JEE Main is an examination conducted for admission to some of the reputed institutions in the country. It is the dream of many students to crack this examination. One of the best ways to prepare for the examination is by practising previous years’ question papers. On this page, JEE Main 2020 Chemistry paper is solved. Practise these questions to get an excellent score in the examination.
1. The number of bonds between sulphur and oxygen atoms in S_{2}O_{8}^{2−} and number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are:
Solution:
Here, we have to count S − O single bonds as well as S = O in S_{2}O_{8}^{2–}, as each double bond also has one sigma bond. The structure of S_{2}O_{8}^{2-} and S_{8} is shown below:
2. The predominant intermolecular forces present in ethyl acetate, a liquid, are:
Solution:
London dispersion forces (also called as induced dipole - induced dipole interactions), exist because of the generation of temporary polarity due to collision of particles and for this very reason, they are present in all molecules and inert gases as well.
Because of the presence of a permanent dipole, there will be dipole-dipole interactions present here.
There is no H that is directly attached to an oxygen atom, so H-bonding cannot be present.
3. For the Balmer series in the spectrum of H-atom,
v̅ = R_{H} [ (1/n_{1}^{2}) – (1/n_{2}^{2})]
The correct statements among (A) to (D) are:
A) The integer n_{1}= 2.
B) The ionization energy of hydrogen can be calculated from the wave number of these lines.
C) The lines of longest wavelength corresponds to n_{2}= 3.
D) As wavelength decreases, the lines of the series converge.
Solution:
Energy of a photon released on transition from n= 100 to n= 2 will have similar energy to that of the photon that gets released on transition from n= 101 to n= 2, because energy of the 100th and the 101th orbit will be very close in value. That means they will also have very close values of wavelengths, which further implies that these two lines will be situated quite close to each other on the photographic plate.
In a similar fashion, we can see that as the nhigher increases, the lines start to converge together. And since, increasing the nhigher will indeed lead to an increase in the energy of the photon released, it will end up releasing photons of shorter wavelengths. Combining these two statements we can easily see that as the wavelength decreases, the spectral lines start to converge.
4. The first ionization energy (in kJ/mol) of Na, Mg, Al and Si, respectively,
Solution:
The expected order is Na < Mg < Al < Si.
But the actual/experimental order turns out to be Na < Al < Mg < Si, because of the fully filled s- sub shell of magnesium and the s^{2}p^{1} configuration of Al which makes it relatively easy for Al to lose its outermost electron.
5. The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:
Solution:
(a) X_{2}Y(s) ⇌ 2X^{+} + Y^{2−}
Ksp = [X^{+}]^{2}[Y^{2−}] = 4 × 10^{−6} × 10^{−3} = 4 × 10^{−9}
(b) XY_{2}(s) ⇌ X^{2+} + 2Y^{−}
Ksp = [X^{2+}][Y^{−}]^{2} = 10^{−3} × 4 × 10^{−6} = 4 × 10^{−9}
(c) XY_{2}(s) ⇌ X^{2+} + 2Y^{−}
Ksp = [X^{2+}][Y^{−}]^{2} = 10^{−3} × ( 2 × 10^{−3})^{2} = 4 × 10^{−9}
(d) XY_{(s)} ⇌ X^{+} + Y^{−}
Ksp = [X^{+}][Y^{−}] = 10^{−3} × 10^{−3} = 10^{−6}
6. The complex that can show fac- and mer-isomers is:
Solution:
Facial and meridional geometrical isomerism is observed only in [MA_{3}B_{3}] type complexes which is given in option a.
7. A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below:
The following inferences are made:
A) X has higher intermolecular interactions compared to Y
B) X has lower intermolecular interactions compared to Y
C) Z has lower intermolecular interactions compared to Y The correct inference(s) is/are:
Solution:
As shown in the plot below, for the same T, the vapour pressure of X is the highest and of Z is the lowest. Now, that means with the same average K.E. of X, Y and Z molecules, the X molecules are able to compensate their respective intermolecular forces better. So, X molecules have the highest vapour pressure. This implies that the intermolecular forces in X are the weakest among the three. The opposite could be said for Z as well.
8. As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:
Solution:
9. The rate of a certain biochemical reaction at physiological temperature (T) occurs 10^{6} times faster with enzyme than without. The change in activation energy upon adding enzyme is:
Solution:
K_{1} = Ae^{−Ea1/RT} ....(1)
K_{2} = Ae^{−Ea2/RT} ....(2)
Dividing equation 1 with equation 2, we get
K_{1}/K_{2} = e ^{(Ea2−Ea1)/RT}
10^{−6} = e ^{(Ea2−Ea1)/RT}
Taking loge on both sides, we get
∆E = E_{a2} − E_{a1} = – 6 × 2.303 RT
10. When gypsum is heated to 393K, it forms:
Solution:
11. The third ionization enthalpy is minimum for:
Solution:
Consider an element E
E^{2+} → E^{3+} would be the 3rd I.E. of the element E.
