1 7 . HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give - (A) BrCH2CHO and CH3OH (B) BrCH2–CH2–OCH3 [AIEEE-2006] (C) H3C–CHBr–OCH3 (D) CH3CHO and CH3Br 1 8 . The structure of the compound that gives tribromo derivative on treatment with bromine water is - [AIEEE-2006] CH2OH CH3 CH3 CH3 (A) OH (C) (D) (B) OH OH 1 9 . Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives [AIEEE-2008] (A) 2, 4, 6– trinitrobenzene (B) –nitrophenol (C) p–nitrophenol (D) nitrobenzene 2 0 . Bakelite is obtained from phenol by reacting with [AIEEE-2008] (A) (CH OH) (B) CH CHO (C) CH COCH (D) HCHO 22 3 33 2 1 . The best reagent to convert pent–3–en–2–ol into pent –3–en –2–one is - [AIEEE-2005] (A) Acidic dichromate (B) Acidic permanganate (C) Pyridinium chloro-chromate (D) Chromic anhydride in glacial acetic acid 2 2 . Rate of the reaction- [AIEEE-2005] OO R–C + Nu R – C + X X Nu is fastest when X is - (A) NH2 (B) Cl (C) OCOR (D) OC2H5 2 3 . In the following sequence of reactions CH3CH2OH P + I2 A Mg B HCHO C H2O D, then compound 'D' is - [AIEEE-2007] Ether (A) butanal (B) n–butyl alcohol (C) n–propyl alcohol (D) propanal 2 4 . End product of the following reaction is - [AIEEE-2002] CH3CH2COOH Cl2 Alc.KOH red P (A) CH3CHCOOH (B) CH2CH2COOH (C) CH2= CHCOOH (D) CH2CH–COOH OH OH Cl OH 2 5 . On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is - [AIEEE-2004] (A) CH3COOC2H5 + NaCl (B) CH3COONa + C2H5OH (C) CH3COCl + C2H5OH + NaOH (D) CH3Cl + C2H5COONa 2 6 . The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :- [AIEEE-2009] (A) Salicylic acid (B) Phthalic acid (C) Benzoic acid (D) Salicyladehyde 2 7 . In Cannizzaro reaction given below :- 2PhCHO : OH PhCH2OH + PhCO2 [AIEEE-2009] the slowest step is :- (A) The abstraction of proton from the carboxylic group (B) The deprotonation of PhCH2OH (C) The attack of : OH at the carboxyl group (D) The transfer of hydride to the carbonyl group
2 8 . A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was :- [AIEEE-2009] (A) CH3COCH3 (B) CH3COOH (C) CH3OH (D) HCHO 2 9 . From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is :- [ AIEEE- 2010] (A) 1–Butanol (B) 2–Butanol (C) 2–Methylpropan–2–ol(D) 2–Methylpropanol 3 0 . Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is :- [ AIEEE- 2011] (A) Ethyl chloride (B) Ethyl ethanoate (C) Diethyl ether (D) 2–Butanone 3 1 . Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is :- [ AIEEE- 2011] (A) 4-Bromophenol (B) 2,4,6-Tribromophenol (C) 2-Bromophenol (D) 3-Bromophenol 3 2 . Trichloroacetaldehyde was subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :- [AIEEE-2011] (A) 2,2,2–Trichloropropanol (B) Chloroform (C) 2,2,2–Trichloroethanol (D) Trichloromethanol 3 3 . Silver Mirror test is given by which one of the following compounds ? [AIEEE-2011] (A) Formaldehyde (B) Benzophenone (C) Acetaldehyde (D) Acetone 3 4 . Which of the following reagents may be used to distinguish between phenol and benzoic acid? [ A I E E E - 2 0 1 1 ] (A) Molisch reagent (B) Neutral FeCl3 (C) Aqueous NaOH (D) Tollen's reagent 3 5 . Consider the following reaction : C2H5OH + H2SO4 Produce Among the following, which one cannot be formed as a product under any conditions ? [ AIEEE- 2011] (A) Ethyl-hydrogen sulphate (B) Ethylene (C) Acetylene (D) Diethyl ether 3 6 . Iodoform can be prepared from all except : [AIEEE-2012] (A) Isobutyl alcohol (B) Ethyl methyl ketone (C) Isopropyl alcohol (D) 3-Methyl–2–butanone 3 7 . Which of the following on heating with aqueous KOH, produces acetaldehyde ? [AIEEE-2009] (A) CH2ClCH2Cl (B) CH3CHCl2 (C) CH3COCl (D) CH3CH2Cl [AIEEE-2010] 3 8 . Consider the following bromides :- Me Br Me Me A Me Br C Br B The correct order of SN1 reactivity is (A) A > B > C (B) B > C > A (C) B > A > C (D) C > B > A
3 9 . A solution of (–) –1–chloro–1–phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of :- [JEE-MAIN 2013] (D) free radical (A) carbanion (B) Carbene (C) carbocation 4 0 . Arrange the following compounds in order of decreasing acidity : OH OH OH OH ;; ; Cl CH3 NO2 OCH3 (I) (II) (III) (IV) (A) II > IV > I > III (B) I > II > III > IV (C) III > I > II > IV (D) IV > III > I > II 4 1 . Compound (A), C8H9Br, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A) : CH2Br (B) C2H5 CH2Br CH2Br Br (A) (C) (D) CH3 CH3 CH3 4 2 . An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is :- (A) CH3COOH (B) CH3CH2CH2COOH (C) CH3–CH–COOH (D) CH3CH2COOH CH3 4 3 . An unknown alcohol is treated with the \"Lucas reagent' to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism :- (A) secondary alcohol by SN1 (B) tertiary alcohol by SN1 (C) secondary alcohol by SN2 (D) tertiary alcohol by SN2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 Ans B C B D D B C A B C B D D Que. 14 15 16 17 18 19 20 21 22 23 24 25 26 Ans B CAD D Bonus D D B C C B A Que. 27 28 29 30 31 32 33 34 35 36 37 38 39 Ans D B C B B C A,C B C A Que. 40 41 42 43 Ans C D D B
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Out of the two compounds shown below, the vapour pressure of B at a particular temperature is expected to be OH OH and O2N NO2 (B) (A) higher than that of A (A) (B) lower than that of A (C) same as that of A (D) can be higher or lower depending upon the size of the vessel 2 . The structure of IF5 can be best described as :- F F F F (A) 72° F IF F F 90° I 90° F (C) 90° I (B) 72° (D) none of these FF F FF F F 3 . The correct order of the bond angles is :- (A) NH3 > H2O > PH3 > H2S (B) NH3 > PH3 > H2O > H2S (C) NH3 > H2S > PH3 > H2O (D) PH3 > H2S > NH3 > H2O 4 . The correct increasing bond angle among BF3, PF3 and ClF3 follows the order :- (A) BF3 < PF3 < ClF3 (B) PF3 < BF3 < ClF3 (C) ClF3 < PF3 < BF3 (D) BF3 = PF3 = ClF3 5 . How many sigma and pi bonds are present in tetracyanoethylene ? (A) nine and nine (B) five and nine (C) nine and seven (D) eight and eight 6 . The types of bond present in N2O5 are :- (A) only covalent (B) only ionic (C) ionic and covalent (D) covalent & coordinate 7 . How many bonded electron pairs are present in IF7 molecule :- (A) 6 (B) 7 (C) 5 (D) 8 8 . When 2s – 2s, 2p – 2p and 2p – 2s orbitals overlap, the bond strength decreases in the order :- (A) p – p > s – s > p – s (B) p – p > p – s > s – s (C) s – s > p – p > p – s (D) s – s > p – s > p – p 9 . The volatility of HF is low as compare to other Hydra acid of Halogen because of :- (A) its low polarizability (B) the weak dispersion interaction between the molecules (C) its small molecular mass (D) its strong hydrogen bonding 1 0 . The shapes of PCl4+, PCl4– and AsCl5 are respectively :- (A) square planar, tetrahedral and see-saw (B) tetrahedral, see-saw and trigonal bipyramidal (C) tetrahedral, square planar and pentagonal bipyramidal (D) trigonal bipyramidal, tetrahedral and square pyramidal
1 1 . The experimental value of the dipole moment of HCl is 1.03 D. The length of the H – Cl bond is 1.275 Å. The percentage of ionic character in HCl is :- (A) 43 (B) 21 (C) 17 (D) 7 1 2 . The shapes of IF5 and IF7 are respectively :- (A) tetragonal pyramidal and pentagonal bipyramidal (B) octahedral and pyramidal (C) trigonal bipyramidal and square antiprismatic (D) distorted square planar and distorted octahedral 1 3 . Amongst LiCl, RbCl, BeCl2 and MgCl2, the compounds with the greatest and the least ionic character, respectively are :- (A) LiCl and RbCl (B) RbCl and BeCl2 (C) RbCl and MgCl2 (D) MgCl2 and BeCl2 1 4 . PCl5, exists but NCl5 does not because :- (A) Nitrogen has no vacant 2d-orbitals (B) NCl5 is unstable (D) Nitrogen is highly inert (C) Nitrogen atom is much smaller than P 1 5 . (C – Cl) bond in CH2 = CH – Cl (vinyl chloride) is stabilised in the same way as in :- (A) benzyl chloride (B) ethyl chloride (C) chlorobenzene (D) allyl chloride 16. In IC l , the shape is square planar. The number of bond pair-lone pair repulsion at 90° are :- 4 (A) 6 (B) 8 (C) 12 (D) 4 1 7 . The structure of diborane (B2H6) contains :- (A) four (2c – 2e–) bonds and two (2c – 3e–) bonds (B) two (2c – 2e–) bonds and two (3c – 2e–) bonds (C) four (2c – 2e–) bonds and four (3c – 2e–) bonds (D) none 1 8 . Among the following species, identify the isostructural pairs : NF3, NO3–, BF3, H3O+, HN3 :- (A) [NF3, NO3–] and [BF3, H3O+] (B) [NF3, HN3] and [NO3–, BF3] (C) [NF3, H3O+] and [NO3–, BF3] (D) [NF3, H3O+] and [HN3, BF3] 1 9 . Which of the following exhibit/s H-bonding ? (A) CH4 (B) H2Se (C) N2H4 (D) H2S 2 0 . Valency expresses generally : (A) total e– in an atom (B) atomicity of an element (C) oxidation number of an element (D) combining capacity of an element 2 1 . Which element do not have valency equals to its group no. : (A) sodium (B) aluminium (C) oxygen (D) carbon 2 2 . Which condition favours the bond formation:- (A) maximum attraction and maximum potential energy (B) minimum attraction and minimum potential energy (C) minimum potential energy and maximum attraction (D) none of the above 2 3 . Number of and bonds present in : CH3 – CH = CH – C CH are - (A) 10 , 3 (B) 10, 2 (C) 9, 2 (D) 8, 3 2 4 . Which is not characteristic of -bond:- (A) - bond is formed when a sigma bond already formed (B) - bond are formed from hybrid orbitals (C) - bond may be formed by the overlapping of p-orbitals (D) -bond results from lateral overlap of atomic orbitals 2 5 . When sodium and chlorine reacts then energy is :- (A) released and ionic bond is formed (B) released and covalent bond is formed (C) absorbed and ionic bond is formed (D) absorbed and covalent bond is formed
2 6 . The electron pair which forms a bond between two similar non-metallic atoms will be :- (A) dissimilar shared between the two (B) by complete transfer from one atom to other (C) in a similar spin condition (D) equally shared in between the two 2 7 . In N2 molecule, the atoms are bonded by :- (B) 1 and 1 - bonds, 1 L.P. (A) 1 and 2 - bonds, 2 L.P. (C) 2 and 1 - bonds, No L.P. (D) 1 , 2 & No L.P. 2 8 . An atom of element A has three electrons in its outer shell and B has six electrons in its outermost shell. The formula of the compound formed between these two will be :- (A) A6B6 (B) A2B3 (C) A3B2 (D) A2B 2 9 . Two element X and Y have following electronic configuration :- X 1s2 ; 2s2, 2p6 ; 3s2, 3p6 ; 4s2 Y 1s2 ; 2s2, 2p6 ; 3s2, 3p5 The expected compound formed by combination of X and Y will be expressed as : (A) XY2 (B) X5Y2 (C) X2Y5 (D) XY5 3 0 . Polarisibility of halide ions increases in the order :- (A) F –, I –, Br –, Cl – (B) Cl–, Br–, I–, F– (C) I–, Br–, Cl–, F– (D) F –, Cl–, Br –, I– 3 1 . The correct order of the O–O bond length in O2, H2O2 and O3 is :- (A) O2 > O3 > H2O2 (B) O3 > H2O2 > O2 (C) H2O2 > O3 > O2 (D) H2O2 > O2 > O3 3 2 . In which of the following the central atom does not use sp3 hybrid orbitals in its bonding :- (A) BeF3– (B) OH3+ (C) NH2– (D) NF3 3 3 . According to Fajjan's rule, covalent bond is favoured by :- (A) large cation and small anion (B) large cation and large anion (C) Small cation and large anion (D) Small cation and small anion 3 4 . Resonance hybrid of nitrate ion is :- (A) O–1/2 N O–1/2 (B) O–2/3 N O–2/3 (C) O–1/3 O–1/2 O–1/3 (D) O–2/3 O–2/3 O–2/3 N + N O–1/3 O–2/3 3 5 . The correct order of bond angle (smallest first) in H2S, NH3, BF3 and SiH4 is :- (A) H2S < NH3 < BF3 < SiH4 (B) NH3 < H2S < SiH4 < BF3 (C) H2S < NH3 < SiH4 < BF3 (D) H2S < SiH4 < NH3 < BF3 CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. A C A C A D B B D B C A B A C B D C C D Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 Ans. C C A B A D A B A D C A C C C
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . Which of the following does not have same shape :- (A) SO 2 , S 2 (B) ICl4, I5 (C) CO2, SO2 (D) N O , N O 4 5 2 2 2 . B3N3H6 + solution of hydrochloric acid ? Select correct about above reaction : (A) no reaction (B) B3N3H6 show substitution reaction & produce B3N3Cl6 (C) B3N3H6 show addition reaction and produce B3N3H9Cl3 in which Cl is bonded to boron (D) B3N3H6 show addition reaction and produce B3N3H9Cl3 in which Cl is bonded to nitrogen 3 . Nodal planes of bonds in benzene are located in : (A) all are in molecular plane (B) one in molecular plane and two in plane perpendicular to molecular plane which contain C – C bonds. (C) two in molecular plane and one in plane perpendicular to molecular plane which contain C – C bond and C – H bond (D) perpendicular to molecular plane which bisect benzene ring in two equal half 4 . Which of the following has fractional bond order :- (A) O22 (B) O22 (C) F22 (D) H 2 5 . Which is correct statement ? As the s-character of a hybrid orbital decreases (I) The bond angle decreases (II) The bond strength increases (III) The bond length increases (IV) Size of orbitals increases (A) (I), (III) and (IV) (B) (II), (III) and (IV) (C) (I) and (II) (D) all are correct 6 . Which of the following compounds have the same no. of lone pairs with their central atom :- (I) XeF5– (II) BrF3 (III) XeF2 (IV) Triple methylene (A) (IV) and (V) (B) (I) and (III) (C) (I) and (II) (D) (II), (IV) and (V) 7 . Select pair of compounds in which both have different hybridization but have same molecular geometry :- (A) BF3, BrF3 (B) ICl , BeCl2 (C) BeCl3, PCl3 (D) PCl3, NCl3 2 8 . The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively. (A) sp3 and sp2 (B) sp2 and sp3 (C) sp2 and sp2 (D) sp3 and sp3 9 . Which of the following option w.r.t. increasing bond order is correct ? (A) NO < C < O – < He +(B) C < NO < He + < O – 22 22 22 (C) He + < O – < NO < C (D) He + < O – < C < NO 22 2 2 22 1 0 . Which is most ionic : (A) P O (B) MnO (C) CrO (D) Mn O 25 3 27 1 1 . The molecular orbital configuration of CN+ is :- (A) KK (2s)2, * (2s)2, (2px)2, (2py)2 (B) KK (2s)2, * (2s)2, (2pz)2, (2px)1, (2py)1 (C) KK (2s)2, * (2s)2, (2pz)2, (2px)2, (2py)1 (D) KK (2s)2, * (2s)2, (2pz)2, (2px)2, (2py)2 1 2 . Among the following orbital bonds, the angle is minimum between :- (A) sp3 bonds (B) px and py orbitals (C) H–O–H in water (D) sp bonds
