Popular Chemistry Online: CHEMISTRY PREBOARD –I Question paper (2020 -21) CLASS- XII With answer

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Tuesday, December 22, 2020

CHEMISTRY PREBOARD –I Question paper (2020 -21) CLASS- XII With answer

 NAVODAYA VIDYALAYA SAMITI, BHOPAL REGION

       PREBOARD –I (2020 -21)

CLASS- XII                           CHEMISTRY THEORY (043)                         

 MM : 70                                                         Time: 3 Hours 

General Instructions. Read the following instructions carefully. 

a) There are 33 questions in this question paper. All questions are compulsory.

b) Section A: Q. No. 1 to 2 are case-based questions having four MCQs or Reason Assertion type based on given passage each carrying 1 mark.

c) Section A: Question 3 to 16 are MCQs and Reason Assertion type questions carrying 1 mark each

d) Section B: Q. No. 17 to 25 are short answer questions and carry 2 marks each.

e) Section C: Q. No. 26 to 30 are short answer questions and carry 3 marks each.

f) Section D: Q. No. 31 to 33 are long answer questions carrying 5 marks each.

g) There is no overall choice. However, internal choices have been provided.

h) Use of calculators and log tables is not permitted.

SECTION A (OBJECTIVE TYPE)

1. Read the passage given below and answer the following questions: (1x4=4)

Alcohols are classified into three types i.e., primary, secondary and tertiary, depending upon the

nature of the carbon atom to which the alcoholic (-OH) group is attached. These have common

physical as well as chemical characteristics. However, they differ in their relative reactivities

towards different reagents. The popular tests which can distinguish the three types of alcohols are

Victor Meyer’s test and Lucas reagent test.

 

The following questions are multiple choice questions. Choose the most appropriate answer:  

(i)              In Lucas test, turbidity appears on heating. Predict the nature of alcohol.

 

a) Primary alcohol

b) Secondary alcohol

c) Tertiary alcohol

d) Butan-2-ol


(ii)            What is the correct order of reactivity of the alcohols involving the cleavage of C-OH bond?

 

a) Methyl alcohol >Ethyl alcohol>Isopropyl alcohol>tertiary butyl alcohol

b) Tertiary alcohol>isopropyl alcohol>ethyl alcohol>methyl alcohol

c) Isopropyl alcohol>tertiary alcohol>ethyl alcohol>methyl alcohol

d) Tertiary alcohol>ethyl alcohol>isopropyl alcohol>methyl alcohol

OR   

 Which of the following is a secondary allylic alcohol?

 

a) But-3-en-2-ol

b) But-2-en-2-ol

c) Prop-2-enol

d) Butan-2-ol

 

(iii) Out of water, tert-butyl alcohol, propan-2-ol and ethyl alcohol, which is most acidic?

a) tert-butyl alcohol

b) propan-2-ol

c) ethyl alcohol

d) water

 

(iv) An organic compound ‘X’ with molecular formula C3H8O on heating with copper

gives compound ‘Y’ which reduces Tollen’s reagent. ‘X’ on reaction with sodium

metal gives ‘Z’ . What is the product of reaction of ‘Z’ with 2- chloro-2-methylpropane?

 

a) CH3CH2CH2OC(CH3)3

b) CH3CH2OC(CH3)3

c) CH2=C(CH3)2

d) CH3CH2CH=C(CH3)2

 

Read the passage given below and answer the following questions: (1x4=4)

 

It is often recommended that the first aid kit to be kept in the houses must have a small lump of alum. In case, bleeding occurs while doing shave in the bathroom or from a knife cut in the kitchen, alum should be immediately rubbed on the affected portion. Bleeding stops and medical aid if required, can be obtained later on.

 

2. In these questions (Q. No 5-8 , a statement of assertion followed by a statement of reason

is given. Choose the correct answer out of the following choices.

 

a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

c) Assertion is correct statement but reason is wrong statement.

d) Assertion is wrong statement but reason is correct statement.

 

(i) Assertion: Bleeding stops when alum is rubbed on the affected portion.

     Reason: Alum can be used to remove colloidal impurities from water.

