?, (a) Carbon, (c) Phosphorus, , Ans. (c), , (b) Nitrogen, (d) Boron, , Among given four elements i.e., carbon, nitrogen, phosphorus and boron. Only, phosphorus has vacant d-orbit so only phosphorus has ability to form pp - dp, bonding., , Q. 5 Which of the following pairs of ions are isoelectronic and isostructural?, (a) CO23 - , NO3-, , Ans. (a), , (b) CIO3- , CO32 -, , (c) SO23 - , NO3-, , (d) CIO3- , SO32 -, , Compounds having same value of total number of electrons are known as, isoelectronic., For CO 23 For NO –3, Total number of electrons, Total number of electrons, = 6+ 8´3+ 2, = 7 + 8´3+ 1, = 6 + 24 + 2, = 7 + 25, = 32, = 32, Hence, CO 23 and NO 3 are isoelectronic. These two ions have similar structure so they, are isostructural., s, s, , O, , O, C==O, O, , s, , N==O, O, , Both have triangular planar structure as in both the species carbon and nitrogen are sp2, hybridised. Hence, (a) is the correct choice.

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Q. 6 Affinity for hydrogen decreases in the group from fluorine to iodine. Which, of the halogen acids should have highest bond dissociation enthalpy?, (a) HF, , Ans. (a), , HF, HCl, HBr, HI, , (b) HCl, , (c) HBr, , (d) HI, , On moving top to bottom, · Size of halogen atom increases, · H–X bond length increases, · Bond dissociation enthalpy decreases, , Q. 7 Bond, , dissociation enthalpy of E—H (E= element) bonds is given below., Which of the compounds will act as strongest reducing agent?, Compound, D diss (E ¾ H) / kJ mol -1, , (a) NH3, , Ans. (d), , NH3, , PH3, , 389, , 322, , (b) PH3, , AsH3, , SbH3, , 297, , 255, , (c) AsH3, , (d) SbH3, , On moving top to bottom, size of central atom increases. Bond length of X—H bond, increases and bond dissociation energy decreases. Hence, reducing nature, increases., NH3, PH3, · Bond length increases, AsH3, · Bond dissociation energy decreases, SbH3, · Reducing character increases, Hence, SbH 3 is act as strongest reducing agent among these., , Q. 8 On, , heating with concentrated NaOH solution in an inert atmosphere of, CO2 , white phosphorus gives a gas. Which of the following statement is, incorrect about the gas?, (a) It is highly poisonous and has smell like rotten fish, (b) It's solution in water decomposes in the presence of light, (c) It is more basic than NH3, (d) It is less basic than NH3, , Ans. (c), , White phosphorous on reaction with NaOH solution in the presence of inert, atmosphere of CO 2 it produces phosphine gas which is less basic than NH3 ., P4 + 3 NaOH + 3H2O ¾® PH3 + 3NaH2PO 2, (sodium hypophosphite), , Q. 9 Which of the following acids forms three series of salts?, (a) H3PO2, , (b) H3BO 3, , (c) H3PO 4, , (d) H3PO 3, , Ans. (c) Structure of H3PO4 is, O, P, HO, , OH, , OH, , H3PO 4 has 3 - OH groups i.e., has three ionisable H-atoms and hence forms three, series of salts. These three possible series of salts for H3PO 4 are as follows, NaH2PO 4 , Na 2HPO 4 and Na 3PO 4

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Q. 10 Strong reducing behaviour of H 3PO2 is due to, (a) low oxidation state of phosphorus, (b) presence of two ¾ OH groups and one P ¾ H bond, (c) presence of one ¾ OH group and two P ¾ H bonds, (d) high electron gain enthalpy of phosphorus, , Ans. (c), , Strong reducing behaviour of H3PO 2 is due to presence of two P—H bonds and one, P ¾ OH bond, O, P, H, , H, OH, , Hypophosphorous following, (Monobasic), , Q. 11 On, , heating lead nitrate forms oxides of nitrogen and lead. The oxides, formed are ......... ., (a) N2O, PbO, , Ans. (b), , (b) NO2 , PbO, , (c) NO, PbO, , (d) NO, PbO2, , On heating lead nitrate it produces brown coloured nitrogen dioxide (NO 2 ) and lead (II), oxide., D, , 2Pb (NO 3 )2 ¾¾® 4NO 2 + 2PbO + O 2, , Q. 12 Which of the following elements does not show allotropy?, (a) Nitrogen, , Ans. (a), , (b) Bismuth, , (c) Antimony, , (d) Arsenic, , Nitrogen does not show allotropy due to its weak N¾N single bond. Therefore, ability, of nitrogen to form polymeric structure or more than one structure or form become, less. Hence, nitrogen does not show allotropy., , Q. 13 Maximum covalency of nitrogen is ......... ., (a) 3, , (b) 5, , (c) 4, , (d) 6, , Ans. (c) Maximum covalency of nitrogen is 4 in which one electron is made available by, s-orbital and 3 electrons are made available by p orbitals. Hence, total four electrons, are available for bonding., , Q. 14 Which of the following statements is wrong?, (a) Single N ¾ N bond is stronger than the single P ¾ P bond., (b) PH3 can act as a ligand in the formation of coordination compound with, transition elements., (c) NO2 is paramagnetic in nature., (d) Covalency of nitrogen in N2O5 is four., , Ans. (a), , True statement is that single N ¾ N bond is weaker than the single P ¾ P bond. This is, why phosphorous show allotropy but nitrogen does not., (i) PH3 acts as a ligand in the formation of coordination compound due to presence, of lone pair of electrons.

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(ii) NO 2 is paramagnetic in nature due to presence of one unpaired electron., Structure of NO 2 is, N, , O, , O, , (iii) Covalency of nitrogen in N2O 5 is 4., O, , O, , O, N, , N, , O, , O, , Q. 15 A, , brown ring is formed in the ring test for NO -3 ion. It is due to the, formation of, , Ans. (a), , (a) [Fe (H2O)5 (NO)]2 +, , (b) FeSO 4 × NO2, , (c) [Fe(H2O)4 (NO)2 ]2 +, , (d) FeSO 4 × HNO 3, , When freshly prepared solution of FeSO 4 is added in a solution containing NO -3 ion, it, leads to formation of a brown coloured complex. This is known as brown ring test of, nitrate., NO -3 + 3Fe 2 + + 4H+ ¾® NO + 3Fe 3 + + 2H2O, [Fe (H2O)6 ]2 + + NO ¾® [Fe (H2O)5 (NO)]2 + + H2O, Brown ring, , Q. 16 Elements of group- 15 form compounds in +5 oxidation state. However,, bismuth forms only one well characterised compound in +5 oxidation, state. The compound is, (a) Bi2O 5, (c) BiCl5, , Ans. (b), , (b) BiF5, (d) Bi2S 5, , Stability of + 5 oxidation state decreases top to bottom and + 3 oxidation state, increases top to bottom due to inert pair effect. Meanwhile compound having + 5, oxidation state of Bi is BiF5 . It is due to smaller size and high electronegativity of, fluorine., , Q. 17 On heating ammonium dichromate and barium azide separately we get, (a) N2 in both cases, (b) N2 with ammonium dichromate and NO with barium azide, (c) N2 O with ammonium dichromate and N2 with barium azide, (d) N2 O with ammonium dichromate and NO2 with barium azide, , Ans. (a), , On heating ammonium dichromate and barium azide it produces N2 gas separately., D, , (NH4 )2 Cr2O 7 ¾¾® N2 + 4H2O + Cr2O 3, Ba(N3 )2 ¾® Ba + 3N2, , Q. 18 In, , the preparation of HNO 3 , we get NO gas by catalytic oxidation of, ammonia. The moles of NO produced by the oxidation of two moles of, NH 3 will be ......... ., (a) 2, , (b) 3, , (c) 4, , (d) 6

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Ans. (a), , Two moles of NH3 will produce 2 moles of NO on catalytic oxidation of ammonia in, preparation of nitric acid., D, , 4NH3 + 5O 2 ¾¾¾¾¾® 4 NO (g ) + 6H2O (l), Pt \Rh gauge catalyst, , Q. 19 The oxidation, , state of central atom in the anion of compound NaH2PO2, will be ......... ., (a) + 3, , (b) + 5, , (c) + 1, , (d) - 3, , Ans. (c) Let oxidation state of P in NaH2PO2 is x., 1+ 2 ´1+ x + 2 ´ - 2 = 0, 1+ 2 + x - 4 = 0, + x - 1= 0, x=+1, , Q. 20 Which of the following is not tetrahedral in shape?, (a) NH+4, , Ans. (c), , (b) SiCl4, , (d) SO 42 -, , (c) SF4, , SF4 has sea-saw shaped as shown below, , S, , F, F, , F, F, , It has trigonal bipyramidal geometry having sp3d hybridisation., , Q. 21 Which of the following are peroxoacids of sulphur?, (b) H2SO 5 and H2S2O7, (d) H2S2O 6 and H2S2O7, , (a) H2SO 5 and H2S2O 8, (c) H2S2O7 and H2S2O 8, , Ans. (a), , Peroxoacids of sulphur must contain one¾O¾O¾bond as shown below, O, , O, , S, , H—O—O, , S, , OH, , O, , H2SO5, , O, , O, , O—O, , OH, , S, , OH, , O, , H2S2O8, , Q. 22 Hot conc. H2SO 4, , acts as moderately strong oxidising agent. It oxidises, both metals and non-metals. Which of the following element is oxidised, by conc. H2SO 4 into two gaseous products?, (a) Cu, , Ans. (c), , (b) S, , (c) C, , (d) Zn, , H2SO 4 is a moderately strong oxidising agent which oxidises both metals and, non-metals as shown below, Cu + 2H2SO 4 (conc) ¾® CuSO 4 + SO 2 + 2H2O, S + 2H2SO 4 (conc) ¾® 3SO 2 + 2H2O, While carbon on oxidation with H2SO 4 produces two types of oxides CO 2 and SO 2 ., C + 2H2SO 4 (conc) ¾® CO 2 + 2SO 2 + 2H2O

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Q. 23 A, , black compound of manganese reacts with a halogen acid to give, greenish yellow gas. When excess of this gas reacts with NH 3 an unstable, trihalide is formed. In this process the oxidation state of nitrogen, changes from ......... ., (a) - 3 to + 3, (c) - 3 to + 5, , Ans. (a), , (b) - 3 to 0, (d) 0 to - 3, , Black coloured compound MnO 2 reacts with HCl to produce greenish yellow coloured, gas of Cl 2, MnO 2 + 4HCl ¾® MnCl 2 + 2H2O + Cl 2, (Black), , (greenish, yellow gas), , Cl 2 on further treatment with NH3 produces NCl 3 ., -3, , +3, , N H3 + 3Cl 2 ¾® N Cl 3 + 3HCl, , NH3 (- 3) changes to NCl 3 (+3) in the above reaction. Hence, (a) is the correct choice., , Q. 24 In the preparation of compounds of Xe, Bartlett had taken O2+ Pt F6-, , as a, , base compound. This is because, , (a) both O2 and Xe have same size., (b) both O2 and Xe have same electron gain enthalpy., (c) both O2 and Xe have almost same ionisation enthalpy., (d) both Xe and O2 are gases., , Ans. (c), , Bertlett had taken O +2 Pt F6- as a base compound because O 2 and Xe both have, almost same ionisation enthalpy. The ionisation enthalpies of noble gases are the, highest in their respective periods due to their stable electronic configurations., , Q. 25 In solid state PCl 5 is a ......... ., (a) covalent solid, (b) octahedral structure, (c) ionic solid with [PCl6]+ octahedral and [PCl4]- tetrahedral, (d) ionic solid with [ PCl4]+ tetrahedral and [PCl6] - octahedral, , Ans. (d), , In solid state PCl 5 exists as an ionic solid with [PCl 4 ]+ tetrahedral and [PCl 6 ]octahedral., s, , Cl, Cl, , Cl, , P, Cl, , +, , Cl, , P, , Cl, , Cl, [PCl6]–, octahedral, , Cl, , Cl, , Cl, +, , [PCl4], tetrahedral

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Q. 26 Reduction, , potentials of some ions are given below. Arrange them in, decreasing order of oxidising power., ClO4-, , Ion, Reduction potential E - /V, , IO4-, , E° = 1.19 V, , BrO4-, , s, , E = 1.65 V, , s, , E = 1.74 V, , (a) CIO -4 > IO -4 > BrO 4-, , (b) IO -4 > BrO -4 > CIO 4-, , (c) BrO -4 > IO -4 > CIO 4-, , (d) BrO -4 > CIO 4- > IO 4-, , K Thinking Process, This problem is based on concept of standard reduction potential of species and, oxidising property., , Ans. (c), , Greater the SRP value of species higher will be its oxidising power., , Species, , E°cell, , ClO -4, , 1.19V, n, , SRP of species increases., , -, , 1.65V, n, , Oxidising power increase., , BrO -4, , 1.74V, , IO 4, , Here, SRP = standard reduction potential., , Q. 27 Which of the following is isoelectronic pair?, Ans. (b), , (a) ICI2 , CIO2, , (b) BrO2- , BrF2+, , (c) CIO2 , BrF, , (d) CN- , O 3, , Isoelectronic pair have same number of electrons, , Total number of electrons, , BrO2-, , BrF2+, , = 35 + 2 ´ 8 + 1= 52, , = 35 + 9 ´ 2 - 1= 52, , Hence, (b) is the correct choice, while in another cases this value is not equal., , ICl2, 53 + 2 ´ 17 = 87, , ClO2, 17 + 16 = 33, , CN = 6 + 7 + 1= 14, Hence, only (b) is the correct choice., , ClO2, 17 + 16 = 33, , BrF, 35 + 9 = 44, , O3, = 8 ´ 3 = 24

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Multiple Choice Questions (More Than One Options), Q. 28 If, , chlorine gas is passed through hot NaOH solution, two changes are, observed in the oxidation number of chlorine during the reaction. These, are ......... and ........., (a) 0 to +5, , (b) 0 to +3, , (c) 0 to -1, , (d) 0 to +1, , Ans. (a, c), When chlorine gas is passed through hot NaOH solution it produces NaCl and NaClO 3 ., 0, , –1, , 6NaOH + 3Cl2, , –5, , 5NaCl + NaClO3 + 3H2O, , Oxidation state varies from 0 to - 1and 0 to + 5., Hence, (a) and (c) are correct choices., , Q. 29 Which of the following options are not in accordance with the property, mentioned against them?, (a) F2 > Cl2 > Br2 > I2, (b) MI > MBr > MCl > MF, (c) F2 > Cl2 > Br2 > I2, (d) HI < HBr < HCI < HF, , Oxidising power, Ionic character of metal halide, Bond dissociation enthalpy, Hydrogen-halogen bond strength, , Ans. (b, c), F2 > Cl 2 > Br2 >I2 As ability to gain electron increases oxidising property increases. Here, F is, the most electronegative element having highest value of SRP hence it has highest oxidising, power., MI > MBr > MCl > MF, This is the incorrect order of ionic character of metal halide., Correct order can be written as, MI < MBr < MCl < MF, As electronegativity difference between metal and halogen increases ionic character, increases., F2 > Cl 2 > Br2 >I2, This is incorrect order of bond dissociation energy. Correct order is Cl 2 > Br2 > F2 >I2 due to, electronic repulsion among lone pairs in F2 molecule., , Q. 30 Which of the following is correct for P4, (a) It has 6 lone pairs of electrons, (c) It has three P ¾ P single bonds, , molecule of white phosphorus?, , (b) It has six P ¾ P single bonds, (d) It has four lone pairs of electrons, , Ans. (b, d), Structure of P4 molecule can be represented as, P, 60°, P, , P, P, , It has total four lone pairs of electrons situated at each P-atom., It has six P—P single bond.

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Q. 31 Which of the following statements are correct?, (a) Among halogens, radius ratio between iodine and fluorine is maximum., (b) Leaving F ¾ F bond, all halogens have weaker X ¾ X bond than X ¾ X ¢ bond, in interhalogens., (c) Among interhalogen compounds maximum number of atoms are present in, iodine fluoride., (d) Interhalogen compounds are more reactive than halogen compounds., , Ans. (a, c, d), (a) Among halogens, radius ratio between iodine and fluorine is maximum because iodine, has maximum radius and fluorine has minimum radius., (b) It can be correctly stated as in general interhalogen compounds are more reactive than, halogen. This is because X ¾ X ¢ bond in interhalogen is weaker than X ¾ X bond in, halogens except F—F bond., (c) Among interhalogen compounds maximum number of atoms are present in iodine, fluoride because radius ratio of iodine and fluorine has maximum value., (d) Interhalogen compounds are more reactive than halogen due to weaker X ¾ X ¢ bond, as compared to X ¾ X of halogen compounds., , Q. 32 Which of the following statements are correct for SO2 gas?, (a) It acts as bleaching agent in moist conditions., (b) Its molecule has linear geometry., (c) Its dilute solution is used as disinfectant., (d) It can be prepared by the reaction of dilute H2SO4 with metal sulphide., , Ans. (a, c), (a) In moist condition SO 2 gas acts as a bleaching agent., e.g., it converts Fe (III) to Fe (II) ion and decolourises acidified KMnO 4 (VII)., 2Fe 3 + + SO 2 + 2H2O ¾® 2Fe 2 + + SO 24 - + 4H+, (b) is incorrect it has bent structure., S, O, , O, , (c) Its dilute solution is used as a disinfectant., (d) It can be prepared by the reaction of O 2 with sulphide ore,, 4FeS2 + 11O 2 ¾® 2Fe 2O 3 + 8SO 2, while metal on treatment with H2SO 4 produces H2S., Hence, options (a) and (c) are correct choices., , Q. 33 Which of the following statements are correct?, (a) All the three N ¾ O bond lengths in HNO3 are equal., (b) All P ¾ Cl bond lengths in PCl5 molecule in gaseous state are equal, (c) P4 molecule in white phosphorus have angular strain therefore white, phosphorus is very reactive, (d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.

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Ans. (c, d), (a) All the three N—O bond lengths in HNO 3 are not equal., (b) All P—Cl bond lengths in PCl 5 molecule in gaseous state are not equal. Axial bond is, longer than equatorial bond., (c) P4 molecule in white phosphorous have angular strain therefore white phosphorous is, very reactive., (d) PCl 5 is ionic in solid state in which cation is tetrahedral and anion is octahedral., Cation — [PCl 4 ]+, Anion — [PCl 6 ]-, , Q. 34 Which, , of the following orders are correct as per the properties, mentioned against each?, (a) As2O 3 < SiO2 < P2O 3 < SO2, (b) AsH3 < PH3 < NH3, (c) S < O < Cl < F, (d) H2O > H2S > H2Se > H2Te, , Acid strength., Enthalpy of vaporisation., More negative electron gain enthalpy., Thermal stability., , Ans. (a, d), As O < SiO < P O < SO, , 2 3, 2, (a) ¾ ¾2¾3 ¾ ¾2 ¾ ¾, ¾¾, ¾, ®, , acidic strength increases, AsH > PH > NH, , 3, 3, (b) Correct order is ¬ ¾ ¾, ¾¾, ¾¾3 ¾, , enthalpy of vaporisation, , (c) S < O < Cl < F More negative electron gain enthalpy, (d) H2O > H2S > H2Se > H2 Te Thermal stability decreases on moving top to bottom due to, increase in its bond length., , Q. 35 Which of the following statements are correct?, (a) S¾S bond is present in H2S2O6, (b) In peroxosulphuric acid (H2SO5) sulphur is in + 6 oxidation state, (c) Iron powder along with Al2O3 and K 2O is used as a catalyst in the preparation of, NH3 by Haber's process, (d) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation, of SO2, , Ans. (a, b), (a) Structure of H2S2O 6 is as shown below, O, , O, , S ——–—— S, O, , OH OH, , O, , It contains one S—S bond., (b) In peroxosulphuric acid (H2SO 5 ) sulphur is in + 6 oxidation state., Structure of H2SO 5 is, O, S, H—O — O, , O, , OH

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Let oxidation state of S = x, 2 ´ ( + 1) + x + 3 ´ (- 2 ) + 2 ´ (-1) = 0, x -6=0, x=6, (c) During preparation of ammonia, iron oxide with small amount of K 2O and Al 2O 3 is used, as a catalyst to increase the rate of attainment of equilibrium., (d) Change in enthalpy is negative for preparation of SO 3 by catalytic oxidation of SO 2 ., , Q. 36 In which of the following reactions conc. H2SO 4, , is used as an oxidising, , reagent?, (a) CaF2 + H2SO 4 ¾® CaSO 4 + 2HF, (b) 2 HI + H2SO 4 ¾® I2 + SO2 + 2H2O, (c) Cu + 2H2SO 4 ¾® CuSO 4 + SO2 + 2H2O, (d) NaCl + H2SO 4 ¾® NaHSO 4 + HCl, , Ans. (b, c), In the above given four reactions, (b) and (c) represent oxidising behaviour of H2SO 4 . As we, know that oxidising agent reduces itself as oxidation state of central atom decreases., Here,, –1, , –6, , 0, , –4, , 2HI + H2SO 4 ¾® I2 + SO 2 + 2H2O, 0, , +6, , +2, , +4, , Cu + 2H2 S O 4 ¾® C uSO 4 + SO 2 + 2H2O, , Q. 37 Which of the following statements are true?, (a) Only type of interactions between particles of noble gases are due to weak, dispersion forces., (b) Ionisation enthalpy of molecular oxygen is very close to that of xenon., (c) Hydrolysis of XeF6 is a redox reaction., (d) Xenon fluorides are not reactive., , Ans. (a, b), (a) Only one type of interactions between particles of noble gases are due to weak, dispersion forces., (b) Ionisation enthalpy of molecular oxygen is very close to that of xenon. This is the reason, for the formation of xenon oxides., +6, , –1, , +1 –2, , +6, , –2, , +1–1, , (c) Hydrolysis of XeF6 ( X eF6 + 3H2 O ¾® X e O 3 + 3 H F ) is not a redox reaction., (d) Xenon fluorides are highly reactive hydrolysis readily even by traces of water.

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Short Answer Type Questions, Q. 38 In the preparation of H2SO 4, , by Contact process, why is SO 3 not absorbed, directly in water to form H2SO 4 ?, , Ans., , In Contact process SO 3 is not absorbed directly in water to from H2SO 4 because the, reaction is highly exothermic, acid mist is formed. Hence, the reaction becomes difficult to, handle., , Note, S, Air, , Burners, , SO2, , Arsenic, purifier, , Cooling, and drying, , Precipator, , Testing, box, , Conc., H2SO4, , Absorption, tower, , H2S2O7, Oleum, , SO3, , Catalyst, chamber, , Preheater, , Flow chart of Contact process, , Q. 39 Write, , a balanced chemical equation for the reaction showing catalytic, oxidation of NH 3 by atmospheric oxygen., , Ans., , Ammonia (NH3 ) on catalytic oxidation by atmospheric oxygen in presence of Rh/Pt gauge, at 500K under pressure of 9 bar produces nitrous oxide., Balanced chemical reaction can be written as, Pt /Rh gauge catalyst, , 4NH3 + 5O 2 ¾¾¾¾¾¾¾® 4NO + 6H2O, 500 K; 9 bar, , From air, , Q. 40 Write the structure of pyrophosphoric acid., Ans., , Molecular formula of pyrophosphoric acid is H4P2O 7 and its structure is as follows, O, , O, , P, , P, OH, , HO, , OH, , O, , OH, , Pyrophosphoric acid (H4P 7O7), , Q., , 41 PH3 forms bubbles when passed slowly in water but NH 3 dissolves., Explain why?, , Ans., , Dissolution of NH3 and PH3 in water can be explained on the basis of H-bonding. NH3, forms H-bond with water so it is soluble but PH3 does not form H-bond with water so it, remains as gas and forms bubble in water.

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Q. 42 In PCl 5 , phosphorus is in sp 3d hybridised state but all its five bonds are, not equivalent. Justify your answer with reason., Ans., , It has trigonal bipyramidal geometry, in which two Cl atoms occupy axial position while, three occupy equatorial positions. All five P¾Cl bonds are not identical. There are two, types of bond lengths (i) Axial bond lengths (ii) Equatorial bond lengths, Cl, 240 pm, , Cl, , P, , 202 pm, , Cl, , Cl, Cl, , Structure of PCl 5, Thus, difference in bond length is due to fact that axial bond pairs suffer more repulsion as, compared to equatorial bond pairs., , Q. 43 Why is nitric oxide paramagnetic in gaseous state but the solid obtained, on cooling it is diamagnetic?, Ans., , In gaseous state, NO 2 exists as a monomer which has one unpaired electron but in solid, state, it dimerises to N2O 4 so no unpaired electron left. Therefore, NO 2 is paramagnetic in, gaseous state but diamagnetic in solid state., O, , O, N, , N—N, , O, , O, , O, , O, , Gaseous state, (Monomer), , Solid state, (Dimer), , Q. 44 Give one reason to explain why ClF 3 exists but FCl 3 does not exist?, Ans., , Existance of ClF3 and FCl 3 can be explained on the basis of size of central atom. Because, fluorine is more electronegative as compared to chlorine and has smaller size. Thus, one, large Cl atom can accomodate three smaller F atoms but reverse is not true., , Q. 45 Out of H2O and H2S, which one has higher bond angle and why?, Ans., , Bond angle of H2O (H ¾ O ¾ H = 104.5° ) is larger than that of H2S (H ¾ S ¾ H = 92 ° ), because oxygen is more electronegative than sulphur therefore, bond pair electron of, O¾H bond will be closer to oxygen and there will be more bond pair¾bond pair repulsion, between bond pairs of two O¾H bonds., , H, , O, , S, , 104.5°, , 92°, H, , due to more b.p–b.p, repulsion, , H, , H

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Q. 46 SF6 is known but SCl 6 is not. Why?, Ans., , Fluorine atom is smaller in size so, six F - ions can surround a sulphur atom. The case is not so, with chlorine atom due to its large size. So, SF6 is known but SCl 6 is not known due to, interionic repulsion between larger Cl - ions., , Q., , 47 On reaction with Cl 2 , phosphorus forms two types of halides ‘A’ and ‘B’., Halide ‘A’ is yellowish-white powder but halide ‘B’ is colourless oily liquid., Identify A and B and write the formulae of their hydrolysis products., , Ans., , Phosphorus on reaction with Cl 2 forms two types of halides A and B., ‘A’ is PCl 5 and ‘B’ is PCl 3 ., P4 + 10 Cl 2 ¾® 4 PCl 5, P4 + 6 Cl 2 ¾® 4 PCl 3, When ‘A’ and ‘B’ are hydrolysed, (a) PCl 5, + 4 H2O ¾® H3PO 4 + 5HCl, [A ], Phosphorus, pentachloride, , (b), , PCl 3, , Phosphoric, acid, , + 3 H2O ¾® H3PO 3 + 3 HCl, , [B ], Phosphorus, trichloride, , Phosphorus, acid, , ion, Fe2+ ion reduces nitrate ion to nitric oxide,, which combines with Fe2+ (aq) ion to form brown complex. Write the, reactions involved in the formation of brown ring., , Q. 48 In the ring test of NO -3, Ans., , NO -3 + 3Fe 2 + + 4H+ ¾® NO + 3Fe 3 + + 2H2O, [Fe (H2O)6 ]2 + + NO ¾® [Fe(H2O)5 NO] 2 + + H2O, Brown ring, , This test is known as brown ring test of nitrates generally used to identify the presence of, nitrate ion in given solution., , Q. 49 Explain why the stability of oxoacids of chlorine increases in the order, given below., HClO < HClO2 < HClO 3 < HClO 4, Ans., , Oxygen is more electronegative than chlorine, therefore dispersal of negative charge, present on chlorine increases from ClO - to ClO -4 ion because number of oxygen atoms, attached to chlorine is increasing.Therefore, stability of ions will increase in the order given, below, ClO - < ClO -2 < ClO -3 < ClO -4, Due to increase in stability of conjugate base, acidic strength of corresponding acid, increases in the same order, HClO < HClO 2 < HClO 3 < HClO 4, , Q. 50 Explain why ozone is thermodynamically less stable than oxygen?, Ans., , Ozone is thermodynamically less stable than oxygen because its decomposition into, oxygen results in the liberation of heat (DH is negative) and an increase in entropy (DS is, positive). These two effects reinforce each other, resulting in large negative Gibbs energy, change (DG ) for its conversion into oxygen.

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Q. 53 Name three oxoacids of nitrogen. Write the disproportionation reaction, of that oxoacid of nitrogen in which nitrogen is in + 3 oxidation state., Ans., , Three oxoacids of nitrogen having oxidation state + 3 are, (a) HNO 2 , nitrous acid, (b) HNO 3 , nitric acid, (c) Hyponitrous acid, H2N2O 2, In HNO 2 , N is in + 3 oxidation state, Disproportionation reaction, Disproportionation, , 3HNO 2 ¾¾¾¾¾¾¾¾® HNO 3 + H2O + 2NO, , Q. 54 Nitric acid forms an oxide of nitrogen on reaction with P4 O 10 . Write the, reaction involved. Also write the resonating structures of the oxide of, nitrogen formed., Ans., , P4O10 being a dehydrating agent, on reaction with HNO 3 removes a molecule of water and, forms anhydride of HNO 3 ., 4HNO 3 + P4O10 ¾® 4HPO 3 + 2N2O 5, Resonating structures of N2O 5 are, O, , O, , N, , O, , O, , O, , N, , N, , O, , O, , O, , N, , O, , O, , Q. 55 (i), , white phosphorus (ii) red phosphorus and (iii) black phosphorus., Write the difference between white red and black phosphorus on the, basis of their structure and reactivity., Phosphorus has three allotropic forms —, , Ans., , White phosphorus, , Red phosphorus, , Black phosphorus, , 1., , It is less stable form of P, , More stable than white P., , It is most stable form of P, , 2., , It is highly reactive., , Less reactive than white P., , It is least reactive., , P, , P, , P, , P, , P, P, , P, , P, , P, , P, P, , P, , P, P, , P, , 99°, , pm, 218 P, P, , P, , P, , P, , P, , P, P, , P, , P, , P, , 3., , It has regular, tetrahedron structure., , P, , P, P, , P, , P, , It has polymeric structure., , P, P, , P, , P, , P, , P, P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , It has a layered structure.

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Q. 56 Give an example to show the effect of concentration of nitric acid on the, formation of oxidation product., Ans., , Effect of concentration of nitric acid on the formation of oxidation product can be, understood by its reaction with conc HNO 3 . Dilute and concentrated nitric acid give, different oxidation products on reaction with copper metal., 3 Cu + 8HNO 3 (Dil.) ¾® 3Cu (NO 3 )2 + 2NO + 4H2O, Cu + 4HNO 3 (Conc.) ¾® Cu (NO 3 )2 + 2NO 2 + 2H2O, , Q. 57 PCl 5, , reacts with finely divided silver on heating and a white silver salt, is obtained, which dissolves on adding excess aqueous NH 3 solution., Write the reactions involved to explain what happens., , Ans., , PCl 5 on reaction with finely divided silver produced silver halide., PCl 5 + 2Ag ¾® 2AgCl + PCl 3, AgCl on further reaction with aqueous ammonia solution produces a soluble complex of, [Ag (NH3 )2 ]+ Cl AgCl + 2NH3 (aq ) ¾® [Ag(NH3 )2 ]+ Cl Soluble complex, , Q., , 58 Phosphorus forms a number of oxoacids. Out of these oxoacids,, phosphinic acid has strong reducing property. Write its structure and, also write a reaction showing its reducing behaviour., , Ans., , Among various forms of oxoacids, phosphinic acid has stronger reducing property., O, P, H, , OH, , H, , Structure of phosphinic acid, Reaction showing reducing behaviour of phosphinic acid is as follows, 4AgNO 3 + 2H2O + H3PO 2 ¾® 4Ag ¯ + 4HNO 3 + H3PO 4, , Matching The Columns, Q. 59 Match, , the compounds given in Column I with the hybridisation and, shape given in Column II and mark the correct option., Column I, , Codes, A B, (a) 1 3, (c) 4 3, , Column II, , A., , Xe F6, , 1., , sp3d 3 -distorted octahedral, , B., , XeO 3, , 2., , sp3d2 -square planar, , C., , XeOF4, , 3., , sp3-pyramidal, , D., , Xe F4, , 4., , sp3d2 -square pyramidal, , C, 4, 1, , D, 2, 2, , A, (b) 1, (d) 4, , B, 2, 1, , C, 4, 2, , D, 3, 3

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Ans. (a), , A. ® (1), , B. ® (3), , S. No., , C. ® (4), , D. ® (2), , Compound, , Hybridisation, 3 3, , F, , sp d -distorted octahedral, F, , F, , A., , Xe, , F, F, , F, , sp3-pyramidal, B., , Xe, , O, , O, , O, O, , F, , sp3d2 -square pyramidal, , F, , Xe, , C., F, , F, , F, , F, , sp3d2 -square planar, , Xe, , D., , F, , F, , Q. 60 Match the formulas of oxides given in Column I with the type of oxide, given in Column II and mark the correct option., Column I, , Codes, A, (a) 1, (c) 3, , Ans. (b), , B, 2, 2, , A. ® (4), , Column II, , A., , Pb 3 O 4, , 1., , Neutral oxide, , B., , N2 O, , 2., , Acidic oxide, , C., , Mn2 O7, , 3., , Basic oxide, , D., , Bi2 O 3, , 4., , Mixed oxide, , C, 3, 4, , D, 4, 1, , B. ® (1), , A, (b) 4, (d) 4, C. ® (2), , D. ® (3), , Formulas of the compound, , Type of oxide, , A., , Pb 3 O 4 (PbO × Pb2 O 3 ), , Mixed oxide, , B., , N2 O, , Neutral oxide, , C., , Mn2 O7, , Acidic oxide, , D., , Bi2 O 3, , Basic oxide, , Mn2O 7 on dissolution in water produces acidic solution., Bi 2O 3 on dissolution in water produces basic solution., , B, 1, 3, , C, 2, 1, , D, 3, 2

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Q. 61 Match the items of Columns I and II and mark the correct option., Column I, A., B., C., D., , Ans. (a), , Column II, 1., 2., 3., 4., , H2 SO 4, CCl 3NO2, Cl2, Sulphur, , Codes, A B, (a) 4 3, (c) 4 1, , C, 1, 2, , A. ® (4), , B. ® (3), , Highest electron gain enthalpy, Chalcogen, Tear gas, Storage batteries, , D, 2, 3, , A, (b) 3, (d) 2, C. ® (1), , B, 4, 1, , C, 1, 3, , D, 2, 4, , D. ® (2), , A. H2SO 4 is used in storage batteries., B. CCl 3NO 2 is known as tear gas., C. Cl 2 has highest electron gain enthalpy., D. Sulphur is a member of chalcogen i.e., ore producing elements., , Q. 62 Match the species given in Column I with the shape given in Column II, and mark the correct option., Column I, , Codes, A B, (a) 3 2, (c) 1 2, , Ans. (b), , A. ® (3), , Column II, , A., B., C., , SF4, BrF3, BrO -3, , 1., 2., 3., , Tetrahedral, Pyramidal, Sea-saw shaped, , D., , NH4+, , 4., , Bent T-shaped, , C, 1, 3, , D, 4, 4, , B. ® (4), , Species, , A, (b) 3, (d) 1, C. ® (2), , B, 4, 4, , C, 2, 3, , D, 1, 2, , D. ® (1), , Shape, , A., , SF4, , Sea-saw shaped, , B., , BrF3, , Bent T-shaped, , Structure, S, , F, , F, F, F, , F, , F, , Br, F, , C., , BrO 3, , Pyramidal, Br, , D., , NH+4, , O, , O, O, , H, , Tetrahedral, , N, H, , H, H, , +

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Q. 63 Match the items of Columns I and II and mark the correct option., Column I, A., , Its partial hydrolysis does not change, oxidation state of central atom., , 1., , He, , B., , It is used in modern diving apparatus., , 2., , XeF6, , C., , It is used to provide inert atmosphere, for filling electrical bulbs., , 3., , XeF4, , D., , Its central atom is in sp3 d2 hybridisation., , 4., , Ar, , Codes, A B, (a) 1 4, (c) 2 1, , Ans. (c), , Column II, , A. ® (2), , C, 2, 4, , D, 3, 3, , A, (b) 1, (d) 1, , B. ® (1), , C. ® (4), , B, 2, 3, , C, 3, 2, , D, 4, 4, , D. ® (3), , (A) Partial hydrolysis of XeF6 does not change oxidation state of central atom., +6, , +6, , XeF6 + 2H2O ¾® XeO 3 + 6 HF, (B) He is used in modern diving apparatus., (C) Ar is used to provide inert atmosphere for filling electrical bulbs, (D) Central atom (Xe ) of XeF4 is in sp3d 2 hybridisation., , F, , F, , Xe, F, , F, , Square planar geometry, , Assertion and Reason, In the following questions a statement of Assertion (A) followed by a, statement of Reason (R) is given. Choose the correct answer out of the, following choices., , (a) Both Assertion and Reason are correct statements, and Reason is the, correct explanation of the Assertion., (b) Both Assertion and Reason are correct statements, and Reason is not, the correct explanation of the Assertion., (c) Assertion is correct, but Reason is wrong statement., (d) Assertion is wrong but Reason is correct statement., (e) Both Assertion and Reason are wrong statements., , Q. 64 Assertion, , (A) N2 is less reactive than P4 ., Reason (R) Nitrogen has more electron gain enthalpy than phosphorus.

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Ans. (c), , Assertion is true, but reason is false., N2 is less reactive than P4 due to high value of bond dissociation energy which is due to, presence of triple bond between two N-atoms of N2 molecule., , Q. 65 Assertion (A) HNO 3 makes iron passive., Reason (R) HNO 3 forms a protective layer of ferric nitrate on the surface, of iron., Ans. (c), , Assertion is true, but reason is false., HNO 3 makes iron passive due to formation of passive form of oxide on the surface., Hence, Fe does not dissolve in conc HNO 3 solution., , Q. 66 Assertion, , (A) HI cannot be prepared by the reaction of KI with, concentrated H2SO 4 ., Reason (R) HI has lowest H¾X bond strength among halogen acids., , Ans. (b), , Both assertion and reason are correct statements, but reason is not the correct, explanation of the assertion., HI cannot be prepared by the reaction of KI with concentrated H2SO 4 because HI is, converted into I2 on reaction with H2SO 4 ., , Q. 67 Assertion, , (A) Both rhombic and monoclinic sulphur exist as S 8 but, oxygen exists as O2 ., Reason (R) Oxygen forms pp - pp multiple bond due to small size and, small bond length but pp - pp bonding is not possible in sulphur., , Ans. (a), , Both assertion and reason are correct statements, and reason is the correct, explanation of the assertion., Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O 2 , because, oxygen forms pp - pp multiple bond due to its small size and small bond length. But pp, -pp bonding is not possible in sulphur due to its bigger size as compared to oxygen., S, , 204, , pm, , S, S, , S, 107, , S, , S, o, , S, S, , pp-pp bond, , O==O, Structure of O2, , Structure of S8, , Q. 68 Assertion, , (A) NaCl reacts with concentrated H2SO 4 to give colourless, fumes with pungent smell. But on adding MnO2 the fumes become, greenish yellow., Reason (R) MnO2 oxidises HCl to chlorine gas which is greenish yellow., , Ans. (a), , Both assertion and reason are correct statements, and reason is the correct, explanation of the assertion., NaCl reacts with concentrated H2SO 4 to give colourless fumes with pungent smell., Pungent smell is due to formation of HCl., NaCl + H2SO 4 ¾® Na 2SO 4 + 2HCl, But on adding MnO 2 the fumes become greenish yellow due to formation of chlorine gas.

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Q. 69 Assertion (A) SF6 cannot be hydrolysed but SF 4, , can be., Reason (R) Six F-atoms in SF6 prevent the attack of H2O on sulphur atom, of SF6 ., , Ans. (a), , Assertion and reason both are true and reason is the correct explanation of assertion., SF4 can be hydrolysed but SF6 can not because six F-atoms in SF6 prevent the attack of, H2O on sulphur atoms of SF6 ., , Long Answer Type Questions, Q. 70 An amorphous solid “A” burns in air to form a gas “B” which turns lime, water milky. The gas is also produced as a by-product during roasting of, sulphide ore. This gas decolourises acidified aqueous KMnO 4 solution, and reduces Fe 3+ to Fe2+ . Identify the solid “A” and the gas “B” and write, the reactions involved., K Thinking Process, This problem is based on concept of properties of sulphur and its oxide., in, A ¾Burn, ¾¾, ® (gas), air, , (amorphous solid), , Amorphous solid A gives B is a gas which turns lime water milky and also produced as a, by product during roasting of sulphide ore. This gas decolourises acidified aqueous, KMnO 4 solution and reduces Fe3+to Fe2+. Hence, compound B (g) must be SO2 ., , Ans., , Since, the by-product of roasting of sulphide ore is SO 2 , so A is S8 ‘ A’ = S8 ; ‘B’ = SO 2, Reactions, D, (i) S8 + 8O 2 ¾¾® 8SO 2, (ii) Ca(OH)2 + SO 2 ¾® CaSO 3 + H2O, (iii) 2MnO -4 + 5SO 2 + 2H2O ¾® 5SO 24 - + 4H+ + 2Mn2+, , (Colourless), , (Violet), , (iv) 2Fe, , 3+, , + SO 2 + 2H2O ¾® 2Fe, , 2-, , +, , SO 24 -, , + 4H+, , Q. 71 On, , heating lead (II) nitrate gives a brown gas “A”. The gas “A” on, cooling changes to colourless solid “B”. Solid “B” on heating with NO, changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions, involved and draw the structures of ‘B’ and ‘C ’., K Thinking Process, This problem is based on preparation and properties of NO2., , Ans., , Pb(NO 3 )2 on heating produces a brown coloured gas which may be NO 2 . Since, on, reaction with N2O 4 and on heating it produces N2O 3 and N2O 4 respectively.

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Structures, (i) N2O 4, , O, , O, , O, , O, , O, , O, , O, , O, , O, , O, , O, , N, , (ii) N2O 3, , N, , O, , N, , N, , N, , N, , O, , N, , N, , O, , Q. 72 On heating compound (A) gives a gas (B) which is a constituent of air., This gas when treated with 3 moles of hydrogen (H2) in the presence of a, catalyst gives another gas (C) which is basic in nature. Gas C on further, oxidation in moist condition gives a compound (D) which is a part of, acid rain. Identify compounds (A) to (D) and also give necessary, equations of all the steps involved., Ans., , The main constituents of air are nitrogen (78%) and oxygen (21%). Only N2 reacts with, three moles of H2 in the presence of a catalyst to give NH3 (ammonia) which is a gas, having basic nature. On oxidation, NH3 gives NO 2 which is a part of acid rain. So, the, compounds A to D are as, A = NH4NO 2 ; B = N2 ; C = NH3 ; D = HNO 3, Reactions involved can be given, as, D, , (i) NH4NO 2 ¾¾® N2 + 2H2O, (A), , ( B), , (ii) N2 + 3H2 s, , 2NH3, , [B ], , [C ], Oxidation, , (iii) 4NH3 + 5O 2 ¾® 4NO + 6H2O, (Iv) 2NO + O 2 ¾® 2NO 2, (v) 3NO 2 + H2O ¾® 2HNO 3 + NO, ( D)