2 2 . The reagent (s) for the following conversion : [IIT-07] Br ? H C C H is /are - Br (A) alcoholic KOH (B) Alcoholic KOH followed by NaNH2 (C) aquesous KOH followed by NaNH2 (D) Zn/CH3OH 2 3 . In the following groups : [IIT-97] (I) –OAc ; (II) –OMe ; (III) –OSO2Me ; (IV) OSO2CF3 the order of leaving group ability is - (A) I > II > III > IV (B) IV > III > I > II (C) III > II > I > IV (D) II > III > IV > I 2 4 . CH3CH2CHCl2 Aq. KOH ? [IIT 1991] 2 5 . Draw the stereochemical structure of the products in the following reaction. [IIT 1994] C2H5 [IIT 2005] CH3 C 2H5OH (aq.) neutral Br H OH/SDN2MSO CH3 CH3 2 6 . Give reasons : H3C Br (a) (i) C 2H5OH (aq.) acidic solution ; (ii) Br— CH3 FF (b) (i) O2N NaOH(aq ) F– (liberated) ; (ii) H3C NaOH(aq ) F– is not liberated CH3 CH2NO2 2 7 . An alkyl halide ,(A) of formula C6H11Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes (B) and (C) C6H10 .Both alkene on hydrogenation give methycyclopentane. Predict the structure of (A), (B) and (C). 2 8 . Which would be the major product in each of the following reactions ? CH3 F (b) (a) H3C C 2H5 OH NaOCH3 CH3 NO2 (c) O base Br
2 9 . The following compound on hydrolysis in aqueous acetone will give: [IIT 2005] CH3 H CH3 MeO NO2 NO2 H Cl CH3 NO2 CH3 H CH3 NO2 MeO (C)L and M (K) H OH CH3 CH3 H CH3 MeO (L) OH H CH3 CH3 H CH3 MeO (M) H CH3 OH It mainly gives (B) Only K (D) Only M (A) K and L 3 0 . Match the following: [IIT 2006] Column I (A) CH3–CHBr–CD3 on treatment with alc. KOH gives Column II CH2=CH-CD3 as a major product. (P) E1 reaction (B) Ph – CHBr - CH3 reacts faster than Ph-CHBr-CD3. (Q) E2 reaction (C) Ph-CD2-CH2Br on treatment with C2H5OD/C2H5O– (R) E1cb reaction gives Ph-CD=CH2 as the major product. (S) First order reaction (D) PhCH2CH2Br and PhCD2CH2Br react with same rate. [IIT 2008] 3 1 . The major product of the following reaction is H3C Br F dim ethPyhlSfoNrmaamide NO2 H3C SPh H3C SPh H3C Br H3C SPh F F SPh SPh (C) (A) (B) (D) NO2 NO2 NO2 NO2
3 2 . In the reaction OCH3 HBr the products are [IIT 2010] (A) Br OCH3 and H2 (B) Br and CH3Br (C) Br and CH3OH (D) OH and CH3Br 3 3 . The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is [IIT-2011] 3 4 . KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as - [IIT 2013] O Cl H3C–Cl Cl Cl PQ R S (A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q 3 5 . Match the chemical conversions in List-I with the appropritate reagents in List-II and select the correct answer using the code given below this lists - [IIT 2013] List-I List-II (P) Cl (1) (i) Hg(OAc)2 (ii) NaBH4 (Q) ONa OEt (2) NaOEt OH (R) (3) Et-Br (S) (4) (i) BH3 (ii) H2O2 /NaOH OH Codes : P QRS (A) 2 3 1 4 (B) 3 2 1 4 (C) 2 3 4 1 (D) 3 2 4 1
PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5 1. (D) 2. (B) 3. (A) 4.(A) 5. (B) 6. (D) 7.(A) 8. (B) 12. (A) 13. (D) 14. (D) 15. (C) 16. (D) 9. (A) 10. (C) 11. (A) 20. (B) 21. (A) 22. (B) 23. (B) 17. (D) 18. (C) 19. (B) 24. CH —CH —CHO 32 25. H C2H5 OH CH3 26. (a) (i) Products are CH3 —C—OC2H5 + HBr (aq.) ; Hence acidic. CH3 (ii) No reaction takes place, as aryl halide is inert to nucleophilic subsititution. (b) (i) —NO group facilitates the nucleophilic attack because it stabilised the anion formed when attack 2 of OH takes place. (ii) Does not show reaction. Cl CH3 CH3 CH2 27. (B) (B) (C) (A) CH3 OCH3 (c) 28. (a) (b) O H2C—C=CH—CH3 NO2 32. D 2 9 . A,C 30. D 31. C 33. A 34. B 35. A
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Tertiary butyl amine is a (A) 1° Amine (B) 2° Amine (C) 3° Amine (D) Quaternary salt 2 . The correct set of the products obtained in the following reactions RCN reduction (A) , RCN iCiiHH3M2OgBr (B), RNC hydrolysis (C), RNH2 HNO2 (D) The answer is : A B CD (A) 2°Amine Methyl ketone 1° Amine Alcohol (B) 1°Amine Methyl ketone 1° Amine Alcohol (C) 2°Amine Methyl ketone 2° Amine Acid (D) 2°Amine Methyl ketone 2° Amine Aldehyde 3 . Which one of the following tests can be used to identify primary amino group in a given organic compound (A) Iodoform test (B) Victor Meyer's test (C) Carbylamine reaction (D) Libermann's reaction CO 4 . CO NH KOH A C2H5Br B HHOH C + D C and D in the above sequence are (A) Benzoic acid + aniline (B) Phthalic acid + ethylamine (C) Phthalic acid + aniline (D) Benzoic acid + ethylamine 5 . Lowest boiling point will be of the compound (A) 2-propanamine (B) Ethylmethylamine (C) 1–propanamine (D) N, N–dimethylmethanamine 6 . How many primary amines can be formulated by C3H9N and how many 1° hydrogen are associated with carbon atoms of each compound (A) Two primary amines [3, 6] (B) One primary amine [3] (C) Three primary amines [3, 6, 6] (D) Two primary amines [5, 6] 7 . Identify B, X and R respectively in the following sequence of reactions C2H5MgBr ClCN A H3O B , CH3COCH3 I2 X Ag Y NaOH C6H5NH2 NHaNCOl 2 P CuCN Q 4H R (A) C2H5COOH, CHI3, C6H5CH2NH2 (B) C2H5COOH, CH3I, C6H5COOH (C) C2H5CH2NH2, CH3I, C6H5COOH (D) C2H5COOH, C2H5I, C6H5CONH2
HNO2 a bC6H5CHO 8. C2H5NH2 NOCl c dC6H5SO2Cl Which product is a Schiff's base :- (A) a (B) b (C) c (D) d 9 . A + CS2 + HgCl2 gives C2H5—N C S. Thus compound A is :- (A) C2H5NH2 (B) C2H5NHC2H5 (C) CH3—CH NOH (D) CH3CH2NO2 1 0 . The end–product in the reaction sequence would be : Ethyl amine HNO2 A PCl5 B NH3 C (A) Ethyl cyanide (B) Ethyl amine (C) Methyl amine (D) Acetamide 1 1 . In the reaction sequence :- A SnCl2 /HCl B NaNO02/ HCl C H2O/ C6H5OH, A, B and C are :- (A) Benzene, nitrobenzene, aniline (B) Nitrobenzene, aniline and azo–compound (C) Nitrobenzene, benzene, aniline (D) Benzene, amino compound, aniline 1 2 . Methylamine on treatment with chloroform and ethanolic caustic alkali gives foul smelling compound, the compound is (A) CH3NCO (B) CH3CNO (C) CH3CN (D) CH3NC 1 3 . Which of the following is used as a solvent in the Friedel–Crafts reaction :- (A) Toluene (B) Nitrobenzene (C) Benzene (D) Aniline 14. CH3CH2CONH2 N aOH A, Br2 Aqueous solution of A (A) Turns blue litmus to red (B) Turns red litmus to blue (C) Does not affect the litmus (D) Decolourise the litmus 1 5 . Hinsberg reagent is used to distinguish between (A) — CHO, C O (B) — CH2OH, CHOH, C OH (C) —O—, —OH (D) —NH2, —NH—, N 1 6 . Reactants of reaction – I are : CH3CONH2, KOH, Br2 Reactants of reaction–II are : CH3NH2, CHCl3, KOH The intermediate species of reaction–I and reaction–II are respectively (A) Carbonium ion, carbene (B) Alkyl isocyanate, carbene (C) Carbene, nitrene (D) Carbocation, carbanion 1 7 . Which of the following amine does not respond to carbylamine reaction :- (A) Ethylamine (B) (CH3)2NH (C) CH3NH2 (D) Phenylamine
1 8 . (a) CH3CONH2 + KOH + Br2 (b) CH3COOH + soda lime (c) CH3COOAg + Br2 All the three reactions show similarity in - (A) Descending of carbon atoms (B) Ascending of carbon atoms (C) Both the above (D) None of the above 1 9 . Which of the following reagents can convert benzenediazonium chloride into benzene ? (A) Water (B) Acid (C) Hypophosphorous acid (D) HCl 2 0 . Hofmann's method to separate amines in a mixture uses the reagent :- (A) Benzenesulphonyl chloride (B) Diethyl oxalate (C) Benzeneisocyanide (D) p–toluenesulphonic acid 2 1 . Which of the following compounds will dissolve in an alkali solution after it has undergone reaction with Hinsberg reagent ? (A) (C2H5)2NH (B) CH3 N C6H5 (C) CH3NH2 (D) C6H5NHC6H5 CH3 2 2 . The number of resonating structures of aniline is :- (A) 2 (B) 3 (C) 4 (D) 5 2 3 . How many isomeric amines with formula C7H9N contain a benzene ring ? (A) Two (B) Three (C) Four (D) Five 2 4 . Reaction of RCONH2 with a mixture of Br2 and KOH gives RNH2 as the main product. The intermediate involved in the reaction is :- O (B) R – NHBr O Br (D) R – C = N = O (A) R C NHBr (C) R C N Br 2 5 . The bromination of aniline in water produces :- (A) 2-bromoaniline (B) 4-bromoaniline (C) 2, 4, 6-tribromoaniline (D) 2, 6-dibromoaniline 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 B C B D D A B A B B D B B D B B A C B Que. 21 22 23 24 25 Ans. C D D A C
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . Carbylamine test is performed in alcoholic KOH by heating a mixture of :- (A) Chloroform and silver powder (B) Trihalogenated methane and a primary amine (C) An alkyl halie and a primary amine (D) An alkyl cyanide and a primary amine 2 . Acetamide is treated separately with following reagents. Which one of these would give methylamine ? (A) PCl5 (B) Sodalime (C) NaOH + Br2 (D) Hot concentrated H2SO4 3 . Maximum basic in gas phase is ? (A) NH3 (B) CH3CH2NH2 (C) (CH3CH2)NH (D) (CH3CH2)3N 4 . Benzenediazonium chloride on reaction with aniline in weakly basic medium gives (A) diphenyl ether (B) p-aminoazobenzene (C) Chlorobenzene (D) benzene 5 . Which of the following statements is correct ? (A) Replacement of halogen by NH2 in alkyl halide is a nucleophilic substitution reaction (B) Aryl halides show more reactivity as compared to alkyl halides in the replacements of halogen by the NH2 group (C) During the replacement of halogen by –NH2 group, ammonia is taken in large excess so as to avoid the formation of 2° and 3° amines (D) Tertiary alkyl halide generally produces alkene instead of the replacement of halogen by NH2 group 6 . Which of the following statements is correct ? (A) Primary amines show intermolecular hydrogen bonding (B) Secondary amines show intermolecular hydrogen bonding (C) Tertiary amines show intermolecular hydrogen bonding (D) Amines have lower boiling points as compared to those of alcohols and carboxylic acid of comparable molar masses. 7 . Which of the following statements is correct ? (A) Aliphatic amines are stronger bases than ammonia (B) Aromatic amines are stronger bases than ammonia (C) The alkyl group in alkyl ammonium ion more stabilizes the ion relative to the amine (D) The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine 8 . Which of the following orders regarding the basic strength of substituted aniline is correct ? (A) p–methylaniline > p–chloroaniline > p–aminoacetophenone (B) p–methylaniline > p–aminoacetophenone > p–chloroaniline (C) p–aminoacetophenone > p–methylaniline > p–chloroaniline (D) p–aminoacetophenone > p–chloroaniline > p–methylaniline
9 . Consider the following reaction, C6H5NH2 + CHCl3 + KOH (A) H / H2O (B) + (C) The compounds (B) and (C) are :- (A) C6H5COOH and NH3 respectively (B) C6H5NH2 and HCOOH respectively (C) C6H5NH2 and H2O respectively (D) None of these 1 0 . CH3CH2NH2 is soluble in :- (A) Dilute HCl (B) CuSO4 solution (C) AgNO3 (D) None of these 1 1 . The end product (Z) of the following reaction is :- N+2Cl– CuCN (X) H / H2O (Y) CNaaOO,H (Z) (A) A cyanide (B) A carboxylic acid (C) An amine (D) An arene 1 2 . The increasing order of basicity of RCN, RCH = NR and RNH2 is :- (A) RCN < RCH = NR < RH2N (B) RNH2 < RCN < RCH = NR (C) RCH = NR < RNH2 < RCN (D) RH2N < RCH = NR < RCN 1 3 . Amongst the following, the most basic compound is :- (A) Benzylamine (B) Aniline (C) Acetanilide (D) p–nitroaniline 1 4 . isocyanide test can be given by : (A) C2H5NH2 (B) C6H5NH2 (C) C6H5NHC2H5 (D) C6H5CH—NH2 C2H5 (D) II is an acceptable canonical structure 1 5 . In the following compounds O N N NN H HH (I) (II) (III) (IV) the order of basicity is (A) IV > I > III > II (B) III > I > IV > II (C) II > I > III > IV (D) I > III > II > IV 1 6 . A positive carbylamine test is not given by : (A) N, N-dimethylaniline (B) 2, 4-hyroxyazobenzene (C) N-methyl-o-methylaniline (D) p-methylbenzylamine 1 7 . Among the following, the strongest base is (A) C6H5NH2 (B) p-NO2C6H4NH2 (C) m-NO2C6H4NH2 (D) C6H5CH2NH2 1 8 . Phenyl cyanide on reduction with Na/C2H5OH yields (A) C6H5CH2NH2 (B) C6H5NHCH3 (C) NH2 (D) C6H5NH2 CH3
