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WBJEE 2016 Solved Physics and Chemistry Question Paper – Part 6

Jan 20, 2017 15:52 IST

    Find WBJEE Solved Physics and Chemistry Question Paper for the year 2016. This article contains ten questions from the paper with their detailed solution. Previous year question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam.

    The Physics and Chemistry section of WBJEE engineering exam has 80 multiple choice questions. The first 40 questions are from Physics and the last 40 questions are from Chemistry.

    The detailed solution from question #51 to #60 is provided in this article.

    51. Which of the following reactions will not result in the formation of carbon–carbon bonds?

    (A) Cannizaro reaction

    (B) Wurtz reaction

    (C) Reimer-Tiemann reaction

    (D) Friedel-Crafts acylation

    Ans : (A)

    Sol.

    Wurtz reaction, Reimer-Tiemann reaction, Friedel-Crafts acylation result in the formation of C-C bond while Cannizaro reaction do not form Carbon-Carbon bond.

    52. Point out the false statement.

    (A) Colloidal sols are homogenous

    (B) Colloids carry +ve or –ve charges

    (C) Colloids show Tyndall effect

    (D) The size range of colloidal particles is 10-1000Å

    Ans : (A)

    Sol.

    Colloidal solution is heterogeneous in nature.

    It consists of two phases-

    (a)dispersed phase

    (b) and dispersion medium.

    All the remaining options about colloidal solution is true.

    So, option (A)  is false and hence it is correct answer.

    53. The correct structure of the drug paracetamol is

    Wbjee 2016 Chemistry

    Ans : (B)

    Sol.

    The compound N-acetyl-para-aminophenol is called paracetamol in India. When mixed with codeine it goes by the tradename Tylenol.

    The structure of paracetamol is

    Wbjee 2016 Chemistry

    So, the correct option is B.

    54. Which of the following statements regarding Lanthanides is false?

    (A) All lanthanides are solid at room temperature.

    (B) Their usual oxidation state is +3

    (C) They can be separated from one another by ion-exchange method.

    (D) Ionic radii of trivalent lanthanides steadily increases with increase in atomic number.

    Ans : (D)

    Sol.

    Lanthanide are reffered as rare earth metals. they are found in 6th period in periodic table. At room temperature all lanthanides are solid. They have usual oxidation state of +3. Also, they can be separated by ion exchange method.

    Ionic radii of trivalent lanthanides steadily decreases with increase in atomic number and the phenomenon is known as Lanthanoid contraction.

    So, D is not the correct option regarding Lanthanoid.

    Sequence and Series: Practice Questions Set 1

    55. Nitrogen dioxide is not produced on heating

    (A)  KNO3 

    (B) Pb(NO3)2

    (C) Cu(NO3)2 

    (D) AgNO3

    Ans : (A)

    Sol.

    Wbjee 2016 Chemistry

    So, KNO3 does not liberate NO2 on heating.

    56. The boiling points of HF, HCl, HBr and HI follow the order

    (A) HF > HCl > HBr > HI

    (B) HF > HI > HBr > HCl

    (C) HI > HBr > HCl > HF

    (D) HCl > HF > HBr > HI

    Ans : (B)

    Sol.

    HF is liquid at or below 19°C so, it has highest boiling point among hydrogen halides. Other hydrogen halides are gaseous and their boiling points depend on the Van Der Waals’ forces. Also, larger the size (or molecular mass), greater is the van der Waals’ forces, hence higher is the boiling point.

    So, the correct order of boiling point is;

    HF > HI > HBr > HCl

    57. In the solid state PCl5 exists as

    (A) [PCl4] and [PCl6]+ ions

    (B) Covalent PCl5 molecules only

    (C) [PCl4]+ and [PCl6] ions

    (D) Covalent P2Cl10 molecules only

    Ans : (C)

    Sol.

    PCl5  in solid state exist as oppositely charged ions as the ionic bonding enhances the crystalline nature. So, in solid state PCl5 exists as a combination of two complex ions. [PCl4]+  and  [PCl6]-

    58. Which statement is not correct for ortho and para hydrogen?

    (A) They have different boiling points.

    (B) Ortho-form is more stable than para-form.

    (C) They differ in their nuclear spin

    (D) The ratio of ortho to para hydrogen changes with change in temperature.

    Ans : (B)

    Sol.

    At the room temperature, the ratio of ortho to para hydrogen is 3 : 1. So, ortho form is more stable than para form at and above room temperature, whereas at low temperature para form is more stable.

    59. The acid in which O – O bonding is present is

    (A) H2S2O3

    (B) H2S2O6

    (C) H2S2O8

    (D) H2S4O6

    Ans : (C)

    Sol.

    In Marshall’s acid or peroxodisulphuric acid (H2S2O8) O – O bonding is present. Hence, the correct option is C.

    60. The metal which can be used to obtain metallic Cu from aqueous CuSO4 solution is

    (A) Na

    (B) Ag

    (C) Hg

    (D) Fe

    Ans : (D)

    Sol.

    When iron (Fe) reacts with copper sulphate solution(CuSO4) then it undergo single displacement reaction known as substitution reaction to form solid copper(Cu) and aqueous iron sulphate. No other metal in the given option undergo single displacement reaction.

    Permutation and Combination: Practice Questions Set 1

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