(held on 20th may sunday 2018) - momentum academy...question paper with solutions of jee advanced -...

H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509

follow us on : momentumacademy www.momentumacademy.com momentumacademy

SECTION 1 (Maximum Marks: 24)

· This section contains SIX (06) questions.

· Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s)is (are) correct option(s).

· For each question, choose the correct option(s) to answer the question.

· Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of which

are correct options.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct

option.

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : –2 In all other cases.

For Example: If first, third and fourth are the ONLY three correct options for a question with second optionbeing an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting onlytwo of the three correct options (e.g. the first and fourth options), without selecting any incorrect option(second option in this case), will result in +2 marks. Selecting only one of the three correct options (eitherfirst or third or fourth option) ,without selecting any incorrect option (second option in this case), will resultin +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of anycorrect option(s) will result in –2 marks.

1. The correct option(s) regarding the complex 33 3 2[ ( )( ) ( )]Co en NH H O +

2 2 2 2( )en H NCH CH NH= is (are)

(A) It has two geometrical isomers(B) It will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide ligands(C) It is paramagnetic

(D) It absorbs light at longer wavelength as compared to 33 4[ ( ) ) ]Co en NH +

Sol.(A,B,D)

(A) 33 3 2[ ( )( ) ( )]Co en NH H O + has 2 geometrical isomers

Our Top class IITian faculty team promises to give you an authentic solutions which will befastest in the whole country.

CHEMISTRY [ PAPER-2 ]

FEEL THE POWER OF OUR KNOWLEDGE & EXPERIENCE

QUESTION PAPER WITH SOLUTIONS OF JEE Advanced - 2018

(HELD ON 20th MAY SUNDAY 2018)



(B) Compound 2 3 3 2[ ( ) ( ) ( )]Co CN NH H O + will have three geometrical isomers.

(C) 33 2[ ( )( )( )]CO en NH H O + is diamagnetic

(D) 33 3 2[ ( )( ) ( )]Co en NH H O + absorbs light at longer wavelength as compared to

33 4[ ( )( ) ]Co en NH + as

2H O is weaker ligand than 3NH .

\ (A,B, D) is correct answer

2. The correct option(s) to distinguish nitrate salts of 2Mn + and 2Cu + taken separately is (are)

(A) 2Mn + show the characteristic green colour in the flame test

(B) Only 2Cu + show the formation of precipitate by passing 2H S in acidic medium

(C) Only 2Mn + show the formation of precipitate by passing 2H S in faintly basic medium

(D) 2 /Cu Cu+ has higher reduction potential than 2 /Mn Mn+ (measured under similar conditions)

(A) (B) (C) (D)

Sol.(D)

\ (D) is correct answer



4. The Fishcher presentation of D-glucose is given below :

The correct structure(s) Lb - - glucopyranose is (are)

(A) (B) (C) (D)

Sol.(D)


5. For a first order reaction ( ) 2 ( ) ( )A g B g C g® + at constant volume and 300 K , the total pressure at

the beginning ( = 0)t and at time t are 0P and tP , respectively. Initially, only A is present with concentration

0[ ]A , and 1/3t is the time required for the partial pressure of A to reach 1/ 3rd of its initial value. TheThe

correct option(s) is (are)(Assume that all these gases behave as ideal gases)

(A) (B) (C)



Sol.(A,D)

First order( ) 2 ( ) ( ) constant, T = 300 KA g B g C g V¾¾¾¾® + =

0t = 0P

1/3t t=0

0

2

3

PPæ ö

-ç ÷è ø

04

3

P 02

3

P

0

3

P=

t t= 0P x- 2x x

So, 0 02 2tP P x x x P x= - + + = +

or 02 tx P P= -

0

0

1ln

( )

Pt

k P x=

-

or 0 0

0 0 00

21 1ln ln

( ) 2

2t t

P pt

P Pk k P P PP

= =- - +-

or0

0 0

0

2ln , ln 2 ln(3 )

3t

t

PKt Kt P P P

P P= = - -

-

or 0 0ln(3 ) ln 2tP P Kt P- = - +

Graph between In 0(3 )tP P- vs ‘t’ is a straight line with negative slope.

So (A) is correct option

01/ 3

0

1 1ln ln 3

( / 3)

Pt

K P K= = Þ It is independent of initial concentration.

So (B) is wrong option

As rate constant is a constant quantity and independent of initial concentration.

So Graph (D) is correct.

\ (A,D) is correct answer



6. For a reaction, A Pˆ †̂‡ˆ̂ , the plots of [ ]A and [ ]P with time at temperature 1T and 2T are given below..

If 2 1T T> , the correct statement(s) is (are)

(Assume H qD and SqD are independent of temperature and ratio of lnK at 1T to lnK at 2T is greater than

2 1/T T . Here , ,H S G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.).)