1 3 . Molecule having dipole moment is :- (A) 2, 2-dimethylpropane (B) trans-2-pentene (C) neopentane (D) 2,2.3,3-tetramethylbutane. 1 4 . The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are :- (A) dx2 – y2, dz2, s, px, py (B) dxy, s, px, py, pz (C) s, px, py, pz,dz2 (D) dx2 – y2, s, px, py 1 5 . Polarisation is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct :- (A) maximum polarisation is brought about by a cation of high charge (B) minimum polarisation is brought about by a cation of low radius (C) a large cation is likely to bring about a large degree of polarisation (D) polarising power of a cation is less than that of anion 1 6 . Amongest NO3–, AsO33–, CO32–, ClO3–, SO32– and BO33–, the non-planar species are :- (A) CO32–, SO32– and BO33– (B) AsO33–, ClO3– and SO32– (C) NO3–, CO32– and BO33– (D) SO32–, NO3– and BO33– 1 7 . The nature of -bonds in perchlorate ion is :- (A) O (d ) – Cl (p ) (B) O (p ) – Cl (d ) (C) O (d ) – Cl (d ) (D) O (p ) – Cl (p ) 1 8 . CaO and NaCl have the same crystal structure and approximately the same ionic radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is :- (A) U/2 (B) U (C) 2 U (D) 4 U 1 9 . The ease of hydrolysis of trichlorides of group 15 elements decreases in the order :- (A) NCl3 > PCl3 > AsCl3 > SbCl3 > BiCl3 (B) PCl3 > NCl3 > AsCl3 > SbCl3 > BiCl3 (C) AsCl3 > NCl3 > PCl3 > SbCl3 > BiCl3 (D) SbCl3 > BiCl3 > PCl3 > NCl3 > AsCl3 2 0 . Which of the following solid sold have highest value of Kp when heated in closed vessel :- (A) Li2CO3 (s) (B) BeCO3 (s) (C) Na2CO3 (s) (D) BaCO3 (s) 2 1 . Type of bonds between calcium and carbon in CaC2 are :- (A) (B) only (C) only (D) ionic bond 2 2 . Ethanol has a higher boiling point than dimethyl ether though they have the same molecular weight. This is due to :- (A) resonance (B) coordinate bonding (C) hydrogen bonding (D) ionic bonding 2 3 . Write order of dipole moment of following compounds :- Cl Cl Cl Cl Cl (iii) Cl (i) (ii) Cl Cl Cl Cl Cl Cl Cl (iv) (v) (vi) Cl Cl Cl (A) (iii) > (ii) > (i) > (iv) > (v) > (vi) (B) (iii) > (i) > (ii) = (vi) > (iv) > (v) (C) (ii) > (i) = (iii) = (iv) > (v) = (vi) (D) (iii) > (i) > (iv) > (v) > (ii) > (vi)
2 4 . The correct order of increasing X – O – X bond angle is (X = H, F or Cl) :- (A) H2O > Cl2O > F2O (B) Cl2O > H2O > F2O (C) F2O > Cl2O > H2O (D) F2O > H2O > Cl2O 2 5 . Out of given reaction which show change in hybridisation of central atom :- (A) H2BO3 dissolve in water (B) H2SO4 dissolve in water (C) N2O5(g) N2O5(s) (D) PBr5 (g) PBr5(s) (E) C2H6 bond cleaHvaogmeoolyftiCc C bond 2 6 . In the cyclo-S8 molecule of rhombic sulphur, all the S–S bond lengths and all the S–S–S bond angles are respectively (give approximate values) :- (A) 204 pm and 105° (B) 102 pm and 120° (C) 204 pm and 180° (D) 102 pm and 60° 2 7 . The structure of the SO3 molecule in the gaseous phase contains :- (A) only -bonds between sulphur and oxygen (B) -bonds and a (p-p) bonds between sulphur and oxygen (C) -bonds and a (d-p) bonds between sulphur and oxygen (D) -bonds, and a (p-p) and a (p-d) bonds between sulphur and oxygen 2 8 . Which of the following allotropic forms of sulphur is the most stable thermodynamically :- (A) Orthorhombic (B) -monoclinic (C) -monoclinic (D) Plastic sulphur 2 9 . Which of the following acids is not a peroxo acid :- (A) CF3CO3H (B) H2S2O8 (C) H2S2O7 (D) H2N2O2 3 0 . The hydrolysis of 1 mol of peroxodisulphuric acid produces :- (A) 2 mol of sulphuric acid (B) 2 mol of peroxomonosulphuric acid (C) 1 mol each of sulphuric acid and peroxomonosulphuric acid (D) 1 mol each of sulphuric acid, peroxomonosulphuric acid and hydrogen peroxide 3 1 . Which of the following statements is correct :- (A) SF6 does not react with water (B) OF6 is d2Sp3-hybridized (C) S2O32– is a linear ion (D) There is no -bonding in SO42– 3 2 . In H2O, the bond angle H–O–H is 104°28' but in H2S, H2Se and H2Te the bond angles are pretty close to 90°. This suggests that :- (A) oxygen uses sp2-hybrid orbitals while S, Se and Te use sp3-hybrid orbitals for bonding with the hydrogen atoms (B) oxygen uses sp3-hybrid orbitals to bond with the two hydrogen atoms while S, Se and Te use almost pure p orbitals (C) oxygen uses sp3-hybrid orbitals while S, Se and Te utilize d orbitals for bonding with the hydrogen atoms (D) all the atoms use pure p orbitals to bond with the hydrogen atoms 3 3 . Which of the following statements are correct for the SO42– ion ? (A) it is tetrahedral (B) all the S–O bond length are equal, and shorter than expected (C) it contains four -bonds between the S and the O atoms, two -bonds delocalized over the S and the four O atoms, and all the S–O bonds have a bond order of 1.5 (D) Oxidation state of sulphur is +6 and all oxygen in –2
3 4 . The colour of halogens progressively deepens from fluorine to iodine because :- (A) halogens of higher atomic number absorb light of longer wavelength since the difference in energy between the ground state and excited state decreases as the atomic number increases. (B) fluorescence and phosphorescence become more intense as the atomic numbers of halogen increases (C) the standard electrode potential increases from I2 to F2 (D) halogens of higher atomic number absorb light of shorter wavelength since the difference in energy between the ground state and excited state increases as the atomic number increases. 3 5 . Which of the following pairs of halogens have approximately identical bond energy ? (A) F2 and Br2 (B) F2 and I2 (C) F2 and Cl2 (D) Cl2 and I2 3 6 . Which of the following is arranged in order of increasing ionic character :- (A) PbCl2 < SnCl4 < KCl < MgCl2 (B) SnCl4 < PbCl2 < KCl < MgCl2 (C) SnCl4 < PbCl2 < MgCl2 < KCl (D) PbCl2 < SnCl4 < MgCl2 < KCl 3 7 . XeOF4 contains :- (A) six electron pairs forming an octahedron with two positions occupied by lone pairs (B) two -bonds and the remaining six electron pairs, forming an octahedron (C) three -bonds and the remaining four electron pairs forming an tetrahedron (D) one -bonds and the remaining six electron pairs forming an octahedron with one position occupied by a lone pair 3 8 . The azide ion has :- (A) 20 outer electrons and is isoelectronic with Br2O (B) 18 outer electrons and is isoelectronic with NO2– (C) 16 outer electrons and is isoelectronic with CO2 (D) 14 outer electrons and is isoelectronic with H2O2 3 9 . Which of the following pairs of ions do not represent cyclic and chain silicates ? (A) Si2O72– and (SiO3)n2n– (B) Si3O96– and (Si4O11)n6n– (C) Si2O72– and (Si2O5)n2n– (D) Si2O77– and (SiO3)n2n– 4 0 . Which of the following statements is/are incorrect ? (A) B2H6 is not an electron-deficient molecule. (B) the dipole moment of BF3 is zero (C) B(OH)3 partially reacts with water to form H3O+ and [B(OH)4]–, and behaves like a weak acid. (D) BF3 and BrF3 molecules have different shapes. 4 1 . Rotation around the bond (between the underlined atoms) is restricted in : (A) C2H4 (B) H2O2 (C) C2H2 (D) C2H6 4 2 . The H bond in solid HF can be best represented as : (A) H – F ...H – F ...H – F (B) H H H H F F F FF HH (C) H H H H (D) F FF F F H 4 3 . Which of the following statements is/are correct ? (A) NH2+ shows sp2 – hybridisation whereas NH2– shows sp3 – hybridisation (B) Al(OH)4– has a regular tetrahedral geometry (C) sp2–hybridized orbitals have equal s-and p-character (D) usually hybridized orbitals form -bonds
4 4 . Which of the following statements is/are true for BaO and MgO ? (A) BaO is more ionic than MgO (B) MgO is more ionic than BaO (C) BaO has a higher melting point than MgO (D) MgO has a higher melting point than BaO 4 5 . Select the correct statement (s) about the compound NO[BF4] : (A) it has 5 and 2 bond (B) nitrogen - oxygen bond length is higher than nitric oxide (NO) (C) it is a diamagnetic species (D) B–F bond length in this compound is lower than in BF3 4 6 . Silane is more reactive than CH4 due to : (A) larger size of Si compared to C which facilitate the attack by nucleophile (B) polarity of Si–H bond is opposite to that of C – H bond (C) availability of vacant 3d orbitals in case of Si to form the reaction intermediate easily (D) Si-H bond energy is lower than that of C–H bond 4 7 . Select correct statement (s) : (A) acidic strength of HBr > HCl but reverse is true for their reducing property (B) basic strength of PH3 > AsH3 but reverse is true for their bond angle (C) dipole moment of CH3Cl > CH3F but reverse is true for their HCˆH bond angle (D) K a1 of fumaric acid is higher than maleic acid but reverse is true for their K a2 4 8 . Nodal planes of -bonds (s) in CH2 = C = C = CH2 are located in : (A) all are in molecular plane (B) two in molecular plane and one in a plane perpendicular to molecular plane which contains C – C -bond (C) one in molecular plane and two in a plane perpendicular to molecular plane which contains C – C -bonds (D) two in molecular plane and one in a plane perpendicular to molecular plane which bisects C – C -bonds at right angle 4 9 . BF3 and NF3 both molecules are covalent, but BF3 is non-polar and NF3 is polar. Its reason is : (A) in uncombined state boron is metal and nitrogen is gas (B) B–F bond has no dipole moment whereas N–F bond has dipole moment (C) the size of boron atom is smaller than nitrogen (D) BF3 is planar whereas NF3 is pyramidal 5 0 . Which of the following has been arranged in order of decreasing bond length :- (A) P – O > Cl – O > S – O (B) P – O > S – O > Cl – O (C) S – O > Cl – O > P – O (D) Cl – O > S – O > P – O 5 1 . Which of the following models best describes the bonding with in a layer of the graphite structure ? (A) metallic bonding (B) ionic bonding (C) non - metallic covalent bonding (D) vander Waals forces 5 2 . Which of the following is tetrabasic acid ? (A) orthophosphoric acid (B) hypophosphorus acid (C) metaphosphoric acid (D) pyrophosphoric acid BRAIN TEASERS ANSWER KEY EXERCISE -2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A,B,C,D C A D A C B C D B A B B C A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 B B D A B D C C B A,C,D, E A D A C,D C 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 A B A,B,C,D A B C D C A,C,D A A,B,C C A,B,D A,D A,C 46 47 48 49 50 51 52 A,B,C,D C B D B C D
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE OR FALSE : 1 . The polarising power of a cation is directly proportional to its size. 2 . The polarisability of an anion is directly proportional to its charge. 3 . For a given cation, greater the polarisability of the anion, more the covalent character. 4 . An element with low ionization potential is most likely to form a covalent bond with an other element having a high electron affinity. 5 . Ionic interactions are stronger than covalent bonds. 6 . Two non-metal atoms are likely to form covalent bonds on combination. 7 . Ionic interactions are directional. 8 . All molecules having polar bonds are polar. 9 . The CH2Cl2 molecule may be polar or nonpolar depending on its geometry. 1 0 . Two isomers of C H Cl are polar. 22 2 1 1 . The net dipole in the water molecules is the resultant of its bond dipoles. 1 2 . SO2 is polar whereas CO2 is non-polar. 1 3 . If all bonds in a molecule are polar, the molecule as a whole must be polar. 1 4 . The bond angle around B in BCl and BF is same. 33 1 5 . NH molecule involve sp3 hybridisation of N-atom. 3 1 6 . The bond length decreases with increase in multiplicity of bonds between two atoms. 1 7 . The geometry of NH and BH is same. 33 1 8 . Dipole moment of NF is less than that of NH . 33 1 9 . A non-polar molecule can have a polar bond in it. 2 0 . In ionic bond formation octet is completed. 2 1 . -bond is formed by the colateral overlapping of atomic orbitals. 2 2 . Normally the covalent bond is non-directional. 2 3 . He having no existence because its bond order is zero. 2 2 4 . O , O – and O + all are paramagnetic. 22 2 2 5 . When bond order increases, bond length decreases. 2 6 . Bond order is the measurement of bond strength. 2 7 . In hybridisation, only orbitals are involved not electrons. 2 8 . SF molecule is octahedral. 6 2 9 . Ethyne is a linear molecule. 3 0 . CO is a polar molecule but not have polar bond. 2 3 1 . C–C sigma bond in ethyne is sp2 –sp2. 3 2 . XeF molecule is square planar in shape. 4 3 3 . Hybridised orbitals have identical shape. 3 4 . The bond angle in Cl O is equal to that of OF . 22 3 5 . The density of water is more than ice. 3 6 . HCl is a gas while HF is high boiling point liquid because H–F bond is stronger. 3 7 . Liquid NH does not contain H–bond. 3
