a) ion-dipole > dipole-dipole > ion-ion
b) dipole-dipole > ion-dipole > ion-ion
c) ion-ion > ion-dipole > dipole-dipole
d) ion-dipole > ion-ion > dipoledipole
Ion-ion interactions are stronger because they have stronger electrostatic forces of attraction whereas dipoles have partial charges and hence the electrostatic forces in their case would be relatively weak.
Answer: c
a) +2, +1, and
b) +1, +2 and +4
c) +1, +1 and +1
d) +1, +4 and +2
Alkali metals always possess a +1 oxidation state, whereas oxygen present in K2O (oxide) is -2, and in K2O2 (peroxide) is -1 and in KO2 (superoxide) is
Answer: c
a) CS2 and acetone are less attracted to each other than to themselves
b) heat must be absorbed in order to produce the solution at 350C
c) Raoult’s law is not obeyed by this system
d) a mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL
PTotal = PT = p0A XA+ p0B XB
The maximum value XA can hold is 1, and hence the maximum value of PT should come out to be 512 mm of Hg, which is less than the value of PT observed (600 mm of Hg). Therefore, positive deviation from Raoult’s law that is observed. This implies that A-A interactions and B-B interactions are stronger than A-B interactions. As we know, for a system not obeying Raoult’s law and showing positive deviation, ΔVmix>0, ΔHmix>0 .
Answer: d
a) Ni
b) Cu
c) Au
d) Hg
Because of Lanthanide contraction, an increase in Zeff is observed and so, the size of Au instead of being greater, as is expected, turns out be similar to that of Ag.
Answer: c
a) CH4 < CCl4 < CHCl3
b) CHCl3 < CH4 = CCl4
c) CH4 = CCl4 < CHCl3
d) CCl4 < CH4 < CHCl3
All the three compounds possess a tetrahedral geometry. In both CCl4 and CH4, µnet= 0, whereas in CHCl3, µnet > 0.
Answer: c
a) less efficient as it exchanges only anions
b) more efficient as it can exchanges only cations
c) less efficient as the resins cannot be regenerated
d) more efficient as it can exchange both cations as well as anions
Answer: d
a) matter consists of indivisible atoms
b) when gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P.
c) Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction
d) all the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
When gases combine or react in a chemical reaction they do so in a simple ratio by volume provided all gases are maintained at the same temperature and pressure- Gay-Lussac’s law.
Answer: b
(i) H2N-CH=NH
(ii)
(iii) CH3NHCH3
a) ii < iii < i
b) iii < i < ii
c) i < ii < iii
d) ii < i < iii
Weaker the conjugate acid, stronger the base. (ii) is the most basic as it has a guanidine like structure. It has a high tendency of accepting a proton, giving rise to a very stable conjugate acid and hence, is a very strong base. In compound (i), the N is sp2 hybridised and its electronegativity is higher as compared to the compound (iii) which is a 20 amine (sp3 hybridised). So compound (ii) is more basic compared to compound (iii).
So the order of basicity is ii > i > iii and thus the order of pKb value will be iii > i > ii
Answer: d
Hex-3-ynal
(i) NaBH4
(ii) PBr3
(iii) Mg/ether
(iv) CO2/H3O+
Answer: b
a) 11
b) 15
c) 25
d) 50
n = 5; l = (n – 1) = 4; hence the possible sub-shells for n=5 are: 5s, 5p, 5d, 5f and 5g.
The number of orbitals in each would be 1,3,5,7 and 9, respectively and summing them up gives the answer as 25.
Answer: c
a) cast iron
b) wrought iron
c) scrap iron and pig iron
d) pig iron
Answer: b
a) Werner’s theory
b) Crystal Field Theory
c) Molecular Orbital Theory
d) Valence Bond Theory
Answer: c
a) Diamminechlorido (methanamine) platinum (II) chloride
b) Bisammine (methanamine) chloridoplatinum (II) chloride
c) Diammine (methanamine) chloridoplatinum (II) chloride
d) Diamminechlorido (aminomethane) platinum (II) chloride
Answer: a
Answer: c
The product ‘X’ is used:
a) in protein estimation as an alternative to ninhydrin
b) as food grade colourant
c) in laboratory test for phenols
d) in acid-base titration as an indicator
X formed is methyl orange.