1 9 . Conisider the following ions (I) Me2N N+ N (II) O2N N+ N (III) H3C O N+ N (IV) H3C N+ N The reactiveities of these ions in azo-coupling reaction (under similar conditions) will be such that (A) (I) < (IV) < (II) < (III) (B) (I) < (III) < (IV) < (II) (C) (III) < (I) < (II) < (IV) < (II) (D) (III) < (I) < (IV) < (II) 2 0 . The strongest base among the following is H2N NH H2N NH2 H2N O H2N (A) C (B) C (C) C (D) CH OH H2N H2N H2N H2N BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A n s . B C D B A,C,D A,B,D A,C,D A B A ,B,C D A A A,B,D D Que. 16 17 18 19 20 A n s . A ,B,C D A B A
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE OR FALSE : 1 . Secondary amines show carbylamine reaction. 2 . The aqueous solution of ethylamine precipitates iron as ferric hydroxide when added to ferric chloride soluiton. 3 . Nitrogen is evolved when ethylamine is treated with nitrous acid. 4 . Primary, secondary and tertiary amines can be separated from their mixture by using ethyl oxalate. 5 . Tertiary amines react with Grignard reagents to form hydrocarbon. 6 . Primary amines are less soluble than tertiary amines. 7 . -Amino acids on heating with Ba(OH)2 shows decarboxylation to produce primary amines. FILL IN THE BLANKS : 1 . (CH3)2CHNH2 is a ................... amine. 2 . Methyl cyanide on reaction with sodium and C2H5OH forms ................... 3 . Acetamide on treatment with ................... forms ethyl amine. 4 . Tertiary amines do not react with acetyl chloride since they to not have ................... on nitrogen. 5 . The reaction RCN SnCl2 /HCl(A ) H2O RCHO + NH4Cl, is known as ................... 6 . nitroalkanes when reduced with zinc dust and ammonium chloride form ................... 7 . Primary amines on heating with ................... and ................... form alkul isocyanides. MATCH THE COLUMN 1 . Match the column I with column II. (p) Column-II Column-I (q) Secondary amine Hofmann's bromamide reaction ( A ) C6H5SO2Cl ( B ) Conversion of amide to amine ( C ) Conversion of primary amine ( r ) Hinsberg's reagent to isocyanide ( D ) Dimethylamine ( s ) Carbylamine reaction 2 . Match the column I with column II. (p) Column-II Column-I Quaternary salt ( A ) Tetraethyl ammonium iodide Tertiary amine ( B ) C–NH2 (q) Schmidt reaction Primary amine (C) RCOOH + N3H + Conc. H2SO4 (r) (D) R3N (s)
3 . Match list I with II and choose the correct answer from the codes given below :- Column-I Column-II ( A ) Aniline ( p ) Used in making azodyes ( B ) Nitrobenzene ( q ) Sulpha drug ( C ) Sulphanilamide ( r ) Solvent in the Friedel Crafts reaction ( D ) Trinitrotoluene ( s ) Used as explosive 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 : Alkyl isocyanide can be prepared by carbyl amine reaction. Because Statement-II : Ethyl amine when heated with chloroform in presence of alcoholic KOH, alkanamide is formed. 2 . Statement-I : CN– ion is an ambident nucleophile. Because Statement-II : Nucleophiles are electron rich species. 3 . Statement-I : Conversion of alkyl halides into alkyl cyanide or isocyanides is a nucleophilic substitution reaction. Because Statement-II : When an alkyl halide is treated with alcoholic solution of AgCN, alkyl cyanides are formed as major product. 4 . Statement-I : Nitrobenzene undergoes electrophilic substitution at m-position, while nucleophilic substitution occurs at o-and p-position. Because Statement-II : Nitro group in benzene ring significantly lowers the electron density at o- and p-positions, while at m-position, the electron density is only slightly lowered. 5 . Statement-I : Nitrobenzene is used as a solvent in Friedel-Craft's reaction. Because Statement-II : Fusion of p-chloronitrobenzene with KOH gives a low yield of a mixture of p-nitro phenoles. 6 . Statement-I : In Hoffmann bromide reaction, the amine formed has one carbon atom less than the parent 1° amide. Because Statement-II : N-methyl acetamide undergoes Hoffmann bromamide reaction. 7 . Statement-I : Nitrobenzene does not undergo Friedel-Craft's alkylation. Because Statement-II : Nitrobenzene is used as solvent in laboratory and industry. 8 . Statement-I : A reaction between a diazo salt and an aromatic amine or a phenol, giving an amino-azo or hydroxy-azo compound is called coupling reaction. Because Statement-II : Condensation of diazonium with phenol is carried out in weakly acidic medium.
9 . Statement-I : Aniline can be prepared by the reaction of chlorobenzene with NaNH2/Liq. NH3. Because Statement-II : NH2 ion is a stronger nucleophile. 1 0 . Statement-I : Carbylamine reaction involves the reaction between 1° amine and chloroform in basic medium. Because Statement-II : In carbylamine reaction, –NH2 group is converted into –NC group via reaction with carbene intermediate. COMPREHENSION BASED QUESTIONS : Comprehension # 1 The origin of acidity and basicity in organic compound is great interest and provides an extensive comparison. Among hydrocarbons % s character is taken in acconut while to decide the acidity, in simple aliphatic acids, more the number of alkyl groups, (+I effect) less is the acidity & more the (– I effect) alkyl groups larger the acidity and vice-versa in the case of simple aliphatic bases. Benzoic acid is more stronge than carboxylic acid as benzoate ion is stabilised more by resonance. Aromatic amines are less basic than aliphatic amine as the electron pair is less available in case of aromatic amines. The presence of solvent also plays a very important role and at times governs the order too. 1 . Amongst the following the most basic compound is - NH2 NH2 NO2 CHO (A) (B) (C) (D) NH2 NH2 NH2 2 . Among the following compounds the strongest acid is (A) HC CH (B) NH3 (C) C2H5NH2 (D) CH3NH2 3 . Which of the following orders regarding acid strength is correct ? ((A) HCOOH < CH3COOH < PhCOOH (B) HCOOH > PhCOOH > CH3COOH (C) HCOOH > CH3COOH > PhCOOH (D) CH3COOH > HCOOH > PhCOOH 4 . In the following compounds, the order of basicity NN O N CH3 (II) N CH3 (I) CH3 (IV) (III) (A) IV > I > III > II (B) III > I > IV > II (C) II > I > III > IV (D) I > III > II > IV 5 . Which of the following is most acidic NH3 NH3 NH2 (A) (B) (C) (D) CH3 NH3
Comprehension # 2 Diazonium salt formation and coupling reactions : When a reaction mixture of phenyl amine and nitrous acid is kept below 10°C, a diazonium salt is formed. This reaction is called diazotization reaction. NH2 + HNO2+ HCl +– NN Cl diazonium chloride The diazonium ion, – N , is rather unstable and decomposes readily to nitrogen. However, delocalization 2 of the diazonium from -bond electron over a benzene ring stabilizes phenyl diazonium sufficiently for it to form at low temperatures. The phenyl diazonium ion behaves as an electrophile, and will attack another arene molecule such as phenol. Electrophilic substitution takes place at the 4 position, producing 4-hydroxy phenyl azobenzene. The reaction is known as coupling reaction. N2 + OH N N OH + H+ The compound formed is an energetically stable, yellow azo dye (the azo group is –N = N–). The stability is due to extensive delocalisation of electrons via the nitrogen- nitrogen double bonds. 1 . The azo dye obtained on reacting 4-aminophenol with nitrous acid (in dilute hydrochloric acid) below 10°C and coupling the resulting diazonium salt with phenol is : (A) N (B) N N N OH (C) HO N (D) H2N N N N OH OH 2 . Benzene diazonium chloride on reaction with phenol in weakly basic medium gives : (A) diphenyl ether (B) p-hydroxy azobenzene (C) chlorobenzene (D) benzene 3 . The product HO HO which is a red azodye obtained on reacting benzene diazonium NN chloride with one of the following compounds : OH (A) OH (B) OH HO OH (C) (D) Comprehension # 3 Amines are derivatives of ammonia and are classified as 1°, 2°, and 3°. Primary and secondary (but not tertiary amines) form intermolecular hydrogen bonds and thus they boil at highter temperatures than expected. Like ammonia, all amines are basic, although they differ in their basic nature.
As amines are considered as derivatives of ammonia, quaternary ammonium salts are considered as derivatives of ammonium salts. Only the quaternary ammonium salts can shown optical activity. 1 . Which of the follwoing statement is correct ? (A) All classes of amines form hydrogen bonds with each other (B) Only primary and secondary amines form hydrogen bonds with water (C) All classes of amines can form hydrogen bonds with water (D) All amines are completely soluble in water 2 . When nitrogen is bounded to three different groups (A) The molecule is optically inactive (B) The molecule is tetrahedral (C) The molecule is not superimposable on its mirror image (D) The amine boils at nearly similar b.p as the 1°, and 2° amine of comparable molecular weight. 3 . Which of the following shows configurational isomerism ? (A) Diethyldimethylammonium iodide (B) Dimethylpropylamine (C) Methylallylphenylbenzyl ammonium bromide (D) None of these 4 . Methylethylpropyl amine is optically inactive because (i) It is not tetrahedral (ii) Its molecule is superimposable on its mirror image (iii) The enantiomers are rapidly interconverted (iv) The nitrogen is sp2 hybridized Which of the above statement(s) is (are) true ? (A) Only (i) (B) Only (iii) (C) (i) and (iii) (D) All the four MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 1. F 2. T 3. T 4. T 5. F 6. F 7. T Fill in the Blanks 1. Primary 2. ethyl amine 3. LiAlH4 4. replaceable hydrogen 5. stephen's reaction 6. hydroxyl amine 7. CHCl3 ,KOH or NaOH Match the Column 1. (A) r ; (B) q ; (C) s ; (D) p 2. (A) p ; B s ; (C) r ; (D) q 3. (A) p ; (B) r ; (C) q ; (D) s Assertion - Reason Questions 1. C 2. B 3. C 4. B 5. B 6. C 7. B 8. C 9. D 10. A 4. (D) Comprehension Based Questions 3. (B) 4. (B) 3. (B) Comprehension #1 : 1. (B) 2. (A) 3. (C) 5. (B) Comprehension #2 : 1. (C) 2. (B) Comprehension #3 : 1. (C) 2. (C)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Arrange the following in decreasing basic order with proper reasoning. O O RNH2, R–C–NH2, NH O 2 . Sulphanilic acid is not soluble in organic solvents. Explain ? 3 . What is the order of basicity of the following compounds ? (I) CH3NH2,(CH3)2NH,(CH3)3 N (in aprotic solvent) (II) C2H5NH2,(C2H5)2NH,(C2H5)3 N (in aprotic solvent) 4 . What is the order of basicity of the following compounds ? (I) CH3NH2,(CH3)2 NH,(CH3)3 N (in protic solvent) (II) C H NH ,(C H ) NH,(C H ) N (in protic solvent) 2 5 2 2 52 2 53 5 . Unlike other aromatic amines, why is the following amine stongly basic ? NO2 NMe2 NO2 6 . In this compound OH NH2 which site acts as an acid and which as a base ? 7 . Alkyl cyanides (CH CN) when treated with hydrogen in presence of Pt or with LiAlH produces same carbon 34 number compound. What is the formula of that compound ? 8 . Write the compound (A) and (B) formed in this NH2 NaNH2 (A) NaOCH3 (B) (major) Cl 9 . Identify the stronger base in each of the following pairs : O NH2 NH2 (I) CH3—C—NH2 or CH3NH2 (II) or 10. NH2 NaN0O–52/CHCl A wHa2rOm B ; what are A & B ?
CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) O 1. RNH2 > RCONH2 > NH O the lone pair of electrons on N is accommodated by sp3 hybrid atomic orbital in RNH2 and does not involve in resonance, whereas in RCONH2 and succinamide the lone pair of –NH2 is delocalised. 2 . Sulphanilic acid exist as Z witterion HO3S NH2 – + O3S NH2 Due to its ionic character it is insoluble in organic solvents. 3 . (I) (CH3)3N > (CH3)2NH >CH3NH2 ; (II) (C2H5)3N > (C2H5)2NH >C2H5NH2 4 . (I) (CH3)2NH > CH3NH2 > (CH3)3N ; (II) (C2H5)2NH > (C2H5)3N >C2H5NH2 5 . Due to the presence of bulky –NO2 groups on its ortho positions, the –NMe2 group goes outside the plane of resonance to avoid steric repulsion. The C–N bond rotates and hence the lone pair of N goes perpendicu- lar to the plane of benzene ring. As a result the resonance is stopped and hence the lone pair is readily available as a base. 6 . OH NH+2 ;—NH2 is a more basic site than OH ; O NH2 ; H–O is more acidic than H–N NH2 NH2 7 . CH3CH2 NH2 8. (A) H3CO (B) Benzyne NH2 N2 OH 9 . (I) CH NH ; (II) ;B: 10. A : 32
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . An organic compound (A) composed of C, H and O gives characteristic colour with ceric ammonium nitrate. Treatement of (A) with PCl5 gives (B), which reacts with KCN to form (C). The reduction of (C) with warm Na/C2H5OH produces (D).which on heating gives (E) with evolution of ammonia. Pyridine is obtained on treatment of (E) with nitrobenzene. Give structure of compounds (A) to (E) with proper reasoning. 2 . Compound (A) having M.F. C8H8O on treatment with NH2OH.HCl gives (B) and (C). (B) and (C) rearrange to give (D) and (E), respectively on treatment with acid. Compounds (B), (C), (D) and (E) are all isomers of molecular formula C8H9NO. When (D) is boiled with alcoholic KOH, an oil (F) C6H7N separated out. (F) reacts rapidly with CH3COCl to give back (D). On the other hand, (E) on boiling with alkali followed by acidification gives a white solid (G), C7H6O2. Identify the compounds (A) to (G). 3 . Two isomeric compounds (A) and (B) have C4H11N as molecular formula. Both on separately treating with HNO2 lose their N2 producing two isomeric alcohols (C) and (D) respectively of molecular formula C4H10O. (C) reacts with Lucas reagent immediately and units oxidation. (D) does not reacts with Lucas reagent in cold but can be easily oxidized. Complete methylation of either (A) or (B) is made which on decomposition does not produce 1-butene. Identify A to D. 4 . A mixture of two aromatic compounds A and B was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound A, when heated with alcoholic solution of KOH produced a compound C (C7H5N) associated with an unplesant odour. The alkaline aqueous layer on the other hand, when heated with chloroform and then acidified gave a mixture of two isomeric compounds D and E of molecular formula C7H6O2. Identify the compound A, B,C,D and E write their structures. 5 . Give structures of (A), (B), (X), (Y) in the given scheme of reactions, if reagents a, b, c, d, e and f are Br2/CH3COOH, NaNO2/H2SO4, CuBr, HNO3/H2SO4, Br2/Fe and Sn/HCl respectively. (A) a (B) (i) b (Z) f ClH3N+ Br Br (ii) c Br (X) d (Y) e 6 . Explain the mechanism of the following change : O OH HONO CH2NH2 CH3 7 . CH3CH2N—CH2CH2CH3 A + B + C. Identify A, B (both alkene) and C. O 8 . Write isomeric amines of the formula C3H9N. ANSWER KEY 9 . Identify A, B, C and D. Acid A N3H, conc. H2SO4 B CHCl3 + KOH C H3O B + HCOOH reduction HNO2 CH3 D CH3—C—NHCH3 CH3
1 0 . The two amines shown differ by a factor of about 40, 000 in their ionisation constants. Which is stronger base ? Explain : tetrahydroisoquinoline H (I) tetrahydroquinoline (II) BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 1. CH2OH CH2Cl CH2CN CH2CH2NH2 CH2 –– CH2 H2C H2C H2C H2C H2C NH CH2OH CH2Cl CH2CN CH2CH2NH2 CH2 –– CH2 (A) (B) (C) (D) (E) C6H5 H3C OH H3C CO CN CN H3C C6H5 C6H5 OH 2 . (A) (B) (C) Acetophenone Acetophenone oxime Z–Acetophenone oxime OO C6H5–NH–C–CH3 C6H5–C–NH–CH3 C6H5–NH2 C6H5–COOH (D) (E) (F) (G) Acetanilide N–methyl benzamide Aniline Benzoic acid 3. A = (CH3)3 C – NH2, B = (CH3)2CH–CH2–NH2 C = (CH3)3C–OH, D = (CH3)2CHCH2OH 4 . A = C6H5NH2 (aniline), B = C6H5OH(phenol), C = C6H5NC = (phenyl isocyanide). OH D= CHO and E = HO CHO salicylaldehyde p-hydroxybenzaldehyde NH2 NH2 Br Br NO2 Br Br Br Br Br 5 . (A) = (B) = (X) = (Y) = (Z) = NO2 NO2 NO2 Br Br Br
H OH O OH HONO OH OH 1,2 alkyl shift –H 6. CH2NH2 CH2N2 –N2 CH2 carbocation conjugate acid of ketone Thus, it involves pinacol-type ring expansion via the carbocation. 7 . A : CH2=CH2 B. CH3CH=CH2 C. CH3—N—OH or CH3—N—OH CH2CH3 CH2CH2CH3 8 . I. CH3CH2CH2NH2 II. CH3CHCH3 III. CH3CH2NHCH3 IV. (CH3)3 N NH2 CH3 CH3 CH3 CH3 9 . A : CH3—C—COOH ; (B) CH3—C—NH2 ; (C) CH3—C—NC ; (D) CH3—C—OH, CH3—C—CH2 CH3 CH3 CH3 CH3 CH3 10. tetrahydroisoquinoline is a stronger base. In tetrahydroquinoline (II), lone-pair on nitrogen is used in delocalisation of -electrons of benzene ring hence, basicity of II is decreased. I resembles aliphatic amine II resembles aniline.
EXERCISE–05[A] PREVIOUS YEARS QUESTIONS 1 . Among the following compounds , which will react with acetone to give a product containing >C=N– ? (A) C6H5NH2 (B) (CH3)3N [IIT-90] [IIT-90] (C) C6H5NHC6H5 (D) C6H5NHNH2 2 . The compound that willl react most readily with NaOH to form methanol is (A) (CH3)4N+I– (B) CH3OCH3 (C) (CH3)3S+I– (D) (CH3)3CCl 3 . Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below : [IIT-93] + NH3 NH3 + (I) (II) (A) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions (B) II is not an acceptable canonical structure because it is non-aromatic (C) II is not an acceptable canonical structure because the nitrogen has ten valence electrons (D) Both (A) & (C) 4 . Allyl isocyanide contains and bonds are - [IIT-95] (A) 9 and 3 (B) 9 and 9 (C) 3 and 4 (D) 5 and 7 5. Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and conc. H2SO4. In the nitrating mixture, nitric acid acts as - [IIT-97] (A) base (B) acid (C) reducing agent (D) catalyst 6 . Among following statements on the nitration of aromatic compounds, the false one is - [IIT-97] (A) the rate of nitration of benzene is almost the same as that of hexadeuterobenzene (B) the rate of nitration of toluene is greater than that of benzene (C) the rate of nitration of benzene is greater than that of hexadeuterobenzene (D) nitration is an electrophilic substitution reaction. 7. p-Chloroaniline and anilinium hydrochloride can not be distinguished by [IIT-98] (A) Sandmeyer reaction (B) NaHCO3 (C) AgNO3 (D) Carbylamine test 8. Read the following statement and explanation and answer as per the option given below : [IIT-98] Statement I : Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide Because Statement II : Cyanide (CN–) is a strong nucleophile (A) If both assertion and reason are correct, and reason is the correct explanation of the assertion (B) If both assertion and reason are correct, but reason is not correct explanation of the assertion (C) If assertion is correct but reason is incorrect (D) If assertion is incorrect but reason is correct 9. Benzenediazonium chloride on reaction with phenol in weakly basic medium gives [IIT-98] (A) Diphenyl ether (B) p–Hydroxyazobenzene (C) Chlorobenzene (D) Benzene
1 0 . A positive carbylamine test is given by [IIT-99] (A) N, N-Dimethylaniline (B) 2, 4-Dimethylaniline (C) N-Methyl-o-methylaniline (D) o-methylbenzylamine 1 1 . Among the following, the strongest base is : [IIT-2000] (A) C6H5NH2 (B) p–NO2–C6H4NH2 (C) m–NO2–C6H4NH2 (D) C6H5CH2NH2 1 2 . Read the following statement and explanation and answer as per the option given below : [IIT-01] Statement I : In strongly acidic solutions, aniline becomes less reactive towards electrophilic reagents. Beacuse Statement II : The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance. (A) If both assertion and reason are correct, and reason is the correct explanation of the assertion (B) If both assertion and reason are correct, but reason is not correct explanation of the assertion (C) If assertion is correct but reason is incorrect (D) If assertion is incorrect but reason is correct 1 3 . The correct order of basicities of the following compounds is - [IIT-01] NH ; (II) CH3CH2NH2 ; (III) (CH3)2NH O (I) CH3–C (B) I > III > II > IV ; (IV) CH3–C–NH2 NH2 (C) III > I > II > IV (D) I > II > III > IV (A) II > I > III > IV 1 4 . When benzamide is treated with POCl3, the product formed is : [IIT-04] (D) Benzylamine (A) Benzonitrile (B) Aniline (C) Chlorobenzene 1 5 . CH3NH2 + CHCl3 KOH Product, Product is- [IIT-06] (B) •• (C) CH3–NH–CH3 (D) CH3–C N (A) CH3–N C•• CH3–N•• C : 1 6 . In the following reaction, O conc. HNO3 X, the structure of the major product 'X' is - N conc. H2SO4 H O2N O [IIT-07] O (B) N (A) NO2 H N H O (D) O (C) N O2N N H H NO2 1 7 . An aromatic compound contains 69.4% carbon and 5.8% hydrogen. A sample of 0.303 g of this com- pound was analysed for nitrogen by Kjeldahl's method. The ammonia evolved was absorbed in 50ml of 0.05 M sulphuric acid. The excess of acid required 25 ml of 0.1 M sodium hydroxide for neutralization. Determine the molecular formula of the compound if its molecular weight is 121. Draw two possible structures for this compound. [IIT-82]
1 8 . Compound ‘X’ containing chlorine, on treatment with NH3 gives a solid ‘Y’ which is free from chlorine. (Y) analysed as C = 49.31%, H = 9.59% and N = 19.18% and reacts with Br2 and caustic soda to give a basic compound (Z), (Z) reacts with HNO2 to give ethanol. Suggest structure for (X), (Y) and (Z). [IIT-88] 1 9 . Show with equations how the following compounds are prepared (equations need not be balanced) : (i) 4-nitroaniline to 1, 2,3-tribromobenzene. [IIT-90] (ii) Aniline Benzylamine (in 3 steps) [IIT-2000] NO2 NO2 [IIT-04] (iii) Convert to in not more than four steps. OH 2 0 . Complete the following with appropriate structures : [IIT-92] —CONH2 P2O5(A ) H,H2O 2 1 . The aqueous solution of a nitrogen and chlorine containing compound (A) is acidic to litmus. (A) on treatement with aqueous NaOH gives a compound (B) containing nitrogen, but not chlorine. Compound (B) on treatment with C6H5SO2Cl in presence of NaOH gives an insoluble product (C), C13H13NO2S. Give structure of (A) and (B). [IIT-93] 2 2 . Complete the following with appropriate structures : [IIT-95] 2,4-Dinitroaniline (I)NaNO(2II)aanndisHolCel at5C C [IIT-99] 2 3 . Complete the following reaction with appropriate reagents : OI —NO2 —N—C—CH3 1 2 3 4 H I 2 4 . Compound (A), C5H11NO is not soluble in cold alkaline or acidic solutions. When (A) is refluxed in NaOH solution, a gas (B) is evolved and salt (C) is formed. Acetyl chloride reacts with (B) to give (D), (C4H9NO). (B) reacts with HNO2 to give a yellow oil (E). Give structures of (A) to (E) with reason. [IIT-99] 25. Identify (A) to (D) in the following series of reactions : [IIT - 04] CH2Cl KDMCNF(A ) C2CH65HO5NCaH,CO2,Hhe5aOt H(B ) Hh3eOat(C ) (ii()iC)SHO3CNlH22(D ) 2 6 . C5H13N (optically active) aq.NaNNO22/ HCl Y (Ter tiary alcohol) + Some other products [IIT-05] (i) Identify (X) and (Y) (ii) Is (Y) optically active ? (iii) Give structure (s) of intermediate (s), if any, in the formation of (Y) from (X).