3 8 . Dipole moment is completely based on ionic nature of bond. 3 9 . BeF containing dipole moment while H O having zero dipole moment. 22 4 0 . In CO molecule, C–O bond is polar but CO molecule is non-polar because the vector sum of two C—O 22 bond is zero. 4 1 . Odd electron molecule is paramagnetic. 4 2 . The reason for resonance, is delocalisation of -electrons. 4 3 . During the formation of covalent bond both shared electron having opposite spin. 4 4 . The bond Hg–Cl is more ionic in HgCl than Hg Cl . 2 22 4 5 . The solubility of ionic solids decreases when hydration energy of ions increases. 4 6 . Between layers of graphite, bond will be covalent. 4 7 . The polarising power of Zn2+ is greater than Ca2+ ions. 4 8 . The colour of PbI is yellow. The reason for this is large size of Pb+2 ion. 2 4 9 . The H – N – H bond angle in NH is greater than H – As – H bond angle in AsH . 33 5 0 . Linear overlap of atomic p-orbitals leads to a sigma bond. 5 1 . The dipole moment of CH F is greater than that of CH Cl. 33 5 2 . sp2 hybrid orbitals have equal s- and p-character. 5 3 . All the Al – Cl bonds in Al Cl are not equivalent. 26 FILL IN THE BLANKS : 1 . Hydrogen bond energy is around ...................... . 2 . The valence atomic orbitals on carbon in silver acetylide are ...................... hybridised. 3 . The hybridisation state of oxygen in water molecule is ...................... . 4 . When N goes to N +, then N–N bond distance ...................... and when O goes to O +, then O–O bond 22 22 distance ...................... . 5 . Among N2O, SO2, I3+ and I3– , the linear species are ...................... and ...................... . MATCH THE COLUMN : 1. Column-I Column-II (A) (p) covalent bond (B) O3 ( q ) Co-ordinate bond (C) H2O() ( r ) Hydrogen bond (D) CuSO4 . 5H2O ( s ) Ionic bond (NH4)2SO4 2. Column-II (A) Column-I (p) planar geometry (B) (q ) non planar geometry (C) B3N3H6 ( r ) no lone pair (D) H2O2 ( s ) non polar molecule B2H6 I2Cl6
3. Column-I Column-II (A) B2H6 + 2NaH (p ) no change in hybridisation of (B) H BO + water under line atom (C) 33 (D) ( q ) sp2 sp3 (change in hybridisation) BeH2 (BeH2)(s) ( r ) breaking of 3C – 2e– bond ( s ) formation of 3C – 2e– bond BF3 + NaF 4. Column-I Column-II (A) N2+ is stable than N – (p) due to one have higher electrons 2 in antibonding than other one have B.O. 3 and other ( B ) NO can easily loss its electron (q) have 2.5 (r) both are paramagnetic with same than N (s) bond order 2 one paramagnetic and other diamagnetic ( C ) NO have large bond length than NO+ ( D ) He2+ exist but less stable than H2+ ASSERTION & REASON QUESTION : These questions contains, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. 1 . Statement-I : FeI3 cannot exist in an aqueous solution. Because Statement-II : Fe3+ oxidizes I– to I2 easily. 2 . Statement-I : SF6 exists but SH6 does not. Because Statement-II : d-p bonding cannot take place in SH6 3 . Statement-I : The stability of peroxides and superoxides increases in passing from Li to Cs. Because Statement-II : The electropositive character of the elements in the periodic table increases on moving down a group. 4 . Statement-I : Borazole is aromatic in nature. Because Statement-II : Nitrogen contributes -electrons to the system. 5 . Statement-I : The first ionization energy of Be is greater than that of B. Because Statement-II : The 2p orbital is lower in energy than the 2s. 6 . Statement-I : Bond order of O and BN is same. 2 Because Statement-II : O and BN are isoelectronic 2 7 . Statement-I : CO is non polar while SO is polar molecule. 22 Because Statement-II : S-O bonds are polar while C-O non polar.
8 . Statement-I : CO2 and SiO2 has same physical state at room temp. Because Statement-II : Carbon has more electronegativity than silicon atom. 9 . Statement-I : In NF3 molecule lone pair resides in sp3 hybrid orbital. Because Statement-II : NF3 has pyramidal shape. 1 0 . Statement-I : N2O, CO2 & I3– are isostructral. Because Statement-II : All three have same hybridise central atom. 1 1 . Statement-I : Dipole moment of Cl is similar as bromo benzene. Because Statement-II : Dipole moment of Cl - C bond is greater than Br- C bond. O 1 2 . Statement-I : PCl3 on hydrolysis gives OH P OH and OH P OH. Because H OH H Statement-II : H3PO3 exists in two tautomeric forms : HO P OH HO P OH. OH O 1 3 . Statement-I : Super oxide ion is paramagnetic whereas peroxide ion is diamagnetic. Because Statement-II : Super oxide ion has one unpaired electron whereas per oxide ion has no unpaired electron. 1 4 . Statement-I : Although PF5, PCl5 and PBr5 are known, the penta halide of nitrogen have not been observed. Because Statement-II : Phosphorus has lower electronegativity than nitrogen. 1 5 . Statement-I : Among alkali metal cations, Li+(aq.) has highest electrical conductance. Because Statement-II : Li+(aq.) is largest alkali metal cation because of greater degree of hydration. 1 6 . Statement-I : HNO3 is a stronger acid than HNO2. Because Statement-II : In HNO3, there are two N–O linkage whereas in HNO2 there is only one. 1 7 . Statement-I : Al(OH)3 is amphoteric in nature. Because Statement-II : Al–O and O–H bonds can be broken with equal case in Al(OH)3. 1 8 . Statement-I : H2SO4 in more viscous than water. Because Statement-II : In H2SO4, S has highest oxidation state. 19. Statement-I : F – ion is known, which has same geometry as X3– (X = Cl, Br, I) 3 Because Statement-II : F is most electronegative element.
2 0 . Statement-I : The p-isomer of dichlorobenzene has higher melting point than o-and m-isomer. Because Statement-II : p-isomer is symmetrical and thus shows more closely packed structure. 2 1 . Statement-I : Boron does not show univalent nature but unipositive nature of thallium is quite stable. Because Statement-II : Inert pair effect predominates in thallium. 2 2 . Statement-I : H3BO3 is monobasic acid in water. Because Statement-II : In water ionise as H3BO3 H2BO – + H+. 3 COMPREHENSION BASED QUESTIONS : Comprehension # 1 In order to explain the shape and geometry of molecules. The valence bond theory was supplemented by the concept of Hybridization. This is a Hypothetical concept and has been introduced by Pauling and Slater. According to this concept any number of atomic orbitals of an atom which differ in energy slightly may mix with each other to form new orbitals called hybrid orbitals. The process of mixing or amalgamation of atomic orbitals of nearly same energy to produced a set of entirely new orbitals of equivalent energy is known as Hybridization. 1 . The hybridization of carbon atoms in C – C single bond of H – C C – CH = CH2 is : (A) sp3 – sp3 (B) sp2 – sp3 (C) sp – sp2 (D) sp3 – sp 2 . In XeF2, XeF4 and XeF6 the number of lone pairs on Xe is respectively : (A) 2, 3, 1 (B) 1, 2, 3 (C) 4, 1, 2 (D) 3, 2, 1 3 . Which of the following is the correct set : (A) H2O ; sp3, angular (B) H2O ; sp2, linear (C) NH ; sp2, pyramidal (D) BF ; sp3, trigonal planar 3 3 4 . In NO3– ion, number of bond pair and lone pairs of electrons on nitrogen atom are : (A) 2, 2 (B) 3, 1 (C) 1, 3 (D) 4, 0 5 . T-shape is exhibited by molecule : (A) ClF3 (B) CHCl3 (C) CCl4 (D) PCl5 Comprehension # 2 Na[BH4] is ionic compound contain B H tetrahydrido borate ion and in solid state NaBH4 has sodium 4 chloride structure. Not all tetrahydridoborates are ionic. The beryllium, aluminium and transition metal borohydrides become increasingly covalent and volatile. In these type of tetrahydrido borate, the B H 4 form covalent bond with metal ion. One or more H atoms in a B H act as a bridging and bond to metal, 4 forming a three centre bond with two electrons shared by three atoms. The B H is usually in that it may form one two or three such three centre bonds to the metal ion when 4 forms covalent bond. Be(BH4)2 and Al(BH4)3 are covalent and structures are given below.
HH HH B HH Be Al H H HH H Be H Be H HH H H H BB HH 1 . Li[AlH4] is used as a reducing agent in many reaction & it is prepared by excess LiH and AlCl3. Select incorrect statement about Li[AlH4]. (A) hybridisation of Al is same as B in Na[BH ] (B) geometry of around Al is same as A lC l 4 4 (C) AlH , B H , A lC l are iso-structral (D) A lH , A lC l , B H ar e iso electronic. 4 4 4 4 4 4 2 . Select correct about Al(BH4)3 : (A) all three tetrahydride borate form two hydrogen bridges (B) two B H form 2 hydrogen bridges and one form one hydrogen bridge. 4 (C) one B H fo rm 2 hydrogen bridge and two form one hydrogen bridge 4 (D) B form only 2c–2e– bond 3 . Total no. of 2c–2e– bond and 3c–2e– bond in Al(BH ) are respectively : 43 (A) 6 , 12 (B) 6 , 6 (C) 12 , 12 (D) 12 , 6 4 . Total 2c–2e and 3c–2e bonds in Be(BH4)2 are respectively : (A) 8, 4 (B) 4, 8 (C) 4, 4 (D) 8, 8 Comprehension # 3 The molecular orbital with the lowest energy is filled first. Thus (1s) is filled first where as * (2p) is filled in the last, also the maximum number of electron in bonding and antibonding molecular orbtitals are according to Pauli and Hund's rule. As an electron in an antibonding molecular orbital cancels out the stability introduced by the electron in a bonding molecular orbital, it means that in order for bonding of atoms to occur there should be an excess of bonding electrons over antibonding electrons. In case where the number of bonding and antibonding electrons are equal, no bond will be formed between the atoms. With the help of above discussion, we can define easily bond order, relative bond length, relative stability and magnetic properties for a molecule. Read the above paragraph carefully and give the answer of following questions :
1 . In an antibonding molecular orbital, there is a point between the two probability contours of hydrogen atoms. This is called. (A) antinode (B) node (C) a plane where electron charge density is maximum (D) A and C both are correct 2 . According to MOT which statement is correct about Boron molecule ? (A) it is diamagnetic in nature (B) it is paramagnetic in nature having magnetic moment 2.8 B.M. by using spin only formula (C) it is paramagnetic but having magnetic moment 1.7 B.M. (D) its bond order is 2 3 . Which of the following are paramagnetic in nature ? (A) B2, N2 and C2 (B) O2, B2, N2– and O – (C) O – –, N + and CO 2 22 (D) B , C and F 2 22 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -5 True / False 1. F 2. T 3. T 4. F 5. F 6. T 7. F 8. F 9. F 10. T 11. F 12. T 13. F 14. T 15. T 16. T 17. F 18. T 19. T 20. T 21. T 22. F 23. T 24. T 25. T 26. T 27.T 28. T 29. T 30. F 31. F 32. T 33. T 34. F 35. T 36. F 37. F 38. F 39. F 40. T 41. T 42. T 43. T 44. F 45. F 46. F 47. T 48. F 49. T 50. T 51. F 52. F 53. T Fill in the Blanks 1. 4.2 – 8.4 kJ mol–1 2. sp 3. sp3 4. increases, decreases 5. N O, I – Match the Column 23 1. A - (p, q), B - (p, r), C - (p, q, r, s), D - (p, q, s) 2. A - (p, r, s), B - (q), C - (q, r, s), D - (p, s) 3. A - (p, r), B - (q), C - (s), D - (q) 4. A - (p, r), B - (p, q, s), C - (p, q, s), D - (p, r) Assertion - Reason Questions 1.A 2. B 3. A 4. B 5. C 6. C 7. C 8. D 9. A 10.C 11. D 12. A 13. A 14. B 15. D 16.C 17.C 18.B 19. D 20. A 21. A 22. C 3. (A) 4. (D) 5. (A) Comprehension Based Questions 3. (B) 4. (C) 3. (B) Comprehension #1 : 1. (C) 2. (D) Comprehension #2 : 1. (D) 2. (A) Comprehension #3 : 1. (B) 2. (B)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE SUBJECTIVE QUESTIONS : 1 . Give reasons for the following : (a) KHF is known whereas KHCl is unknown. 22 (b) (CH ) N is pyramidal but (SiH ) N is trigonal planar. 33 33 (c) CO is a gas but SiO is a solid. 22 2 . (a) The structures of N O and P O are different. Explain. 23 23 (b) Among H–I, H–Br and H–Cl bonds which is weakest? 3 . Suppose that the stability of carbonates when heated depends on the ability of the metal cation to polarize the carbonate ion and remove an oxide ion from it, so releasing CO . 2 (a) Predict the order of thermal stability of the group 1 and 2 metal carbonates ? (b) Comment on the likely stability of aluminium carbonate ? 4 . (a) BF has less lewis acidic property than BBr . Why ? 33 (b) (CH ) C–OH is less acid than (CH ) Si–OH. Why ? 33 3 3 (c) Why in the presence of ethyne HF exhibit more acidic character. (d) Why SF can be hydrolysed easily but SF can not ? 46 5 . In what hybridization state is the beryllium atom in BeCl molecule? How will the type of hybridization change 2 when BeCl transform to the solid state. 2 6 . Give a suitable reason for the high mobility of H+ ions through the ice than through liquid water. 7 . (a) Example the following : (i) Boron trichloride has triangular planar geometry whereas aluminium trichloride has tetrahedral dimeric structure. (ii) Barium sulphate is sparingly soluble in water whereas beryllium sulphate is freely soluble. (iii) Dioxygen (O ) molecule is stable whereas disulphur (S ) is unstable at the room temperature. 22 8 . Assuming Z-axis as molecular axis, lable the molecular orbitals formed by the following combination of atomic orbitals : (i) 1s + 1s (ii) 2p – 2p yy (iii) 2p + 2p zz (iv) 2s + 2s (v) 2p + 2p xx 9 . When a magnet is dipped in a jar of liquid O , some O clings to it. Why ? 22 1 0 . (a) What type of bonding holds the layer in graphite ? Why will graphite conduct electricity well in a direction parallel to the planes of hexagons, but not at all well in a direction perpendicular to the planes ? (b) PCl exists as solid in the form of [PCl ]+ [PCl ]–, yet it is a non conductor of electricity. Why? 5 46 1 1 . A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 0.1 Å, what fraction of an electron charge, e exist on each atom? 1 2 . Through the electronegativities of nitrogen and chlorine are same, NH exists as liquid whereas HCl as gas. 3 Why?