 

(ii) Assertion: Excess of electrolyte can bring about coagulation of colloidal solution.

      Reason: The flocculating ion of the electrolyte removes the charge from the sol particles.

 

(iii) Assertion: Coagulation power of Al+3 is more than that of Na+

       Reason: Greater the valency of the flocculating ion added, greater is its power to cause coagulation           (Hardy-Schulze rule)

 

(iv) Assertion: Addition of gelatin to a lyophobic sol may result in to its coagulation

       Reason: Smaller the gold number of protective colloid, greater will be its protective power.

 

OR

Assertion: persistent dialysis may also result in to coagulation of a lyophobic colloid.

Reason: Electrophoresis involves the migration of dispersion medium under the influence of  electric field when the dispersed phase particles are prevented from moving.

 

Following questions (No. 3 -11) are multiple choice questions carrying 1 mark each:

 

3   Which of the following option will be the limiting molar conductivity of CH3COOH if the limiting molar     conductivity of CH3COONa is 91 Scm2mol-1? Limiting molar conductivity for individual ions are given in the following table.

 

S.No

Ions

limiting molar conductivity / Scm2mol-1

1

H+

349.6

2

Na+

50.1

3

K+

73.5

4

OH-

199.1


a) 350 Scm2mol-1                                 b) 375.3 Scm2mol-1

c) 390.5 Scm2mol-1                              d) 340.4 Scm2mol-1

4. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet        structure, α-helix structure of protein is stabilized by:

 

a) Peptide bonds

b) Van der Waal’s forces

c) Hydrogen bonds

d) Dipole-dipole interactions. 

OR

 

   Curdling of milk is an example of:

a) breaking of peptide linkage

b) hydrolysis of lactose

c) breaking of protein into amino acids

d) denauration of proetin

 

5. A binary solution is prepared by mixing n-heptane and ethanol. Which one of the following

    statements is correct regarding the behaviour of the solution?

a) The solution formed is an ideal solution

b) The solution is non-ideal showing positive deviation from Raoult’s law.

c) The solution is non-ideal showing negative deviation from Raoult’s law.

d) n-heptane shows positive deviation while ethanol shows negative deviation from Raoult’s law.

 

6. Because of lanthanoid contraction:

 

a) separation of the lanthanoid elements become difficult.

b) there is a very small difference in the atomic size of the transition metals of 5th and 6th period

   in the same group.

c) there is a gradual decrease in the basic strength of the hydroxides of lanthanoids

d) all are correct

OR

 

Which of the following is a diamagnetic ion:

(Atomic numbers of Cr, Mn, Ni and Cu are 24, 25, 28 and 29 respectively)

a) Cr2+

b) Cu+

c) Ni2+

d) Mn2+

7. Method by which aniline cannot be prepared is

a) degradation of benzamide with bromine in alkaline solution

b) reduction of nitrobenzene with H2/Pd in ethanol

c) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous

sodium hydroxide solution

d) hydrolysis of phenylisocyanide with acidic solution

OR

IUPAC name of product formed by reaction of methyl amine with two moles of ethyl chloride

a) N,N-Dimethylethanamine

b) N,N-Diethylmethanamine

c) N-Methyl ethanamine

d) N-Ethyl - N-methylethanamine

8. What is the correct electronic configuration of the central atom in [Co(H2O)6]Cl3 based on crystal

    field theory?                                                                                                                                           

a) e4t22

b) t2g4eg2

c) t2g6eg0

d) e2t24

OR

Which is true for the complex [Ni(en)2]2+ ?

a) paramagnetic, dsp2, square planar, CN of Ni = 2

b) diamagnetic, dsp2, square planar, CN of Ni = 4

c) diamagnetic, sp3, tetrahedral, CN of Ni = 4

d) paramagnetic, sp3, square planar, CN of Ni = 4

 

9. The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(ox)(SO4)2] respectively are

a) +3, +2 and +4

b) +3, 0 and +6

c) +3, +4 and +6

d) +3, 0 and +4

10. Identify A,B,C and D:

 

a) A = C2H4, B= C2H5OH, C= C2H5NC, D= C2H5CN

b) A= C2H5OH, B= C2H4, C = C2H5CN, D=C2H5NC 

c) A = C2H4, B= C2H5OH, C= C2H5CN, D= C2H5NC

d) A= C2H5OH, B= C2H4, C= C2H5NC, D= C2H5CN

 

11.The crystal showing Frenkel defect is :

 

(a)



(b)



( c)



(d )



                                                                                                         


In the following questions (Q. No. 12 - 16) a statement of assertion followed by a statement of

reason is given. Choose the correct answer out of the following choices.