Answer: d
|
List I |
List II |
|
i) Riboflavin |
p) Beri beri |
|
ii) Thiamine |
q) Scurvy |
|
iii) Ascorbic acid |
r) Cheliosis |
|
iv) Pyridoxine |
s) Convulsions |
|
i |
ii |
iii |
iv |
|
|
a) |
s |
q |
p |
r |
|
b) |
r |
p |
q |
s |
|
c) |
p |
r |
q |
s |
|
d) |
s |
r |
q |
p |
|
Vitamins |
Deficiency diseases |
|
i) Riboflavin (Vitamin B2) |
Cheilosis |
|
ii) Thiamine (Vitamin B1) |
Beri beri |
|
iii) Ascorbic acid (Vitamin C) |
Scurvy |
|
iv) Pyridoxine (Vitamin B6) |
Convulsions |
Answer: b
a) +0.158 V
b) -0.158 V
c) 0.182 V
d) -0.182 V
Cu2+ + 2e– Cu
E° = 0.340 V
Cu – Cu+ + e–
E°= -0.522 V
Cu2+ + e– ⇾ Cu+
E° = ?
Applying ΔG = nFE° We get,
(-1 × F × E°) = -2 × F × 0.340 + (-1 × F × -0.522)
Solving, we get, E°= 0.158 V
Answer: a
a) m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline
b) m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline
c) m-chloroaniline, m-chlorobenzoic acid and m-chlorophenol
d) m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol
m-chlorobenzoic acid being the most acidic can be separated by a weak base like NaHCO3 and hence will be labelled fraction A.
m-chlorophenol is not as acidic as m-chlorobenzoic acid, and can be separated by a stronger base like NaOH, and hence can be labelled as fraction B.
m-chloroaniline being a base, does not react with either of the bases and hence would be labelled as fraction C.
Answer: a
a) -333, -325, -349 and -296
b) -333, -349, -325 and -296
c) -296, -325, -333 and -349
d) -349, -333, -325 and -296
Cl > F > Br > I
Answer: b
Which of these reaction(s) will not produce Saytzeff product?
a) b and d
b) d only
c) a, c and d
d) c only
Answer: d
Molarity of NaOH (4 g in 100 L) = 10-3 M
Molarity of H2SO4 (9.8 g in 100 L) = 10-3 M
Equivalents of NaOH= M × V × nf = 10-3 × 40 × 1= 0.04
Equivalents of H2SO4= M × V × nf = 10-3 × 10 × 2= 0.02
MNaOH . VNaOH . (nf)NaOH – MH2SO4 . VH2SO4. VH2SO4 . (nf)H2SO4 = M. Vtotal
10-3 × 40 × 1 – 10-3 × 10 × 2 = M. 50
M = 4 × 10-4
pOH = -log M
= 4 – 2log2
= 3.4
pH = 14 – 3.4 = 10.6
Answer: 10.6
All nuclear processes follow first order kinetics, and hence
T1/2 = 0.693 / λ
λ = 0.1(year)-1
t = (2.303/ λ) (log (a0)/at)
t = (2.303/ 0.1) (log (a0)/0.1a0)
on solving, t = 23.03 years.
Answer: 23.03
Bond order = total no. of bonds / total resonating structures = 5/3 = 1.67.
Answer: 1.67
Answer: 2
∆U= 2.1 kcal, ∆S= 20 calK-1 at 300 K, Hence ∆G in kcal is
∆H= ∆U + ∆ngRT
∆H= 2100 + (2 × 2 × 300) (R=2 calK-1mol-1)
= 3300 cal
∆G= ∆H – T∆S
∆G= 3300 – (300 × 20) = -2.7 kcal.
Answer: -2.7
Video Lessons – January 7 Shift 1 Chemistry
JEE Main 2020 Chemistry Paper January 7 Shift 1
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