Transcript
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PART-I
1. The major product of the following reaction aqueous NaOH⎯⎯⎯⎯⎯→ are:
(a) Br3C–OH and
(b) and CHBr3
(c) and NaBr (d) PhH and CBr3CO2Na
2. Among the following,
The compounds which can undergo an SN1 reaction in an aqueous solution, are
(a) I and IV only (b) II and IV only
(c) II and III only (d) II, III and IV only
3. The major product of the following reaction
3
Excess DIBAL-H
Toluene, –78°C then H O+⎯⎯⎯⎯⎯⎯⎯→ is
(a) (b)
(c) (d)
HO
O
H
O
EtO
O
H
O
H
O
CN H
O
H
O
EtO
O
CN
Br
I
Br
II
Br
III MeO Br
IV
Ph CHBr2
O
Ph ONa
O
Ph H
O
Ph CBr3
O
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4. Permanent hardness of water can be removed by
(a) heating
(b) treating with sodium acetate (CH3CO2Na)
(c) treating with Ca(HCO3)2
(d) treatment with sodium hexametaphosphate (Na6P6O18)
5. Alkali metals (M) dissolve in liquid NH3 to give
(a) MNH2 (b) MH
(c) [M(NH3)x]+ + [e(NH3)y]– (d) M3N
6. The absolute configurations of the following compounds
Respectively are
(a) R and R (b) S and S (c) R and S (d) S and R
7. The diamagnetic species among the following is
(a) 2O+ (b) –2O (c) O2 (d) 2–
2O
8. Among the following transformations, the hybridization of the central atom remains
unchanged in
(a) CO2⎯→ HCOOH (b) BF3⎯→ –4BF
(c) NH3⎯→ 4NH+ (d) PCl3⎯→ PCl5
9. For an octahedral complex MX4Y2 (M = a transition metal, X and Y are monodentate
achiral ligands), the correct statement, among the following, is
(a) MX4Y2 has 2 geometrical isomers one of which is chiral
(b) MX4Y2 has 2 geometrical isomers both of which are achiral
(c) MX4Y2 has 4 geometrical isomers all of which are achiral
(d) MX4Y2 has 4 geometrical isomers two of which are chiral
CH2SH
CH2OH
H
H3C CH2SH
HO
H3C
H
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10. The values of the Henry's law constant of Ar, CO2, CH4 and O2 in water at 25°C are
40.30, 1.67, 0.41 and 34.86 kbar, respectively. The order of their solubility in water at
the same temperature and pressure is
(a) Ar > O2> CO2> CH4 (b) CH4> CO2> Ar > O2
(c) CH4> CO2> O2> Ar (d) Ar > CH4> O2> CO2
11. Thermal decomposition of N2O5 occurs as per the equation below
2N2O5⎯→ 4NO2 + O2
The correct statement is
(a) O2 production rate is four times the NO2 production rate
(b) O2 production rate is the same as the rate of disappearance of N2O5
(c) rate of disappearance of N2O5 is one-fourth of NO2 production rate
(d) rate of disappearance of N2O5 is twice the O2 production rate
12. For a 1st order chemical reaction.
(a) the product formation rate is independent of reactant concentration
(b) the time taken for the completion of half of the reaction (t1/2) is 69.3% of the rate
constant (k)
(c) the dimension of Arrhenius pre-exponential factor is reciprocal of time
(d) the concentration vs time plot for the reactant should be linear with a negative
slope
13. The boiling point of 0.001 M aqueous solutions of NaCl, Na2SO4, K3PO4 and CH3COOH
should follows the order
(a) CH3COOH < NaCl < Na2SO4< K3PO4
(b) NaCl < Na2SO4< K3PO4< CH3COOH
(c) CH3COOH < K3PO4< Na2SO4< NaCl
(d) CH3COOH < K3PO4< NaCl < Na2SO4
14. An allotrope of carbon which exhibits only two types of C–C bond distance of 143.5
pm and 138.3 pm, is
(a) charcoal (b) graphite (c) diamond (d) fullerene
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15. Nylon-2-nylon-6 is a co-polymer of 6-aminohexanoic acid and
(a) glycine (b) valine (c) alanine (d) leucine
16. A solid is hard and brittle. It is an insulator is solid state but conducts electricity in
molten state. The solid is a
(a) molecular solid (b) ionic solid (c) metallic solid (d) covalent solid
17. The curve that best describes the adsorption of a gas (x g) on 1.0 g of a solid
substrate as a function of pressure (p) at a fixed temperature
is:
(a) 1 (b) 2 (c) 3 (d) 4
18. The octahedral complex CoSO4Cl.5NH3 exists in two isomeric forms X and Y. Isomer X
reacts with AgNO3 to give a white precipitate, but does not react with BaCl2, Isomer Y
gives white precipitate with BaCl2 but does react with AgNO3.