 

a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

c) Assertion is correct statement but reason is wrong statement.

d) Assertion is wrong statement but reason is correct statement.

 

12. Assertion: Deoxyribose, C5H10O4 is not a carbohydrate.

 

Reason: Carbohydrates are optically active polyhydroxy aldehyde or ketone or the compound which produce such units on hydrolysis.

 

13. Assertion: Sulphur exhibits paramagnetic behaviour in the vapour state.

Reason: In vapour state, sulphur partly exists as S2 molecules which have two unpaired electrons in       antibonding π* orbital.

 

14. Assertion: Osmotic pressure is a colligative property.

Reason: Osmotic pressure is directly proportional to molarity. 

                                                                             OR

Assertion: Acetone-aniline mixtures shows negative deviation from Raoult’s law

Reason: H-bonding between acetone and aniline is stronger than that between acetone- acetone

              and aniline-aniline.

 

15. Assertion: The pKa of acetic acid is lower than that of phenol.

Reason: Phenoxide ion is more resonance stabilized than acetate ion.

 

16. Assertion: tert-Butyl methyl ether on treatment with HI at 373 K gives a mixture of methyl alcohol and          tert-butyl iodide.

Reason: The reaction occurs by SN2 mechanism.

 

SECTION B

 

The following questions, Q.No 17 25 are short answer type and carry 2 marks each.

 

17. With the help of resonating structures explain the effect of presence of nitro group at ortho position     in chlorobenzene.

 

OR

Carry out the following conversions in not more than 2 steps:                                                                                     (i)  Aniline to chlorobenzene

(ii) 2-bromopropane to 1- bromopropane

 

18.   5% aqueous solution of a non-volatile solute was made and its vapour pressure at 373 K was

        found to be 745 mm. Calculate the molar mass of solute.

 

19.  (i) Using the valence bond approach, write the hybridization of iron ion in the following

       complex ion. Also predict its magnetic behaviour :[Fe(H2O)6]Cl3

       (ii)Write the IUPAC name of the coordination complex: [CoCl2(en)2]NO3

OR

(i) What is meant by chelate effect? Give an example.

(ii) Predict the geometry and magnetic moment of [NiCl4]2- ion.

 

20. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of

       production of N2 and H2 if k = 2.5 × 10-4  mol L-1 s-1 ?

OR

                                                                                                                                                                                          The following results have been obtained during the kinetic studies of the reaction:

 

P + 2Q à R + 2S Exp.

Initial P(mol/L)

Initial Q (mol/L)

Init. Rate of Formation of R (M min-1)

1

0.10

0.10

3.0 x 10-4

2

0.30

0.30

9.0 x 10-4

3

0.10

0.30

3.0 x 10-4

4

0.20

0.40

6.0 x 10-4


Determine the rate law expression for the reaction.

21. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees.          How old is that piece of wood ?  (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years ) 

 22. Give a mechanism for following  reaction:

  (CH3)3CBr  +  OH -   à  (CH3)3COH  + Br-

23.  Draw the structure of the following compounds:

       (a) H2S2O8

       (b) XeOF4

 

24. (a) Write chemical equations for Reimer Tiemann reaction?

      (b) Which compound out of the following pairs will react faster in SN2 reaction and why ?

      CH2=CHBr  or  CH2=CHCH2Br.

25.  A compound is consist of three element  P, Q  and R . Atoms of element P form ccp lattice and those of the element Q occupy 1/3rd of tetrahedral voids and those of the element R occupy2/3rd of octahedral voids. What is the formula of the compound formed by the elements P, Q and R?

SECTION C

 

Q.No 26 - 30 are Short Answer Type II carrying 3 mark each.