2 7 . Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II . Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I Column II [IIT 2008] +- (P) sodium fusion extract of the compound gives (A) H2N – NH3Cl Prussian blue colour with FeSO4 +- NH3I (B) HO (Q) gives positive FeCl3 test COOH (C) HO +- (R) gives white precipitate with AgNO3 NH3Cl (S) reacts with aldehydes to form the corresponding hydrazone derivative (D) O2N +- NH – NH3Br NO2 2 8 . Match each of the compound in Column I with its characteristic reaction(s) in Column II. Column I Column II [IIT 2009] (A) CH3CH2CH2CN (P) Reduction with Pd–C / H2 (B) CH3CH2OCOCH3 (Q) Reduction with SnCl2 / HCl (C) CH3 – CH = CH – CH2OH (R) Development of foul smell on treatment with chloroform and alcoholic KOH (D) CH3CH2CH2CH2NH2 (S) Reduction with diisobutylaluminium hydride (DIBAL – H) (T) Alkaline hydrolysis 2 9 . The major product of the following reaction is [IIT 2011] O C (i) KOH CH2Cl (ii) Br NH C O O O C C (B) N CH2Cl (A) N–CH2 Br C C O O O O C C (D) N (C) N O CH2Cl O–CH2– Br
EXERCISE–05[B] PREVIOUS YEARS QUESTIONS 1 . Reaction - [AIEEE-2002] Primary amine + CHCl3 + KOH product, here product will be - (1) Cyanide (2) Isocyanide (3) Amine (4) Alcohol 2 . The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is- [AIEEE-2004] (1) Fe4[Fe(CN)6]3 (2) Na3[Fe(CN)6] (3) Fe(CN)3 (4) Na4[Fe(CN)5)NOS] 3 . Which one of the following methods is neither meant for the synthesis nor for separation of amines- [AIEEE-2005] (1) Hofmann method (2) Hinsberg method (3) Curtius reaction (4) Wurtz reaction 4 . In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH (A) + (B) + 3H2O, the compounds (A) and (B) are respectively - [AIEEE-2007] (1) C2H5CN and 3KCl (2) CH3CH2 CONH2 and 3KCl (3) C2H5NC and K2CO3 (4) C2H5NC and 3KCl 5 . In the chemical reactions, [AIEEE-2010] NH2 NaNO2 A HBF4 B HCl, 278 K the compounds 'A' and 'B' respectively are :- (2) Nitrobenzene and fluorobenzene (1) Nitrobenzene and chlorobenzene (4) Benzene diazonium chloride and fluorobenzene (3) Phenol and benzene 6 . In the chemical reactions NH2 [AIEEE-2011] HNCal,N2O782K A CuCN B, the compounds A and B respectively are : (1) Fluorobenzene and phenol (2) Benzene diazonium chloride and benzonitrile (3) Nitrobenzene and chlorobenzene (4) Phenol and bromobenzene 7 . 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): [ A I E E E - 2 0 1 3 ] CH2Br (2) C2H5 CH2Br CH2Br Br (1) (3) (4) CH3 CH3 CH3 8 . 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 :- [AIEEE-2013] (1) CH3COOH (2) CH3CH2CH2COOH (3) CH3–CH–COOH (4) CH3CH2COOH CH3
PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5[A] 1. (D) 2. (C) 3. (C) 4.(A) 5. (A) 6. (C) 7.(C) 8. (D) 9. (B) 10. (B,D) 11. (D) 12. (A) 13. (B) 14. (A) 15. (A) 16. (B) 1 7 . Calculation of % nitrogen 50 ml of 0.05 M H2SO4 = 50 ml of 0.1 NH2SO4 ( Normality of H2SO4) = 2 × molarity) Excess of acid requires 25 ml of 0.1 M or 0.1 N NaOH ( Normality of NaOH = molarity of NaOH) 25 ml of 0.1 N NaOH 25 ml of 0.1 N H2SO4 vol. of 0.1 NH2SO4 used for the neutralisation of NH3 = 50 – 25 = 25 ml Now we know that, % of nitrogen = 1.4 Normality of acid Vol.of acid Wt.of compound 18. = 1.4 0.1 25 11.55% ; Hence % of oxygen = 100 –(69.4 + 5.8 + 11.55) = 13.25 19. (i) 0.303 Empirical formula of (Y) is C H NO 37 CH3CH2NH2 CH3.CH2COCl + NH3 2 CH3CH2CONH2 NH2 NH2 N2Br Br Br Br Br Br Br Br CHuBBrr Br2 NaNO2/ HBr NO2 NO2 NO2 NO2 4-nitroaniline Br Br Br Br rZendu/cHtioCnl Br Br Br Br NaNO2 Br H3 P O2 / Diazotisation N2+Cl 1,2,3-tribromobenzene NH2 NH2 N2 CN CH2NH2 (ii) NaNO2 (HCl) CuCN SnHCl NO2 NO2 NO2 NO2 NO2 (iii) 1H0N0OC3 NH4SH H0 O5NCO H2O 20. NO2 NH2 OH N2 CONH2 CN COOH P2O5 H, H2O 21. Aminhydrochloride Amine . HCl NaOH Amine +H2O +NaCl (A) (B) A = [C6H5NH2CH3] Cl– and B = C6H5NHCH3
NH2 22. NO2 NO2 —OCH3 (ii()iA)NnaisNoOle2,C/6HHC5lO,5CHC3 O2N— —N=N— (C) NO2 2,4-dinitroaniline 23. O O —NH2 24. —NH—C—CH3 (1c)ocnocn.cH.H2SNOO43 O2N— —NH—C—CH3 (2)H2O /H O2N— 25. ((3H)ICCll) O2N— I I —NH2 ( 4)( i()iNi)HaN3OPO2 2HCl O2N— I I (A) = CH3CH2CON(CH3)2, B = (CH3)2 NH, C = CH3CH2COONa, D = CH3CON(CH3)2 and E = (CH3)2N – N = O CH2Cl CH2CN CN COOH DKCMNF C2CH65HO5NCaH,CO2,Hhe5aOt H C=CHC6H5 C=CHC6H5 (A) H3 O ,h eat (C) (B) ( ii()iC) SHO3CNlH22 CONHCH3 C=CHC6H5 (D) NH2 OH 26. (i) (CH3)2CH—CH—CH3 aq.NHaCNlO2 (CH3)2C—CH2CH3+NaCl+N2+H2O (X) optically active (Y) 3°alcohol (ii) [Y], a 3° alcohol is optically inactive (iii) Formation of [Y] from [X] NH2 N2 (CH3)2CH—CHCH3 aq.NHaCNlO2 (CH3)2CH—CHCH3 (CH3)2C—CHCH3 rearrangement (X) H 2°carbocation OH2 OH (CH3)2C—CH2CH3 H2O (CH3)2C—CH2CH3 H (CH3)2C—CH2CH3 3°carbocation (Y) 2 7 . (A) R,S ;(B) P,Q; (C) P,Q,R ;(D) P,S 2 8 . (A) P,Q,S,T (B) P,S,T (C) P (D) R 29. A
PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5[B] 1. 2 2. 1 3. 4 4. 4 5. 4 6. 2 7. 4 8. 4 1 . Carbylamine Test Na 3 . Wurtz reaction R × dry ether R – R + NaX 4 . Isocyanide test (carbylamine test). Pungent smelling isocyanide is formed as a major product. NH2 +– N+ 2BF4– F N2Cl 5 . NaNO2 HBF4 HCl, 278K Aniline Benzenediazonium Chloride (B) (A) Sandmeyer's Reaction +– CN N2Cl NH2 NaNO2 CuCN / + N2 HCl, 278K 6. Aniline Benzenediazonium Chloride (B) (A) CH2–Br 7 . alc Ag NO3 Write ppt of AgBr. CH3 [O] P2O5 / O COOH COOH C O C O OO O 8. CH3CH2 – C – OH CH3CH2 – C – ONH4 –H2O CH3CH2 – C – NH2 KOH Br2 CH3 – CH2 – NH2
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The element having no neutron in the nucleus of its atom is - (A) hydrogen (B) nitrogen (C) helium (D) boron 2 . The particles present in the nucleus of an atom are - (A) the proton and the electron (B) the electron and the neutron (C) the proton and the neutron (D) none of these 3 . The fraction of volume occupied by the nucleus with respect to the total volume of an atom is - (A) 10–15 (B) 10–5 (C) 10–30 (D) 10–10 4 . Which of the following is iso-electronic with neon - (A) O2– (B) F+ (C) Mg (D) Na 5. The approximate size of the nucleus of 64 N i is - 28 (A) 3 fm (B) 4 fm (C) 5 fm (D) 2 fm 6 . Which is true about an electron - (A) rest mass of electron is 9.1 × 10–28 g (B) mass of electron increases with the increase in velocity (C) molar mass of electron is 5.48 × 10–4 g/mole (D) e/m of electron is 1.7 × 108 coulomb/g 7. An isotone of 7 6 G e is - 3 2 (A) 77 G e (B) 77 A s (C) 77 S e (D) 78 Se 32 33 34 34 8 . When alpha particles are sent through a thin metal foil, most of them go straight through the foil because - (A) alpha particles are much heavier than electrons (B) alpha particles are positively charged (C) most part of the atom is empty space (D) alpha particles move with high speed 9 . Many elements have nonintegral atomic masses because - (A) they have isotopes (B) their isotopes have non-integral masses (C) their isotopes have different masses (D) the constituents, neutrons, protons and electrons combine to give fractional masses 1 0 . The MRI (magnetic resonance imaging) body scanners used in hospitals operate with 400 MHz radio frequency energy. The wavelength corresponding to this radio frequency is - (A) 0.75 m (B) 0.75 cm (C) 1.5 m (D) 2 cm 1 1 . Photon of which light has maximum energy - (A) Red (B) Blue (C) Violet (D) Green 1 2 . The value of Planck's constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 8 × 1015 sec–1 - (A) 5 × 10–18 (B) 4 × 101 (C) 3 × 107 (D) 2 × 10–25 1 3 . Bohr's theory is not applicable to - (A) He (B) Li2+ (C) He2+ (D) the H-atom
1 4 . What is likely to be principal quantum number for a circular orbit of diameter 20 nm of the hydrogen atom if we assume Bohr orbit be the same as that represented by the principal quantum number ? (A) 10 (B) 14 (C) 12 (D) 16 1 5 . Which is the correct relationship - (A) E of H = 1/2 E of He+ = 1/3 E of Li2+ = 1/4 E of Be3+ 12 3 4 (B) E (H) = E (He+) = E (Li2+) = E (Be3+) 12 3 4 (C) E (H) = 2E (He+) = 3E (Li2+) = 4E (Be3+) 12 3 4 (D) No relation 1 6 . If the value of E = – 78.4 kcal/mole, the order of the orbit in hydrogen atom is - (A) 4 (B) 3 (C) 2 (D) 1 1 7 . If velocity of an electron in 1st orbit of H atom is V, what will be the velocity of 3rd orbit of Li+2 - (A) V (B) V/3 (C) 3 V (D) 9 V 1 8 . In a certain electronic transition in the hydrogen atoms from an initial state (1) to a final state (2), the difference in the orbit radius (r – r ) is 24 times the first Bohr radius. Identify the transition - 12 (A) 5 1 (B) 25 1 (C) 8 3 (D) 7 5 1 9 . Match the following - (i) –6.04 eV (a) Energy of ground state of He+ (b) Potential energy of I orbit of H-atom (ii) –27.2 eV (c) Kinetic energy of II excited state of He+ (iii) 8.7 × 10–18 J (d) Ionisation potential of He+ (iv) –54.4 eV (A) A – (i), B – (ii), C – (iii), D – (iv) (B) A – (iv), B – (iii), C – (ii), D – (i) (C) A – (iv), B – (ii), C – (i), D – (iii) (D) A – (ii), B – (iii), C – (i), D – (iv) 2 0 . The energy of hydrogen atom in its ground state is –13.6 eV. The energy of the level corresponding to n=5- (A) –0.54 eV (B) –5.40 eV (C) –0.85 eV (D) –2.72 eV 2 1 . Total no. of lines in Lyman series of H spectrum will be - (where n = no. of orbits) (A) n (B) n – 1 (C) n – 2 (D) n (n + 1) (D) Na 2 2 . The spectrum of He+ is expected to be similar to that of - (A) Li+ (B) He (C) H 2 3 . What possibly can the ratio be of the de Broglie wavelengths for two electrons having the same initial energy and accelerated through 50 volts and 200 volts ? (A) 3 : 10 (B) 10 : 3 (C) 1 : 2 (D) 2 : 1 2 4 . The uncertainty in the momentum of an electron is 1.0 × 10–5 kg m s–1. The uncertainty of its position will be (h = 6.626 × 10–34 kg m2 s–1) - (A) 1.05 × 10–28 m (B) 1.05 × 10–26 m (C) 5.27 × 10–30 m (D) 5.25 × 10–28 m 2 5 . An -particle is accelerated through a potential difference of V volts from rest. The de-Broglie's wavelength associated with it is - (A) 150 A (B) 0.286 A (C) 0.101 A (D) 0.983 A V V V V 2 6 . The orbital with zero angular momentum is - (A) s (B) p (C) d (D) f