1 3 . The percent ionic character in HCl is 18.08. The observed dipole moment is 1.08 D. Find the internuclear distance in HCl. 1 4 . HBr has dipole moment 2.6 × 10–30 CM. If the ionic character of the bond is 11.5 %, calculate the interatomic spacing. 1 5 . Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate percentage ionic character in LiF molecule Li – F bond length is 0.156 nm. 1 6 . Based upon M.O. theory state reason for the paramagnetic characater of CN, the diamagnetic character of CN–, the stability of CN– and calculate their respective bond orders. 1 7 . Draw the structure of following compound (i) S 8 (ii) N H 24 (iii) P H 4 10 (iv) POCl 3 (v) XeOF 4 (vi) C3O2 (vii) BrF 5 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 4 . 1.4 Å 15 . 84.35 % 1 16. 2 , 3 2
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE SUBJECTIVE QUESTIONS : 1 . Draw the structure of following compound (i) Na[B O (OH) ] 33 4 (ii) Na[B O (OH) ].8H O 45 42 (iii) Na P O 6 6 18 (iv) S O –2 36 (v) S O 39 (vi) (CN) 2 2 . What is the increasing order of the bond angle for the following compounds ? OO O H H, Cl Cl and F F 3 . How many S – S linkage present in H2SnO6 4 . Draw the geometry of following three molecules and explain with proper reasoning : (i) N(CH3)3 (ii) N(SiH3)3 (iii) P(SiH3)3 5 . The dipole moment of LiH is 1.964 × 10–29 CM and the intermolecular distance between Li and H in this molecule is 1.596 Å. What is percent ionic character in molecule ? 6 . The dipole moment of KCl is 3.336 × 10–29 coulomb metre which indicates that it is a highly polar molecule. The interatomic distance between K+ and Cl– in this molecule is 2.6 × 10–10 m. Calculate the dipole moment of KCl molecule, if these were opposite charges of one fundamental unit located at each nucleus. Calculate percentage ionic charcter of KCl. 7 . Assuming covalent radii to be additive property ; calculate the iodine - iodine distance in o–, m–, p– di- iodobenzene. The benzene ring is regular hexagon and each C – I bond lies on a line passing through the centre of hexagon. The C – C bond length C6H6 are 1.40 Å and covalent radius of iodine and carbon atom are 1.33 Å and 0.77 Å. Also neglect different overlapping effect. 8 . Assuming that all the four valency of carbon atom in propane pointing towards the corners of a regular tetrahedron. Calculate the distance between the terminal carbon atoms in propane. Given, C – C single bond length is 1.54 Å. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 5. 76.82% 6 . 4.165 × 10–29 coulomb metre. 7 . 3.50 Å, 6.06 Å, 7.0 Å 8 . 2.514 Å
EXERCISE–05[A] PREVIOUS YEARS QUESTIONS 1 . In which of the following species is the underlined carbon having sp3 - hybridisation ? [AIEEE 2002] (1) CH3–COOH (2) CH3CH2OH (3) CH3COCH3 (4) CH2=CH–CH3 2 . Which of the following statements is true ? [AIEEE 2002] (1) HF is less polar than HBr (2) Water does not contain any ions (3) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion (4) In covalent bond, transfer of electrons takes place 3 . A square planar complex is formed by hybridisation of which atomic orbital ? [AIEEE 2002] (1) s, p , p , d (2) s, p x , p y , d x2 y2 (3) s, p , p , d z2 (4) s, p p , d x y yz xy x y xy 4 . The reason for double helical structure of DNA is operation of : [AIEEE 03] (1) dipole-dipole interaction (2) hydrogen bonding (3) electrostatic attractions (4) vander Walls' forces 5 . Which one of the following pairs of molecules will have permanent dipole moments for both members [AIEEE 03] (1) NO and CO (2) NO and O (3) SiF and CO (4) SiF and NO 22 23 42 42 6 . The pair of species having identical shapes for molecules of both species is [AIEEE 03] (1) XeF , CO (2) BF , PCl (3) PF , IF (4) CF , SF 22 33 55 44 7 . The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is :– [AIEEE–2004] (1) H2S < NH3 < SiH4 < BF3 (2) NH3 < H2S < SiH4 < BF3 (3) H2S < SiH4 < NH3 < BF3 (4) H2S < NH3 < BF3 < SiH4 8 . The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species ? [AIEEE–2004] (1) Bond length in NO+ is equal to that in NO (2) Bond length in NO is greater than in NO+ (3) Bond length in NO+ is greater than in NO (4) Bond length is unpredictable 9 . The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively (1) sp3 and sp2 (2) sp2 and sp3 (3) sp2 and sp2 [AIEEE–2004] (4) sp3 and sp3 1 0 . Which one of the following has the regular tetrahedral structure ? [AIEEE–2004] (1) BF4– (2) SF4 (3) XeF4 (4) [Ni(CN)4]2– (Atomic nos.: B = 5, S = 16, Ni = 28, Xe = 54) 1 1 . The maximum number of 90° angles between bond pair-bond pair of electrons is observed in :- (1) dsp2 hybridization (2) sp3d hybridization (3) dsp3 hybridization [AIEEE–2004] (4) sp3d2 hybridization 1 2 . Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in (1) Forming covalent halides (2) Forming polymeric hydrides [AIEEE–2004] (3) Exhibiting maximum covalency in compounds (4) Exhibiting amphoteric nature in their oxides
1 3 . Which one of the following species is diamagnetic in nature ? [AIEEE-2005] (1) He2+ (2) H2 (3) H2+ (4) H2– 1 4 . lattice energy of an ionic compound depends upon [AIEEE-2005] (1) charge on the ion only (3) packing of the ion only (2) size of the ion only (4) charge and size of the ion 1 5 . The molecular shapes of SF4, CF4 and XeF4 are [AIEEE-2005] (1) the same with 2, 0 and 1 lone pair of electrons on the central atom, respecitvely [AIEEE-2005] [AIEEE-2006] (2) the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively (3) different with 0, 1 and 2 lone pair of electrons on the central atoms, respectively (4) different with 1, 0 and 2 lone pair of electrons on the central atoms, respectively 1 6 . Of the following sets which one does not contain isoelectronic species ? (1) PO43–, SO42–, ClO4– (2) CN–, N2, C22– (3) SO32–, CO32–, NO3– (4) BO33–, CO32–, NO–3 1 7 . Which of the following molecules\\ions does not contain unpaired electrons? (1) N (2) O2 (3) O 2 (4) B2 2 2 1 8 . Among the following mixtures, dipole-dipole as the major interaction, is present in [AIEEE-2006] (1) KCl and water (2) benzene and carbon tetrachloride (3) benzene and ethanol (4) acetonitrile and acetone 1 9 . A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct? [AIEEE-2006] (1) MCl2 is more ionic than MCl4 (2) MCl2 is more easily hydrolysed than MCl4 (3) MCl2 is more volatile than MCl4 (4) MCl2 is more soluble in anhydrous ethanol than MCl4 2 0 . In which of the following molecules/ions are all the bonds not equal? [AIEEE-2006] (1) XeF4 (2) BF4– (3) SF4 (4) SiF4 2 1 . The decreasing values of bond angles from NH3 (106°) to SbH3 (91°) down group-15 of the periodic table is due to [AIEEE-2006] (1) decreasing lp – bp repulsion (2) increasing electronegativity (3) increasing bp – bp repulsion (4) increasing p-orbital character in sp3 2 2 . In which of the following ionizion processes, the bond order has increased and the magnetic behaviour has changed [AIEEE-2007] (1) NO NO+ (2) O2 O (3) N2 N (4) C2 C 2 2 2
2 3 . Which of the following hydrogen bonds is the strongest [AIEEE-2007] (1) F–H.....F (2) O–H.....O (3) O–H.....F (4) O–H.....N 2 4 . Which of the following species exhibits the diamagnetic behaviour [AIEEE-2007] (1) O (2) O2 (3) NO (4) O 2 2 2 2 5 . The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca+2, Mg+2, Be+2 [AIEEE-2007] (1) Be+2 < K+ < Ca+2 < Mg+2 (2) K+ <Ca+2 < Mg+2 < Be+2 (3) Ca+2 < Mg+2 < Be+2 < K+ (4) Mg+2 < Be+2 <K+ <Ca+2 2 6 . Using MO theory predict which of the following species has the shortest bond length? [AIEEE-2009] (1) O2 (2) O 2 (3) O 2 (4) O2 2 2 2 7 . The hybridisation of orbitals of N atom in NO3–, NO2+ and NH4+ are respectively:- [AIEEE-2011] (1) sp, sp3, sp2 (2) sp2, sp3, sp (3) sp, sp2, sp3 (4) sp2, sp, sp3 2 8 . The structure of IF7 is :- [AIEEE-2011] (1) octahedral (2) pentagonal bipyramid (3) square pyramid (4) trigonal bipyramid 2 9 . Among the following the maximum covalent character is shown by the compound :- [AIEEE-2011] (1) AlCl3 (2) MgCl2 (3) FeCl2 (4) SnCl2 3 0 . Which of the following has maximum number of lone pairs associated with Xe ? [AIEEE-2011] (1) XeO3 (2) XeF4 (3) XeF6 (4) XeF2 3 1 . The number of types of bonds between two carbon atoms in calcium carbide is :- [AIEEE-2005, 2011] (1) One sigma, two pi (2) One sigma, one pi (3) Two sigma, one pi (4) Two sigma, two pi 3 2 . Ortho-Nitrophenol is less soluble in water than p– and m– Nitrophenols because :- [AIEEE-2005, 2012] (1) Melting point of o–Nitrophenol is lower than those of m– and p– isomers (2) o–Nitrophenol is more volatile in steam than those of m– and p– isomers (3) o–Nitrophenol shows Intramolecular H–bonding (4) o–Nitrophenol shows Intermolecular H–bonding 3 3 . Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ? [AIEEE-2012] (1) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. (2) Ferrous oxide is more basic in nature than the ferric oxide. (3) Ferrous compounds are relatively more ionic than the corresponding ferric compounds. (4) Ferrous compounds are less volatile than the corresponding ferric compounds. 3 4 . The molecule having smallest bond angle is :- [AIEEE-2012] (1) PCl3 (2) NCl3 (3) AsCl3 (4) SbCl3 3 5 . In which of the following pairs the two species are not isostructural ? [AIEEE-2012] (1) AlF63– and SF6 (2) CO32– and NO3– (3) PCl4+ and SiCl4 (4) PF5 and BrF5 PREVIOUS YEAR QUESTIONS CHEMICAL BONDING EXERCISE-05(A) Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 A. 2 3 2 2 2 1 1 2 3 1 4 3 2 4 4 3 3 Q. 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 A. 4 1 3 4 1 1 4 2 3 4 2 1 4 1 3 1 4 Q. 35 A. 4
EXERCISE–05[B] PREVIOUS YEARS QUESTIONS Choose the correct alter-native (only one correct answer). 1 . The geometry & the type of hybrid orbitals present about the central atom in BF is : [JEE '98] 3 (A) linear, sp (B) trigonal planar, sp2 (C) tetrahedral, sp3 (D) pyramidal, sp3 2 . The correct order of increasing C–O bond length of, CO, CO2–, CO is [JEE '99] 32 (A) CO2– < CO < CO (B) CO < CO2– < CO (C) CO < CO2– < CO (D) CO < CO < CO2– 32 23 32 23 3 . In the dichromate anion [JEE '99] (A) 4Cr – O bonds are equivalent (B) 6Cr – O bonds are equivalent (C) all Cr – O bonds are equivalent (D) all Cr – O bonds are non equivalent 4 . The geometry of H S and its dipole moment are [JEE '99] 2 (A) angular & non zero (B) angular & zero (C) linear & non zero (D) linear & zero 5 . In compounds type ECl , where E = B, P, As or Bi, the angles Cl – E – CI for different E are in the order 3 [JEE '99] (A) B > P = As = Bi (B) B > P > As > Bi (C) B < P = As = Bi (D) B < P < As < Bi 6 . The most likely representation of resonance structure of p-nitrophenoxide is: [JEE '99] —O O —O + O OO —O + O N+ N N+ N (A) (B) (C) (D) O— O O— O— 7 . Amongst H O, H S , H Se and H Te, the one with the highest boiling point is [JEE 2000] 22 2 2 (A) H O because of hydrogen bonding (B) H Te because of higher molecular weight 2 2 (C) H S because of hydrogen bonding (D) H Se because of lower molecular weight 2 2 8 . The hybridization of atomic orbitals of nitrogen in NO+, NO – and NH+ are [JEE 2000] 23 4 (A) sp2, sp3 and sp2 respectively (B) sp, sp2 and sp3 respectively (C) sp2, sp, and sp3 respectively (D) sp2, sp3 and sp respectively 9 . The correct order of hybridization of the central atom in the following species NH , PtCl –2, PCl and BCl 34 5 3 is [JEE 2001] (A) dsp2, sp3d, sp2 and sp3 (B) sp3, dsp2, sp3d, sp2 (C) dsp2, sp2, sp3, sp3d (D) dsp2, sp3, sp2, sp3d 1 0 . The common features among the species CN–, CO and NO+ are [JEE 2001] (A) Bond order three and isoelectronic (B) Bond other three and weak field ligands (C) Bond order two and –acceptors (D) Isoelectronic and weak field ligands 1 1 . Specify hybridization of N and B atoms in a 1 : 1 complex of BF and NH [JEE 2002] 33 (A) N : tetrahedral, sp3; B : tetrahedral, sp3 (B) N : pyramidal, sp3; B : pyramidal, sp3 (C) N : pyramidal, sp3; B : planar, sp2 (D) N : pyramidal, sp3; B : tetrahedral, sp3