 

26. Give reasons for the following:

      i. Transition elements exhibit variable oxidation states .

      ii. Zirconium and hafnium have almost similar atomic radii.

      iii. Cu+ ion is not known in aqueous solution.

 

OR

 

Observed and calculated values for the standard electrode potentials of elements from Ti to Zn in

the first reactivity series are depicted in figure (1):

                  FIGURE 1 (source NCERT)

Explain the following observations:

i. The general trend towards less negative Eo values across the series

ii. The unique behaviour of Copper

iii. More negative Eo values of Mn and Zn

27.  Arrange the following in increasing order of property specified:

       i. triethylamine, ethanamine, diethylethanamine (solubility in water)

       ii. C4H9NH2, C2H5N(CH3)2 , (C2H5)2NH (boiling point)

       iii. Ethanamine, N-Ethyl ethanamine, N,N-Diethyl ehanamine (Basic strength)

 

OR

      i. Give a chemical test to distinguish between aniline and N,N-dimethylaniline.

      ii. Write chemical equations for Hoffmann bromamide degradation reaction.

      iii. Methyl amine is soluble in water but not aniline. Explain.

 

28. An element ‘X’ (atomic mass = 40 g mol-1) having fcc structure, has unit cell edge length of 400 pm.      Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’.

 

29.  (i) what is the effect of denaturation on the structure of proteins?

       (ii) What is the difference between a nucleoside and nucleotide?

       (iii) Represent Alanine in the zwitter ionic form.

 

30.  i. Explain why N does not form pentahalides while phosphorus does.

       ii. . Arrange the following in decreasing order of bond dissociation enthalpy                                                                                                                             F2 ,  Cl2 ,   Br2 ,  I2

      iii. Electron gain enthalpy of fluorine is less negative than chlorine. Justify

 

SECTION D

 

Q.No 31 to 33 are long answer type carrying 5 marks each.

 

31.  Answer the following questions:                                                                                          (2+3)

  (i) Write the balanced chemical reaction for reaction of Cl2 with hot and concentrated NaOH.

  (ii) Complete the following chemical equations:

        (a)  PbS + O3 à

        (b) H2SO4 + Cu à

  (iii) Give reason

(a)    Amongst all noble gases only xenon is known to form compounds with oxygen and fluorine.

(b)    I-Cl is more reactive than I2 .

 

 

OR

 

Answer the following questions:                                                                                                   (1+4)

(i) Arrange the following in the increasing order of basic strength: (Give reason)

     NH3, PH3, AsH3, SbH3, BiH3

(ii) Account for the following observation

 (a) Acidity of oxo –acids of chlorine is HOCl < HOClO < HOClO2 < HOClO3.

 (b) Nitrogen can’t form compounds like R3N=O, while phosphorus can form R3P=O .

 (c) Fluorine never acts as the central atom in polyatomic interhalogen compounds.

(iii) Complete the following reaction:

     XeF6 + 3H2O     à

 

 

32. An organic compound ‘A’ C8H6 on treatment with dilute H2SO4 containing mercuric sulphate gives compound ‘B’. This compound ‘B’ can also be obtained from a reaction of benzene with acetyl chloride in presence of anhy AlCl3. ‘B’ on treatment with I2 in aq. KOH gives ‘C’ and a yellow compound ‘D’. Identify A, B, C and D. Give the chemical reactions involved.                                                                (5)

                                                                                 OR

 

(i) Give a chemical reaction to distinguish between methanol and ethanol.                               (1+3+1)

(ii) How will you carry out the following conversions:

a) Benzaldehyde to α-hydroxyphenylacetic acid

b) Propanone to propene

(iii) Arrange the following in increasing order of acidic character:

  CH3COOH,  HCOOH,  C6H5OH,  2,4,6-trinitrophenol

 

33. (i) How conductivity is affected with dilution?                                                                    (1+3+1)

(ii) Calculate the emf of the following cell at 298 K:

Al(s)/Al3+ (0.15M)//Ni2+ (0.025M) /Ni(s)

(Given Eo(Al3+ Al) = -1.66 V, Eo(Ni2+ /Ni) = 0.25V, log 0.15 = -0.8239, log 0.025 = -1.6020)

(iii)           Calculate ∆Go for above cell reaction.

 

OR

 

(i) Why on dilution the molar conductivity of CH3COOH increases drastically, while that of

       CH3COONa increases gradually?

(ii) A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of silver metal is placed on a 1-molar solution of AgNO3. An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltmeter.

Calculate the cell potential, E, at 25oC for the cell if the initial concentration of  Ni(NO3)2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar.

[EoNi2+/Ni = -0.25 V,   EoAg+/Ag = 0.80 V,  log 10-1 = -1

 (iii) ) The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1.                              Calculate its molar conductivity.                                                                      

 (1+3+1)



NAVODAYA VIDYALAYA SAMITI BHOPAL REGION

PREBOARD - I   (2020 – 2021) CHEMISTRY – XII                                                          

 MARKING SCHEME

Q.No

VALUE POINT

MARKs

            1 i

a

1

ii

b

1

        or ii

a

1

            iii

d

1

             iv

a

1

                    2 i

b

1

                     ii

a

1

                    iii

a

1

                     iv

d

1

                    Or iv

c

1

     3

c

1

             4

c

1

      Or 4

d

1

              5

b

1

            6

d

1

        Or 6 or or 

b

1

             7

c

1

           Or 

      7

d

1

             8

b

1

      Or 8  or or 

b

1

             9

b

1

            10

a

1

    11

a

1

    12

d

1

    13

a

1

    14

a

1

    Or 14

a

1

    15

c

1

    16

c

1

            17


2

                 Or

     17

(I)  C6H5NH2   -------- NaNO2 +HCl /273 -278 K ---à C6H5N2Cl  -----CuCl/HCl   ---à C6H5Cl

(ii)  CH3CH(Br)CH3  ---alc KOH à CH3CH=CH2   -----  HBr/ peroxide à CH3CH2CH2Br     

1+1

                  18

PoA -PA/ PoA=WBXMA/MBX WA

760-745/760 =  5X 18/ MB X 95

MB  = 24 g/mol

 

½+1 +1/2

                 19

(i)d2sp3 hybridisation, magnetic moment = 5.9 BM                                               (ii)Dichloridobis(ethane -1,2-diammine)cobalt (III) nitrate

1/2 +1/2 +1      

 

       Or

      19

(i)Whenever a central metal ion is surrounded by poly dentate ligand to form ring type /cyclic structure is called chelate ex  [Co(en)3]+3

(ii) tetrahedral , magnetic moment = 2.83 BM

 

1 +1

                 2O


 

 

 

 

 

 

Or 20 or

For the reaction ,  2NH3  à N2  +3H2

Reaction rate = -1/2x d[NH3]/dt  =  d[N2]/dt  = 1/3x d[H2]/dt  = k[NH3]O  =K

Then , this relationship,

Rate of production of N= d[N2]/dt =k = 2.5 x 10-4 mol L -1S-1

Rate of production of H= d[H2]/dt = 3k =  3X2.5 x 10-4 = 7.5 X 10-4 mol L -1S-1

         Or

Let the rate law expression be

 Rate   Ro  = K [P]m [Q] n  

Using the given data

  3.0x 10-4  = K[0.10]m [0.10] n     eqn  (i)                                                                                                    9.0x 10-4  = K[0.30]m [0.30] n     eqn  (ii)

  3.0x 10-4  = K[0.10]m [0.30] n     eqn  (iii)

  6.0x 10-4  = K[0.20]m [0.40] n     eqn  (iv)

From  eqn  (iii)/ eqn  (i)    3.0x 10-4  / 3.0x 10-4   = K[0.10]m [0.30] n / K[0.10]m [0.10] n    

1  = (3)n ,  (3)0  =  (3)n   , n = 0    

From  eqn  (ii)/ eqn  (iii)    9.0x 10-4  / 3.0x 10-4   = K[0.30]m [0.30] n / K[0.10]m [0.30] n    

3 = (3)m ,  (3)1  =  (3)m   , m = 1

Thus the rate law expression for the reaction is     

Rate  = K [P]1 [Q] 0   

½ x4

    

 

 

 

 

 

 

 

 

½ x4

     21   

k = 0.693/t1/2

k = 0.693/5730 years-1

t = 2.303/ k log Co/Ct

 

let Co = 1, Ct = 3/10 so Co/Ct = 1/ (3/10) = 10/3

t = 2.303 x 5730/0.693 log 10/3

 

t  = 9957 years

½

 

 

1

 

 

1/2

                  22



       1+1

       23

   (a)


 (b) XeOF4 

1+1

      24

(a)    


(b)   CH2=CHCH2Br will react faster in SN2 reaction than CH2=CHBr  because in CH2=CHCH2Br , Br is attached sp3 hybridised carbon and in CH2=CHBr ,Br is attached sp2 hybridised carbon

 

1+1

      25

 Atoms of Element  P  forms ccp lattice or fcc lattice

Number of atoms of element P in ccp /fcc unit cell = 4

Total number of tetrahedral voids in unit cell  = 2x4 =8

Atoms of element Q occupy 1/3 rd of tetrahedral voids

Number of atoms of element Q in unit cell  = 1/3X8 = 8/3

Total number of octahedral voids in unit cell  =  4

Atoms of element R occupy 2/3 rd of octahedral voids

Number of atoms of element R in unit cell  =2/3X8 = 16/3

Formula of the compound =P4 Q8/3 R16/3

Multiplying by 4 ,           =P12Q8 R16 

dividing by 4 ,                  =P3Q2R4       

formula of the compound  is P3Q2R4

½ x4

       26

(i)                  (n-1)d electrons also participate in bond formation because (n-1)d electrons and ns  electrons has comparable energies

(ii)                Due to lanthanoid contraction.

(iii)               Cu+ easily disproportionates in aqueous solution as follows

2 Cu+ à  Cu+2  +   Cu

1+1+1

                   Or

      26

(i) The general trend towards less negative Eo V values across the series is

related to the general increase in the sum of the first and second

ionisation enthalpies.

(ii) The high energy to transform Cu(s) to Cu2+ (aq) is not balanced by its

hydration enthalpy.

(iii) The stability of the half-filled d sub-shell in Mn2+ and the completely

filled d10 configuration in Zn2+ are related to their more negative Eo V

values

 

 

 

 

1+1+1

                  27

i.    triethylamine<diethylethanamine < ethanamine ( increasing order of solubility in water)

ii. C2H5N(CH3)2 < (C2H5)2NH< C4H9NH2 (increasing order of boiling point)

iii. Ethanamine< N ,N-Diethyl ehanamine < N-Ethyl ethanamine (increasing order of Basic strength)

1+1+1

                   

                 Or  

       27

(i)                  Carbylamine test or Hinsberg test

(ii)                RCONH2 + Br2 + 4NaOH  à  RNH2 + Na2CO3 + 2NaBr + 2H2O

(iii)               Due to large hydrophobic -C6H5  group , aniline is insoluble in water.

1+1+1

       28

Density, d =z.M/a3 .NA

 d  =4 x 40/(400x10-10)3 x 6.022x1023

 d =4.15gcm-3

40 g (1 mol) element contain 6.022x1023 atoms or 6.022x1023/ 4  unit cell

So 4 g  element contain  6.022x1023x 4 /40x4 = 1.5x1023 unit cell

2+1

                 29

(i)            Proteins loses its biological activities on denaturation /secondary and tertiary structure is destroyed; primary structure remain intact

(ii)                The base – sugar unit in any nucleic acid chain is called a nucleosides whereas the base – sugar- phosphate unit in any nucleic acid chain is called  a nucleotides

 

(iii)               


  (Zwitter ion )

 

1+1+1

      30

(i)                  N does not have vacant d orbital  for sp3d hybridization

(ii)                Cl2 >  Br2  >  F2> I2

(iii)               Due to too small size and high electron density on fluorine it repel the incoming electron thus incoming electron does not experience much attraction force.

1+1+1

     31

 (i)3Cl2 + 6NaOH   à 5NaCl + NaClO3 +3H2O

(ii (a)  PbS + 4Oà PbSO4 +4O2

(b) 2H2SO4 + Cu  à   CuSO4+ SO2 +2H2O

(iii) (a) Larger atomic size and lower ionization energy of Xe, only xenon is known to form compounds with oxygen and fluorine

(b)    Interhalogen compound has lower bond enthalpy than parent halogen due to improper overlapping.

1+2+2

      

      31

     (i)BiH3  <  SbH3 < AsH3< PH3 < NH3

Smaller the size greater the electron density available to accept the proton.

 (ii)(a) HOCl < HOClO < HOClO2 < HOClO3.

Because order of stability of conjugate base is as follows on the basis  of number of resonating structure /delocalization of electron cloud.

OCl- < OClO- < OClO2 - < OClO3-

(b)Nitrogen does not vacant d orbital and can not form dπ-pπ bond like phosphorus.

(c)     Fluorine never acts as the central atom in polyatomic interhalogen compounds due to following reason i) very small size ii)most electronegative atom iii) absence of vacant d orbital

(iii) XeF6 + 3H2O     à XeO3 + 6HF

1+3+1

32



 

 

½ X4

 

 

 

 

 

 

 

 

1

1

1

Or 32

(i)                  Iodoform test

(ii)                (a)C6H5CHO  -------HCN---à C6H5CH(OH)CN -----H3O+ ---à C6H5CH(OH)COOH

(b) CH3COCH3 ---NaBH4---à CH3CH(OH)CH3 --conc.H2SO4/heat ---à CH3CH=CH2

       (iii)          C6H5OH<2,4,6-trinitrophenol <  CH3COOH <HCOOH

1+3+1

      33

(i)                  Conductivity decreases on  dilution because number of ion per unit volume decreases.

(ii)            Al(s)/Al3+ (0.15M)//Ni2+ (0.025M) /Ni(s)

(Given Eo(Al3+/ Al) = -1.66 V, Eo(Ni2+ /Ni) = 0.25V

2Al   à 2Al3+  + 6e

3Ni2++ 6e à 3 Ni

---------------------------------------

Net cell reaction: 2Al  3Ni2+ à 2Al3+  + 3 Ni

Here number of electron  exchange n= 6

Eocell = Eocathode -Eoanode = 0.25 -(-1.66) = 1.91 V

 

                         Ecell = Eocell – 0.059/n log [Al3+]2/ [Ni2+]3

                            =1.91 - 0.059/6 log[ 0.15]2/ [0.025]3

                             =1.91 - 0.059/6{ 2log[ 0.15]-3log [0.025]}

 

                             =1.91 - 0.059/6{ 2x-0.8239 -3 x-1.6020]}

                                      =1.91 - 0.059/6( -1.6478+ 4.806 )

                             =1.91 - 0.059/6x3.1582

                              = 1.91 – 0.031056

                    Ecell  =  1.87894V

   (iii)  ∆G=  -  nFEo cell   

            =  -6x 96500x 1.91

           =  - 1105898 j

           = - 1105.89 kj

   

1+3+1

                 Or 

     33

(i) On dilution the molar conductivity of CH3COOH increases drastically, while that of CH3COONa increases gradually because CH3COOH is weak electrolyte and dissociation of weak electrolyte increases on dilution whereas CH3COONa is strong electrolyte which is completely ionised at any concentration

 

1

 

 

(ii)             Ni   à  + Ni2++2e

2Ag++ 2e à 2Ag

---------------------------------------

Net cell reaction: Ni +2Ag+ à Ni+2+ 2Ag

Here number of electron  exchange n= 2

Eocell = Eocathode -Eoanode = 0.80 -(-0.25) = 1.05V

 

                         Ecell = Eocell – 0.059/n log [Ni2+]/ [Ag+]2

                            =1.05 - 0.059/2 log[ 0.1]/ [1.0]2

                               =1.05 - 0.059/2 log 10 -1

                            =1.05 - 0.059/2x(-1)

                           =1.05 + 0.0295

                            = 1.0795 V

3

 

(iii)          λm = k x 1000/C

0.025 x 1000/0.20

25/0.20 = 125Scm2mol-1

1

 

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