(A) 0, 0H Sq qD < D < (B) 0, 0G Hq qD < D <

(C) 0, 0G Sq qD < D < (D) 0, 0G Sq qD < D >

Sol.(A,C)

As temperature increases concentration of product decreases

so reaction is exothermic 0HÞD ° <

1

1 2

2

ln1 ln ln

ln

T

T

KK KT

KT> Þ > So,

1 2T TK K>

Also, 1 2

2 1

ln

ln

TK T

KT T>

or1 2 21 2 1 1 2ln ln ln lnT T TT K T K R T KT RT K> Þ - < -

or1 2T TG G° °D < D

or 1 2H T S H T SD °- D ° < D ° - D °

As 1 2T TG G° °D < D , since as temperature increase GD increases this is possible only when 0SD ° <

\ (A,C) is correct answer




· This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE.

· For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to thesecond decimal place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the onscreenvirtual numeric keypad in the place designated to enter the answer.

· Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 If ONLY the correct numerical value is entered as answer.

Zero Marks : 0 In all other cases.7. The total number of compounds having at least one bridging oxo group among the molecules given below

is _________.

2 3 2 5 4 6 4 7 4 2 5 5 3 10 2 2 3 2 2 5, , , , , , ,N O N O PO PO H PO H PO H S O H S O

Sol.(6)

\ (6) is correct answer



8. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, thepassage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo

self-reduction. The weight (in kg) of Pb produced per kg of 2O consumed is _____.

(Atomic weights in g mol–1 : 16, 32, 207O S Pb= = =

Sol. 6.47 kg.

2 2PbS O Pb SO+ ¾¾® +

Mole

310

32

Moles of Pb formed

310

32=

\ Mass of Pb formed

310207 6468.75

32gm= ´ =

6.46875kg=

6.47 kg=

\ 6.47 kg is correct answer

9. To measure the quantity of 2MnCl dissolved in an aqueous solution, it was completely converted to

4KMnO using the reaction.

2 2 2 8 2 4 2 4MnCl K S O H O KMnO H SO HCl+ + ® + + (equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid

( 225mg ) was added in portions till the colour of the permanganate ion disappeared. The quantity of

2MnCl (in mg) present in the initial solution is _______ .

(Atomic weights in 1g mol - : 55, 35.5Mn Cl= = )

Sol. 126 mg

2 2 2 8 2 4 2 4 2 42 5 8 2 4 6 4MnCl K S O H O KMnO K SO H SO HCl+ + ¾¾® + + + ....(1)

4 2 2 4 2 4 2 4 4 2 22 5 3 2 8 10KMnO H C O H SO K SO MnSO H O CO+ + ¾¾® + + + ....(2)

Mass of oxalic acid added 225mg=

Milimoles of oxalic acid added 225

2.590

= =

From equation (2)

Milimoles of 4KMnO used to react with oxalic acid =1

and milimoles of 2MnCl required initially =1

\ Mass of 2MnCl required initially 1 126 126mg= ´ =

\ 126 mg is correct answer



10. For the given compound X , the total number of optically active stereoisomers is .......

Sol.(7) \ (7) is correct answer

11. In the following reaction seuquence, the amount of D (in g) formed from 10 moles of acetophenone is .....

(Atomic weights in g mol-1 : 1, 12, 14, 16, 80H C N O Br= = = = = . The yield (%) corresponding to the

product in each step is given in the parenthesis)

Sol. 495 g

moles of D formed 10 0.6 0.5 0.5 1 1.5= ´ ´ ´ ´ =

Mass of D formed 1.5 330 495= ´ = gram

\ 495 g is correct answer



12. The surface of copper gets tarnished by the formation of copper oxide. 2N gas was passed to prevent the

oxide formation during heating of copper at 1250K . However, the 2N gas contains 1 mole % of water

vapour as impurity. The water vapour oxidises copper as per the reaction given below :

2 2 22 ( ) ( ) ( ) ( )Cu s H O g Cu O s H g+ ¾¾® +

2pH is the minimum partial pressure of 2H (in bar) needed to prevent the oxidation at 1250 K . TheThe

value of 2

ln ( )Hp is _______ .

(Given : total pressure = 1 bar, R (universal gas constant) 1 18J K mol- -= , ln(10) 2.3, ( )Cu s= and

2 ( )Cu O s are mutually immiscible.