2 7 . Which of the following is electronic configuration of Cu2+ (Z = 29) - (A) [Ar]4s1 3d8 (B) [Ar]4s2 3d10 4p1 (C) [Ar]4s1 3d10 (D) [Ar] 3d9 (D) 3d34s0 2 8 . The electronic configuration of the Mn4+ ion is - (A) 3d44s0 (B) 3d24s1 (C) 3d14s2 2 9 . Which of the following ions has the maximum number of unpaired d-electrons - (A) Zn2+ (B) Fe2+ (C) Ni3+ (D) Cu+ 3 0 . The total spin resulting from a d7 configuration is - (A) 1 (B) 2 (C) 5/2 (D) 3/2 3 1 . Given K L M N 2 8 11 2 The number of electrons present in = 2 is - (A) 3 (B) 6 (C) 5 (D) 4 3 2 . The configuration 1s2 2s2 2p5 3s1 shows the - (A) ground state of the fluorine atom (B) excited state of the fluorine atom (C) excited state of the neon atom (D) excited state of O – ion 2 3 3 . The value and m for the last electron in the Cl – ion are - (A) 1 and 2 (B) 2 and +1 (C) 3 and –1 (D) 1 and –1 3 4 . In which transition, one quantum of energy is emitted - (A) n = 4 n = 2 (B) n = 3 n = 1 (C) n = 4 n = 1 (D) n = 2 n = 1 3 5 . Choose the correct relation on the basis of Bohr's theory - 1 Z2 (A) velocity of electron n (B) frequency of revolution n3 (C) radius of orbit n2Z Z3 (D) force on electron n 4 3 6 . The magnitude of the spin angular momentum of an electron is given by - (A) S = s(s 1) h (B) S = s h (C) S = 3 h (D) S = ± 1 h 2 2 2 2 2 2 3 7 . The change in orbital angular momentum corresponding to an electron transition inside a hydrogen atom can be - h h h h (A) (B) (C) (D) 4 2 8 3 8 . In which of these options do both constituents of the pair have the same magnetic moment - (A) Zn2+ and Cu+ (B) Co2+ and Ni2+ (C) Mn4+ and Co2+ (D) Mg2+ and Sc+ 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 C A A C A,B,C,D B,D A,C A,C A C B A,C B B Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. C A A C A B C D C C A D D B D Que. 31 32 33 34 35 36 37 38 An s . A C D A,B,C,D A,B,D A,C B,C A,C
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . The maximum energy is present in any electron at :- (A) Nucleus (B) Ground state (C) First excited state (D) Infinite distance from the nucleus 2 . Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon :- (A) 3s (B) 2p (C) 2s (D) 1s 3 . The third line in Balmer series corresponds to an electronic transition between which Bohr's orbits in hydrogen :- (A) 5 3 (B) 5 2 (C) 4 3 (D) 4 2 4 . Correct set of four quantum numbers for valence electron of rubidium (Z = 37) is :- 1 1 1 1 (A) 5, 0, 0, + (B) 5, 1, 0, + (C) 5, 1, 1, + (D) 6, 0, 0, + 2 2 2 2 5 . The orbital diagram in which the Aufbau's principle is violated is :- 2s 2px 2py 2pz 2s 2px 2py 2pz 2s 2px 2py 2pz (D) 2s 2px 2py 2pz (A) (B) (C) 6 . The total number of neutrons in dipositive zinc ions with mass number 70 is :- (A) 34 (B) 40 (C) 36 (D) 38 7 . Which of the following sets of quantum numbers represent an impossible arrangement :- nl m m s (A) 3 2 –2 1 2 (B) 4 0 0 1 2 (C) 3 2 –3 1 2 (D) 5 3 0 1 2 8 . The explanation for the presence of three unpaired electrons in the nitrogen atom can be given by :- (A) Pauli's exclusions principle (B) Hund's rule (C) Aufbau's principle (D) Uncertainty principle 9 . The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d5 4s1. This represents its :- (A) Excited state (B) Ground state (C) Cationic form (D) None 1 0 . Which of the following has maximum number of unpaired electron (atomic number of Fe 26) :- (A) Fe (B) Fe (II) (C) Fe (III) (D) Fe (IV) 1 1 . Which quantum number is not related with Schrodinger equation :- (A) Principal (B) Azimuthal (C) Magnetic (D) Spin 1 2 . If is the threshold wavelength for photoelectric emission, wavelength of light falling on the surface of 0 metal, and m mass of electron, then de Broglie wavelength of emitted electron is :- 1 1 1 1 h(0 ) 2 h(0 ) 2 h( 0 ) 2 h0 2 2mc(0 2mc0 ) 2 m c0 ) 2mc (A) ) (B) (C) (D)
1 3 . It is known that atom contain protons, neutrons and electrons. If the mass of neutron is assumed to half of its original value where as that of proton is assumed to be twice of its original value then the atomic mass of 14C will be :- 6 (A) same (B) 25 % more (C) 14.28 % more (D) 28.5 % less 1 4 . Give the correct order of initials T (true) or F (false) for following statements :- (I) If an ion has 2 electrons in K shell, 8 electrons in L shell and 6 electrons in M shell, then number of S electrons present in that element is 6. (II) The maximum number of electrons in a subshell is given by 2n2 (III) If electron has magnetic quantam number –1, then it cannot be present in s-orbital. (IV) Only one radial node is present in 3p orbital. (A) TTFF (B) FFTF (C) TFTT (D) FFTF 1 5 . The shortest wavelength of He+ in Balmer series is x, then longest wavelength in the Paschene series of Li+2 is :- 36x 16x 9x 5x (A) (B) 7 (C) (D) 5 5 9 1 6 . An electron in a hydrogen atom in its ground state absorbs energy equal to ionisation energy of Li+2. The wavelength of the emitted electron is :- (A) 3.32 × 10–10 m (B) 1.17 Å (C) 2.32 × 10–9 nm (D) 3.33 pm 1 7 . In compound FeCl the orbital angular momentum of last electron in its cation & magnetic moment (in Bohr 2 Magneton) of this compound are :- (A) ( 6 ), 35 (B) ( 6 ), 24 (C) 0, 35 (D) none of these 1 8 . An electron, a proton and an alpha particle have kinetic energy of 16 E, 4E and E respectively. What is the qualitative order of their de Broglie wavelengths :- (A) > = (B) = > ep pe (C) > > (D) < >> p pe e 1 9 . Question : Is the specie paramagnetic ? STATE-1 : The atomic number of specie is 29. STATE-2 : The charge on the specie is +1. (A) Statements (1) alone is sufficient but statement (2) is not sufficient (B) Statement (2) alone is sufficient but statement (1) is not sufficient (C) Both statement together are sufficient but neither statement alone is sufficient (D) Statement (1) & (2) together are not sufficient 2 0 . Given H for the process Li (g) Li+3 (g) + 3e– is 19800 kJ/mole & IE for Li is 520 then IE & IE of 1 21 Li+ are respectively (approx value) :- (A) 11775, 7505 (B) 19280, 520 (C) 11775, 19280 (D) Data insufficient 2 1 . The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :- (A) 2.5 : 1 (B) 3.5 : 1 (C) 4.5 : 1 (D) 5.5 : 1
2 2 . Which of the following statement is INCORRECT. e (A) ratio for canal rays is maximum for hydrogen ion. m e (B) ratio for cathode rays is independent of the gas taken. m (C) The nature of canal rays is dependent on the electrode material. e E2 ,when the cathode rays go undeflected under the (D) The ratio for electron is expressed as m 2B2V influence of electric field (E), magnetic field (B) and V is potential difference applied across electrodes. 2 3 . The quantum numbers of four electrons (e1 to e4) are given below :- n m s e1 3 0 0 +1/2 e2 4 0 1 1/2 e3 3 2 2 –1/2 e 4 3 1 –1 1/2 The correct order of decreasing energy of these electrons is : (A) e4 > e3 > e2 > e1 (B) e2 > e3 > e4 > e1 (C) e3 > e2 > e4 > e1 (D) none 2 4 . If radius of second stationary orbit (in Bohr's atom) is R. Then radius of third orbit will be :- (A) R/3 (B) 9R (C) R/9 (D) 2.25 R 2 5 . The wavelength associated with a gold weighing 200 g and moving at a speed of 5 m/h is of the order :- (A) 10–10 m (B) 10–20 m (C) 10–30 m (D) 10–40 m 2 6 . If the nitrogen atom had electronic configuration 1 s7, it would have energy lower that of normal ground state configuration 1s2 2s2 2p3, because the electrons would be closer to the nucleus. Yet 1 s7 is not observed because it violates :- (A) Heisenberg uncertainty principle (B) Hunds rule (C) Pauli's exclusion principle (D) Bohr postulate of stationary orbits 2 7 . From the following observations predict the type of orbital : Observation 1 : x y plane acts as nodal plane Observation 2 : The angular function of the orbital intersect the three axis at origin only. Observation 3 : R2(r) v/s r curve is obtained for the orbital is R2(r) r (A) 5p (B) 6d (C) 6dx2 – y2 (D) 6 d z xy yz 2 8 . Question : Is the orbital of hydrogen atom 3p ? x STATE 1 : The radial function of the orbital is R(r) = 1 (4 – ) e–/2, = r 2 9 6 a 3 / 2 0 STATE 2: The orbital has 1 radial node & 0 angular node. (A) Statement (1) alone is sufficient. (B) Statement (2) alone is sufficient (C) Both together is sufficient (D) Neither is sufficient