Isomers X and Y are
(a) Ionization isomers (b) Linkage isomers
(c) Coordination isomers (d) Solvate isomers
19. The correct order of basicity of the following amines
is:
(a) I > II > III > IV (b) I > III > II > IV (c) III > II > I > IV (d) IV > III > II > I
20. Electrolysis of a concentrated aqueous solution of NaCl results in
(a) increase in pH of the solution (b) decrease in pH of the solution
(c) O2 liberation at the cathode (d) H2 liberation at the anode
I II
NH2
III H3C
IV
NH
2
NH
2
NH2
O2N
x
p
1
2 3
4
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PART-II
21. The product of which of the following reaction forms a reddish brown precipitate
when subjected to Fehling's test?
(a)3
CO, HCl
anhy. AlCl , CuCl⎯⎯⎯⎯⎯→ (b) + (CH3CH2)2Cd ⎯→
(c) 5
2 4
1. PCl
2. H , Pd-BaSO⎯⎯⎯⎯→ (d) 3
2
1. O
2. Zn/H O⎯⎯⎯→
22. The major products X, Y and Z in the following sequence of transformations
are:
(a)
(b)
(c)
(d)
H
O X = Y =
NO2
Z = N
H
O
N
NO2
NH2
NH2
O
X = Y = Z = NH2
O
O2N
NH2
O
O2N OH
H
O X = Y =
O2N
NH2 Z =
N H
O
N
O2N
NH2
O
X = NH2
O
Y = NO2
NH2
O
Z = NO2
HO
NH2 O
O
O
X conc. HNO3
conc. H2SO4
15°C
Y aq. NaOH
Z
CO2H
O
Cl
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23. In the following reaction, P gives two products Q and R, each in 40% yield.
If the reaction is carried out with 420 mg of P, the reaction yields 108.8 mg of Q.
The amount of R produced in the reaction is closest to
(a) 97.6 mg (b) 108.8 mg (c) 84.8 mg (d) 121.6 mg
24. Solubility products of CuI and Ag2CrO4 have almost the same value (~4 × 10–12).
The ratio of solubilities of the two salts (CuI: Ag2CrO4) is closest to
(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.10
25. Given that the molar combustion enthalpy of benzene, cyclohexane, and hydrogen
are x, y, and z, respectively, the molar enthalpy of hydrogenation of benzene to
cyclohexane is
(a) x – y + z (b) x – y + 3z (c) y – x + z (d) y – x + 3z
26. Among the following, the pair of paramagnetic complexes is
(a) K3[Fe(CN)6] and K3[CoF6] (b) K3[Fe(CN)6] and [Co(NH3)6]Cl3
(c) K4[Fe(CN)6] and K3[CoF6] (d) K4[Fe(CN)6] and [Co(NH3)6]Cl3
OMe 1. O3 Q
40%
2. Zn, H2O + R
40% P (MW = 210)
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27. The major products X and Y in the following sequence of transformations
are:
(a)
(b)
(c)
(d)
28. 3.0 g of oxalic acid [(CO2H)2.2H2O] is dissolved in a solvent to prepare a 250 mL
solutions. The density of the solution is 1.9 g/mL. The molality and normality of the
solution, respectively, are closest to
(a) 0.10 and 0.38 (b) 0.10 and 0.19 (c) 0.05 and 0.19 (d) 0.05 and 0.09
29. In a titration experiment, 10 mL of a FeCl2 solution consumed 25 mL of a standard
K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is
prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the
FeCl2 solution is closest to [Given: molecular weight of K2Cr2O7 = 294 g mol–1]
(a) 0.25 N (b) 0.50 N (c) 0.10 N (d) 0.04 N
30. Atoms of an element Z form hexagonal closed pack (hcp) lattice and atoms of element
X occupy all the tetrahedral voids. The formula of the compound is