At 1

2 2

11250 : 2 ( ) ( ) ( ); 78,000

2K Cu s O g Cu O s G J mol -+ ¾¾® D ° = -

12 2 2

1( ) ( ) ( ); 1,78,000 ;

2H g O g H O g G J mol G-+ ¾¾® D ° = - is the Gibbs energy)

Sol. -14.6

(i) 2 2

12 ( ) ( ) ( ) : 78000 /

2Cu s O g Cu O s G J mol+ ¾¾® D ° = -

(ii) 2 2 2

1( ) ( ) ( ), 178000 /

2H g O g H O g G J mol+ ¾¾® D ° = -

(i) - (ii) then

2 2 22 ( ) ( ) ( ) ( )Cu s H O g Cu O s H g+ ¾¾® +

78000 178000 100000 /G J molD ° = - + =

Now for the above reaction

2

2

nH

H O

PG G RT

P

æ öD = D °+ ç ÷ç ÷

è øl

To prevent the above reaction :

0GD ³

2

2

n 0H

H O

PG RT

P

æ öD °+ ³ç ÷ç ÷

è øl

2 2

4 510 ( n n ) 10H H OP P- ³ -l l

2 2n 10 nH H OP P³ - +l l

10 2.3log(0.01)³ - +

2n 10 4.6HP ³ - +l

2n 14.6HP ³ -l

\ Minimum 2

n 14.6HP = -l

\ -14.6 is correct answer



13. Consider the following reversible reaction, ( ) ( ) ( )A g B g AB g+ ˆ †̂‡ˆ̂

The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J mol–1).

If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute

value of GqD (in J mole–1) for the reaction at 300 K is _____.

(Given : ln (2) 0.7= , 2500RT J= mol–1 at 300 K and G is the Gibbs energy)

Sol. -8500

( ) ( ) ( )A g B g AB g+ ˆ †̂‡ˆ̂

2b fa aE E RT= + & 4f bA A=

Now, Rate constant of forward reaction /fEa RT

f fk A e-

=

Rate constant of reverse reaction /bEa RT

b bK A e-=

Equilibrium constant

( ) /af abE E RTf f

eq

b b

K AK e

K A

--= = 2 24 4e e+= =

Now,2ln 2500 ln(4 )eqG RT K eD ° = - = -

22500 (ln 4 ln )e= - +

2500 (1.4 2) 2500 3.4= - + = - ´

8500 / moleJ= -

\ -8500 J/mole is correct answer

14. Consider an electrochemical cell : 2( ) | ( , 2 ) || ( ,1 ) | ( )n nA s A aq M B aq M B s++ . The value of HD ° for

the cell reaction is twice that of GD ° at 300 K . If the emf of the cell is zero, the SD ° (in 1 -1molJ K - ) of

the cell reaction per mole of B formed at 300 K is ______.

(Given : In (2) 0.7, R= *universal gas constant) 1 -18.3 molJ K -= . ,H S and G are enthalpy, entropy

and Gibbs energy, respectively.)Sol.-11.62

2( ) | ( , 2 || ( ,1 ) | ( )n nA s A aq M B aq M B s++

Reactions

Anode ( ) 2nA A ne+¾¾® + ´

Cathode 2 2nB ne B+ + ¾¾®_______________________________

Overall reaction :

22 ( ) 2n nA s B A B+ ++ ¾¾® +

n2

RTE E Q

nF= ° - l

2

2

[ ]n

2 [ ]

n

n

RT AO E

nF B

+

+= ° - l



n 42

RTE

nF° = l

Now,2

2 n 4 n 42

nFRTG nFE RT

nF

-D ° = - ° = = -l l

2G H T S G T SD ° = D °- D ° = D ° = - D °

T S GD ° = D °

n 4n 4

G RTS R

T T

D ° -D ° = = = -

ll

1 -18.3 2 0.7 11.62 molJK -= - ´ ´ = -

\ -11.62 is correct answer


• This section contains FOUR (04) questions.

• Each question has TWO (02) matching lists: LIST-I and LIST-II.

• FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of thesefour options corresponds to a correct matching.

• For each question, choose the option corresponding to the correct matching.

• For each question, marks will be awarded according to the following marking scheme:

Full Marks : 3 If ONLY the option corresponding to the correct matching is chosen.

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : 1 In all other cases.15. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II

LIST-I LIST-II

(P) 2dsp (1) 46[ ]FeF -

(Q) 3sp (2) 2 3 3[ ( ) ]Ti H O Cl

(R) 3 2sp d (3) 33 6[ ( ) ]Cr NH +

(S) 2 3d sp (4) 24[ ]FeCl -

(5) 4( )Ni CO

(6) 24[ ( ) ]Ni CN -

The correct option is

(A) P®5 ; Q®4, 6 ; R®2, 3 ; S®1

(B) P®5, 6 ; Q®4 ; R®3 ; S®1, 2

(C) P®6 ; Q®4, 5 ; R®1 ; S®2, 3

(D) P®4, 6 ; Q®5, 6 ; R®1, 2 ; S®3Sol. (C)