2 9 . Consider the following nuclear reactions involving X & Y. X Y + 4 He 2 Y O18 + H1 81 If both neutrons as well as protons in both the sides are conserved in nuclear reaction then identify period number of X & moles of neutrons in 4.6 g of X (A) 3, 2.4 N (B) 3, 2.4 (C) 2, 4.6 (D) 3, 0.2 N A A 3 0 . Electromagnetic radiations having = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic Energy.... (A) 0, no emission will occur (B) 2.18 × 106 m/s (C) 2.18 2 × 106 m/s (D) 8.72 × 106 m/s 3 1 . If in Bohr's model, for unielectronic atom, time period of revolution is represented as T where n repre- n,z sents shell no. and z represents atomic number then the value of T : T will be :- 1,2 2,1 (A) 8 : 1 (B) 1 : 8 (C) 1 : 1 (D) None of these 3 2 . Column I & Column II contain data on Schrondinger Wave-Mechanical model, where symbols have their usual meanings. Match the columns :- Column I Column II (Type of orbital) r (p) 4s (A) r r2 4r2 (q) 5p (B) x (C) (, ) = K (independent of ) (r) 3s (D) atleast one angular node is present (s) 6d xy 3 3 . Which orbital is non-directional :- (A) s (B) p (C) d (D) All 3 4 . A hydrogen - like atom has ground state binding energy 122.4 eV. Then : (A) its atomic number is 3 (B) an electron of 90 eV can excite it to a higher state (C) an 80 eV electron cannot excite it to a higher state (D) an electron of 8.2 eV and a photon of 91.8 eV are emitted when a 100 eV electron interacts with it 3 5 . Uncertainty in position is twice the uncertainty in momentum Uncertainty in velocity is :- h 1h 1 h (A) (B) (C) (D) 2m 2m 4 3 6 . For which orbital angular probability distribution is maximum at an angle of 45° to the axial direction :- (A) dx2 – y2 (B) dz2 (C) d (D) P xy x
3 7 . Which orbit would be the first to have 'g' subshell :- (A) 3rd (B) 4th (C) 5th (D) 6th 3 8 . The decreasing order of energy of the 3d, 4s, 3p, 3s orbitals is :- (A) 3d > 3s > 4s > 3p (B) 3s > 4s > 3p > 3d (C) 3d > 4s > 3p > 3s (D) 4s > 3d > 3s > 3p 3 9 . If n and are respectively the principle and azimuthal quantum numbers, then the expression for calculating the total number of electrons in any orbit is :- n n 1 n 1 n 1 (A) 2(2 1) (B) 2(2 1) (C) 2(2 1) (D) 2(2 1) 1 1 0 0 4 0 . If wavelength is equal to the distance travelled by the electron in one second, then :- h h h (D) h (A) (B) (C) p m p m 4 1 . According to Schrodinger model nature of electron in an atom is as :- (A) Particles only (B) Wave only (C) Both simultaneously (D) Sometimes waves and sometimes particle 4 2 . Which describes orbital :- (A) (B) 2 (C) |2| (D) All 4 3 . In order to have the same wavelength for the electron (mass m ) and the neutron (mass m ) their velocities en should be in the ratio (electron velocity/neutron velocity) :- (A) m /m (B) m × m (C) m /m (D) one ne ne en 4 4 . The quantum numbers +1/2 and –1/2 for the electron spin represent :- (A) Rotation of the electron in clockwise and anticlockwise direction respectively. (B) Rotation of the electron in anticlockwise and clockwise direction respectively. (C) Magnetic moment of the electron pointing up and down respectively. (D) Two quantum mechanical spin states which have no classifical analogue. 4 5 . Which is true about :- (A) represents the probability of finding an electron around the nucleus (B) represent the amplitude of the electron wave (C) Both A and B (D) None of these 4 6 . Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Broglie wavelength of the electron as :- (A) (0.529) n (B) n (C) (13.6) (D) n 4 7 . A particle X moving with a certain velocity has a debroglie wave length of 1A°. If particle Y has a mass of 25% that of X and velocity 75% that of X, debroglies wave length of Y will be :- (A) 3 A° (B) 5.33 A° (C) 6.88 A° (D) 48 A° 4 8 . What are the values of the orbital angular momentum of an electron in the orbitals 1s, 3s, 3d and 2p :- (A) 0, 0, 6, 2 (B) 1, 1, 4, 2 (C) 0, 1, 6, 3 (D) 0, 0, 20, 6 4 9 . If m = magnetic quantum number and = azimuthal quantum number then :- (A) m = + 2 (B) m = 22 + 1 (C) = m 1 (D) = 2m + 1 2
5 0 . The number of unpaired electrons in Mn4+ (Z = 25) is :- (A) Four (B) Two (C) Five (D) Three 5 1 . After np orbitals are filled, the next orbital filled will be :- (A) (n + 1) s (B) (n + 2) p (C) (n + 1) d (D) (n + 2) s 5 2 . The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :- (A) Fe2+ (B) Ni2+ (C) Mn2+ (D) Co3+ 5 3 . In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit of n = 1 to the period of the revolution of the electron in the orbit n = 2 is :- (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 1 : 8 5 4 . Let be the frequency of the series limit of the Lyman series, be the frequency of the first line of the 12 lyman series, and be the frequency of the series limit of the Balmer series :- 3 (A) – = (B) – = (C) = 1/2 ( – ) (D) + = 123 213 3 13 123 5 5 . The energies of energy levels A, B and C for a given atom are in the sequence E < E < E . If the radiations ABC of wavelengths , and are emitted due to the atomic transitions C to B, B to A and C to A respectively 12 3 then which of the following relations is correct :- (A) + + = 0 (B) = + 2 (C) = + (D) = 12 123 312 312 3 1 2 5 6 . The wavelengths of photons emitted by electron transition between two similar levels in H and He+ are 1 and respectively. Then :- 2 (A) = (B) = 2 (C) = /2 (D) = /4 21 21 21 21 5 7 . If first ionization potential of an atom is 16 V, then the first excitation potential will be :- (A) 10.2 V (B) 12 V (C) 14 V (D) 16 V 5 8 . In which transition minimum energy is emitted :- (A) 1 (B) 2 1 (C) 3 2 (D) n (n – 1) (n 4) 5 9 . No. of visible lines when an electron returns from 5th orbit to ground state in H spectrum :- (A) 5 (B) 4 (C) 3 (D) 10 BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. D D B A B B C B B C D A C C B Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. B B A C A B C C D C C D B B C Que. 31 32(A) (B) (C) (D) 33 34 35 36 37 38 39 40 41 42 Ans. D p p,q,s p,r q,s A A,C,D C C C C D D B B Que. 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 Ans. A D B D B A C D A B D A D D B Que. 58 59 Ans. D C
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . The electron density in xy plane of 3dx2 y2 orbital is zero. 2 . 3d6 configuration is more stable than 3d5. 3 . The potential energy of an electron in an orbit is twice in magnitude as compared to its kinetic energy. 4 . The increasing order for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle () is n < < p < e. 5 . The orbital 3dxy has no probability of finding electron along x and y-axis. FILL IN THE BLANKS 1 . Nitrogen has an atomic number of 7 and oxygen has an atomic number of 8. The total number of electrons in the nitrate ion ( NO – ) is ......................... . 3 2 . h/ is the angular momentum of the electron in the ......................... orbit of He+. 3 . An emission spectrum has electromagnetic radiation of definite ......................... . 4 . The maximum number of electrons that can be accommodated in all the orbitals for which = 3 is ......................... . 5 . The values of n1 and n2 in the Paschen spectral series of hydrogen atom are ......................... and ......................... respectively. MATCH THE COLUMN Column-II 1. Column-I (p) Line spectrum in visible region (A) Aufbau principle (q) Orientation of an electron in an orbital (B) de broglie (r) Photon (C) Angular momentum (s) = h/mv (D) Hund's rule (t) Electronic configuration (E) Balmer series (u) mvr (F) Planck's law 2. Column-I Column-II (A) Cathode rays (p) Helium nuclei (B) dumb-bell (q) Uncertainty principle (C) Alpha particles (r) Electromagnetic radiation (D) Moseley (s) p-orbital (E) Heisenberg (t) Atomic number (F) X-ray (u) Electrons 3 . Frequency = f, Time period = T, Energy of nth orbit = E , radius of nth orbit = r , Atomic number = Z, nn Orbit number = n : Column-I Column-II (A) f (p) n3 (B) T (q) Z2 (C) E 1 n (r) n2 (s) Z 1 (D) rn
4. Column-I Column-II (A) Lyman series (B) Balmer series (p) maximum number of spectral line observed = 6 (C) In a sample 5 2 (q) maximum number of spectral line observed = 2 (r) 2nd line has wave number 8R (D) In a single isolated H-atom for 3 1 transition 9 (s) 2nd line has wave number 3R 4 (t) total number of spectral line is 10 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 : Nodal plane of p atomic orbital is yz plane. x Because Statement-II : In p atomic orbital electron density is zero in the yz plane. x 2 . Statement-I : No two electrons in an atom can have the same values of four quantum numbers. Because Statement-II : No two electrons in an atom can be simultaneously in the same shell, same subshell, same orbitals and have same spin. 3 . Statement-I : p-orbital has dumb-bell shape. Because Statement-II : Electrons present in p-orbital can have one of three values for 'm', i.e. 0, +1, –1 4 . Statement-I : The ground state configuration of Cr is 3d5 4s1. Because Statement-II : A set of exactly half filled orbitals containing parallel spin arrangement provide extra stability. 5 . Statement-I : Mass numbers of most of the elements are fractional. Because Statement-II : Mass numbers are obtained by comparing with the mass number of carbon taken as 12. 6 . Statement-I : Limiting line in the balmer series has a wavelength of 36.4 m. Because Statement-II : Limiting lines is obtained for a jump of electron from n = to n = 2 for Balmer series. 7 . Statement-I : The electronic configuration of nitrogen atom is represented as : not as Because Statement-II : The configuration of ground state of an atom is the one which has the greatest multiplicity.