(a) XZ (b) XZ2 (c) X2Z (d) X4Z3
OH X = Y =
CO2H
OH
SO3H X = Y =
OH
CO2H
OH
OH X = Y =
CO2H
SO3H X = Y =
HO SO3H HO2C
X 1. oleum
2. molten NaOH, 3. H3O+
1. NaOH
2. CO2 3. H3O+
Y
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ANSWER KEY
1. (b) 2. (c) 3. (a) 4. (d) 5. (c)
6. (d) 7. (d) 8. (c) 9. (b) 10. (c)
11. (d) 12. (c) 13. (a) 14. (d) 15. (a)
16. (b) 17. (b) 18. (a) 19. (b) 20. (a)
21. (d) 22. (b) 23. (c) 24. (b) 25. (b)
26. (a) 27. (d) 28. (c) 29. (a) 30. (c)
SOLUTIONS
PART-I
1. (b)
O
Ph C Br Br
Br
NaOH
O
Ph C Br Br
Br
–
OH
O
Ph OH + – C–Br
Br
Br
O
Ph O + H – C–Br
Br
Br
– Na+
O
+ Ph O Na+ – H – C–Br
Br
Br
+
–
OH–
Nucleophile
Good leaving group
Proton transfer
Resonance stabilised
Option (b) is correct.
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2. (c)
Condition for SN1 reaction is whether type of carbocation formed is stable or not.
(I)
⎯→ not stable carbocation due to –I effect of alkene.
(II) ⎯→stable carbocation +I effect of three methyl group.
(III)
MeO ⎯→stable due to resonance and delocalization of +ve charge
(IV)
⎯→according to Bredt’s rule, carbocation at bridge position is unstable.
So, option (c) is correct.
3. (a)
[DIBAL–H] → (i–Bu)2 Al–H → Di isobutyl aluminium hydride
RCN or ester 2
DIBAL–H
H O
⎯⎯⎯⎯⎯→ R–CHO
EtO
O DIBAL–H
Toulene, –78°C then H3O CN
Ester R–CN
H
O H
O
4. (d)
The permanent hardness of water can be removed by using calgon process. The
hardness of water causes by salts of Ca and Mg.
Na2[Na4(PO3)6] + 2Ca2+/Mg2+⎯→ Na2[Ca2/Mg2(PO3)6] + 4Na+
(soluble complex)
Sodium hexametaphosphate
(Calgon)
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5. (c)
All alkali metals like lithium, sodium, potassium etc. dissolves in liquid ammonia to
give deep blue coloured solution. The blue colour is due to presence of solvated
electrons.
M + (x + y)NH3⎯→[M(NH3)x]+ + [e(NH3)y]–
(Ammoniated cation) (Ammoniated anion)
(Blue colour)
6. (d)
Draw the Fischer project i.e. 2–D representation and then assign priorities.
CH2SH H
H3C CH2OH
(I)
H HO
H3C CH2SH
(II)
According to Cahn Ingold Prelog (CIP) priority rules –
CH3
CH2SH
CH2OH
H (1)
(2)
(4)
(3)
Clockwise
CH3
H
CH2SH
HO (4)
(2)
(1)
(3)
Anticlockwise
So, it should be ‘R’ but here, It should be ‘S’ lower priority group is on
lower priority group is on horizontal line i.e. ‘R’
horizontal line. Therefore,
Configuration will be reversed.
i.e. correct configuration ‘S’
So, option (d), S, R is correct.
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7. (d)
According to MOT –
(a) ( ) ( )( )+ = = = 2 z x y x y
– 2 *2 2 *2 2 2 2 *1 *01s 1s 2s 2s 2p 2p 2p 2p 2pO 15e
Number of unpaired e–s = 1 Paramagnetic
(b) ( ) ( )( )= = = 2 z x y x y
– – 2 *2 2 *2 2 2 2 *2 *11s 1s 2s 2s 2p 2p 2p 2p 2pO 17e
Number of unpaired e–s = 1
(c) ( ) ( )( )= = = z x y x y
– 2 *2 2 *2 2 2 2 *1 *12 1s 1s 2s 2s 2p 2p 2p 2p 2pO 16e
Number of unpaired e–s = 2 paramagnetic
(d) ( ) ( )( )= = = z x y x y
2– – 2 *2 2 *2 2 2 2 *2 *22 1s 1s 2s 2s 2p 2p 2p 2p 2pO 18e
Number of unpaired e–s = 0 diamagnetic
Option (d) is correct.
