2017-Jee-Advanced
Question Paper-1_Key & Solutions PART-2:CHEMISTRY SECTION -1 (Maximum Marks: 28)
This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : -2 In all other cases. For example, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also darkened.
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Q19.
The colour of the
X 2 molecules of group 17 elements changes gradually from yellow to violet down the group.
This is due to A) The physical state of
X 2 at room temperature changes from gas to solid down the group
B) Decrease in HOMO-LUMO gap down the group C) Decrease in
* O * gap down the group
D) Decrease ionization energy down the group Key:
BC
Sol:
IF Consider,
X 2 as F2
F2 MoT config is / s 2 * /s 2 2s 2 * 2s 2 * 2 pz 2 2 px 2 = 2 p y 2
*2 px2 *2 py2 HOMO * 2 pz0 LuMo Q20.
if gap between HOMO & LUMO Due to absorbed energy decreases complimentary colour is seen light energy is seen both ‘B’ & ‘C’ are thus applicable
Addition of excess aqueous ammonia to a pink coloured aqueous solution of
MCl2 .6H 2O (X) and NH 4Cl
gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1: 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. 2
A) The hybridization of the central metal ion in Y is d sp
3
B) When X and Z are in equilibrium at 0C , the colour of the solution is pink C) Z is a tetrahedral complex D) Addition of silver nitrate to Y gives only two equivalent of silver chloride ABC
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Key:
2017-Jee-Advanced Sol:
Question Paper-1_Key & Solutions
X Co H 2O 6 Cl2 pink sp3d 2 Y Co NH 3 6 Cl3 d 2 sp3
Z CoCl4 blue sp3 Option (B) is correct CBSE book equilibrium chapter. Hence ‘A’ is correct Z is tetrahedral , ‘C’ is correct Q21.
An ideal gas is expanded from
p1,V1,T1 to p2 ,V2 , T2 under different conditions. The correct statement(s)
among the following is (are) A) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic B) The work done by the gas is less when it is expanded reversibly from V1 to V2 under adiabatic conditions as compared to that when expanded reversibly from V1 to
V2 under isothermal conditions
C) The work done on the gas is maximum when it is compressed irreversibly from constant pressure
p2 ,V2 to p1 ,V1 against
p1
D) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with if it is expanded reversibly under adiabatic conditions with Key:
BC
Sol:
in free expansion the heat supplied in zero (z=0)
T1 T2 , and (ii) positive,
T1 T2
And work done is zero in both isothermal and adiabatic process But T1 T2
isothermal
P
Adiabatic process
V In adiabatic process
PV cons tan t
is always greater than 1 so, opp. Pressure is less for same exp is Case of adiabatic process compare to isothermal so work done is less.
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2017-Jee-Advanced
Question Paper-1_Key & Solutions
compression
PV 1 1
P2 ,V2
P
V Expansion
PV 1 1 P
PV 2 2
V Work done in compression is more than expansion or (Area under curve is more) wirr wrev comp
Exp
opp., pressure is more in compression Pext V In case of adiabatic expansion E And temperature also . Q22.
For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here
xL and xM represent mole fractions of L and M, respectively,
in the solution. The correct statement(s) applicable to this system is (are) Z
pL
1
xM
0
A) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mixed in solution
Key:
B) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when
xL 0
C) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from
xL 0 to xL 1
D) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when
xL 1
AD actual given behaviour
Z
Sol:
Xm 1 XL 0
xM
expected ideal behaviour
XM 0 XL 1
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Q
pL
2017-Jee-Advanced
Question Paper-1_Key & Solutions
As actual is above expected ideal
+ve deviation L – L attn > L – M attn Also m –mattn > L – M attm Pt. z X L 1 only ‘L’ is present PL PL , PM 0
@ Z PL PL0 From pt Q to Z the graph (actual) is very similar to expected ideal behaviour where X L 1 it mm ideal behaviour . Q23.