8 . Statement-I : The configuration of B atom cannot be 1s2 2s3. Because Statement-II : Hund's rule demands that the configuration should display maximum multiplicity. 9 . Statement-I : 2p orbitals do not have spherical nodes. Because Statement-II : The number of spherical nodes in p-orbitals is given by (n – 2). 1 0 . Statement-I : In Rutherford's gold foil experiment, very few particles are deflected back. Because Statement-II : Nucleus present inside the atom is heavy. 1 1 . Statement-I : Each electron in an atom has two spin quantum numbers. Because Statement-II : Spin quantum numbers are obtained by solving Schrodinger wave equation. 1 2 . Statement-I : There are two spherical nodes in 3s-orbital. Because Statement-II : There is no angular node in 3s-orbital. COMPREHENSION BASED QUESTIONS Comprehension # 1 Read the following rules and answer the questions at the end of it. Electrons in various suborbits of an orbit are filled in increasing order to their energies. Pairing of electrons in various orbitals of a suborbit takes place only after each orbital is half-filled. No two electrons in an atom can have the same set of quantum number. 1 . Cr (Z = 24), Mn+ (Z = 25), Fe2+ (Z = 26) and Co3+ (Z = 27) are isoelectronic each having 24 electrons. Thus, (A) all have configurations as [Ar] 4s1 3d5 (B) Cr and Mn+ have configurations as [Ar] 4s1 3d5 while Fe2+ and Co3+ have configurations as [Ar]3d5. (C) all have configurations as [Ar] 3d6 (D) all have configurations as [Ar] 4s2 3d6 2 . A compound of vanadium has a magnetic moment of 1.73 BM. Electronic configuration of the vanadium ion in the compound is : (A) [Ar] 4s0 3d1 (B) [Ar] 4s2 3d3 (C) [Ar] 4s1 3d0 (D) [Ar] 4s0 3d5 3 . Which of these ions are expected to be paramagnetic and coloured in aqueous solution ? (A) Fe3+, Ti3+, Co3+ (B) Cu+, Ti4+, Sc3+ (C) Fe3+, Ni2+, V5+ (D) Cu+, Cu2+, Fe2+ 4 . While writing the following electronic configuration of Fe some rules have been violated : I : Aufbau rule, II : Hund's rule III : Pauli's exclusion principle Ar 3d 4s (A) I, II (B) II, III (C) I, III (D) I, II, III
5 . How many elements would be in the second period of the periodic table if the spin quantum number 11 (m ) could have the value of ,0, ? s 22 (A) 8 (B) 10 (C) 12 (D) 18 6 . The sub-shell that arises after f sub-shell is called g sub-shell. (A) it contains 18 electrons and 9 orbitals (B) it corresponds to = 4 and first occurs in 5th energy level (C) a g-orbital can have maximum of two electrons (D) all the above statements are true. MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 3. T 4. T 5. T 5. 3, (4, 5, 6,....) 1. F 2. F 6. A Fill in the Blanks 12. B 6. (D) 1. 32 2. 2nd 3. frequency or wavelength 4. 14 Match the Column 1. (A) t; (B) s ; (C) u ; (D) q ; (E) p; (F) r 2. (A) u ; (B) s ; (C) p ; (D) t ; (E) q ; (F) r 3. (A) q ; (B) p ; (C) q,r ; (D) r,s 4. (A) r ; (B) s; (C) p; (D) q Assertion - Reason Questions 1. A 2. A 3. B 4. A 5. E 10. B 11. E 7. A 8. B 9. A Comprehension Based Questions Comprehension #1 : 1. (B) 2. (A) 3. (A) 4. (D) 5. (C)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . How long would it take a radio wave of frequency 6 × 103 sec–1 to travel from mars to the earth, a distance of 8 × 107 km ? 2 . The energy levels of hypothetical one electron atom are shown below. 0 eV —— n = –0.50 eV —— n = 5 –1.45 eV —— n = 4 –3.08 eV —— n = 3 –5.3 eV —— n = 2 –15.6 eV —— n = 1 (a) Find the ionisation potential of atom? (b) Find the short wavelength limit of the series terminating at n = 2 ? (c) Find the wave no. of photon emitted for the transition made by the electron from third orbit to first orbit ? (d) Find the minimum energy that an electron will have after interacting with this atom in the ground state, if the initial kinetic energy of the electron is (i) 6eV (ii) 11 eV ? 3 . Suppose 10–17 J of light energy is needed by the interior of the human eye to see an object. How many photons of green light ( = 550 nm) are needed to generate this minimum amount of energy? 4 . Find the number of photons of radiation of frequency 5 × 1013 s–1 that must be absorbed in order to melt one g ice when the latent heat of fusion of ice is 330 J/g. 5 . The eyes of certain member of the reptile family pass a single visual signal to the brain when the visual receptors are struck by photons of wavelength 850 nm. If a total energy of 3.15 × 10–14 J is required to trip the signal, what is the minimum number of photons that must strike the receptor? 6 . The wavelength of a certain line in the Paschen series is 1093.6 nm. What is the value of nhigh for this line [RH = 1.0973 × 10+7 m–1]. 7 . Wavelength of the Balmer H line (first line) is 6565 Å. Calculate the wavelength of H (second line). 8 . Calculate the Rydberg constant R if He+ ions are known to have the wavelength difference between the first (of the longest wavelength) lines of Balmer and Lyman series equal to 133.7 nm. 9 . Calculate the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum. 1 0 . A photon having = 854 Å causes the ionization of a nitrogen atom. Give the I.E. per mole of nitrogen in KJ. 1 1 . Calculate energy of electron which is moving in the orbit that has its radius, Sixteen times the radius of first Bohr orbit for H-atom. 12. The electron energy in hydrogen atom is given by En = 21.7 1012 the energy required to remove an e– completely from n = 2 orbit. What is n2 ergs. Calculate of light that can be the largest wavelength in cm used to cause this transition.
1 3 . Calculate the wavelength in angstrom of photon that is emitted when an e– in Bohr orbit n = 2 returns to the orbit n = 1. The ionization potential of the ground state of hydrogen atom is 2.17 × 10–11 erg/atom. 1 4 . The velocity of e– in a certain Bohr orbit of the hydrogen atom bears the ratio 1 : 275 to the velocity of light. What is the quantum no. \"n\" of the orbit and the wave no. of the radiation emitted for the transition form the quantum state (n + 1) to the ground state. 1 5 . A doubly ionised lithium atom is hydrogen like with atomic number Z = 3. Find the wavelength of the radiation required to excite the electron in Li2+ from the first to the third Bohr orbit. 1 6 . Estimate the difference in energy between I and II Bohr Orbit for a hydrogen atom. At what minimum At. no. a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with = 3.0 × 10–8 m? Which hydrogen like species does this At. no. correspond to: 1 7 . 1.8 g atoms of hydrogen are excited to radiations. The study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atoms in 2n d energy level and the rest in ground state. If I.P. of H is 21.7 × 10–12 erg. Calculate. (i) No. of atoms present in III & II energy level. (ii) Total energy evolved when all the atoms return to ground state. 1 8 . One mole He+ ions are excited. Spectral analysis showed existence of 50% ions in 3rd orbit, 25% in 2nd and rest in ground state. Calculate total energy evolved when all the ions return to the ground state. 1 9 . The energy of an excited H-atom is –3.4 eV. Calculate angular momentum of e–. 2 0 . The vapours of Hg absorb some electrons accelerated by a potential difference of 4.5 volt as a result of it light is emitted. If the full energy of single incident e– is supposed to be converted into light emitted by single Hg atom, find the wave no. of the light. 2 1 . The hydrogen atom in the ground state is excited by means of monochromatic radiation of wavelength x A0. The resulting spectrum consists of 15 different lines. Calculate the value of x. 2 2 . If the average life time of an excited state of H atom is of order 10–8 sec, estimate how many orbits an e– makes when it is in the state n = 2 and before it suffers a transition to n = 1 state. 2 3 . Calculate the frequency of e– in the first Bohr orbit in a H-atom. 2 4 . A single electron orbits around a stationary nucleus of charge +Ze where Z is a constant from the nucleus and e is the magnitude of the electric charge. The hydrogen like species required 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find - (i) the value of Z give the hydrogen like species formed. (ii) the kinetic energy and potential energy of the electron in the first Bohr orbit. 2 5 . A stationary He+ ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H atom in ground state. What is the velocity of photoelectron ? 2 6 . To what series does the spectral lines of atomic hydrogen belong if its wave number is equal to the difference between the wave number of the following two lines of the Balmer series 486.1 and 410.2 nm. What is the wavelength of this ?
2 7 . A particle of charge equal to that of an electron and mass 208 times the mass of the electron moves in a circular orbit around a nucleus of charge +3e. Assuming that the Bohr model of the atom is applicable to this system, (a) derive an expression for the radius of the nth bohr orbit, (b) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom, and (c) find the wavelength of the radiation emitted when the revolving particle jumps from the third orbit to the first. 2 8 . A neutrons breaks into a proton and an electron. This decay of neutron is accompanied by release of energy. Assuming that 50% of the energy is produced in the form of electromagnetic radiation, what will be the frequency of radiation produced. Will this photon be sufficient to cause ionization of Aluminium. In case it is able to do so what will be the energy of the electron ejected from the Aluminum atom. IE1 of Al = 577 kJ/mol. 2 9 . Calculate the threshold frequency of metal if the binding energy is 180.69 kJ mol–1 of electron. 3 0 . Calculate the binding energy per mole when threshold wavelength of photon is 240 nm. 3 1 . A metal was irradiated by light of frequency 3.2 × 1015 s–1. The photoelectron produced had its KE, 2 times the KE of the photoelectron which was produced when the same metal was irradiated with a light of frequency 2.0 × 1015 s–1. What is work function ? 3 2 . U.V. light of wavelength 800 A° & 700 A° falls on hydrogen atoms in their ground state & liberates electrons with kinetic energy 1.8 eV and 4 eV respectively. Calculate planck's constant. 3 3 . A potential difference of 20 kV is applied across an X-ray tube. Find the minimum wavelength of X-ray generated. 3 4 . The K.E. of an electron emitted from tungsten surface is 3.06 eV. What voltage would be required to bring the electron to rest. 35. What is de-Broglie wavelength of a He-atom in a container at room t e mp er a t u re. U s e U 8kT avg. m 3 6 . Through what potential difference must an electron pass to have a wavelength of 500 A°. 3 7 . A proton is accelerated to one tenth of the velocity of light. If its velocity can be measured with a precision ± 1%. What must be its uncertainty in position ? 3 8 . To what effective potential a proton beam be subjected to give its protons a wavelength of 1 × 10–10m. 3 9 . Calculate the number of exchange pairs of electrons present in configuration of Cu according to Aufbau principle considering 3d orbitals. 4 0 . He atom can be excited to 1s1 2p1 by = 58.44 nm. If lowest excited state for He lies 4857 cm–1 below the above. Calculate the energy for the lower excitation state. 4 1 . A certain dye absorbs 4530 A° and fluoresence at 5080 A° these being wavelengths of maximum absorp- tion that under given conditions 47% of the absorbed energy is emitted. Calculate the ratio of the no. of quanta emitted to the number absorbed. 4 2 . The reaction between H and Br to form HBr in presence of light is initiated by the photo decomposition 22 of Br2 into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of Br2 is 192 kJ/mole. What is the longest wavelength of the photon that would initiate the reaction? 4 3 . The quantum yield for decomposition of HI is 0.2. In an experiment 0.01 moles of HI are decomposed. Find the number of photons absorbed.