8. (c)
(a)
O=C=O H–C–O–H
O
sp(2B.P.) sp2(3B.P.)
(b)
F–B–F F–B–F
sp2 (3B.P.)
sp3(4B.P.)
F
F
– F
(c)
N
sp3(3 B.P. + 1 .p.) sp3(4 B.P.)
+
H H H
N
H H H
H
(d)
P
sp3(3 B.P. + 1 .p.) sp3d(5 B.P.)
Cl Cl Cl
P Cl
Cl Cl
Cl
Cl
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9. (b)
MX4Y2⎯→ Octahedral complex
Note → If there is plane of symmetry (POS) or centre of symmetry (COS) in a
complex. Then it will be achiral and optically inactive. Geometrical isomers exist is cis
and trans forms.
M
X X
X X
Y
Y
POS & COS present M
Y X
Y X
trans (Achiral)
POS present
Cis (Achiral)
X
X
i.e. two geometrical isomers both of which are achiral.
So, option (b) is correct.
10. (c)
According to Henry’s law, the partial pressure of a gas in vapour phase (P) is
proportional to the mole fraction of the gas (X) in the solution.
P X
P = KHX
KH = Henry’s constant
P = Partial pressure of gas above liquid surface
so, according to Henry’s law solubility of gas in liquid partial pressure of gas above
liquid surface.
KHP solubility
KH Ar O2 CO2 CH4 (given)
40.30 34.86 1.67 0.41
Solubility order Ar < O2< CO2< CH4
11. (d)
Reaction: 2N2O5⎯→ 4NO2 + O2
Rate =
= =2 5 2 2d N O d NO d O1 1
–2 dt 4 dt dt
Rate of disappearance of N2O5 =
= =2 5 2 2d N O 2d O d NO1
–dt dt 2 dt
Rate of disappearance of N2O5= twice the O2 production rate
So, option (d) is correct.
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12. (c)
For Ist order reaction –
t =
o
10
t
A2.303log
K A
And =1
2
0.693t
K
Arrhenius equation: a–E /RTK Ae=
[A = pre-exponential factor]
In a first order reaction, unit of rate constant is s–1. So pre-exponential factor are also
reciprocal seconds. Because exponential factor depends on frequency of collision.Its
related to collision theory and transition statetheory.
13. (a)
We know that, Tb = i ×Kb × m
molarity or molality = 0.001 m {at very diluted solution}
So, Tb i ;{molality is constant for all}
i Boiling point
For NaCl strong electrolyte i = 2 {Na+ + Cl–}
For Na2SO4 strong electrolyte i = 3 + + 2–42Na SO
For K3PO4 strong electrolyte i = 4 + + 3–43K PO
For CH3COOH weak acid i < 2 (calculated from i= 1+ (n-1)α , where α<1 )
So, order of Boiling Point: K3PO4> Na2SO4> NaCl > CH3COOH
So, option (a) is correct.
14. (d) Fullerene a soccer ball shaped molecule has 60 vertices with a carbon atom at each
vertex. It contains both single and double bond with C–C at a distance of 143.5 pm and 138.3 pm.
Where as in diamond C–C bond length 154 pm and in graphite 140 pm.
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15. (a)
Nylon–2–Nylon–6 is a copolymer of glycine and amino caproic acid (6-amino
hexanoic acid)
nH2N–CH2–COOH + nNH2 –(CH2)5–COOH
Glycine Amino caproic acid
–nH2O
C–CH2–NH–C–(CH2)5–NH
O O
Nylon-2-Nylon-6
n
16. (b)
Ionic solid is hard and brittle eg. NaCl
It is an insulator in solid state but in molten state it conducts electricity as it has free
ions in molten state.