The IUPAC name(s) of the following compound is (are)
Cl
H 3C
A) 1-chloro-4-methybenzene B) 4-chlorotoluene C) 1-methy-4-chlorobenzene D) 4-methylchlorobenzene Key:
AB
Cl 1
6
2
5
3 4
CH 3
Sol:
Alphabetical order is the priority order “C” of chloro has greater priority over “W” of methyl 1-chloro -4-methyl benzene A 4- Chlorotoluene B According to NCERT BOOK The correct statement(s) for the following addition reactions is (are)
H 3C H (i)
H 3C (ii)
H
H
Br2 CHCl3
M and N
CH 3 CH 3
Br2 CHCl3
O and P
H
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Q24.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
A) O and P are identical molecules B) Bromination proceeds through trans-addition in both the reactions C) (M and O) and (N and P) are two pairs of enantiomers D) (M and O) and (N and P) are two pairs of diastereomers Key:
BD
Sol: CH 3
CH 3
Br2 / CHCl3
(i )
H
OH
H
OH
HO
H
HO
H
CH 3
CH 3 M
N
CH 3
CH 3
(ii )
OH
H
Br2 CHCl3
H
H
H
HO
HO
OH CH 3
CH 3 O
P
Bromination is anti addition . M & N are identical (M& O) & (N& P) two pairs of diastereomere The correct statement(s) about the oxoacids, A) The conjugate base of
HClO4 and HClO , is(are)
HClO4 is weaker base than H 2O
B) The central atom in both
HClO4 and HClO is sp 3 hybridized
C)
HClO4 is formed in the reaction between Cl2 and H 2O
D)
HClO4 is more acidic than HClO because of the resonance stabilization of its anion
Key:
ABD
Sol:
HClO4 and HClO O O
Cl
O
O
is resonance stabilized, so charge density is less, it is a weak base than mo
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Q25.
2017-Jee-Advanced O
Question Paper-1_Key & Solutions
sp
Cl
3
Cl
OH
O
O
sp
O
3
so HClO4 is strong acid due to resonance stabilization of anion
ClO4 SECTION -2 (Maximum Marks: 15)
26.
This section Contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Negative Marks : -2 In all other cases. **************************************************************************************************
The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is 120 cm with an area 7
of cross section of 1cm . The conductance of this solution was found to be 5 10 S . The pH of the solution is 2
4. The value of limiting molar conductivity m of this weak monobasic acid in aqueous solution is 0
Z 102 S cm1 mol 1 . The value of Z is Key: Sol:
6
120 cm 7 1 K G cell constant k 5 107 S 2 600 10 S cm 1 cm 7 7 600 10 1000 600 10 1000 m 40 C 0.0015M pH 4 HT C 104
1 15
M 0m 40 0m m 600 1/ 15 600S cm2 mol 1 6 102 2 2 10 The sum of the number of lone pairs of electrons on each central atom in the following species is
TeBr6 , BrF2 2
Key:
, SNF3 , and XeF3
(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54) 6
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27.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
F F
F 3 3 F sp d
Te F Sol:
F
Br
2 lonepair F
F
N
1 lonepair
F
S
F
0 zero-lone pair
F
Xe
3 lonepair
Total 6 lone pair 28.
Among the following, the number of aromatic compounds(s) is
Key: Sol:
5
Aromatic
Aromatic
ions
Aromatic
Aromatic
Compounds
Aromatic
* It is clearly mentioned in the question word “compounds” then answer is “2” only
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Question Paper-1_Key & Solutions
A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density 3
of the substance in the crystal is 8 g cm , then the number of atoms present in 256 g of the crystal is N 10 . The value of N is
Key: Sol:
24
2 FCC: Z=4
d 8
Z M a 3 .N O 4 ( Mol.Wt )
(400)3 1012 NO 3
mol.mass 2 400 1012 N0 3
2 400 1012 N 0 3
no.of moles
256
2 400 1012 N 0 3
No.of atoms
256
2 N0
H 2 , He2 , Li2 , Be2 , B2 , C2 , N 2 , O2 , and F2 the number of diamagnetic species is (Among numbers: H 1, He 2 , Li 3 , Be 4 , B 5 , C 6 , N 7 , O 8 , F 9 )
30.
Among
Key: Sol:
5
H 2 diamagnetic He2 paramagnetic
Li2 Diamagnetic Be2 Does not exist B2 Paramagnetic due to s p mixing C2 diamagnetic due to s p mixing N2 diamagnetic(14e ) O2 paramagnetic(17e )
F2 diamagnetic
SECTION -3 (Maximum Marks: 18) This section Contains SIX questions of matching type. This section Contains TWO tables (each having 3 columns and 4 rows) Based on each table, there are THREE questions Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks :0 If none of the bubbles is darkened Negative Marks : -1 In all other cases.