4 4 . Calculate the wavelength of the radiation that would cause photo dissociation of chlorine molecule if the Cl – Cl bond energy is 243 kJ/mol. 4 5 . The dissociation energy of H2 is 430.53 kJ/mol. If H2 is exposed to radiant energy of wavelength 253.7 nm, what % of radiant energy will be converted into K.E ? 4 6 . Iodine molecule dissociates into atoms after absorbing light of 4500 A0 If one quantum of radiation is absorbed by each molecule, calculate the K.E. of iodine atoms. (Bond energy of I2 = 240 kJ/mol) 4 7 . X-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm attributed to the K line of an impurity element. The K lines of a copper (Z = 29) and molybdenum (Z = 42) have wavelength 15.42 nm and 7.12 nm respectively. Using Moseley's law, 1/2 = a (Z – b). Calculate the atomic number of the impurity element. 4 8 . What is de-Broglie wavelength associated with an e– accelerated through P.D. = 100 kV ? 4 9 . Calculate the de-broglie wavelength associated with motion of earth (mass 6 × 1024 kg) orbiting around the sun at a speed of 3 × 106 m/s. 5 0 . A base ball of mass 200 g is moving with velocity 30 × 102 cm/s. If we can locate the base ball with error equal in magnitude to the of the light used (5000 Å), how will the uncertainty in momentum compared with the total momentum of base ball ? 5 1 . An electron has a speed of 40 m/s, accurate up 99.99 %. What is the uncertainty in locating position? CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1. 2.66 × 102sec 2. (a) 15.6 eV (b) 233.9 nm, (c) 1.008 × 107 m–1 (d) (i) electron will not interact (ii) 0.7 eV 3. 28 photons 4. 1022 5. 1.35 × 105 6. 6 7. 4863 A° 8. 1.096 × 107 m–1 9. 1.827 × 105 J/mol 10. 1403 kJ/mol 11. –1.36 × 10–19 Joules 12. 5.425 × 10–12 ergs, 3.7 × 10–5 cm 13. 1220 A° 14. 2 ; 9.75 × 104 cm–1 15. 113.74 A° 16. 10.2 eV, Z=2 17. 292.68 × 1021 atoms, 162.60 × 1021 atoms, 832.50 kJ 18. 331.13×104J 19. h/ 20. 3.63×106m–1 2 1 . 938 A° 22. 8 × 106 23. 6530 × 1012 Hz 24. (i) Z = 5, (ii) 340 eV, –680 eV 26. Brackett ; 2.63 × 10–4 cm 25. 3.09 × 108 cm/sec n2h2 28. 9.15 × 1019 Hz, yes, 58.5 × 10–15 J 27. rn = 4K2 3e2 208m e ; n = 25 ; 55.2 pm 29. 4.5 × 1014 s–1 30. 497 kJ/mol 31. 319.2 kJ/mol 32. 6.57 × 10–34 Js 33. 0.62 A° 3 4 . 3.06 V 35. 0.79 A° 36. 6.03×10–4volt 37. 1.05 × 10–13m 3 8 . 0.0826 volts 39. 16 40. 3.3 × 10–18 J 41. 0.527 4 2 . 6235 A° 4 3 . 3 × 1022 44. 4.9 × 10–7 m 45. 8.68 % 4 6 . 2.186 × 10–20 Joules 47. 24 4 8 . 3.88 pm 49. 3.68 × 10–65 m 5 0 . 1.75 × 10–29 51. 0.0144 m
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . To what series does the spectral lines of atomic hydrogen belong if its wave number is equal number is equal to the difference between the wave numbers of the following two lines of the Balmer series 486.1 and 410.2 nm? What is the wavelength of this line ? 2 . Energy required for the excitation of H-atom its ground state to the 2nd excited state is 2.67 times smaller than dissociation energy of H2(g). If H2 (g) placed in 1.0 litre flask at 27°Cand 1.0 bar is to be excited to their 2nd excited state, what will be the total energy consumption ? 3 . Find the quantum number 'n' corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm. 4 . A gas of identical H-like atom has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV. Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photons have energy 2.7 eV. Some have more and some have less than 2.7 eV. (a) Find the principal quantum number of initially excited level B. (b) Find the ionisation energy for the gas atoms. (c) Find the maximum and the minimum energies of the emitted photons. 5 . A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of n and z (ionisation energy of hydrogen atom = 13.6 eV). 6 . Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 A°. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. 7 . An alpha particle after passing through a potential difference of 2 × 106 volt falls on a silver foil. The atomic number of silver is 47. Calculate (i) the K.E. of the alpha-particle at the time of falling on the foil. (ii) K.E. of the -particle at a distance of 5 × 10–14 m from the nucleus, (iii) the shortest distance from the nucleus of silver to which the -particle reaches. ke2 8 . Suppose the potential energy between electron and proton at a distance r is given by – 3r3 . Use Bohr's theory to obtain energy of such a hypothetical atom. 9 . An energy of 68 eV is required to excite a hydrogen like atom from its second Bohr orbit to the third. The nuclear charge is Ze. Find the value of Z, the kinetic energy of the electron in the first Bohr orbit and the wavelength of the radiation required to eject the electrons from the first Bohr orbit to infinity . 1 0 . The ionisation energy of a H-like Bohr atom is 4 Rydbergs. (i) What is the wavelength of radiation emitted when the e– jumps from the first excited state to the ground state? (ii) What is the radius of first Bohr orbit for this atom? [1 Rydberg = 2.18 × 10–18 J] 1 1 . Photon having wavelength 12.4 nm was allowed to strike a metal plate having work function 25 eV. Calculate the -
(a) Maximum kinetic energy of photoelectrons emitted in eV. (b) Wavelength of electron with maximum kinetic energy in A°. (c) Calculate the uncertainity in wavelength of emitted electron, if the uncertainity in the momentum is 6.62 × 10–28 kg m/sec. 1 2 . Electron present in single electron species jumps from energy level 3 to 1. Emitted photons when passed through a sample containing excited He+ ion causes further excitation to some higher energy level (Given Z2 En = 13.6 n2 ) : Determine . (i) Atomic number of single electron species. (ii) Principal quantum number of initial excited level & higher energy of He+ 1 3 . The angular momentum of an electron in a Bohr's orbit of H-atom is 3.1652 × 10–34 kg-m2/sec. Calculate the wave number in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from this level to the ground state. (Use h = 6.626 × 10–34 Js). BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . n = 4, n = 6, 2.63 × 10–4 cm 2 . 21.8 kJ 12 4 . (a) n = 2, (b) 14.4 eV, (c) 13.5eV, 0.7eV 3. n= 5 5 . n = 6, Z = 3 6 . six, 18800 A° 7. 6.4 × 10–13 J, 2.1 × 10–13 J, 3.4 × 10–14m 8 . n6h6 E = 384m 3K 2e4 6 9 . 6 ; 489.6 eV, 25.28 A° 1 0 . 300.89 A°, 2.645 × 10–9 cm 1 1 . (a) 75 eV ; (b) 1.414 A° ; (c) 2 × 10–14 m 1 2 . (i) Z = 1 (ii) For He+ ion this energy corresponds to excitation from 2 to 6. 8 1 3 . R 9
EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . An atom has a mass of 0.02 kg and uncertainty in its velocity is 9.218 × 10–6 m/s then uncertainty in position is (h = 6.626 × 10–34 Js) [AIEEE 2002] (1) 2.86× 10–28 m (2) 2.86 × 10–32 cm (3) 1.5 × 10–27 m (4) 3.9 × 10–10 m 2 . Energy of H- atom in the ground state is –13.6 eV, Hence energy in the second excited state is- [AIEEE 2002] (1) –6.8 eV (2) –3.4 eV (3) –1.51 eV (4) –4.3 eV 3 . Uncertainty in position of a particle of 25 g in space is 10–5 m. Hence uncertainty in velocity (ms–1) is (Planck's constant h = 6.6 × 10–34 Js) [AIEEE-2002] (1) 2.1 × 10–28 (2) 2.1 × 10–34 (3) 0.5 × 10–34 (4) 5.0 × 20–24 4. The orbital angular momentum for an electron revolving in an orbit is given by ( 1) h . This momentum . 2 for an s-electron will be given by [AIEEE-2003] h 1h (3) zero h (1) 2. (2) + . (4) 2 2 2 2 5 . The number of d-electrons retained in Fe2+ (At. no. of Fe = 26) ion is : [AIEEE-2003] (1) 6 (2) 3 (3) 4 (4) 5 6 . The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately : [AIEEE 2003] (1) 10–25 metres (2) 10–33 metres (3) 10–31 metres (4) 10–16 metres 7 . In Balmer series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ? [AIEEE-2003] (1) 2 5 (2) 3 2 (3) 5 2 (4) 4 1 8 . Which of the following sets of quantum number is correct for an electron in 4f orbital ? [AIEEE-2004] 1 1 (1) n = 3, l = 2, m = – 2, s = + 2 (2) n = 4, l = 4, m = – 4, s = – 2 1 1 (3) n = 4, l = 3, m = + 1, s = + 2 (4) n = 4, l = 3, m = + 4, s = + 2 9 . Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively [AIEEE-2004] (1) 16 and 5 (2) 12 and 5 (3) 16 and 4 (4) 12 and 4 1 0 . The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be [AIEEE-2004] (Rydberg constant = 1.097× 107 m–1) : (1) 9.1 × 10–8 nm (2) 192 nm (3) 406 nm (4) 91 nm 1 1 . Which one of the following sets of ions represents the collection of isoelectronic species ? [AIEEE-2004] (1) Na+, Mg2+, Al3+, Cl– (2) Na+, Ca2+, Sc3+, F– (3) K+, Cl–, Mg2+, Sc3+ (4) K+, Ca2+, Sc3+, Cl– 1 2 . In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields ? [AIEEE-2005] (A) n = 1, l = 0, m = 0 (B) n = 2, l = 0, m = 0 (C) n = 2, l = 1, m = 1 (D) n = 3, l = 2, m = 1 (E) n = 3, l = 2, m = 0 (1) (D) and (E) (2) (C) and (D) (3) (B) and (C) (4) (A) and (B)
1 3 . Of the following sets which one does not contain isoelectronic species ? [AIEEE-2005] (1) B O 3, CO 2, N O (2) SO 2, CO 2, NO (3) CN , N , C 2 (4) PO 3, SO 2, ClO 333 333 22 44 4 1 4 . Which of the following statements in relation to the hydrogen atom is correct ? [AIEEE-2005] (1) 3s, 3p and 3d orbitals all have the same energy (2) 3s and 3p orbitals are of lower energy than 3d orbitals (3) 3p orbital is lower in energy than 3d orbital (4) 3s orbitals is lower in energy than 3p orbital 1 5 . According to Bohr's theory angular momentum of electron in 5th shell is :- [AIEEE-2006] (1) 1.0 h/ (2) 10 h/ (3) 2.5 h/ (4) 25 h/ 1 6 . Uncertainty in the position of an electron (mass = 9.1 × 10–31 Kg) moving with a velocity 300 ms–1, accurate upto 0.001%, will be :- (h = 6.63 × 10–34 Js) [AIEEE-2006] (1) 5.76 × 10–2 m (2) 1.92 × 10–2 m (3) 3.84 × 10–2 m (4) 19.2 × 10–2 m 1 7 . Which of the following sets of quantum numbers represents the highest energy of an atom ? [ AIE E E-2 0 0 7 ] (1) n = 3, l = 1, m =1, s = +½ (2) n = 3, l = 2, m =1, s = +½ (3) n = 4, l = 0, m =0, s = +½ (4) n = 3, l = 0, m =0, s = +½ 1 8 . Which one of the following constitutes a group of the isoelectronic species? [AIEEE-2008] (1) C 2 , O , C O, N O (2) N O , C 2 , C N , N (3) C N , N , O 2 , C 2 (4 ) N , O , N O , C O 2 2 2 2 2 2 2 2 2 1 9 . The ionziation enthalpy of hydrogen atom is 1.312 x 106 J mol–1. The energy required to excite the electron in the atom from n =1 to n =2 is [AIEEE-2008] (1) 8.51 x 105 J mol–1 (2) 6.56 x 105 J mol–1 (3) 7.56 x 105 J mol–1 (4) 9.84 x 105 J mol–1 2 0 . In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 × 10–34 kg m2 s–1, mass of electron, em = 9.1 × 10–31 kg):- [AIEEE-2009] (1) 1.92 × 10–3 m (2) 3.84 × 10–3 m (3) 1.52 × 10–4 m (4) 5.10 × 10–3 m 2 1 . Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms–1 (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 Js) :- [AIEEE-2009] (1) 2.5 nm (2) 14.0 nm (3) 0.032 nm (4) 0.40 nm 2 2 . The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl–Cl bond is (C = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1) [AIEEE-2010] (1) 494 nm (2) 594 nm (3) 640 nm (4) 700 nm 2 3 . Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationar y state (n = 1) of Li2+ is:- (1) 8.82 × 10–17 J atom–1 (2) 4.41 × 10–16 J atom–1 [AIEEE-2010] (3) –4.41 × 10–17 J atom–1 (4) –2.2 × 10–15 J atom–1 2 4 . A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :- [AIEEE-2011] (1) 743 nm (2) 518 nm (3) 1035 nm (4) 325 nm 2 5 . The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following :- [AIEEE-2011] (1) n = 3 to n = 1 (2) n = 2 to n = 1 (3) n = 3 to n = 2 (4) n = 4 to n = 3 2 6 . The increasing order of the ionic radii of the given isoelectronic species is :- [AIEEE-2012] (1) K+, S2–, Ca2+ , Cl– (2) Cl–, Ca2+, K+, S2– (3) S2–, Cl–, Ca2+, K+ (4) Ca2+, K+, Cl–, S2–
2 7 . The electrons identified by quantum numbers n and :- [AIEEE-2012] (a) n = 4 , = 1 (b) n = 4, = 0 (c) n = 3, = 2 (d) n = 3, = 1 Can be placed in order of increasing energy as (1) (a) < (c) < (b) < (d) (2) (c) < (d) < (b) < (a) (3) (d) < (b) < (c) < (a) (4) (b) < (d) < (a) < (c) PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 1 3 1 3 1 2 3 3 2 4 4 1 2 1 3 Que. 16 17 18 19 20 21 22 23 24 25 26 27 Ans 2 2 2 4 1 4 1 3 1 2 4 3
Search
Read the Text Version
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346
- 347
- 348
- 349
- 350
- 351
- 352
- 353
- 354
- 355
- 356
- 357
- 358
- 359
- 360
- 361
- 362
- 363
- 364
- 365
- 366
- 367
- 368
- 369
- 370
- 371
- 372
- 373
- 374
- 375
- 376
- 377
- 378
- 379
- 380
- 381
- 382
- 383
- 384
- 385
- 386
- 387
- 388
- 389
- 390
- 391
- 392
- 393
- 394
- 395
- 396
- 397
- 398
- 399
- 400
- 401
- 402
- 403
- 404
- 405
- 406
- 407
- 408
- 409
- 410
- 411
- 412
- 413
- 414
- 415
- 416
- 417
- 418
- 419
- 420
- 421
- 422
- 423
- 424
- 425
- 426
- 427
- 428
- 429
- 430
- 431
- 432
- 433
- 434
- 435
- 436
- 437
- 438
- 439
- 440
- 441
- 442
- 443
- 444
- 445
- 446
- 447
- 448
- 449
- 450
- 451
- 452
- 453
- 454
- 455
- 456
- 457
- 458
- 459
- 460
- 461
- 462
- 463
- 464
- 465
- 466
- 467
- 468
- 469
- 470
- 471
- 472
- 473
- 474
- 475
- 476
- 477
- 478
- 479
- 480
- 481
- 482
- 483
- 484
- 485
- 486
- 487
- 488
- 489
- 490
- 491
- 492
- 493
- 494
- 495
- 496
- 497
- 498
- 499
- 1 - 50
- 51 - 100
- 101 - 150
- 151 - 200
- 201 - 250
- 251 - 300
- 301 - 350
- 351 - 400
- 401 - 450
- 451 - 499
Pages:
