JEE (MAIN) 2018 PAPER CHEMISTRY

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JEE (MA) 018 PAPER EMSTRY 1. Which of the following salts is the most basic in aqueous solution? (1) 3 K () Fel 3 (3) Pb( 3 ) (4) Al() 3 (1) 3 K is a salt of weak acid ( 3 ) and a strong base (K).

1 JEE (MA) 018 PAPER EMSTRY 1. Which of the following salts is the most basic in aqueous solution? (1) 3 K () Fel 3 (3) Pb( 3 ) (4) Al() 3 (1) 3 K is a salt of weak acid ( 3 ) and a strong base (K). t undergoes complete dissociation in aqueous solution. K(aq) ( aq) ( aq) 3 3 The acetate ions undergo hydrolysis in water to form acetic acid and hydroxide ions. ( aq) l () (aq) ( aq) 3 3 Acetic acid being a weak acid will remain unionized in water and the concentration of hydroxide ions will make the solution basic.. Which of the following compounds will be suitable for Kjeldahl s method for nitrogen estimation? (1) () (3) l (4) (1) Kjeldahl s method is suitable for aniline. n this method, the organic compound containing nitrogen is heated with concentrated sulphuric acid in presence of u which converts the nitrogen in the com pound to ammonium sulphate. 3. Which of the following are Lewis acids? (1) All 3 and Sil 4 () P 3 and Sil 4 (3) Bl 3 and All 3 (4) P 3 and Bl 3 (3) Lewis acid is any ionic or molecular species that can accept a pair of electrons in the formation of a coordinate covalent bond. These substances have an incomplete valence shells octet like Bl 3 and All Phenol on treatment with in the presence of a followed by acidification produces compound X as the major product. X on treatment with ( 3 ) in the presence of catalytic amount of produces (1) () (3) (4) JEE Main 018_hemistry.indd :4:

2 P- JEE (Main) 018 Paper hemistry (4) Kolbe s reaction is usually carried out by allowing sodium phenoxide to absorb carbon dioxide and then heating the product to 15 under a pressure of several atmospheres of carbon dioxide. 1., a. Acidif ication Salicylic acid (X) Reaction of salicylic acid with acetic anhydride yields the widely used pain reliever aspirin: Salicylic acid (X) conc. 3 Acetic anhydride 3 Acetylsalicylic acid (aspirin) 3 5. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination? Base Acid End point (1) Strong Strong Pinkish red to yellow () Weak Strong Yellow to pinkish red (3) Strong Strong Pink to colorless (4) Weak Strong olorless to pink () Methyl orange is a very weak base. t is ionized to give Me and ions. Me Me Yellow Red [ Me ][ ] The ionization constant is K = [ Me] n the presence of alkali, the concentration of ions increases, and the equilibrium shifts to the left imparting yellow color to the solution. By adding acid, the ions are removed and the equilibrium shifts to the right thereby increasing the concentration of Me and imparting pinkish red to the solution. 6. An aqueous solution contains 0.10 M S and 0.0 M l. f the equilibrium constants for the formation of S from S is and that of S from S ions is then the concentration of S ions in aqueous solution is (1) () (3) (4) JEE Main 018_hemistry.indd :4:

3 JEE (Main) 018 Paper hemistry P-3 (1) We have [ S] = 0.10 and [l] = 0.0 M [ ] = 0. M The reactions involved are as follows: - S S ; K1 = S S ; K = So, for reaction S S ; equilibrium constant is given by K = K1 K = = [ S ][ ] We know K = [ S ] [ S ] = = 3 10 M The combustion of benzene (l) gives (g) and (1). Given that heat of combustion of benzene at constant volume is kj mol 1 at 5 ; heat of combustion (in kj mol 1 ) of benzene at constant pressure will be (R = JK 1 mol 1 ) (1) () 360 (3) (4) (3) The reaction involved is 6 6() l ( g) 6 ( g) 3(l) We know that = U nrt g (1) where n = n n g p r = 6 15 = Substituting DU = kJ, R = kj K mol, T = 98Kand Dng =-15. in Eq. (1), we get -3 D = (- 1.) = kj 8. The compound that does not produce nitrogen gas by the thermal decomposition is (1) ( 4 ) r 7 () 4 (3) ( 4 ) (4) Ba( 3 ) (3) The thermal decomposition reactions are as follows: ( 4 ) r 7 4 ( 4 ) Ba( 3 ) r Ba 3 JEE Main 018_hemistry.indd :4:4

4 P-4 JEE (Main) 018 Paper hemistry 9. ow long (approximate) should water be electrolyzed by passing through 100 amperes current so that the oxygen released can completely burn 7.66 g of diborane? (Atomic weight of B = 10.8 u) (1) 0.8 h () 3. h (3) 1.6 h (4) 6.4 h () The reaction involved is B 6 3 B 3 3 umber of moles of required = 3 o. of moles of B = 3 = For reaction, the n-factor is 4. Therefore, number of equivalents = 3 4 = 1 F t = t = s 100 = 3. h 10. Total number of lone pair of electrons in 3 ion is (1) 6 () 9 (3) 1 (4) 3 () 3 is a linear molecule with sp 3 d hybridization. As we can see from the above structure, total number of lone pair in 3 is When metal M is treated with a, a white gelatinous precipitate X is obtained, which is soluble in excess of a. ompound X when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal M is (1) a () Al (3) Fe (4) Zn () The metal M is Al. The reactions are as follows: a excess a Al 3¾ ¾¾ Al() ¾ ¾¾¾ aal() ( M) ( X) D Al() ¾ ¾ Al ( X) Al 3 is used in chromatography as an adsorbent. 3 4 JEE Main 018_hemistry.indd :4:5

5 JEE (Main) 018 Paper hemistry P-5 1. According to molecular orbital theory, which of the following will not be a viable molecule? (1) e () (3) (4) e (3) The electronic configuration of is s1s, s * 1s Bond order (B) of can be calculated as B = b a = = 0 where b and a are the number of bonding and antibonding electrons respectively. ence, does not exist, due to zero bond order. 13. The increasing order of basicity of the following compounds is () () () (V) (1) () < () < () < (V) () () < () < (V) < () (3) (V) < () < () < () (4) () < () < () < (V) 3 () More stable the conjugate base, the more will be the acidic strength. n compound (), all resonating structures have equal contribution in resonance hybrid. This makes it the strongest base. 3 3 (i) 3 (ii) effect increases the basicity of compound. 3 is an electron donating group hence, it increases the basicity of compound (V). More the s character of an orbital, more strongly it holds its electrons. Therefore, lone pair on sp 3 hybridized atom is more protonating than the lone pair on sp. -sp 3 -sp > () () 14. Which type of defect has the presence of cations in the interstitial sites? (1) Vacancy defect () Frenkel defect (3) Metal deficiency defect (4) Schottky defect JEE Main 018_hemistry.indd :4:7

6 P-6 JEE (Main) 018 Paper hemistry () When an ion is displaced from its regular position to an interstitial position, it creates a vacancy; this pair of vacancy interstitial is called Frenkel defect. This defect consists of a vacant lattice site (a hole ), and the ion which ideally should have occupied the site now occupies an interstitial position. Metal ions are generally smaller than the anions. Thus it is easier to squeeze A into alternative interstitial positions, and consequently it is more common to find the positive ions occupying interstitial positions. 15. Which of the following compounds contain(s) no covalent bond(s)? Kl, P 3,, B 6, (1) Kl, () Kl (3) Kl, B 6 (4) Kl, B 6, P 3 () Among the given compounds except Kl all are covalent compounds. 16. The oxidation states of r in [r( ) 6 ]l 3,[r( 6 6 ) ], and K [r() () ( )( 3 )] respectively are (1) 3,, and 4 () 3, 0, and 6 (3) 3, 0, and 4 (4) 3, 4, and 6 () [r( ) 6 ]l 3 Let the oxidation state of r be x [r( 6 6 ) ] Let the oxidation state of r be y K [r() () ( )( 3 )] x 6 0 ( 1) 3 x = 3 y 0 = 0 y = 0 1 z ( 1) ( ) ( ) 0 = 0 z = ydrogen peroxide oxidizes [Fe() 6 ] 4 to [Fe() 6 ] 3 in acidic medium but reduces [Fe() 6 ] 3 to [Fe() 6 ] 4 in alkaline medium. The other products formed are, respectively (1) ( ) and ( ) () and ( ) (3) and ( ) (4) ( ) and 4 3 () [ Fe( )] [ Fe ( )] [ Fe( )] [ Fe( )] Glucose on prolonged heating with gives (1) 1-hexene. () hexanoic acid. (3) 6-iodohexanal. (4) n-hexane. JEE Main 018_hemistry.indd :4:8

7 JEE (Main) 018 Paper hemistry P-7 (4) Glucose reacts with to form n-hexane indicating that the six carbons are linked in a straight chain. () 4 eat 3 3 n-exane 19. The predominant form of histamine present in human blood is (pk a, istidine = 6.0) (1) 3 () 3 (3) (4) (3) The structure of histamine is n the above structure, the terminal is slightly more basic than the other two nitrogens in the ring. The p of blood is approximately Therefore, terminal gets protonated and the predominate structure will be 3 0. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3a 3 (P 4 ) a() ] to (1) [3(aF ) a() ] () [3(a 3 (P 4 ) af ] (3) [3(a() ] af ] (4) [af ] () The concentration of fluoride is maintained up to 1 ppm in drinking water by addition of soluble fluoride ions. The enamel of the teeth is made harder by the fluoride ions. These convert hydroxyapatite [3a 3 (P 4 ) a() ] to much harder fluorapa tite [3a 3 (P 4 ) af ]. 1. onsider the following reaction and statements: [o( 3 ) 4 ] [o( 3 ) 3 3 ] 3 JEE Main 018_hemistry.indd :4:9

