May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
IIT-JEE 2011PAPER-I (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 7]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Passage based single correct type questions. +3 marks will be awarded for correct answer and -1 markfor wrong answer. [2 passage, No. of Ques. : 5]Section – IV : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 7]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 7 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Geometrical shapes of the complexes formed bythe reaction of Ni 2+ with Cl – , CN – and H 2 O,respectively, are(A) octahedral, tetrahedral and square planar(B) tetrahedral, square planar and octahedral(C) square planar, tetrahedral and octahedral(D) octahedral, square planar and octahedralAns. [B]Sol. Complexes are : [NiCl 4 ] –2 , [Ni(CN) 4 ] –2 &[Ni(H 2 O) 6 ] +2Ni +2 = 3d 8 4s 0[NiCl 4 ] –2 : Now Since, Cl – is a weak legand so nopairing of electron take place and geometry istetrahedral[Ni(CN) 4 ] –2 : Since, CN – is a strong legand sopairing of electron will take place & geometry issquare planar.[Ni(H 2 O) 6 ] +2 : It will formed octahedral complexsince C.N. = 62. AgNO 3 (aq.) was added to an aqueous KClsolution gradually and the conductivity of thesolution was measured. The plot of conductance( Λ ) versus the volume of AgNO 3 is -Ans.Sol.ΛΛvolume(P)volume(R)ΛΛvolume(Q)volume(S)(A) (P) (B) (Q) (C) (R) (D) (S)[D]Because in the beginning of the reaction no ofions remain constant so conductivity remainsconstant but after complete precipitation of Cl –the no. of ions increases in solution. Soconductivity increases.3. Bombardment of aluminum by α-particle leads toits artificial disintegration in two ways, (i) and (ii)as shown. Products X, Y and Z respectively are -Ans.(i)273013 Al15 P + Y3014 Si +X(ii)3014 Si +(A) proton, neutron, positron(B) neutron, positron, proton(C) proton, positron, neutron(D) positron, proton, neutron[A]ZXtraEdge for IIT-JEE 62 MAY 2011
Sol.27134He30152Al ⎯⎯→ P +42 He3014 Si + H1 301 Si0 114 + + e4. Extra pure N 2 can be obtained by heating -(A) NH 3 with CuO (B) NH 4 NO 3(C) (NH 4 ) 2 Cr 2 O 7 (D) Ba(N 3 ) 2Ans. [D]Sol. Ba(N 3 ) 2 ⎯⎯→∆ Ba + 3N 2 ↑5. Among the following compounds, the most acidicis -(A) p-nitrophenol(B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) p-toluic acidAns. [C]Sol. o-hydroxy benzoic acid is stronger acid due toortho effect6. The major product of the following reaction is -OC(i) KOHNHC(ii) Br CH 2 ClOO(A)(B)(C)(D)CN–CH 2COOCNCOOCNO–CH 2OCNO10nCH 2 ClBrCH 2 ClBrSol.OCNHCOOC Θ ⊕NKCOCl–CH 2–KOH– BrOCN – CH 2 –C7. Dissolving 120 g of urea (mol. wt. 60) in 1000 gof water gave a solution of density 1.15 g/mL.The molarity of the solution is -(A) 1.78 M(B) 2.00 M(C) 2.05 M(D) 2.22 MAns.[C]Sol. M =x × d × 10 10 .7 × 1.15×10== 2.05 Mmol wt 60x = percentage by weightO120x = × 100120 + 1000SECTION – IIMultiple Correct Choice TypeThis section contains 4 multiple choice questions. Eachquestions has 4 choices (A), (B), (C) and (D), out ofwhich ONE OR MORE is/are correct.8. Extraction of metal from the ore cassiteriteinvolves -(A) carbon reduction of an oxide ore(B) self-reduction of a sulphide ore(C) removal of copper impurity(D) removal of iron impurityAns. [A,C,D]Sol. Cassiterite is SnO 2 .To reduce SnO 2 into Sn, carbon reduction processis used.Sn has iron impurity.SnO 2 + C → Sn + CO 2–BrAns.[A]XtraEdge for IIT-JEE 63 MAY 2011
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IIT-JEE <strong>2011</strong>PAPER-I (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 7]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Passage based single correct type questions. +3 marks will be awarded for correct answer and -1 markfor wrong answer. [2 passage, No. of Ques. : 5]Section – IV : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 7]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 7 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Geometrical shapes of the complexes formed bythe reaction of Ni 2+ with Cl – , CN – and H 2 O,respectively, are(A) octahedral, tetrahedral and square planar(B) tetrahedral, square planar and octahedral(C) square planar, tetrahedral and octahedral(D) octahedral, square planar and octahedralAns. [B]Sol. Complexes are : [NiCl 4 ] –2 , [Ni(CN) 4 ] –2 &[Ni(H 2 O) 6 ] +2Ni +2 = 3d 8 4s 0[NiCl 4 ] –2 : Now Since, Cl – is a weak legand so nopairing of electron take place and geometry istetrahedral[Ni(CN) 4 ] –2 : Since, CN – is a strong legand sopairing of electron will take place & geometry issquare planar.[Ni(H 2 O) 6 ] +2 : It will formed octahedral complexsince C.N. = 62. AgNO 3 (aq.) was added to an aqueous KClsolution gradually and the conductivity of thesolution was measured. The plot of conductance( Λ ) versus the volume of AgNO 3 is -Ans.Sol.ΛΛvolume(P)volume(R)ΛΛvolume(Q)volume(S)(A) (P) (B) (Q) (C) (R) (D) (S)[D]Because in the beginning of the reaction no ofions remain constant so conductivity remainsconstant but after complete precipitation of Cl –the no. of ions increases in solution. Soconductivity increases.3. Bombardment of aluminum by α-particle leads toits artificial disintegration in two ways, (i) and (ii)as shown. Products X, Y and Z respectively are -Ans.(i)273013 Al15 P + Y3014 Si +X(ii)3014 Si +(A) proton, neutron, positron(B) neutron, positron, proton(C) proton, positron, neutron(D) positron, proton, neutron[A]ZXtraEdge for IIT-JEE 62 MAY <strong>2011</strong>
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