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PAPER - 2 : CODES

Q.No. 0 1 2 3 4 5 6 7 8 9

01. C D A C D C D C C D

02. A A C B C A C B A C

03. D C D C B D A D C B

04. B C D A D B D D D A

05. C B C D A C C A B D

06. D D B D C D B C D C

07. 7 2 7 2 7 2 2 3 2 3

08. 2 6 2 7 2 6 2 7 2 6

09. 2 3 2 6 3 2 7 6 7 7

10. 6 2 6 3 2 3 6 2 3 2

11. 3 7 3 2 6 7 3 2 6 2

12. B B B B B B B B B B

13. A C A C A C A C A C

14. D B D B D B D B D B

15. B B B B B B B B B B

16. C A C A C A C A C A

17. B D B D B D B D B D

18. A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s)

B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t)

C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q)

D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p)

19. A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s)

B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t)

C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q)

D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r)

20. A B C B D B D A B B

21. C C A D B C A D C B

22. B A B A D D C D A D

23. D B D C B D B B D C

24 B D B D C A B C D D

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Solutions

toIIT-JEE 2010

Time : 3 hrs. Max. Marks: 252

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472

Instructions :

1. The question paper consists of 3 parts (Chemistry, Mathematics and Physics) and each Partconsists of four Sections.

2. For each question in Section I , you will be awarded 5 marks if you have darkened only the bubblecorresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,minus two (2) mark will be awarded.

3. For each question in Section II , you will be awarded 3 marks if you have darkened the bubblecorresponding to the correct answer and zero mark if no bubbles are darkened. No negative markswill be awarded for incorrect answers in this Section.

4. For each question in Section III , you will be awarded 3 marks if you have darkened only the bubblecorresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,minus one (1) mark will be awarded.

5. For each question in Section IV , you will be awarded 2 marks for each row in which you havedarkened the bubble(s) corresponding to the correct answer. Thus, each question in this sectioncarries a maximum of 8 marks . There is no negative marks awarded for incorrect answer(s) in thisSection.

DATE : 11/04/2010

PAPER - 2 (Code - 5)
http://www.icbse.com/

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IIT-JEE 2010 (Paper-2)

PARTI : CHEMISTRY

SECTION - I

Single Correct Choice TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

1. In the reaction H C3 CO

NH2

(1) NaOH/Br 2

(2) CO

Cl

T , the structure of the product T is

(A)

H C3 CO

O C

O

(B)NH

C

O

CH 3

(C)NH

C

O

H C3(D)

H C3 CO

NH C

O

Answer (C)

Hints :

H C3 CO

NH2

NaOH + Br 2H C3 NH2

CO

Cl

H C3

NH

O

C

2. Assuming that Hunds rule is violated, the bond order and magnetic nature of the diatomic molecule B 2 is

(A) 1 and diamagnetic (B) 0 and diamagnetic

(C) 1 and paramagnetic (D) 0 and paramagnetic

Answer (A)

Hints :

Molecular orbital diagram for B 2 where Hunds rule is violated.

2X2 2 2 20Y

2p1 * 1 2 * 2

2ps s s s

Bond order = 1and diamagnetic.

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3. The species having pyramidal shape is

(A) SO 3 (B) BrF 3 (C)2 3SiO (D) OSF 2

Answer (D)

Hints :

SO F

F

Pyramidal

4. The complex showing a spin-only magnetic moment of 2.82 B.M. is

(A) Ni(CO) 4 (B) [NiCl4]2 (C) Ni(PPh 3)4 (D) [Ni(CN)4]

2

Answer (B)

Hints : Ni2+ is sp 3 hybridized and metal ion is connected with weak ligands.

Ni2+ d 8

sp 3two unpaired electrons

s = 2 (2 2) B.M.+

= 8 B.M. = 2.83 B.M.

5. The compounds P, Q and S

COOH

HOP

OCH 3

H C3Q

C

S

O

O

were separately subjected to nitration using HNO 3/H2SO 4 mixture. The major product formed in each caserespectively, is

(A)

COOH

HO

NO 2

OCH 3

H C3NO 2

CO

O

O N2

(B)

COOH

HO NO2

OCH 3

H C3NO 2

CO

O

NO 2

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(C)

COOH

HO

NO 2

OCH 3

H C3 NO 2

CO

O NO 2

(D)

COOH

HO

NO 2

OCH 3

H C3 NO 2

CO

O

NO 2

Answer (C)

Hints :

COOH

HO

HNO /H SO3 2 4COOH

HONO 2

OCH 3

H C3

HNO /H SO3 2 4 OCH 3

H C3 NO 2

C

O

O

HNO /H SO3 2 4C

O

O NO 2

6. The packing efficiency of the two-dimensional square unit cell shown below is

L

(A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54%

Answer (D)

Hints :

Packing efficiency =Area covered by particle

Total area

=

2

22 r

a

= ( )

2

2

2 r42 2 r

=

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SECTION - IIInteger Type

This section contains a group of 5 questions. The answer to each of the questions is a Single-digit Integer , ranging from0 to 9. The correct digit below the equation number in the ORS is to be bubbled.

7. Among the following, the number of elements showing only one non-zero oxidation state is

O, Cl, F, N, P, Sn, Tl, Na, Ti

Answer (2)

Hints : F will only exhibit 1 oxidation state except zero.and Na will exhibit +1 oxidation state.

8. The total number of diprotic acids among the following is

H3PO 4 H2SO 4 H3PO 3 H2CO3 H2S2O7H3BO3 H3PO 2 H2CrO 4 H2SO 3

Answer (6)

Hints : H2SO 4, H 3PO 3, H 2CO 3, H 2CrO 4 and H 2SO 3 and H 2S 2O7 will behave as dibasic acid.

9. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines asshown in the graph below. If the work done along the solid line path is W s and that along the dotted line path isWd, then the integer closest to the ratio W d/Ws is

0.50.0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.00.0

0.5

1.0

1.52.0

2.5

3.0

3.5

4.0

4.5

P(atm.)

V(lit.)

a

b

Answer (2)Hints :

Solid line path work done (W s) is isothermal beacuse PV is constant at each point & dash line path work done(Wd) is isobaric.

Total work done on solid line path (W s) = 2.303 nRT2

1

Vlog

V

= 2.303 PV 21

Vlog

V

= 2.303 5.5

2log0.5

4.6 l atmTotal work done on dash line path (W d) = 4 1.5 + 1 1 + 0.5 2.5

= 8.255 l atm.

d

s

W 8.252(closest integer)

W 4.6=

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10. Total number of geometrical isomers for the complex [RhCl(CO)(PPh 3)(NH3)] is

Answer (3)

Hints :Rh Cl(CO)(PPh 3)(NH3)] is a square planar complex with four different ligands and hence it will have threegeometrical isomers

a b

cd

Rh

a b

dc

Rh

a c

bd

Rh

11. Silver (atomic weight = 108 g mol 1 ) has a density of 10.5 g cm 3 . The number of silver atoms on a surface ofarea 10 12 m2 can be expressed in scientific notation as y 10 x. The value of x is

Answer (7)

Hints :

Volume of 1 Ag atom =34

r3

3 3

23

4 108r cm

3 6.023 10 10.5 =

r = 1.6 10 8 cm

r = 1.6 10 10 m

Number of Ag atoms in 10 12 m2

r2 n = 10 12 m2

12

10 2

10n 3.14 (1.6 10 )=

710 10n

8=

n = 1.25 10 7

The value of x = 7

SECTION - III

Paragraph TypeThis section contains 2 paragraphs . Based upon each of the paragraphs 3 multiple choice questions have to beanswered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions 12 to 14

The hydrogen-like species Li 2+ is in a spherically symmetric state S 1 with one radial node. Upon absorbing light theion undergoes transition to a state S 2. The state S 2 has one radial node and its energy is equal to the ground stateenergy of the hydrogen atom.

12. The state S 1 is

(A) 1s (B) 2 s (C) 2 p (D) 3 s

Answer (B)

Hints :

S 1 state is 2 s

In 2 s , one radial node is present

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13. Energy of the state S 1 in units of the hydrogen atom ground state energy is

(A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50

Answer (C)

Hints :

Energy in S 1 state2

2

13.6 3 9E 13.6eV

42= =

11( H )

E = 13.6 eV [Ground state]

So energy = 2.25 energy of e in ground state in 1H1.

14. The orbital angular momentum quantum number of the state S 2 is

(A) 0 (B) 1 (C) 2 (D) 3

Answer (B)

Hints :

S 2 state is 3 p orbital

Orbital angular momentum of 3 p is 1.


Two aliphatic aldehydes P and Q react in the presence of aqueous K 2CO 3 to give compound R, which upon treatmentwith HCN provides compound S. On acidification and heating, S gives the product shown below :

H C3

H C3OH

OO15. The compounds P and Q respectively are

(A) H C3

CH 3

C

CH H and H C3

O

C

O

H

(B) H C3

CH 3

C

CH H and H

O

C

O

H

(C)

H C3

CH 3

CHH C3

C

O

HCH 2C

O

Hand

(D)

H C3

CH 3

CHH

C

O

HCH 2C

O

Hand

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SECTION - IV

Matrix TypeThis section contains 2 questions . Each question has four choices (A), (B), (C) and (D) given in Column I and fivestatements (p), (q), (r), (s) and (t) in Column II . Any given statement in Column I can have correct matching withone or more statement(s) given in Column II . For example, if for a given question, statement B matches with the

statements given in (q) and (r), then for that particular question, against statement B, darken the bubbles correspondingto (q) and (r) in ORS.

18. All the compounds listed in Column I react with water. Match the result of the respective reactions with theappropriate options listed in Column II .

Column I Column II

(A) (CH 3)2SiCl2 (p) Hydrogen halide formation

(B) XeF 4 (q) Redox reaction

(C) Cl 2 (r) Reacts with glass

(D) VCl5 (s) Polymerization

(t) O 2 formationAnswer : A(p, s), B(p, q, r, t), C(p, q), D(p)

Hints :

(A) (CH ) SiCl + H O3 2 2 2 SiOH

CH 3

CH 3OH

Polymerise SiO

CH 3

CH 3O

n

(B) 4 2 3 26XeF 12H O 4Xe 2XeO 24HF 3O+ + + +

(C) 2 2Cl H O HCl HOCl+ +

(D) 5 2 3VCl H O VOCl HCl+ +

Note : Vanadium in (+V) oxidation state from only fluoride. Existence of VCl 5 is doubtful.

19. Match the reactions in Column I with appropriate options in Column II.

Column I Column II

(A) N Cl +2 OH N=NNaOH/H O

0C2 OH (p) Racemic mixture

(B) H C3 C C CH 3

OH

CH 3

OH

CH 3

H SO2 4 H C3CO

C

CH 3

CH 3

CH 3

(q) Addition reaction

(C) C

O

CH 3

1. LiAlH

2. H O4

3+ CH

OH

CH 3(r) Substitution reaction

(D)

Base

HS ClS

(s) Coupling reaction

(t) Carbocation intermediate

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PARTII : MATHEMATICS

SECTION - I

(Single Correct Choice Type)

This section contains 5 multiple choice questions . Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

20. Let f be a real-valued function defined on the interval (1, 1) such that ( ) 40

2 1x

x e f x t dt = + + , for allx (1, 1), and let f 1 be the inverse function of f . Then ( f 1) (2) is equal to

(A) 1 (B)13

(C)12

(D)1e

Answer (B)

Hints :

We have,

( ) 40

2 1x

x e f x t dt = + + x (1, 1)

Differentiating w.r.t. x , we get

( ) ( )( ) 4' 1x e f x f x x = +

( ) ( ) 4' 1 x f x f x x e = + +

f 1 is the inverse of f

f 1 (f (x )) = x

f 1 (f (x )) f (x ) = 1

( )( ) ( )1 1'

'f f x

f x =

( )( )( )

1

4

1'

1 x f f x

f x x e

=+ +

at x = 0, f (x ) = 2

( )1 1 1' 22 1 3

f = =+

21. A signal which can be green or red with probability45

and15

respectively, is received by station A and then

transmitted to station B . The probability of each station receiving the signal correctly is34

. If the signal receivedat station B is green, then the probability that the original signal was green is

(A)3

5(B)

6

7(C)

20

23(D)

9

20Answer (C)

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Hints :

From the tree-diagram it follows that

B G B R B G B R B G

AG AR AR AG

RG

S45

34

34

34

34

34

34

14

14 1

4

14

14

15

14

46( )

80=G P B

( ) 10 5|16 8G

P B G = =

( ) 5 4 18 5 2 = =G P B G

( )1

1 80 202|( ) 2 46 23G G

P G B P B

= = =

22. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to

(A) 25 (B) 34 (C) 42 (D) 41

Answer (D)

Hints : Let ,A B = ,A B S

The number of elements in A : 0 Choices for B : 2 4

The number of elements in A : 1 Choices for B : 38

The number of elements in A : 2 Choices for B : 3

Total number of possible subsets is 41

23. For r = 0, 1,..., 10, let Ar , B r and C r denote, respectively, the coefficient of x r in the expansions of

(1 + x )10 , (1 + x )20 and (1 + x )30 . Then ( )10

10 101

r r r r

A B B C A=

is equal to

(A) B 10 C 10 (B) ( )210 10 10 10A B C A (C) 0 (D) C 10 B 10

Answer (D)

Hints : Ar = Coefficient of x r in (1 + x )10 = 10C r

B r = Coefficient of x r in (1 + x )20 = 20C r

C r = Coefficient of x r in (1 + x )30 = 30C r

( )10 10 10

10 10 10 101 1 1

r r r r r r r r r r

A B B C A A B B A C A= = =

= 10 10

10 20 20 10 30 1010 10

1 1r r r r

r r C C C C C C

= ==

10 1010 20 20 10 30 10

10 10 10 101 1

r r r r r r

C C C C C C = =

= 10 10

20 10 20 30 10 1010 10 10 10

1 1.r r r r

r r C C C C C C

= ==

( ) ( )20 30 30 20

10 10 10 101 1C C C C =

30 2010 10 10 10C C C B = =

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24. If the distance of the point P (1, 2, 1) from the plane x + 2 y 2 z = , where > 0, is 5, then the foot of theperpendicular from P to the plane is

(A)8 4 7

, ,3 3 3 (B)

4 4 1, ,

3 3 3 (C)

1 2 10, ,

3 3 3 (D)

2 1 5, ,

3 3 2

Answer (A)Hints : Distance of point P from plane = 5

1 4 25

3

=

= 10

P (1, 2, 1)

Foot of perpendicular

1 2 1 51 2 2 3

x y z + = = =

8 4 7, ,3 3 3

x y z = = =

Thus the foot of the perpendicular is

8 4 7, ,

3 3 3f

25. Two adjacent sides of a parallelogram ABCD are given by

2 10 11AB i j k = + +

and 2 2AD i j k = + +

.

The side AD is rotated by an acute angle in the plane of the parallelogram so that AD becomes AD . If AD makes a right angle with the side AB , then the cosine of the angle is given by

(A)89

(B)179

(C)19

(D)4 5

9

Answer (B)

Hints :

2 10 11AB i j k = + +

2 2AD i j k = + +

Angle ' ' between AB

and AD

A B

C D

= ( ) ( )( ). 2 20 22 8

cos15 3 9

AB AD

AB AD

+ + = = =

( ) 17sin9

=

90 + =

( ) ( ) ( ) 17cos cos 90 sin9

= = =

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SECTION - II

(Integer Type)This Section contains 5 questions . The answer to each question is a single-digit integer , ranging from 0 to 9. The correctdigit below the question no. in the ORS is to be bubbled.

26. Let f be a function defined on R (the set of all real numbers) such thatf '(x ) = 2010 ( x 2009) ( x 2010) 2 (x 2011) 3 (x 2012) 4, for all x R .

If g is a function defined on R with values in the interval (0, ) such that f (x ) = ln ( g (x )), for all x R, then thenumber of points in R at which g has a local maximum is

Answer (1)

Hints : ( ) ( ),f x g x e x R =

( ) ( ) ( )' . 'f x g x e f x =

f (x ) changes its sign from positive to negative in the neighbourhood of x = 2009

f (x ) has local maxima at x = 2009

So, the number of local maximum is one.

27. Two parallel chords of a circle of radius 2 are at a distance 3 1+ apart. If the chords subtend at the center,

angles ofk

and2k

, where k > 0, then the value of [ k ] is

[Note : [k ] denotes the largest integer less than or equal to k ]

Answer (3)

Hints : Let 2 k

=

cos2x =

2

2

2

x

C

3 + 1 x

3 1cos22

x + =

2 3 12cos 1

2x + =

2 3 12 1

4 2x x + =

2 3 3 0x x + =

1 1 12 4 32

x + +=

1 13 4 32

+=

1 2 3 13

2 + += =

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3cos

2 6

= =

Required angle 23k

= = =

k = 3

28. Let k be a positive real number and let

2 1 2 2 0 2 1

2 1 2 and 1 2 0 2 .

2 2 1 2 0

k k k k k

A k k B k k

k k k k

= =

If det (adj A) + det(adj B ) = 10 6, then [ k ] is equal to

[Note : adj M denotes the adjoint of a square matrix M and [ k ] denotes the largest integer less than or equal tok ]

Answer (4)

Hints :

|A| = (2 k + 1) 3, | B | = 0

det (adj A) det (adj B ) = (2 k + 1) 6 = 10 6 9

.2

k =

[k ] = 4

29. Consider a triangle ABC and let a , b and c denote the lengths of the sides opposite to vertices A, B and C

respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3. If ACB is obtuse and if r denotesthe radius of the incircle of the triangle, then r 2 is equal to

Answer (3)

Hints :

3sin and

2C C = is given to be obtuse.

23

C =

30. Let a 1, a 2, a 3,......, a 11 be real numbers satisfying

a 1 = 15, 27 2 a 2 > 0 and a k = 2 a k 1 a k 2 for k = 3, 4, ...., 11.

If2 2 21 2 11.... 90,

11a a a + + + =

then the value of 1 2 11....

11a a a + + +

is equal to

Answer (0)

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SECTION-III (Paragraph Type)

This section contains 2 paragraphs . Based upon each of the paragraphs 3 multiple choice questions have to beanswered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Tangents are drawn from the point P (3, 4) to the ellipse2 2

19 4x y + = touching the ellipse at points A and B .

31. The coordinates of A and B are

(A) (3, 0) and (0, 2) (B)8 2 161 9 8

, and ,5 15 5 5

(C)

8 2 161, and (0, 2)

5 15

(D)

9 8

(3, 0) and ,5 5

Answer (D)

Hints :

Figure is self explanatory

B D

P (3, 4)

A(3, 0)

F

32. The orthocentre of the triangle PAB is

(A)8

5,7

(B)

7 25,

5 8 (C)

11 8,

5 5 (D)

8 7,

25 5

Answer (C)

Hints :

Equation of AB is

8850 ( 3) ( 3)

9 2435

y x x = =

95

, 85

P (3, 4)

(3, 0)

AB

1

( 3)3

y x =

x + 3 y = 3 ...(i)

Equation of the straight line perpendicular to AB through P is 3 x y = 5

Equation of PA is x 3 = 0

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The equation of straight line perpendicular to PA through 9 8,5 5

B

is y = 85

.

Hence the orthocentre is 11 8,5 5

33. The equation of the locus of the point whose distances from the point P and the line AB are equal, is

(A) 9x 2 + y 2 6 xy 54 x 62 y + 241 = 0

(B) x 2 + 9 y 2 + 6 xy 54 x + 62 y 241 = 0

(C) 9 x 2 + 9 y 2 6 xy 54 x 62 y 241 = 0

(D) x 2 + y 2 2 xy + 27 x + 31 y 120 = 0

Answer (A)

Hints :

Equation of 10 ( 3)3

AB y x = =

x + 3 y 3 = 0

|x + 3 y 3| 2 = 10[( x 3) 2 + ( y 4) 2]

(Look at coefficient of x 2 & y 2 in the answers)

Paragraph for question 34 to 36

Consider the polynomial f (x ) = 1 + 2 x + 3 x 2

+ 4 x 3. Let s be the sum of all distinct real roots of f (x ) = 0 and lett = | s |.

34. The real number s lies in the interval

(A) 1 , 04

(B) 311,4

(C) 3 1,4 2

(D) 10,4

Answer (C)

Hints :

f (x ) = 2(6 x 2 + 3 x + 1) = 9 24 < 0

Hence f (x ) = 0 has only one real root.

1 3 41 1 0

2 4 8f = + >

3 6 27 1081

4 4 16 64f = +

64 96 108 108

064

+ = 2, as ray bends away from normal2 > 3, for similar reasons

(r) Similar to (p)

(s) Similar to (q)

(t) 1 > 2, as ray bends away from normal

2 = 3, as there is no deviation

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chemistry 2010