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JEE Main 2020 Chemistry (Shift 2-5th Sept) Paper With Solutions

To succeed in the JEE examination candidates need to prepare well and effectively. One way to study productively is by solving previous years’ question papers. IIT JEE aspirants can refer to this page for the JEE Main 2020 (Shift 2-5th Sept) Chemistry Question Paper with Solutions. These questions are solved by the experts at BYJU’S in the simplest possible method.
September 5 Shift 2 – Chemistry

1. The major product formed in the following reaction is:

JEE Main 2020 Solved Che Paper Shift 2 5th Sept Ques 1


1) CH3CH(Br)CH2CH(CH3)2
2) CH3CH2CH2C(Br)(CH3)2
3) CH3CH2CH(Br)CH(CH3)2
4) Br(CH2)3CH(CH3)2

Answer: (1)

JEE Main 2020 Solved Che Paper Shift 2 5th Sept Sol 1

2. Hydrogen peroxide, in the pure state, is:


1) Linear and blue in color
2) Linear and almost colorless
3) Non-planar and almost colorless
4) Planar and blue in color

Answer: (3)

H2O2 has open book structure it is non planar

3. Boron and silicon of very high purity can be obtained through:


1) Liquation
2) Electrolytic refining
3) Zone refining
4) Vapour phase refining

Answer: (3)

Fact

4. The following molecule acts as an:

JEE Main 2020 Question Paper with Solutions Chemistry Shift 2- 5th Sept


1) Anti-histamine
2) Antiseptic
3) Anti-depressant
4) Anti-bacterial

Answer: (1)

Anti-histamine

5. Among the following compounds, geometrical isomerism is exhibited by:

JEE Main 2020 Questions with Solutions Chemistry Shift 2- 5th Sept


Answer: (1 and 2)

JEE Main 2020 Chemistry Shift 2- 5th Sept Solutions

6. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the gas adsorbed on mass m of the adsorbent, the correct plot of x/m versus p is:

Solutions of JEE Main 2020 Chemistry Shift 2- 5th Sept


Answer: (2)

As temp. increases extent of Adsorption decreases

Therefore correct option (2)

(x/m)= KP 1/n

(x/m) v/s P is non linear curve

7. The compound that has the largest H–M–H bond angle (M=N, O, S, C) is:


1) CH4
2) H2S
3) NH3
4) H2O

Answer: (1)

Solved Paper of JEE Main 2020 Chemistry Shift 2- 5th Sept

8. The correct statement about probability density (except at infinite distance from nucleus) is :


1) It can be zero for 3p orbital
2) It can be zero for 1s orbital
3) It can never be zero for 2s orbital
4) It can negative for 2p orbital

Answer: (1)

Sample Paper of JEE Main 2020 Chemistry Shift 2- 5th Sept

9. The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol–1 is: (R is gas constant)

Practice Paper of JEE Main 2020 Chemistry Shift 2- 5th Sept


1) R
2) 2/R
3) 1/R
4) 2R

Answer: (4)

\(\begin{array}{l}In(k)=In(A)-\frac{Ea}{R}\left [ \frac{1}{T} \right ]\end{array} \)

In(A)= 10

\(\begin{array}{l}Slope = \frac{-Ea}{R}\times 10^{-3}=-10/5\end{array} \)

Ea= 2000R J/mol

Ea= 2R KJ/mol

10. The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure.

JEE Main 2020 Chemistry Shift 2- 5th Sept Practice Paper

The electrolyte X is:


1) HCl
2) CH+COOH
3) NaCl
4) KNO3

Answer: (2)

JEE Main 2020 Chemistry Shift 2- 5th Sept Practice Question Paper

Such type of variation is always for weak electrolyte. Hence Ans. (2) CH3COOH

11. The final major product of the following reaction is:

Practice Question Paper of JEE Main 2020 Chemistry Shift 2- 5th Sept


Answer: (3)

Question Paper with Answers for JEE Main 2020 Chemistry Shift 2- 5th Sept

12. The major product of the following reaction is :
Question Paper with Solutions for JEE Main 2020 Chemistry Shift 2- 5th Sept

13. Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol–1, and 4 kJ mol–1, respectively. The hydration enthalpy of NaCl is:


1) –780kJ mol–1
2) 784kJ mol–1
3) –784kJ mol–1
4) 780kJ mol–1

Answer: (3)