1 2 . The nodal plane in the -bond of ethene is located in [JEE 2002] (A) the molecular plane (B) a plane parallel to the molecular plane (C) a plane perpendicular to the molecular plane which bisects, the carbon-carbon a bond at right angle. (D) a plane perpendicular to the molecular plane which contains, the carbon-carbon bond. 1 3 . Identify the least stable ion amongst the following : [JEE 2002] (A) Li– (B) Be– (C) B– (D) C– 1 4 . Which of the following molecular species has unpaired electron(s) ? [JEE 2002] (A) N (B) F (C) O – (D) O2– 2 2 2 2 1 5 . Which of the following are isoelectronic and isostructural ? [JEE 2003] NO – . CO 2–, ClO– , SO 33 33 (A) NO –, CO2– (B) SO , NO– (C) ClO–, CO2– (D) CO2–, SO 33 33 33 33 1 6 . According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O+ [JEE 2004] 2 (A) Paramagnetic and Bond order < O (B) Paramagnetic and Bond order > O 2 2 (C) Diamagnetic and Bond order < O (D) Diamagnetic and Bond order > O 2 2 1 7 . Which species has the maximum number of lone pair of electrons on the central atom? (A) ClO– (B) XeF (C) SF (D) I – [JEE 2005] 3 4 4 3 1 8 . The percentage of p-character in the orbitals forming P–P bonds in P is (D) 75 4 (A) 25 (B) 33 (C) 50 1 9 . Among the following, the paramagnetic compound is [JEE 2007] (A) Na O (B) O (C) N O (D) KO 22 3 2 2 2 0 . The species having bond order different from that in CO is [JEE 2007] (A) NO– (B) NO+ (C) CN– (D) N (C) pyramidal 2 2 1 . The structure of XeO is [JEE 2007] 3 (A) linear (B) planar (D) T-shaped 2 2 . Statement-1 : p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. a n d [JEE 2007] Statement-2 : o-Hydroxybenzoic acid has intramolecular hydrogen bonding. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 2 3 . Statement-1 : In water, orthoboric acid behaves as a weak monobasic acid. [JEE 2007] and Statement-2 : In water, orthoboric, acid acts as a proton donor. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.
2 4 . Statement-1 : Pb+4 compounds are stronger oxidizing agents than Sn4+ compounds [JEE 2008] and Statement-2 : The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect' . (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. 2 5 . Match each of the diatomic molecules in Column I with its property / properties in Column II. [JEE 2009] Column I Column II (A) B (P) Paramagnetic 2 (Q) undergoes oxidation (R) Undergoes reduction (B) N (S) Bond order 2 2 (T) Mixing of 's' and `p' orbitals (C) O – 2 (D) O 2 2 6 . The nitrogen oxide(s) that contain(s) N–N bond(s) is (are) [JEE 2009] (A) N2O (B) N2O3 (C) N2O4 (D) N2O5 2 7 . In the reaction [JEE 2009] 2X + B2H6 [BH2(X)2]+ [BH4]– the amine(s) X is (are) (A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) (CH3)3N [JEE 2010] 2 8 . The species having pyramidal shape is (A) SO (B) BrF (C) SiO2– (D) OSF 3 3 3 2 2 9 . Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B is 2 [JEE 2010] (A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic 3 0 In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) [JEE 2012] (A) sp and sp3 (B) sp and sp2 (C) only sp2 (D) sp2 and sp3 3 1 Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen- (A) HNO3, NO, NH4Cl, N2 (B) HNO3, NO, N2, NH4Cl [JEE 2012] (C) HNO3, NH4Cl, NO, N2 (D) NO, HNO3, NH4Cl, N2 [JEE 2012] (D) see-saw 3 2 The shape of XeO F molecule is : 22 (A) Trigonal bipyramidal (B) Square planar (C) tetrahedral Explain the following. 1 . Give reason carbon oxygen bond lengths in formic acid are 1.23Å & 1.36 Å and both the carbon oxygen bonds in sodium formate have the same value i.e. 1.27Å. [JEE '88] 2 . Explain the first I.E. of carbon atom is greater than that of boron atom whereas the reverse is true for the second I.E. [JEE '89]
3 . The experimentally determined N–F bond length in NF3 is greater than the sum of single bond covalent radii of N & F. Explain. [JEE '95] 4 . Explain the difference in the nature of bonding in LiF & LiI. [JEE '96] 5 . Give reasons for the following in one or two sentences only. [JEE '99] (a) BeC1 can be easily hydrolyed (b) CrO is an acid anhydride 2 3 Arrange as directed. 1 . N , O , F , Cl in increasing order of bond dissociation energy. [JEE '88] 2 22 2 [JEE '88] [JEE '88] 2 . CO , N O SiO , SO is the increasing order of acidic character. [JEE '97] 2 25 2 3 3 . HOCl, HOClO , HOClO , HOClO in increasing order of thermal stability. 23 4 . The decreasing order of acid strength of ClOH, BrOH, IOH. 5 . Arrange in order of increasing radii , Li+, Mg2+, K+, Al3+ [JEE '97] 6 . Arrange BeSO , MgSO , CaSO , SrSO in order of decreasing thermal stability. [JEE '97] 4 4 44 [JEE 2004] 7 . Decreasing order of the O-O bond length present in them O , KO and O [AsF ] 22 24 Subjective 1 . The number of water molecule(s) directly bonded to the metal centre in CuSO .5H O is [JEE 2009] 42 2 . Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF is [JEE 2010] 5 3 . The value of n in the molecular formula Be Al Si O is [JEE 2010] n 2 6 18 4 . The total number of diprotic acids among the following is [JEE 2010] H PO H SO H PO H CO HSO 34 24 33 23 22 7 H BO H PO H CrO H SO 33 32 23 24 5 . Among the following, the number of elements showing only one non-zero oxidation state is [JEE 2010] O, Cl, F, N , P, Sn, Tl, Na , Ti 6 . The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is.[JEE 2011] PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1. (B) 2. (D) 3. (B) 4. (A) 5. (B) 6. (A) 7. (A) 14. (C) 8. (B) 9. (B) 10. (A) 11. (A) 12. (A) 13. (B) 21. (C) 15. (A) 16. (B) 17. (D) 18. (D) 19. (D) 20. (A) 32. (D) 22. (D) 23. (C) 24. (C) 25. (A) P,Q,R,T ; B Q,R,S,T ; (C) P,Q,R ; (D) P,Q,R,S 26. A,B,C 27. B,C 28. (D) 29. (A) 30. (B) 31. (B) Arrange as directed : 1. F < Cl < O < N 2. SiO < CO < N O < SO 3 2 22 2 2 2 25 3. HOCl < HClO2 < HClO3 < HClO4 4. HOCl > HOBr > HOI 5. Al3+ < Mg2+ < Li+ < K+ 6. SrSO4 > CaSO4 > MgSO4 > BeSO4 7. KO > O > O [AsF ] 222 4 Subjective Questions 1. 4 2. 0 3. 3 4. 6 5. 2 6. 5
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . In a reaction PCl5 PCl3 + Cl2 degree of dissociation is 30%. If initial moles of PCl5 is one then total moles at equilibrium is (A) 1.3 (B) 0.7 (C) 1.6 (D) 1.0 2. For reaction HI ½ H +½ I value of K is 1/8 then value of K for H +I 2HI. 2 c c 22 2 1 (B) 64 1 (D) 8 (A) (C) 64 8 3. In a equilibrium reaction H (g) + I (g) 2HI(g) 22 H = – 3000 calories, which factor favours dissociation of HI :- (A) Low temp. (B) High Pressure (C) High temp. (D) Low pressure. 4. N+ 3H 2NH 2 2 3 If temp. of following equilibrium reaction increase then - (A) Shift Right side (B) Shift left side (C) Unchanged (D) Nothing say. 5. C(s) + H O (g) H (g) + CO(g) 2 2 by increasing pressure following equilibrium (A) Unaffected (B) Proceed in backward direction (C) Proceed in forward direction (D) Unfixed 6 . Unit of equilibrium constant K for following homogenous reaction :- c 4NH + 5O 4NO + 6H O is 32 2 (A) (Concn)-1 (B) (Concn)+1 (C) (Concn)+10 (D) Have no unit 7. Which of the following factor shifted the reaction PCl + Cl PCl at left side. 3 2 5 (A) Adding PCl5 (B) Increase pressure (C) Constant temp. (D) Catalyst. 8 . In which of the following process reaction is fastest complete. (A) K = 10 (B) K = 1 (C) K = 103 (D) K = 10–2 9 . At 298 K equilibrium constant K and K of 12 following reaction SO (g) + ½ O (g) SO (g) ----- (1) 2 2 3 2SO (g) 2SO (g) + O (g) -------- (2) 3 2 2 The relation between K1 and K2 is (A) K = K (B) K = K 2 (C) K = 1/K 2 (D) K = 1/K 12 21 21 21 10. In the following reaction PCl (g) PCl (g) + Cl (g) at constant temp. rate of backward reaction is increase by : 5 32 (A) Inert gas mixed at constant volume (B) Cl gas mixed at constant volume 2 (C) Inert gas mixed at constant pressure (D) PCl mixed in constant volume. 5 1 1 . Some gaseous equilibrium are following : CO + H O K CO + H 2 22 K 2CO + O 1 2CO 22 K 2 2H + O 2H O 2 2 2 then find out the relation between equilibrium constants :- (A) K = K K (B) K = (K K )2 (C) K = (K K )–½ (D) K = (K /K )½ 12 12 12 12
1 2 . For the equilibrium process x + y xy. If the concn. of x and y is doubled, then equilibrium constant. (A) Become twice (B) Become half (C) Unchanged (D) Become thrice 1 3 . Two moles of PCl were heated in closed vessel of 2 litre capacity, at equilibrium 40% of PCl was 55 dissociated into PCl3 and Cl2. The value of equilibrium constant is : (A) .267 (B) .53 (C) 2.67 (D) 5.3 1 4 . The reaction A(g) + B(g) 2C(g) is occur by mixing of 3 moles of A and 1 mole of B in one litre Container. if of B is 1 , then K for this reaction is :- 3C (A) 0.12 (B) 0.25 (C) 0.50 (D) 0.75 15. Reaction 2BaO (s) 2BaO(s) + O (g) ; H = + ve. At equilibrium condition, Pressure of O is 2 2 2 depends on :- (A) Increase mass of BaO (B) Increase mass of BaO 2 (C) Increase temp. at Eqm. (D) Increase mass of BaO and BaO both 2 1 6 . Consider the following equilibrium in a closed container N O (g) 2NO (g). At a fixed temperature, 24 2 the volume of the reaction container is halved. For this change, which of the following statements held true regarding the equilibrium constant (K ) and degree of dissociation () :- p (A) Neither K nor changes (B) Both K and - changes p p (C) K changes, but does not change (D) K does not change, but - changes p p 17. C(s) + CO (g) 2CO(g) 2 According to above reaction, partial Pressure of CO & CO are 4 & 8 respectively then find out K of the 2p above reaction :- (A) 6 (B) 2 (C) 16 (D) 32 1 8 . For the reaction, A+B C + D, K = 9. If A and B are taken in equal amounts, then amount of C at c equilibrium is :- (A) 1 (B) 0.25 (C) 0.75 (D) None of these 1 9 . At equilibrium 500mL vessel contains 1·5 M of each A, B, C, D. If 0·5M of C and D expelled out then what would be the K :- C (A) 1 1 4 5 (B) (C) (D) 9 9 9 2 0 . The following equilibrium are given N + 3H 2NH -------------- K 2 2 3 1 N + O 2NO -------------- K 2 22 H2 + 1 H2O -------------- K3 2 O2 The equilibrium constant of the reaction 2NH + 5 2NO + 3 H O, in terms of K , K and K is : 3 O 2 12 3 22 K1K2 (B) K 1 K 2 (C) K 2 K 3 (D) K K K (A) K 3 3 3 123 K2 K1 2 1 . The reaction A + B C + D is studied in a one litre vessel at 250°C. The initial concentration of A was 3n and that of B was n. When equilibrium was attained, equilibrium concentration of C was found to the equal to the equilibrium concentration of B. What is the concentration of D at equilibrium : (A) n/2 (B) (3n – 1/2) (C) (n – n/3) (D) n
2 2 . In a reversible reaction A k1 B , the initial concentration of A and B are a and b in moles per litre and k2 the equilibrium concentration are (a – x) and (b + x) respectively ; express x in terms of k , k , a and b : 12 k1a k2b k1a k2b k1a k2b k1a k2b (A) k1 k2 (B) k1 k2 (C) k1k2 (D) k1 k2 2 3 . The value of K for the reaction p 2H O (g) + 2Cl O (g) 4HCl (g) + O (g) 2 22 is 0.03 atm at 427°C, when the partial pressure are expressed in atmosphere then the value of K for C the same reaction is : (A) 5.23 × 10–4 (B) 7.34 × 10–4 (C) 3.2 × 10–3 (D) 5.43 × 10–5 24. The equilibrium constant of the reaction SO (g) 1 SO (g) is 4 × 10–3 atm–1/2. The equilibrium 2 + O (g) 3 22 constant of the reaction 2 SO (g) 2 SO (g) + O (g) would be : 3 2 2 (A) 250 atm (B) 4 × 103 atm (C) 0.25 × 104 atm (D) 6.25 × 104 atm 2 5 . When alcohol (C H OH) and acetic acid are mixed together in equimolar ratio at 27°C, 33% is converted 25 into ester. Then the K for the equilibrium : C C H OH () + CH COOH () CH COOC H () + HO () 25 3 3 22 2 (A) 4 (B) 1/4 (C) 9 (D) 1/9 26. 2 moles each of SO , CO, SO and CO is taken in a one lit. vessel. If K for SO + CO SO + CO 3 2 2 C 3 2 2 is 1/9 then : (A) total no. of moles at equilibrium are less than 8 (B) n (SO ) + n(CO ) = 4 32 (C) [n(SO )/n (CO)] < 1 2 (D) both (B) and (C) 2 7 . An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol N , 2 mol O and 3 mol NO. No. 22 of moles of O to be added so that at new equilibrium the conc. of NO is found to be 0.04 mol/lit : 2 (A) (101/18) (B) (101/9) (C) (202/9) (D) None of these 2 8 . Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of NH will be : 3 (A) 0.6 (B) 0.4 (C) unpredictable (D) none of these 29. The degree of dissociation of SO is at equilibrium pressure P: 3 0 Kp for 2SO (g) 2SO (g) + O (g) 3 22 (A) [(P 3)/2(1 – )3] (B) [(P 3)/(2 + )(1–)2] (C) [(P 2)/2(1 – )2] (D) none of these 0 00 30. For the reaction CO(g) + HO (g) CO (g) + H (g) at a given temperature the equilibrium amount 2 2 2 of CO (g) can be increased by : 2 (A) adding a suitable catalyst (B) adding an inert gas (C) decreasing the volume of container (D) increasing the amount of CO (g) 3 1 . For the reaction : PCl (g) PCl (g) + Cl (g) 5 3 2 The forward reaction at constant temperature is favoured by (A) introducing an inert gas at constant volume (B) introducing chlorine gas at constant volume (C) introducing an inert gas at constant pressure (D) increasing the volume of the container (E) introducing PCl at constant volume 5