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2017-Jee-Advanced
Question Paper-1_Key & Solutions **************************************************************************************************
Answer Q.31,Q.32 Q33 by appropriately matching the information given in the three columns of the following table. The wave function n,l ,ml is a mathematical function whose value depends upon spherical polar coordinates (r,
, ) of the electron and characterized by the quantum numbers n, l and ml . Here r is distance from nucleus, is colatitude and is azimuth. In the mathematical functions given in the Table, Z is atomic number and a0 is Bohr radius. Column 1
Column 2
(I) Is orbital
Column3 3
(i) n,l ,ml
Zr
Z 2 e a0 a0
n,l ,m r 1
0
r / a0
(P) (II) 2S orbital
(ii) one radial node
(Q) Probability density at nucleus
5
(III) 2 Pz orbital
(iii) n,l ,ml
Zr
Z 2 re 2 a0 COS a0
(iv) xy-plane is a nodal plane
(IV) 3d z 2 orbital
1 a03
(R) Probability density is maximum at nucleus (S) Energy needed to excite electron from n=2 state to n= 4 state is
27 32
times the energy needed to excite electron from n=2 state to n=6 state
Key: Sol:
For He ion, the only INCORRECT combination is A) (I) (i) (R) B) (II) (ii) (Q) C) (I) (i) (S) D 1s 0 radial nodes , no angular component ( no cos ) 2s 1 radial node, , no angular component ( no cos )
D) (I) (iii) (R)
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos )
a)1s
(i) ‘R’ 1s orbital ’0’ radial node probability max@nucleas correct b)2s 1 radial node of correct c) 1s (i) ’s’ correct
1s (i ) ' s ' correct
(2 4)
3 16
(2 6)
8 36
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Q31.
2017-Jee-Advanced
Question Paper-1_Key & Solutions
3 27 16 8 32 36 d ) 1s no angular component cos should not be present wrong. Q32. Key: Sol:
For the given orbital in column I, the only CORRECT combination for any hydrogen-like species is A) (I) (ii) (S) B) (IV) (iv) (R) C) (III) (iii) (P) D) (II) (ii) (P) D 1s 0 radial nodes , no angular component ( no cos ) 2s 1 radial node, , no angular component ( no cos )
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos )
a) 1s 0 radial node wrong as (ii),(i) 1 radial node b) 3d z2 dz 2 ' R ' is wrong po int
Q33. Key: Sol:
c)for 2 pz graph ' p ' iswrong d) correct For hydrogen atom, the only CORRECT combination is A) (II) (i) (Q) B) (I) (iv) (R) C) (I) (i) (P) D 1s 0 radial nodes , no angular component ( no cos ) 2s 1 radial node, , no angular component ( no cos )
D) (I) (i) (S)
2 pz 0 radial node, angular component is present ( cos ) 3d z2 0 radial node, angular component is present ( cos )
a) 2s 1radial node / coloumn(2) (i) no radial node wrong b) 1s no node plane wrong c) 1s no radial node coloumn (3) ‘p’ 1 radial node wrong d) correct Answer Q.34,Q.35 and Q.36 by appropriately matching the information given in the three columns of the following table. Columns 1, 2 and 3, contain starting materials, reaction conditions, and type of reactions, respectively. Column-1
Key:
Column-3
(I) Toluene
(i) NaOH / Br2
(P) Condensation
(II) Acetophenone
(ii) Br2 / hv
(Q) Carboxylation
(III) Benzaldehyde
(iii) CH 3CO 2 O / CH 3COOK
(R) Substitution
(IV) Phenol
(iv) NaOH / CO2
(S) Haloform
The only CORRECT combination in which the reaction proceeds through radical mechanism is A) (II) (iii) (R)
B) (III) (ii) (P)
C) (IV) ii) (Q)
D) (I) (ii) (R)
D
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Q34.
Column-2
2017-Jee-Advanced CH 3
Question Paper-1_Key & Solutions CH 2 Br
Br2 / h
Sol:
Free radical substitution Ans: (I) (II) (R) Q35.
For the synthesis of benzoic acid, the only CORRECT combination is A) (III) (iv) (R)
Key:
B) (IV) (ii) (P)
C) (II) (i) (S)
D) (I) (iv) (Q)
C
Sol: COOH
0
NaoH / Br
C CH 3
afterH
+
CHBr3
Acetophenone
Haloform r n
Q36.
The only CORRECT combination that gives two different carboxylic acids is A) (IV) (iii) (Q)
Key:
S
B) (I) (i) (S)
C) (III) (iii) (P)
D) (II) (iv) (R)
C
Sol: CH O
AC2O
CH CH COOH ACOH
CH 3COOK Cis & trans After acidification
Condensation R n P
(III)(iii)(P)
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(III)