Electronic configuration of Mn is [Ar]4s^{2}3d^{5}, Co is [Ar]4s^{2}3d^{7}, Fe is [Ar]4s^{2}3d^{6}, Ni is [Ar]4s^{2}3d^{8}
Electronic configuration of Mn^{2+} is [Ar]3d^{5}, Co^{2+} is [Ar]3d^{7}, Fe^{2+} is [Ar]3d^{6}, Ni^{2+} is [Ar]3d^{8}
As it is evident from the above configurations of the E^{2+} for the given elements, Fe^{2+} would require the least amount of energy for removal of electron as it has the configuration 3d^{6} 4s^{0}. That means that its E^{3+} form is the most stable among the four elements provided in their respective E^{3+} states, i.e., when compared, the next electron removal will require least amount of energy.
12. The strength of an aqueous NaOH solution is most accurately determined by titrating: (Note: consider that an appropriate indicator is used)
Solution:
13. The decreasing order of reactivity towards dehydrohalogenation (E_{1}) reaction of the following compounds is:
Solution:
14. Major product in the following reaction is:
Solution:
15. Arrange the following compounds in increasing order of C—OH bond length: methanol, phenol, p-ethoxyphenol
Solution:
16. Among the gases (i) – (v), the gases that cause greenhouse effect are:
i. 𝐶𝑂_{2}
ii. 𝐻_{2}𝑂
iii. 𝐶𝐹𝐶
iv. 𝑂_{2}
v. 𝑂_{3}
Solution:
17. The major products A and B in the following reactions are:
Solution:
18. A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63 °C while the other boils at 60 °C. What is the best way to separate the two liquids and which one will be distilled out first?
Solution:
19. Which of the given statement is not true for glucose?
Solution:
Glucose exists in two crystalline forms alpha and beta which are anomers of each other.
Glucose does not react with Schiff’s reagent because after the internal cyclisation, it forms either alpha- anomer or beta-anomer. In these forms, free aldehydic group is not present.
Glucose forms open chain structure in aqueous solution which contains aldehyde at chain end. This aldehydic group reacts with NH_{4}OH to form oxime. On the other hand, glucose penta acetate being a cyclic structure even in aqueous form does not have terminal carbonyl group. Therefore it will not react with 𝑁𝐻_{4}𝑂𝐻.
20. The reagent used for the given conversion is:
Solution:
21. The volume (in mL) of 0.125 M 𝐴𝑔𝑁𝑂_{3} required to quantitatively precipitate chloride ions in 0.3 g of [𝐶(𝑁𝐻_{3} )_{6} ]𝐶𝑙_{3} is _____.
[𝐶𝑜(𝑁𝐻_{3} )_{6} ]𝐶𝑙_{3} = 267.46 𝑔/𝑚𝑜𝑙
𝑀𝐴𝑔𝑁𝑂_{3} = 169.87 g/mpl
Solution:
To react completely with one mole of [𝑀𝐿_{6}]𝑙_{3} , 3 moles of 𝐴𝑔𝑁𝑂_{3} is required.
0.3 g [𝑀𝐿_{6} ]𝑙_{3} means (0.3/267.46) moles of [𝑀𝐿_{6} ]𝐶𝑙_{3}.
So, moles of 𝐴𝑔𝑁𝑂3 required will be (0.3×3)/267.46 moles
To find the volume, (0.3×3)/267.46 = 0.125× (𝐿)
(𝐿) = 0.02692
(𝑚𝐿) = 26.92
22. What will be the electrode potential for the given half cell reaction at pH= 5?
2𝐻_{2}𝑂 → 𝑂_{2} + 4𝐻^{+} + 4𝑒^{–}; 𝐸° = – 1.23 𝑉
(R=8.314 J mol^{-1}K^{-1}; temp.=298 K; oxygen under std. atm. Pressure of 1 bar.)
Solution:
2𝐻_{2}𝑂 → 𝑂_{2} + 4𝐻^{+} + 4𝑒^{–}; 𝐸° = – 1.23 𝑉
23. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is _______. Atomic weight: Fe=55.85; S=32.00; O=16.00)
Solution:
24. The magnitude of work done by gas that undergoes a reversible expansion along the path ABC shown in figure is
Solution:
Work done by the gas
= The area under the curve
= (Area of the square) + (Area of the triangle)
= 48 J
25. The number of chiral centres in Penicillin is ____.
Solution:
The structure of penicillin is shown below:
So, the number of chiral centers= 3