8 P-8 JEE (Main) 018 Paper hemistry () Two isomers are produced if the reactant complex ion is a cis-isomer. () Two isomers are produced if the reactant complex ion is a trans-isomer. () nly one isomer is produced if the reactant complex ion is a trans-isomer. (V) nly one isomer is produced if the reactant complex ion is a cis-isomer. The correct statements are (1) () and () () () and (V) (3) () and (V) (4) () and () (1) 3 o o 3 3 Fac-isomer 3 (cis) 3 o 3 3 Mer-isomer 3 3 o (trans) 3 o 3 3 Mer-isomer. The trans-alkenes are formed by the reduction of alkynes with (1) ab 4 () a/liq. 3 (3) Sn l (4) Pd/, Ba () Anti addition of hydrogen to the triple bond of alkynes occurs when they are treated with lithium or sodium metal in ammonia or ethylamine at low temperatures. a/liq The ratio of mass percent of and in an organic compound ( X Y Z ) is 6 : 1. f one molecule of the above compound ( X Y Z ) contains half as much oxygen as required to burn one molecule of compound X Y completely to and. The empirical formula of compound X Y Z is (1) 4 () 3 4 (3) 4 3 (4) (3) For compound X Y Z, we have 1 X = 6 X = Y Y 1 JEE Main 018_hemistry.indd :4:30

9 JEE (Main) 018 Paper hemistry P-9 The reaction involved is X Y(g) X Y Y ( g) X( g) l () 4 umber of oxygen atom in X Y Z = Z umber of oxygen atom required for X Y combustion is X Y X Y = 4. Therefore, 1 Z = X Y X Y X 3X = = X = 4 4 3X Thus, X : X : X : 4X : 3X or : 4: 3 ence, the empirical formula is Phenol reacts with methyl chloroformate in the presence of a to form product A. A reacts with to form product B. A and B are respectively (1) and () and (3) (4) 3 3 and 3 and 3 () Me - a Me a l Me Phenol (A) (B) 5. The major product of the following reaction is ame Me JEE Main 018_hemistry.indd :4:31

10 P-10 JEE (Main) 018 Paper hemistry Me (1) () (3) (4) Me (1) With primary halides, however, a strong base favors elimination because steric hindrance in the substrate makes substitution more difficult. ence, the reaction will proceed by E mechanism. ame Me 6. Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic reaction? n K (0, 0) A B 1 T(K) D (1) B and () and D (3) A and D (4) A and B (4) We know S ln K = R RT The graph of lnk vs 1 T can be plotted for exothermic and endothermic reactions. For an exothermic reaction, D = ve slope = = ve R ence, from the given graph lines A and B represent the exothermic reactions. 7. The major product formed in the following reaction is eat (1) () (3) (4) JEE Main 018_hemistry.indd :4:33

11 JEE (Main) 018 Paper hemistry P-11 (3) The steps involved in the reaction are as follows: - S 1 (- 3 ) (- 3 ) - S 8. An aqueous solution contains an unknown concentration of Ba. When 50 ml of a 1 M solution of a is added, Ba just begins to precipitate. The final volume is 500 ml. The solubility product of Ba is What is the original concentration of Ba? (1) 10 9 M () M (3) M (4) M () The concentration of S 4 in Ba solution can be calculated as MV 1 1= MV 1 50 = M M = 10 For just precipitation.p = K sp [ Ba ][ S 4 ] = Ksp[BaS4 ] 1 10 [ Ba ] = [ Ba ] = 10 M in 500 ml solution For calculation of [Ba ] in original solution (450 ml) where M 1 is the molarity of Ba in original solution (450 ml). M = = =. M 9 9 M 1 9. At 518, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 torr, was 1.00 torr s 1 when 5% had reacted and 0.5 torr s 1 when 33% had reacted. The order of the reaction is (1) 3 () 1 (3) 0 (4) (4) The decomposition reaction is 3 4 Let r 1 = 1 torr s 1, when 5% reacted (95% unreacted) r = 0.5 torr s 1, when 33% reacted (67% unreacted) JEE Main 018_hemistry.indd :4:34

12 P-1 JEE (Main) 018 Paper hemistry We know r µ (a x) m where a is the initial concentration, x is the amount reacted at time t and m is the order of reaction. r 1 a x 1 = r a x = m m m m = (. 141) = ( ) m = 30. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? (1) [o( ) 5 l]l () [o( ) 4 l ]l (3) [o( ) 3 l 3 ] 3 (4) [o( ) 6 ]l 3 - (3) [ o( ) l] l [ o( ) l] l (i = 3) [ o( ) l ] l [ o( ) 4 l ] l (i = ) 4 [ o( ) 3l3] 3 [ o( ) 3l3] (i = 1) 3 [ o( ) ] l [ o( ) ] 3l (i = 4) We know that Tf = i Kf m. Since i value of [ o( ) 3l3] 3 is highest (i = 1), so the depression in freezing point is minimum, and hence freezing point is maximum. JEE Main 018_hemistry.indd :4:35

CHEMISTRY. ion is : 62. Which of the following salts is the most basic in aqueous solution?

CHEMISTRY. ion is : 62. Which of the following salts is the most basic in aqueous solution? JEE MAIN-018 CHEMISTRY This section contains 0 multiple choice questions. Each question has choices, (), () and for its answer, out of which Only One is correct. 61. Total number of lone pair of electrons