32. Given the following reaction at equilibrium N (g) + 3H (g) 2 NH (g). Some inert gas at constant 2 2 3 pressure is added to the system. Predict which of the following facts will be affected : (A) more NH (g) is produced (B) less NH (g) is produced 3 3 (C) no affect on the equilibrium (D) K of the reaction is decreased P 33. For an equilibrium HO (s) H O () which of the following statement is true : 2 2 (A) the pressure changes do not affect the equilibrium (B) more of ice melts if pressure on the system is increased (C) more of liquid freezes if pressure on the system is increase (D) the pressure changes may increase or decrease the degree of advancement of the reaction depending upon the temperature of the system. 3 4 . When a bottle of cold drink is opened, the gas comes out with a fizze due to : (A) decrease in temperature (B) increase in pressure (C) decrease in pressure suddenly which results in decrease of solubility of CO2 gas in water (D) none 35. The equilibrium, SO Cl (g) SO (g) + Cl (g) is attained at 25°C in a closed container and an inert 22 2 2 gas, helium, is introduced. Which of the following statements are correct : (A) concentrations of SO , Cl and SO Cl are changed 22 22 (B) no effect on equilibrium (C) concentration of SO is reduced 2 (D) K of reaction is increasing p 3 6 . For the reaction H (g) + I (g) 2HI (g) 22 K = 66.9 at 350°C and K = 50.0 at 448°C. The reaction has : CC (A) H = +ve (B) H = –ve (C) H = Zero (D) H = Not found the signs 1 3 7 . Variation of log K with is shown by the following graph in which straight line 10 T is at 45°, hence H° is : log10K A (A) +4.606 cal (B) –4.606 cal (C) 2 cal (D) – 2cal 1/T CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. A B C B B B A C C B D C A B C Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. D C C A C A A A D B D A B B D Que. 31 32 33 34 35 36 37 Ans . C,D,E B B C BBB
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . Consider following reaction in equilibrium with equilibrium concentration 0.01 M of every species (I) PCl (g) PCl (g) + Cl (g) 5 32 (II) 2HI (g) H (g) + I (g) 22 (III) N (g) + 3H (g) 2NH (g) 2 2 3 Extent of the reactions taking place is : (A) I > II > III (B) I < II < III (C) II < III < I (D) III < I < II 2. For the reaction 3 A (g) + B (g) 2 C (g) at a given temperature, K = 9.0. What must be the volume c of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium? (A) 6 L (B) 9 L (C) 36 L (D) None of these 3 . Sulphide ion in alkaline solution reacts with solid sulphur to form polysulphide ions having formulae S 2–, S 2–, S 2– and so on. The equilibrium constant for the formation of S 2– is 12 (K ) & for the formation 2 3 4-- 21 of S 2– is 132 (K ), both from S and S2–. What is the equilibrium constant for the formation of S 2– from 32 3 S 2– and S ? 2 (A) 11 (B) 12 (C) 132 (D) None of these 4 . For the following gases equilibrium N O (g) 2 NO (g) 24 2 K is found to be equal to K . This is attained when : pc (A) 0°C (B) 273 K (C) 1 K (D) 12.19 K 5 . 1 mole N and 3 mol H are placed in a closed container at a pressure of 4 atm. The pressure falls to 22 3 atm at the same temperature when the following equilibrium is attained N (g) + 3H (g) 2NH (g). The equilibrium constant K for dissociation of NH is : 2 2 3 P 3 (A) 1 (1.5)3 atm 2 (B) 0.5 × (1.5)3 atm2 (C) 0.5 (1.5)3 atm2 33 0.5 33 (D) 0.5 (1.5)3 atm–2 6 . One mole of N O (g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 24 20% by mass of N O (g) decomposes to NO (g). The resultant pressure is : 24 2 (A) 1.2 atm (B) 2.4 atm (C) 2.0 atm (D) 1.0 atm 7. For the reaction : 2 HI (g) H (g) + I (g), the degree of dissociation () of Hl (g) is related to equilibrium 2 2 constant K by the expression : P (A) 1 2 Kp (B) 1 2Kp 2K p 2 Kp 2 2 (C) 1 2KP (D) 1 2 Kp 8 . The vapour density of N O at a certain temperature is 30. What is the % dissociation of N O at this 24 24 temperature : (A) 53.3% (B) 106.6% (C) 26.7% (D) None 9. For the reaction PCl (g) PCl (g) + Cl (g), the forward reaction at constant temperature is favoured 5 3 2 by : (A) introducing an inert gas at constant volume (B) introducing chlorine gas at constant volume (C) introducing an inert gas at constant pressure (D) introducing PCl at constant volume 5
1 0 . When N O is heated at temp. T, it dissociates as N O N O + O , K = 2.5. At the same time 25 25 23 2C N O also decomposes as : NO NO + O. If initially 4.0 moles of N O are taken in 1.0 litre 23 23 2 2 25 flask and allowed to attain equilibrium, concentration of O was formed to be 2.5 M. Equilibrium concen- 2 tration of N O is : 2 (A) 1.0 (B) 1.5 (C) 2.166 (D) 0.334 1 1 . Densities of diamond and graphite are 3.5 and 2.3 g/mL. C (diamond) C (graphite) rH = –1.9 kJ/mole favourable conditions for formation of diamond are : (A) high pressure and low temperature (B) low pressure and high temperature (C) high pressure and high temperature (D) low pressure and low temperature 1 2 . When NaNO is heated in a closed vessel, oxygen is liberated and NaNO is left behind. At equilibrium: 32 (A) addition of NaNO favours reverse reaction 2 (B) addition of NaNO favours forward reaction 3 (C) increasing temperature favours forward reaction (D) increasing pressure favours reverse reaction 13. The equilibrium SO Cl (g) SO (g) + Cl (g) is attained at 25°C in a closed rigid container when 22 2 2 an inert gas, helium is introduced. Which of the following statement is / are correct : (A) concentrations of SO , Cl and SO Cl do not change 22 22 (B) more chlorine is formed (C) concentration of SO is reduced 2 (D) more SO Cl is formed 22 14. For the gas phase reaction, CH +H CH (H = –32.7 kcal), carried out in a closed vessel, 24 2 26 the equilibrium moles of C H can be increased by : 24 (A) increasing the temperature (B) decreasing the pressure (C) removing some H (D) adding some C H 2 26 1 5 . An exothermic reaction is represented by the graph : (A) lnKp (B) lnKp (C) lnKp (D) lnKp 1/T 1/T 1/T 1/T 1 6 . The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K is : (A) –G° = RT ln K (B) G = RT ln K (C) –G = RT ln K (D) G° = RT ln K 1 7 . The value of G°f of gaseous mercury is 31 kJ/mole. At what external pressure mercury start boiling 25°C. [R = 8.3 J/K mole] (A) 10–5.44 (B) 10–12.5 (C) 10–6.52 (D) 10–3.12 1 8 . What is G (kJ/mole) for synthesis of ammonia at 298 K at following sets of partial pressure : r N2(g) + 3H2(g) 2NH3(g) ; rG° = –33 kJ/mole. [Take R = 8.3 J/K mole, log 2 = 0.3 ; log 3 = 0.48] Gas N H NH 3 22 Pressure (atm) 1 3 0.02 (A) +6.5 (B) –6.5 (C) +60.5 (D) –60.5
1 9 . In a 7.0 L evacuated chamber, 0.50 mol H and 0.50 mol I react at 427°C. 22 H (g) + I (g) 2HI (g). At the given temperature, K= 49 for the reaction. 2 2 C (i) What is the value of K ? p (A) 7 (B) 49 (C) 24.5 (D) None (ii) What is the total pressure (atm) in the chamber ? (A) 83.14 (B) 831.4 (C) 8.21 (D) None (iii) How many moles of the iodine remain unreacted at equilibrium ? (A) 0.388 (B) 0.112 (C) 0.25 (D) 0.125 (iv) What is the partial pressure (atm) of HI in the equilibrium mixture? (A) 6.385 (B) 12.77 (C) 40.768 (D) 646.58 20. N +O 2NO, K; 1 N + 1 O NO, K 22 1 2 2 2 2 2 2NO N + O , K ; NO 1 N + 1 O, K 2 23 2 2 2 2 4 Correct relation between K1, K2, K3 and K4 is : (A) K × K = 1 (B) K1 K4 = 1 (C) K3 K2 = 1 (D) None 13 D d 2 1 . The equation, a = (n 1)d is correctly matched for : (A) A nB/2 + nC/3 (B) A nB/3 + (2n/3)C (C) A (n/2)B + (n/4)C (D) A (n/2)B + C 2 2 . Variation of equilibrium constant K for the reaction ; 2A (s) + B (g) C(g) + 2D(g) is plotted against absolute temperature T in figure as - nK Vs (1/T) : nK 1/T (nK vs 1/T diagram) (A) the forward reaction is exothermic (B) the forward reaction is endothermic (C) the slope of line is proportional to H (D) adding 'A' favours forward reaction (E) removing C favours forward reaction 2 3 . The equilibrium of which of the following reactions will not be disturbed by the addition of an inert gas at constant volume? (A) H (g) + I (g) 2HI (g) (B) N O (g) 2NO (g) 2 2 24 2 (C) CO (g) + 2H (g) CH OH (g) (D) C(s) + HO (g) CO (g) + H (g) 2 3 2 2 2 4 . An industrial fuel, 'water gas', which consists of a mixture of H and CO can be made by passing steam 2 over red-hot carbon. The reaction is : C(s) + H2O (g) CO(g) + H2 (g), H = +131 kJ The yield of CO and H at equilibrium would be shifted to the product side by 2 (A) raising the relative pressure of the steam (B) adding hot carbon (C) raising the temperature (D) reducing the volume of the system
2 5 . The dissociation of ammonium carbamate may be represented by the equation : NH CO NH (s) 2NH (g) + CO (g) 422 3 2 H0 for the forward reaction is negative. The equilibrium will shift from right to left if there is (A) a decrease in pressure (B) an increase in temperature (C) an increase in the concentration of ammonia (D) an increase in the concentration of carbon dioxide 2 6 . The percentage of ammonia obtainable, if equilibrium were to be established during the Haber process, is plotted against the operating pressure for two temperatures 400°C and 500°C. Which of the following correctly represents the two graphs ? % NH3 at equilibrium40 400°C % NH3 at equilibrium40 500°C (A) 20 500°C (B) 20 400°C 0 10 20 0 10 20 Pressure/103 kPa Pressure/103 kPa 40% NH3 at equilibrium 400°C 40% NH3 at equilibrium 400°C (C) 20 (D) 20 500°C 500°C 0 10 20 0 10 20 Pressure/103 kPa Pressure/103 kPa BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 345 6 7 8 9 10 11 12 13 14 15 ADB Ans. B A 18 19(i) (ii) B D A C,D D C C,D A A,B,C,D A DBC Que. 16 17 (iii) (iv) 20 21 22 23 24 25 26 Ans. A A B A A,B,C B A,C,E A,B,C,D A,C B,C,D A
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . Van't Hoff's equation gives the quantitative relation between change in value of K with change in tempera- ture. 2 . The larger value of K indicates that the product is more stable relative to reactants. 3 . The value of equilibrium constant changes with change in the initial concentration of the reactants. 4 . Extent of a reaction can always be increased by increasing the temperature. 5 . K is related to K as K = K (RT)n. P C PC 6 . Introduction of inert gas in a gaseous reaction (ng 0) at equilibrium keeping pressure constant has no effect on equilibrium state. 7. For the reaction, N O (g) 2NO (g), K = K (RT). 24 2 PC 8 . For a reaction the value of Q greater than K indicates that the net reaction is proceeding in backward direction. 9 . Solubilities of all solids in water increase with increase in temperature. 1 0 . Dissolution of all gases in water is accompained by evolution of heat. 11. For the reaction, N + 3H 2NH , the equilibrium expression may be written as K = [NH3 ]2 . 2 2 3 [N2 ][H2 ]3 12. For the reaction, CaCO (s) CaO (s) + CO (g), K= p. 3 2 P CO2 1 3 . A catalyst increases the value of the equilibrium constant for a reaction. 1 4 . If concentration quotient of reaction is less than K, the net reaction is proceeding in the backward direction. 1 5 . In case of endothermic reaction, the equilibrium shifts in backward direction on increasing the temperature. 1 6 . The value of K increases with increase in pressure. 17. For the reaction, H + I 2HI, the equilibrium constant, K is dimensionalless. 22 18. The reaction 2SO (g) + O (g) 2SO (g), H = –X kJ, is favoured by high pressure and high 2 2 3 temperature. 1 9 . A very high value of K indicates that at equilibrium most of the reactants are converted into products. 2 0 . The value of K for the reaction, N + 3H 2NH , can be increased by applying high pressure or 22 3 by using a catalyst. FILL IN THE BLANKS 1 . K for the reaction 2 A + B 2C is 1.5 × 1012. This indicates that at equilibrium the concentration of ...................... would be maximum. 2 . The reaction N + O 2NO – Heat, would be favoured by ...................... temperature. 22 3. K for the reaction X +Y 2XY is 100 K for this reaction XY 1 X 1 would be ..................... . +Y 22 2 22 2 4 . Compared to K for the dissociation, 2H S 2H+ + 2HS–, then K' for the H+ + HS– H S 22 would have ...................... . 5 . The equilibrium constant for a reaction decreases with increase in temperature, the reaction must be ...................... . 6. For the reaction, PCl (g) PCl (g) + Cl (g), K and K are related as ...................... . 5 3 2 P C 7. For the reaction, NO (g) 2NO (g), at equilibrium, increase in pressure shifts the equilibrium in 24 2 ...................... direction.
8 . G° is related to K by the relation ..................... . 9 . Vant Hoff's equation is ..................... . 0 . When the reaction is at equilibrium, the value of G is ..................... . 11. Dimensions of equilibrium constant, K for the reaction 2NH N + 3H , are ..................... . c 3 2 2 1 2 . The value of K for a reaction can be changed by changing ..................... . 1 3 . The law of mass action was proposed by ..................... . 14. The degree of dissociation of PCl [PCl (g) PCl (g) + Cl (g)], .................... with increase in pressure 5 5 3 2 at equilibrium. 1 5 . If concentration quotient, Q is greater than K , the net reaction in taking place in ...................... direction. C 16. The reaction, N + 3H 2NH would be favoured by ...................... pressure. 2 2 3 1 7 . K is related to K as ..................... . PC 1 8 . Solubility of a gas in water ...................... with increase in temperature. 1 9 . Introduction of inert gas at constant volume to a gaseous reaction at equilibrium results in formation of ...................... product. 2 0 . The product is more stable than reactants in reaction having ...................... K. MATCH THE COLUMN 1. Column-I Column-II (Reactions) (Favourable conditions) (A) Oxidation of nitrogen (p) Addition of inert gas at constant pressure (q) Decrease in pressure N (g) + O (g) + 180.5 kJ 2NO(g) (r) Decrease in temperature 22 Increase in temperature (B) Dissociation of N O (g) 24 N2O4(g) + 57.2 kJ 2NO2(g) (C) Oxidation of NH (g) 3 4NH (g) + 5O (g) 4NO(g) + 6H O(g) 32 2 + 905.6 kJ (D) Formation of NO (g) (s) 2 NO(g) + O (g) NO (g) + O (g) 3 22 + 200 kJ 2. Column-I Column-II (Reaction) (If is negligiable w.r.t. 1) (A) 2X(g) Y(g) + Z(g) (p) = 2 × K c (B) X(g) Y(g) + Z(g) (C) 3X(g) Y(g) + Z(g) (q) = 3 × K c (D) 2X(g) Y(g) + 2Z(g) (r) = (2K )1/3 c (s) = K c
ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : The melting point of ice decreases with increase of pressure. Because Statement-II : Ice contracts on melting. 2 . Statement-I : The equilibrium of A(g) B(g) + C(g) is not affected by changing the volume. Because Statement-II : Kc for the reaction does not depend on volume of the container. 3 . Statement-I : For the reaction A(g) B(g) + C(g), Kp = 1 atm. If we start with equal moles of all gases at 9 atm of initial pressure, then at equilibrium partial pressure of A increases. Because Statement-II : Reaction quotient Q > K hence equilibrium shifts in backward direction. pp 4 . Statement-I : For a reaction at equilibrium, the Gibb's free energy of reaction is minimum at constant temp. and pressure. Because Statement-II : The Gibb's free energy of both reactants and products increases and become equal at equilibrium. 5 . Statement-I : Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction. Because Statement-II : Equilibrium constant depends upon the way in which the reaction is written. 6 . Statement-I : For the reaction H2(g) + I2(g) 2HI(g) if the volume of vessel is reduced to half of its original volume, equilibrium concentration of all gases will be doubled. Because Statement-II : According to Le Chatelier's principle, reaction shifts in a direction that tends to minimized the effect of the stress. 7 . Statement-I : For the reaction at certain temperature A(g) + B(g) C(g) there will be no effect by addition of inert gas at constant volume. Because Statement-II : Molar concentration of all gases remains constant. 8 . Statement-I : The catalyst does not alter the equilibrium constant. Because Statement-II : For the catalysed reaction and uncatalysed reaction H remains same and equilibrium constant depends on H.
COMPREHENSION BASED QUESTIONS Comprehension # 1 On July, 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial Island and a bridge from the island to Malmo in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials. Concrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarly of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production may lead to formation of unwanted hemihydrate, CaSO4 . 1 H2O. Consider 2 the following reaction : 11 CaSO4 . 2H2O(s) CaSO4 . 2 H2O(s) + 1 H2O(g) 2 The following thermodynamic data apply at 25°C, standard pressure : 1.00 bar : Compound H°/(kJ mol–1) ( H ° ) S°/(JK–1 mol–1) CaSO4 . 2H2O(s) 194.0 f 1 130.5 CaSO4 . 2 H2O(s) –2021.0 188.6 H O(g) –1575.0 2 –241.8 Gas constant ; R = 8.314 J mol–1 K–1 1 1 . H° for the transformation of 1.00 kg of CaSO4.2H2O(s) to CaSO4. 2 H2O(s) is : (A) +446 kJ (B) +484 kJ (C) –446 kJ (D) –484 kJ 2 . Equilibrium pressure (in bar) of water vapour in a closed vessel containing CaSO4 . 2H2O(s), CaSO4(s) .1 H2O(s) and H2O (g) at 25°C is : 2 (A) 7.35 × 10–4 bar (B) 2.15 × 10–4 bar (C) 8.10 × 10–3 bar (D) 7.00 × 10–4 bar 3 . Temperature at which the equilibrium water vapour pressure is 1.00 bar. (A) 107°C (B) 380°C (C) 215°C (D) 240°C Comprehension # 2 Questions are based on the manufacture of Na CO by Solvay process : 23 In the manufacture of Na2CO3(s) by Solvay process, NaHCO3(s) is decomposed by heating : 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) Kp = 0.23 at 100°C H° = 136 kJ 1 . If a sample of NaHCO3 (s) is brought to a temperature of 100°C in a closed container total gas pressure at equilibrium is : (A) 0.96 atm (B) 0.23 atm (C) 0.48 atm (D) 0.46 atm
2 . A mixture of 1.00 mol each of NaHCO3(s) and Na2CO3(s) is introduced into a 2.5 L flask in which PCO2 = 2.10 atm and PH2O = 0.94 atm.When equilibrium is established at 100°C, then partial pressure of : (A) CO2(g) and H2O(g) will be greater than their initial pressure (B) CO2(g) and H2O(g) will be less than their initial pressure (C) CO2(g) will be larger and that of H2O(g) will be less than their initial pressure (D) H2O(g) will be larger and that of CO2(g) will be less than their initial pressure MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 2. T 3. F 4. F 1. T 6. F 7. T 8. T 5. T 10. T 11. T 12. T 9. F 14. F 15. F 16. F 13. F 18. F 19. T 20. F 17. T 1 1 Fill in the Blanks 3. 4. 1. C 2. high 10 K 5. exothermic 6. K = K (RT) 7. backward 8. G°=–RT lnK PC 9. log K2 H T2 T1 10. zero 11. mol2 L–2 K1 2.303R T2 T1 12. temperature 13. Guldberg and Waage 14. decreases 15. backward 16. high 19. same amount of 20. large value of 17. K = K (RT)n 18. decreases PC Match the Column 1. A - (s), B - (p,q,s), C - (p,q,r), D - (r) 2. A - (p), B - (s), C - (q), D - (r) Assertion - Reason Questions 1. A 2. D 3. A 4. C 5. A 6. B 7. A 8. A Comprehension Based Questions Comprehension #1 : 1. B 2. C 3. A Comprehension #2 : 1. A 2. B
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE Reaction quotient and equillibrium constant 1 . The initial concentration or pressure of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the directions in which each system will shift to reach equilibrium. (a) 2NH3 (g) N2 (g) + 3H2 (g) K = 17 [NH ] = 0.20 M ; [N ] = 1.00 M ; [H ] = 1.00 M 32 2 (b) 2NH (g) N (g) + 3H (g) K = 6.8 × 104 atm2 3 2 2 p Initial pressure : NH = 3.0 atm ; N = 2.0 atm ; H = 1.0 atm 32 2 (c) 2SO (g) 2SO (g) + O (g) K = 0.230 atm 3 2 2 [SO ] = 0.00 M ; [SO ] = 1.00 M ; [O ] = 1.00 M 3 22 (d) 2SO (g) 2SO (g) + O (g) K = 16.5 atm 3 2 2 p Initial pressure : SO = 1.0 atm ; SO = 1.0 atm ; O = 1.0 atm 3 22 (e) 2NO (g) + Cl (g) 2NOCl (g) K = 4.6 × 104 2 [NO] = 1.00 M ; [Cl2] = 1.00 M ; [NOCl] = 0 M 2 . Among the solubility rules, the statement that all chlorides are soluble except Hg Cl , AgCl, PbCl , and 22 2 CuCl. (a) Write the expression for the equilibrium constant for the reaction represented by the equation AgCl (s) Ag+ (aq) + Cl– (aq) Is K greater than 1, less than 1, or about equal to 1? Explain your answer. (b) Write the expression for the equilibrium constant for the reaction represented by the equation Pb2+ (aq) + 2Cl– (aq) PbCl2 (s) Is K greater than 1, less than 1, or about equal to 1? Explain your answer. 3 . Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene. 3C H C6H6 22 Would this reaction be most useful commercially if K were about 0.01, about 1, or about 10? Explain your answer. 4 . For which of the following reactions will the equilibrium mixture contain an appreciable concentration of both reactants and products? (a) Cl (g) 2Cl (g) ; K = 6.4 × 10–39 2 c (b) Cl (g) + 2NO (g) 2NOCl (g) ; K = 3.7 × 108 2 c (c) Cl (g) + 2NO (g) 2NO Cl (g) ; K = 1.8 2 2 2 c 5. The value of K for the reaction 3 O (g) 2O (g) is 1.7 × 10–56 at 25°C. Do you expect pure air c 2 3 at 25°C to contain much O (ozone) when O and O are in equilibrium ? If the equilibrium concentration 3 23 of O in air at 25°C is 8 × 10–3 M, what is the equilibrium concentration of O . 23
6. At 1400 K, K = 2.5 × 10–3 for the reaction CH (g) + 2H S CS (g) + 4H (g). A 10.0 L reaction c 42 22 vessel at 1400 K contains 2.0 mol of CH , 3.0 mol of CS , 3.0 mol of H and 4.0 mol of H S. Is this reaction 4 22 2 mixture at equilibrium ? If not, in which direction does the reaction proceed to reach equilibrium ? 7 . An equilibrium mixture of N , H and NH at 700 K contains 0.036 M N and 0.15 M H . At this tem- 22 3 22 perature, K for the reaction N (g) + 3H (g) 2NH (g) is 0.29. What is the concentration of c2 2 3 NH ? 3 8 . The air pollutant NO is produced in automobile engines from the high temperature reaction N (g) + O (g) 2NO (g) ; K = 1.7 × 10–3 at 2300 K. If the initial concentrations of N and O 2 2 c 2 2 at 2300 K are both 1.40 M, what are the concentration of NO, N2 and O2 when the reaction mixture reaches equilibrium? 9. At a certain temperature, the reaction PCl (g) PCl (g) + Cl (g) has an equilibrium constant 5 3 2 K = 5.8 × 10–2. Calculate the equilibrium concentrations of PCl , PCl and Cl if only PCl is present initially, c 53 2 5 at a concentration of 0.160 M. 10. At 700 K, K= 0.140 for the reaction ClF (g) ClF (g) + F (g). Calculate the equilibrium partial p 3 2 pressure of ClF , ClF and F if only ClF is present initially ; at a partial pressure of 1.47 atm. 3 23 Homogeneous equilibria degree of dissociation, vapour density and equilibrium constant 1 1 . The degree of dissociation of N O into NO at 1.5 atmosphere and 40°C is 0.25. Calculate its K at 40°C. 24 2 p Also report degree of dissociation at 10 atmospheric pressure at same temperature. 12. At 46°C, K for the reaction N O (g) 2NO (g) is 0.667 atm. Compute the percent dissociation p 24 2 of N O at 46°C at a total pressure of 380 Torr. 24 1 3 . When 36.8 g N O (g) is introduced into a 1.0 - litre flask at 27°C. The following equilibrium reaction 24 occurs : N O (g) 2NO (g) ; K = 0.1642 atm. 24 2 p (a) Calculate K of the equilibrium reaction. c (b) What are the number of moles of N O and NO at equilibrium? 24 2 (c) What is the total gas pressure in the flask at equilibrium? (d) What is the percent dissociation of N O ? 24 1 4 . At some temperature and under a pressure of 4 atm, PCl is 10% dissociated. Calculate the pressure at 5 which PCl will be 20% dissociated, temperature remaining same. 5 1 5 . In a mixture of N and H in the ratio of 1 : 3 at 64 atmospheric pressure and 300°C, the percentage 22 of ammonia under equilibrium is 33.33 by volume. Calculate the equilibrium constant of the reaction using the equation. N (g) + 3H (g) 2NH (g) 2 2 3 16. The system NO 2 NO maintained in a closed vessel at 60°C & pressure of 5 atm has an average 24 2 (i.e. observed) molecular weight of 69, calculate K . At what pressure at the same temperature would the p observed molecular weight be (230/3) ? 1 7 . The vapour density of N2O4 at a certain temperature is 30. Calculate the percentage dissociation of N2O4 at this temperature. N O (g) 2NO (g). 24 2 18. In the esterification C H OH () + CH COOH () CH COOC H () + H O () an equimolar mixture 25 3 3 25 2 of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. Calculate the equilibrium constant.