P®6 ; Q®4, 5 ; R®1 ; S®2, 3

1. 4 66 3FeF d- & weak field ligand

\ Hybridization is 3 2sp d



2.1

2 3 3[ ( ) ]3Ti H O Cl d & weak field ligand

\ Hybridization is 2 3d sp

3. 3 33 6[ ( ) ] ,3Cr NH d+ & strong field ligand

\ Hybridization is 2 3d sp

4. 2 64[ ] 3FeCl d- & weak field ligand

\ Hybridization is 3sp

5. 104( ) , 3Ni CO d

\ Hybridization is 3sp

6. 2 84[ ( ) ] , 3Ni CN d-

\ 2dsp hybridization

\ (C) is correct answer

16. The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one ormore appropriate reagents in LIST-II.(given, order of migrarylory aptitude: ary > alkyl > hydrogen)

LIST-I LIST-II

(P) 2 4H SO+ (1) 2 ,I NaOH

(Q) 2HNO+ (2) 3 2[ ( ) ]Ag NH OH

(R) 2 4H SO+ (3) Fehling solution

(S) 3AgNO+ (4) ,HCHO NaOH

(5) NaOBr



The correct option is

(A) P®1 ; Q®2, 3 ; R®1, 4 ; S®2, 4

(B) P®1, 5 ; Q®3, 4 ; R®4, 5 ; S®3

(C) P®1, 5 ; Q®3, 4 ; R®5 ; S®2, 4

(D) P®1, 5 ; Q®2, 3 ; R®1, 5 ; S®2, 3Sol.(D)

(P)

(S)




17. LIST-I contains reactions and LIST-II contains major products.LIST-I LIST-II

(P) (1)

(Q) (2)

(R) (3)

(S) (4)

(5)

Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option

Sol.(B)

(Q)

(R)

(S)

\ (B) is correct answer



18. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of

the solution on [ ]H + are given in LIST-II.

(Note : degree of dissociation ( )a of weak acid and weak base is <<1; degree of hydrolysis of salt <<1;

[ ]H + represents the concentration of H + ions)

LIST-I LIST-II

(P) (10mL of 0.1M 20NaOH mL+ of (1) The value of [ ]H + does not change on dilution

0.1M acetic acid) diluted to 60mL

(Q) ( 20mL of 0.1 20M NaOH mL+ of (2) The value of [ ]H + changes to hafd of itss

0.1M acetic acid) diluted to 80mL initial value on dilution

(R) ( 20mL of 0.1 20M HCl mL+ of (3) The value of [ ]H + changes to two times of itss

0.1M ammonia solution) diluted to 80mL initial value on dilution

(S) 10mL saturated solution of 2( )Ni OH in (4) The value of [ ]H + changes to 1

2 times of itss

equilibrium with excess solid 2( )Ni OH initial value of dilution

is diluted to 20mL (solid 2( )Ni OH is

still present after dilution)

(5) The value of [ ]H + changes to 2 times of itss

initial value on dilution

Match each process given in LIST-1 with one or more effect(s) in LIST-II the correct option is

(A) P®4 ; Q®2 ; R®3 ; S®1

(B) P®4 ; Q®3 ; R®2 ; S®3

(C) P®1 ; Q®4 ; R®5 ; S®3

(D) P®1 ; Q®5 ; R®4 ; S®1

Sol.(D) P®1 ; Q®5 ; R®4 ; S®1

(P) 3 3 2NaOH CH COOH CH COONa H O+ ¾¾® +

M.Mole 1 2

Now solution contains 1 m. mole 3CH COOH & 1 m.mole 3CH COONa in 30ml solution.

It is a Buffer solution

\ [ ]H + does not change with dilution.



(Q)3 3 2NaOH CH COOH CH COONa H O+ ¾¾® +

M.Mole 2 2

Now solution contain 2 m.mole of 3CH COONa in 40 ml solution (salt of weak acid strong base)

initial[ ] w aK KH

C+ =

Now on dilution upto 80ml , now can. Becomes 2

C.

\ new initial[ ] [ ] 2/ 2

w aK KH H

C+ += = ´

(R)3 4HCl NH NH Cl+ ¾¾®

M.Mole 2 2

Now solution contain 2 m.mole of 4NH Cl in 40ml solution (salt of SA & WB)

initial[ ] w

b

K CH

K+ =

Now on dilution upto 80ml , new conc. becomes 2

C

\initial

new

[ ][ ]

2 2

w

b

K C HH

K

++ = =

(S) 22( ) ( ) 2Ni OH s Ni OH+ -¾¾® +

Q it is sparingly soluble salt

\ on dilution [ ]OH - conc. in saturated solution of 2( )Ni OH remains const.

\ new initial[ ] [ ]H H+ +=

(held on 20th may sunday 2018) - momentum academy...question paper with solutions of jee advanced -...

Documents