17. (b)
Given: mass = 1g
According to freundlich adsorption isotherm –
=1
nx
KPm
(n > 1)
=1
nx KP
=1
nx
Pk
=
nx
Pk
On comparing parabola equation yn = x. If n = 2
x
So, graph (b) is correct.
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18. (a)
CoSO4Cl.5NH3
Isomers of CoSO4Cl.5NH3 are (I) [Co(NH3)5SO4]Cl gives white ppt with AgNO3 but not
with BaCl2 and (II) [Co(NH3)5Cl]SO4 gives white ppt with BaCl2 but not with AgSO4.
(I) [Co(NH3)5SO4]Cl + AgNO3⎯→ AgCl + [Co(NH3)5SO4]NO3
[white ppt]
(II) [Co(NH3)5Cl]SO4 + BaCl2⎯→ BaSO4 + [Co(NH3)5Cl]Cl2
[white ppt]
So, isomer (I) and (II) are ionization isomers.
19. (b)
(I) NH2
⎯→ lone pair of ‘N’ are not delocalised so, easily donate e–.
(II) NH2
⎯→No effect, lone pair of ‘N’ are in resonance. So, e– not donate
easily.
(III)
NH2
H3C ⎯→ +I effect of CH3 and resonance of lone pair of ‘N’. So,
easily donate e– as compared to (II).
(IV)
NH2
O2N ⎯→ –I effect of –NO2 and resonance of lone pair of ‘N’. So hard
to remove electrons.
So, order of basicity – I > III > II > IV
20. (a)
Electrolysis of aqueous NaCl leads to the formation of NaOH, which is a base. SO, its
pH will be increases.
NaCl + H2O ⎯⎯⎯⎯→Electrolysis NaOH + Cl2 + H2
(base)
H2 liberation will be at cathode due to (H+⎯→ H2) gain of electron at cathode.
So, option (a) is correct.
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PART-II
21. (d)
Fehling test: Fehling solution A + Fehling solution B
(Cu2+) (OH–)
Fehling reagent react with aldehyde which has –H such aldehyde form an enolate
and thus give a positive Fehling test and a reddish brown precipitate is obtained.
Reaction: R–CHO + 2Cu2+ + 5OH–⎯→RCOO– + Cu2O + 3H2O
(reddish brown)
Benzaldehyde gives tollen’s as well as schiff’s test but not gives fehling's solution
test because benzaldehyde does not contain 𝐴𝑙𝑝ℎ𝑎 hydrogen and cannot form
intermediate enolate to proceed for further and hence it does not react with fehling's
solution test however aliphatic aldehydes gives fehling's solution test.
(a)
CO, HCl
C–H
anhy. AlCl3, CuCl
O
(Gattermann Koch Rxn) (No –H)
do not give Fehling test
(b)
O –CdCl2 2
CH2–CH3 Cl
+ (CH3CH2)2Cd O
(c)
PCl5 C–Cl
O
C–OH
O
H2, Pd-BaSO4 CH2–OH
(d) 1. O3 CH3–C–CH2–CH2–CH2–CH2–C–H give Fehling test
O 2. Zn/H2O
O aldehyde
(–H)
So, option (d) product gives Fehling test.
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22. (b)
(CH3CO)2O
H–N–C–CH3
O
Conc. HNO3
NH2
(x)
Conc. H2SO4
H–N–C–CH3
O
NO2 (Acetanilide) (>51%)
(y)
aq. NaOH
NH2
NO2 (z)
Pyridine (15°C)
(Aniline)
So, option (b) is correct.
23. (c)
1. O3 OMe 2. Zn, H2O
OMe
CHO
+
OHC
(P) (40%) (40%)
mw = 210 Given: w = 420 mg
Moles = =w 420
Mw 210 = 2 m moles =
40 42
100 5 m moles =
40 42
100 5 m moles
Molecular weight for 108.8 mg = mass
mole =
108.8 mg
4m mole
5
= 136 g/mol
So, 136 g/mol is forC8H8O2
OMe OHC Q
Then CHO
R C7H6O mol. wt = 106 g mol–1
Mass of R = mole × m.wt. = 4
5 m mole × 106 g mol–1 = 84.4 mg
So, option (c) is correct.
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24. (b)
Given: ( ) ( )= = 2 4
–12sp spCuI Ag CrO
K K 4 10
CuI(aq) ⎯→ Cu+(aq) + I–(aq)
S1 S1
1spK = S1 × S1 4 × 10–12 = (S1)2
S1 = 2 × 10–6 mol/L
Ag2CrO4⎯→ 2Ag+ + 2–4CrO
2S2 S2
2spK = (2S2)2 (S2) 4 × 10–12 = 4(S2)3
(S2)3 = 10–12 S2 = 10–4 mol/L
= = = –2
–61
–42 4 2
Ssolubility of CuI 2 102 10
solubility of Ag CrO S 10
=1
2
S0.02
S
So, option (b) is correct.
25. (b)
6 6Benzene
C H + 15
2O2⎯→ 6CO2 + 3H2O; H1 = x .....(1)
6 12Cyclohexane
C H + 9O2⎯→ 6CO2 + 6H2O; H2 = y .....(2)
H2 + 1
2O2⎯→ H2O; H3 = z .....(3)
C6H6 + 3H2⎯→ C6H12; H =? .....(4)
So, equation (1) + equation (3) × 3 – equation (2); to get equation (4)
C6H6 + 3H2⎯→ C6H12
x + 3z – y = x – y + 3z
So, molar enthalpy of hydrogenation of benzene to cyclo hexane = x – y + 3z
Option (b) is correct.
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26. (a)
(a) K3[Fe(CN)6] CN– is a strong field ligand so, pairing possible
Fe3+
26Fe [Ar]3d64s2
Fe3+ [Ar]3d54s04p0
3d 4s 4p
Fe3+
K2[Fe(CN)6] xx
CN–
d2sp3
xx xx xx xx xx
CN– CN– CN– CN– CN–
Number of unpaired e–s = 1 paramagnetic
K3[CoF6] F– weak field ligand, so no pairing
Co3+
27Co [Ar]3d74s2
Co3+ [Ar]3d64s04p04d0
3d 4s 4p
Co3+
4d
K3[CoF6] xx xx xx xx xx xx
F– F– F– F– F– F–
sp3d2 Number of unpaired e–s = 4 paramagnetic
So, option (a) is correct.
27. (d)
1. oleum 2. NaOH
3. H3O+
SO3H
1. NaOH
2. CO2 3. H3O+
OH OH COOH
(x) (y)
(salicylic acid) (Kolbe schmidt reaction)
Option (d) is correct.
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28. (c)
Given: Mass of oxalic acid = 3.0 g
Mol. wt. of oxalic acid (COOH)2.2H2O = 126 g mol–1
Density = 1.9 g/mL and volume of solution = 250 mL
d = m
v m = d × v
Molality = ( )
mass of solute
m.mass of solute mass of solvent kg
( )
=
mmolality
m.m d v –3
3
126 10.9 250 10=
Molality = 0.05
We know that, Normality = Molarity × nfactor( )
molemolarity
volume L
=
nfactorof oxalic acid = 2 (Because it lose 2 H+)
( ) factornn
NV L
=
Normality = –3
3126 2 0.19
250 10 =
So, option (c) is correct.
29. (a)
Reaction: FeCl2 + K2Cr2O7
+6 Fe3+ + Cr3+
nf = 6
So, by law of equivalence,
Number of eq. of FeCl2 = number of eq. of K2Cr2O7
N1V1 = N2V2
Given: volume of FeCl2 = 10 mL
Volume of K2Cr2O7 = 25 mL
Wt. of K2Cr2O7 = 1.225 g
Volume of solution = 250 mL
m. wt. of K2Cr2O7 = 294 g mol–1
N1 × 10 = M × nf × 25
=
1 –3
1.225/ 294N 10 25 6
250 10
( )mole
molarityvolume L
=
=
1 –3
1.225 25 6N
294 250 10 10
N1 = 0.25 N
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30. (c)
Let, number of atom in HCP = a
We know, the number of tetrahedral voids will be twice as that of number of atoms
present in a unit cell.
So, no. of tetrahedral void = 2a
Z = a, X = 2a
Z : X
a : 2a
1 : 1
So, formula of compound = X2Z
Therefore, option (c) is correct.
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