Hetrogeneous equilibrium 19. Solid Ammonium carbamate dissociates as : NH COONH (s) 2NH (g) + CO (g). In a closed vessel 24 3 2 solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of NH at new equilibrium now equals the original total pressure. Calculate 3 the ratio of total pressure at new equilibrium to that of original total pressure. 2 0 . A sample of CaCO (s) is introduced into a sealed container of volume 0.821 litre & heated to 1000 K until 3 equilibrium is reached. The equilibrium constant for the reaction CaCO (s) CaO (s) + CO (g) is 3 2 4 × 10–2 atm at this temperature. Calculate the mass of CaO present at equilibrium. 2 1 . Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of CaCl , the amount 2 of the water taken up is governed by K = 6.4 × 1085 for the following reaction at room temperature, p CaCl (s) + 6H O (g) CaCl . 6H O (s). What is the equilibrium vapour pressure of water in a closed 22 22 vessel that contains CaCl (s) ? 2 2 2 . 20.0 grams of CaCO (s) were placed in a closed vessel, heated & maintained at 727°C under equilibrium 3 CaCO (s) CaO (s) + CO (g) and it is found that 75% of CaCO was decomposed. What is the 3 2 3 value of K ? The volume of the container was 15 litres. p Changes in concentration at equilibrium Le Chatelier's principle 2 3 . Suggest four ways in which the concentration of hydrazine, N H , could be increased in an equilibrium 24 described by the equation N2 (g) + 2H2 (g) N2H4 (g) H = 95 kJ 2 4 . How will an increase in temperature and increase in pressure affect each of the following equilibria? (a) 2NH (g) N (g) + 3H (g) H = 92 kJ 3 2 2 (b) N (g) + O (g) 2NO (g) H = 181 kJ 2 2 H = –285 kJ H = –176 kJ (c) 2O (g) 3 O (g) 32 (d) CaO (s) + CO (g) CaCO (s) 2 3 2 5 . (a) Water gas, a mixture of H and CO, is an important industrial fuel produced by the reaction of steam 2 with red-hot coke,essentially pure carbon. Write the expression for the equilibrium constant for the reversible reaction. C (s) + HO (g) CO (g) + H (g) H = 131.30 kJ 2 2 (b) Assume that equilibrium has been established and predict how the concentration of each reactant and product will differ at a new equilibrium if (1) more C is added. (2) H O is removed. (3) CO is added 2 (4) the pressure on the system is increased. (5) the temperature of the system is increased. 2 6 . Ammonia is weak base that reacts with water according to the equation NH (aq) + HO (l) NH + (aq) + OH– (aq) 3 2 4 Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water? (a) Addition of NaOH. (b) Addition of HCl. (c) Addition of NH Cl. 4 2 7 . Suggest two ways in which equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl–, Ag+ and NO –, in contact with solid AgCl. 3 Na+ (aq) + Cl– (aq) + Ag+ (aq) + NO – (aq) AgCl (s) + Na+ (aq) + NO – (aq) H = –65.9 kJ 3 3
Kinetics and equilibrium constant 2 8 . Consider a general single-step reaction of the type A + B C. Show that the equilibrium constant is equal to the ratio of the rate constant for the forward and reverse reaction, K = K /K . c fr 2 9 . Consider the reaction of chloromethane with OH– in aqueous solution CH Cl (aq) + OH– (aq) kf CH OH (aq) + Cl– (aq) 3 kr 3 At 25°C, the rate constant for the forward reaction is 6 × 10–6 M–1 s–1, and the equilibrium constant K c is 1 × 1016. Calculate the rate constant for the reverse reaction at 25°C. 3 0 . The progress of the reaction A nB with time, is presented in figure, Determine 0.5 (i) the value of n. 0.3 (ii) the equilibrium constant K. 0.1 (iii) the initial rate of conversion of A. 13 57 Time/Hour Temperature dependence of equi librium consta nt 31. Listed in the table are forward and reverse rate constants for the reaction 2NO (g) N (g) +O (g) 2 2 Temperature (K) k (M–1s–1) k (M–1s–1) 1400 f r 0.29 1.1 × 10–6 1500 1.3 1.4 × 10–5 Is the reaction endothermic or exothermic? 3 2 . Rate of disappearance of the reactant A at two different temperature is given by A B d[A ] = (2 × 10–2 S–1) [A] –4 × 10–3 S–1 [B] ; 300 K dt d[A ] = (4 × 10–2 S–1) [A] –16 × 10–4 S–1 [B] ; 400 K dt Calculate heat of reaction in the given temperature range. When equilibrium is set up. 3 3 . The K for reaction A + B C + D is 1.34 at 60°C and 6.64 at 100°C. Determine the free energy P change of this reaction at each temperature and H° for the reaction over this range of temperature? Equilibrium expressions and equilibrium constants 3 4 . If K = 7.5 × 10–9 at 1000 K for the reaction N (g) + O (g) 2NO (g), what is K at 1000 K forc c 22 the reaction 2NO (g) N (g) + O (g)? 22 3 5 . A sample of HI (9.30 × 10–3 mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I was 6.29 × 10–4 M. Calculate the value of K at 1000 K for the reaction 2c H (g) + I (g) 2HI (g). 22 3 6 . The vapour pressure of water at 25°C is 0.0313 atm. Calculate the values of Kp and Kc at 25°C for the equilibrium HO () H O (g). 2 2
3 7 . For each of the following equilibria, write the equilibrium constant expression for K . Where appropriate, c also write the equilibrium constant expression for K . p (a) Fe O (s) + 3CO (g) 2Fe () + 3 CO (g) 23 2 (b) 4Fe (s) + 3 O (g) 2Fe O (s) 2 23 (c) BaSO (s) BaO (s) + SO (g) 4 3 (d) BaSO (s) Ba2+ (aq) + SO 2– (aq) 4 4 General problems 3 8 . When 0.5 mol of N2O4 is placed in a 4.00 L reaction vessel and heated at 400 K, 79.3 % of the N2O4 decomposes to NO . 2 Calculate K and K at 400 K for the reaction N O (g) 2 NO (g). c p 24 2 39. At 100 K, then value of K for the reaction C(s) + HO (g) CO (g) + H (g) is 3.0 × 10–2. Calculate c 2 2 equilibrium concentrations of H O, CO and H in the reaction mixture obtained by heating 6.0 mole of 22 2 steam and an excess of solid carbon in a 5.0 L container. What is the molar composition of the equilibrium mixture? 4 0 . When 1.0 mol of PCl is introduced into a 5.0 L container at 500 K, 78.5 % of the PCl dissociates to 55 given an equilibrium mixture of PCl , PCl and Cl . 53 2 PCl (g) PCl (g) + Cl (g) 5 3 2 (a) Calculate the values of K and K . cp (b) If the initial concentrations in a particular mixture of reactants and products are [PCl ] = 0.5 M, 5 [PCl ] = 0.15 M, and [Cl ] = 0.6 M, in which direction does the reaction proceed to reach equilibrium? 32 What are the concentrations when the mixture reaches equilibrium? Thermodynamic and equilibrium constant 4 1 . -D-Glucose undergoes mutarotation to -D-Glucose in aqueous solution. If at 298 K there is 60% conversion. Calculate G° of the reaction. -D-Glucose -D-Glucose 4 2 . For the reaction at 298 K A (g) + B (g) C (g) + D (g) H° = – 29.8 kcal ; S° = – 0.1 kcal/K Calculate G° and K. 43. The equilibrium constant of the reaction 2 C H (g) CH (g) + CH (g) is found to fit the expression 36 24 48 nK = – 1.04 – 1088 T Calculate the standard reaction enthalpy and entropy at 400 K. 44. PCl dissociates according to the reaction PCl (g) PCl (g) + Cl (g). At 523 K, K =1.78 atm. Find 5 5 3 2 p the density of the equilibrium mixture at a total pressure of 1 atm. 4 5 . The following data for the equilibrium composition of the reaction 2 Na (g) Na (g) 2 at 1.013 MP pressure and 1482.53 K have been obtained. a mass % Na (monomer gas) = 71.3 mass % Na (dimer gas) = 28.7 2 Calculate the equilibrium constant K . p
46. A certain gas A polymerizes to a small extent at a given temperature & pressure, nA A. Show n that the gas obeys the approx equation PV = 1 (n 1)Kc where Kc = A n & V is the volume of the RT V n 1 A n container. Assume that initially one mole of A was taken in the container. 4 7 . When 1 mole of A (g) is introduced in a closed rigid 1 litre vessel maintained at constant temperature the following equilibria are established. A (g) B(g) + C(g) : K C (g) D(g) + B(g) : C1 K C2 The pressure at equilibrium is twice the initial pressure. Calculate the value of K C2 if [C]eq 1 K C1 [B]eq 5
CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1. (a) 25, shifts left, (b) 0.22, shifts right, (c) , shifts left, (d) 1, shifts right, (e) 0, shifts right 2. (a) K = [Ag+] [Cl–] is less than 1. AgCl is insoluble thus the concentration of ions are much less than 1 M (b) K = 1/[Pb2+] [Cl–]2 is greater than one because PbCl is insoluble and formation of the solid will reduce the 2 concentration of ions to a low level. 3. K about 10 4. c 5. ~ 9 × 10–32 mol/L 6. The reaction is not in equilibrium because Q > K . The reaction will proceed from right to left to reach equilibrium. cc 7. 5.9 × 10–3 M 8. [NO] = 0.056 M, [N ] = [O ] = 1.37 M 22 9. [PCl ] = [Cl ] = 0.071 M, [PCl ] = 0.089 M 10. P = P = 0.389 atm, P = 1.08 atm 32 5 CIF F2 ClF3 11.K = 0.4, ~ 0.1 12. 50% P 13. (a) 6.667 × 10–3mol L–1 ; (b) n (N O ) = 0.374 mol ; n (NO ) = 0.052 mol ; (c) 10.49 atm (d) 6.44% 24 2 14. 0.97 atm 15. K = 1.3 × 10–3 atm–2 P 16. K = 2.5 atm , P = 15 atm 17. 53.33% 18. K = 4 19. 31/27 p 20. 22.4 mg 21. P = 5 × 10–15atm 22. 0.821 atm H2O 23. add N , add H , increase the pressure, heat the reaction 22 24. (a) shift right, shift left, (b) shift right, no effect, (c) shift left, shift left, (d) shift left, shift right 25. (a) K = [CO][H ] / [H O] ; 22 (b) in each of the following cases the mass of carbon will change, but its concentration (activity) will not change. 1. [H O] no change, [CO] no change, [H ] no change ; 2. [H O] decrease, [CO] decrease, [H ] decrease ; 22 22 3. [H O] increase, [CO] increase, [H ] decrease ; 4. [H O] increase, [CO] decrease, [H ] decrease ; 22 22 5. [H O] decrease, [CO] increase, [H ] increase ; 22 26.b 27. Add NaCl or some other salt that produces Cl– in the solution. Cool the solution 28. k [A][B] = k [C] ; kf [C] kc 29. 6 × 10–22 30. (i) 2 ; (ii) 1.2 mol/L ; (iii) 0.1 moles/hr fr kr [A ][B] 32. 16.06 kJ 31. The reaction is exothermic 33. –810 J/mol ; –5872 J/mol and 41.3 kJ/mol 34. 1.3 × 108 35. 29.0 36. K = 0.0313 atm, K = 1.28 × 10–3 pc 37. (a) K = [CO2 ]3 , K = 3 PCO2 1 PCO 3 , (b) K 1 , K = (PO2 )3 (d) K = [Ba2+] [SO 2–] c = [O2 ]3 p ; (c) K = [SO ], K = P c4 c [CO]3 p c 3p SO3 38. K = 1.51, K = 49.6 39. [CO] = [H ] = 0.18 M ; [H O] = 1.02 M cp 22 40. (a) K = 0.573 and K = 23.5 ; (b) to the right, [PCl ] = 0.365 M ; [PCl ] = 0.285 M ; [Cl ] = 0.735 M cp 5 32 41. –1.005 kJ/mol 42. G° = 0 ; K = 1 43. H° = 9.04 kJ/mol ; S° = – 8.64 J / mol–1 K–1 44. 2.71 g/L 45. P = 0.843 M Pa, P = 0.170 M Pa, k = 0.293 46. To be proved 47. 4 Na Na2 P
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