ΔHsol=L.E + ΔHhyd

4 = 788 + ΔHhyd

ΔHhyd = – 784 KJ/mol Ans

14. Reaction of ammonia with excess Cl2 gives:


1) NH4Cl and N2
2) NH4Cl and HCl
3) NCl3 and HCl
4) NCl3 and NH4Cl

Answer: (3)

JEE Main 2020 Chemistry Shift 2- 5th Sept Question Paper with Solutions

15. Which one of the following polymers is not obtained by condensation polymerisation?


1) Bakelite
2) Nylon 6
3) Buna-N
4) Nylon 6, 6

Answer: (3)

JEE Main 2020 Chemistry Shift 2- 5th Sept Question Paper

16. Consider the comples ions, trans-[Co(en)2Cl2]+ (A) and cis-[Co(en)2Cl2]+ (B). The correct statement regarding them is:


1) Both (A) and (B) can be optically active.
2) (A) can be optically active, but (B) cannot be optically active.
3) Both (A) and (B) cannot be optically active.
4) (A) cannot be optically active, but (B) can be optically active.

Answer: (4)

Due to presence of Pos (A) cannot be optically active, but (B) can be optically active

17. An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is:


1) a
2) a/2
3) √2a
4) a/√2

Answer: (4)

Nearest octahedral voids

One along edge center & other at Body centre

\(\begin{array}{l}Distance=\sqrt{\left [ \frac{a}{2} \right ]^{2}+\left [ \frac{a}{2} \right ]^{2}}=\sqrt{2}\frac{a}{2}=\frac{a}{\sqrt{2}}\end{array} \)

18. The correct order of the ionic radii of O2–, N 3–, F, Mg2+, Na+ and Al3+ is:


1) N3– < O2– < F < Na+ < Mg2+ < Al3+
2) N3– < F < O2– < Mg2+ < Na+ < Al3+
3) Al3+ < Na+ < Mg2+ < O2– < F < N3–
4) Al 3+ < Mg2+ < Na+ < F < O2– < N3–

Answer: (4)

All are Isoelectronic

JEE Main 2020 Chemistry Shift 2- 5th Sept Solution Paper

19. The increasing order of boiling points of the following compounds is:

JEE Main 2020 Chemistry Shift 2- 5th Sept Solved Paper


1) I < III < IV < II
2) IV < I < II < III
3) I < IV < III < II
4) III < I < II < IV

Answer: (3)

JEE Main 2020  Solved Paper of Chemistry Shift 2

20. The one that is NOT suitable for the removal of permanent hardness of water is:


1) Ion-exchange method
2) Calgon’s method
3) Treatment with sodium carbonate
4) Clark’s method

Answer: (4)

Clark’s method is used for Removal of Temporary hardness

JEE Main 2020 Solved Che Paper Shift 2 5th Sept Ques 20

21. For a reaction

\(\begin{array}{l}X+Y\rightleftharpoons 2Z\end{array} \)
, 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L–1. The equilibrium constant of reaction is ________ x/15. The value of x is _________.


Answer: (16)

JEE Main 2020 Solved Che Paper Shift 2 5th Sept Ques 21

22. The volume, in mL, of 0.02 M K2Cr2O7 solution required to react with 0.288 g of ferrous oxalate in acidic medium is ________. (Molar mass of Fe= 56 g mol–1)


Answer: (50 ml)

K2Cr2O7+ FeC2O4 → Cr3+ + Fe3+ + CO2

\(\begin{array}{l}\frac{0.02\times vol\times 6}{1000}=3 \times \frac{0.288}{144}\times 100\end{array} \)

Vol. = 200/4 = 50 ml Ans.

23. Considering that ∆0 > P, the magnetic moment (in BM) of [Ru(H2O)6] 2+ would be ________.


[Ru(H2O)6)2+

Ru 2+ = 3d6(∆0 > P)

= t2g6 eg0

n = 0, u = 0

JEE Main 2020 Chemistry (Shift 2-5th Sept) Paper With Solutions q-24
JEE Main 2020 Chemistry (Shift 2-5th Sept) Paper With Solutions answer 24

25. The number of chiral carbons present in sucrose is ______.


Answer: (9)

JEE Main 2020 Solved Che Paper Shift 2 5th Sept Ques 25

Video Lessons – September 5 Shift21 Chemistry

JEE Main 2020 Chemistry Paper With Solutions Shift 2 September 5

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