JEE Main 2021 February 25 Shift 2 Chemistry Question Paper with Solutions

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JEE Main 2021 February 25 – Shift 2 Chemistry Question Paper with Solutions

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JEE Main 2021 25th February Shift 2 Chemistry Question Paper

SECTION A

Question 1. Given below are two statements:

Statement I: The identification of Ni²⁺ is carried out by dimethylglyoxime in the presence of NH₄OH

Statement II: The dimethylglyoxime is a bidentate neutral ligand.

In the light of the above statements, choose the correct answer from the options given below:

a. Both statement I and statement II are true
b. Both statement I and statement II are false
c. Statement I is false but statement II is true
d. Statement I is true but statement II is false

Answer: (d)

JEE Main 25th Feb Shift 2 Chemistry Paper Question 1 solution

Dimethylglyoxime is a negative bidentate legend.

Question 2. Carbylamine test is used to detect the presence of a primary amino group in an organic compound. Which of the following compounds is formed when this test is performed with aniline?

JEE Main 25th Feb Shift 2 Chemistry Paper Question 2

Answer: (b)

Question 3. The correct order of bond dissociation enthalpy of halogen is:

a. F₂>Cl₂>Br₂>I₂
b. Cl₂>F₂>Br₂>I₂
c. Cl₂>Br₂>F₂>I₂
d. I₂>Br₂>Cl₂>F₂

Answer: (c)

Fact based

F₂ has F — F, F₂involves repulsion of non-bonding electrons & more over its size is small and hence due to high repulsion its bond dissociation energy is very low.

Question 4. Which one of the following statements is FALSE for hydrophilic sols?

a. These sols are reversible in nature
b. The sols cannot be easily coagulated
c. They do not require electrolytes for stability
d. Their viscosity is of the order of that of H₂O

Answer: (d)

Fact based

Question 5. Water does not produce CO on reacting with

a. C₃H₈
b. C
c. CH₄
d. CO₂

Answer: (d)

H₂O + CO₂ H₂CO₃

Question 6. What is ‘X’ in the given reaction?

JEE Main 25th Feb Shift 2 Chemistry Paper Question 6

Answer: (a)

Question 7. If which of the following orders the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?

(i) [FeF₆]^3-

(ii) [Co(NH₃)₆]³⁺

(iii) [NiCl₄]^2-

(iv) [Cu(NH₃)₄]²⁺

a. (ii)>(i)>(iii)>(iv)
b. (iii)>(iv)>(ii)>(i)
c. (ii)>(iii)>(i)>(iv)
d. (i)>(iii)>(iv)>(ii)

Answer: (d)

[FeF₆]^3-: Fe³⁺ 3d⁵ → 5-unpaired electrons as F^– is weak field legend

[Co(NH₃)₆]³⁺: Co³⁺ 3d⁶→ No-unpaired electron as NH₃ is strong field light and causes pairing

[NiCl4]^2-: Ni²⁺ 3d⁸ → 2-unpaired electrons

[Cu(NH₃)₄]²⁺: Cu²⁺ 3d⁹→ 1-unpaired electrons

Question 8. The major product of the following reaction:

Answer: (d)

JEE Main 25th Feb Shift 2 Chemistry Paper Question 8 solution

Question 9: The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is:

a. CH₃COCl/AlCl₃, Br₂/AlBr₃, HNO₃/H₂SO₄, Zn/HCl
b. CH₃COCl/AlCl₃, Zn-Hg/HCl, Br₂/AlBr₃, HNO₃/H₂SO₄
c. Br₂/AlBr₃, CH₃COCl/AlCl₃, HNO₃/H₂SO₄, Zn/HCl
d. HNO₃/H₂SO₄, Br₂/AlCl₃, CH₃COCl/AlCl₃, Zn-Hg/HCl

Answer: (b)

Question 10. The major components of German Silver are:

a. Cu, Zn and Mg
b. Ge, Cu and Ag
c. Zn, Ni and Ag
d. Cu, Zn and Ni

Answer: (d)

Fact

German silver is alloy which does not have silver.

Cu-50%; Ni-30%; Zn-20%

Question 11. The method used for the purification of Indium is:

a. Van Arkel method
b. Vapour phase refining
c. Zone refining
d. Liquation

Answer: (c)

Fact

Ga, In, Si, Ge are refined by zone refining or vacuum refining.

Question 12. Which of the following is correct structure of α-anomer of maltose

Answer: (d)

Question 13. The major product of the following reaction is:

a. CH₃CH₂CH₂CHO
b. CH₃CH₂CH=CH–CHO
c. CH₃CH₂CH₂CH₂CHO
d.

Answer: (c)

Question 14. The correct order of acid character of the following compounds is:

a. II>III>IV>I
b. III>II>I>IV
c. IV>III>II>I
d. I>II>III>IV

Answer: (a)

Acidity of carboxylic acid α-R>-H>-I

1 / α + R > + H > + I

Question 15. Which among the following species has unequal bond lengths?

a. XeF₄
b. SiF₄
c. BF₄^–
d. SF₄

Answer: (d)

JEE Main 25th Feb Shift 2 Chemistry Paper Question 15 solution

Question 16. Given below are two statements:

Statement I: α and β forms of sulphur can change reversibly between themselves with slow heating or slow cooling.

Statement II: At room temperature the stable crystalline form of sulphur is monoclinic sulphur.

In the light of the above statements, choose the correct answer from the options given below.

a. Both statement I and statement II are false
b. Statement I is true but statement II is false
c. Both statement I and statement II are true
d. Statement I is false but statement II is true

Answer: (b)

Question 17. Correct statement about the given chemical reaction is:

a. Reaction is possible and compound (A) will be a major product.
b. The reaction will form a sulphonated product instead of nitration.
c. NH₂ group is ortho and para directive, so product (B) is not possible.
d. Reaction is possible and compound (B) will be the major product.

Answer: (a)

Question 18. Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?

a. Nitric acid
b. Sodium hydroxide
c. Hydrochloric acid
d. Ammonia

Answer: (a)

NaCN + HNO₃ → NaNO₃ + HCN ↑

Na₂S + HNO₃ → NaNO₃ + H₂S ↑

Nitric acid decomposes NaCN & Na₂S, else they precipitate in test & misguide the resolve.

Question 19: Given below are two statements:

Statement I: The pH of rain water is normally ~5.6.

Statement II: If the pH of rain water drops below 5.6, it is called acid rain.

In the light of the above statements, choose the correct answer from the option given below.

a. Statement I is false but Statement II is true
b. Both statement I and statement II are true
c. Both statement I and statement II are false
d. Statement I is true but statement II is false

Answer: (b)

Both statements are correct

Question 20. The solubility of Ca(OH)₂ in water is:

[Given: The solubility product of Ca(OH)₂ in water = 5.5×10^-6]

a. 1.11 ×10^-6
b. 1.77 ×10^-6
c. 1.77×10^-2
d. 1.11×10^-2

Answer: (d)

Ca(OH)₂ ⇌ Ca⁺² + 2OH^–

s (2s + 10^-7)

s(2s+10^-7)² = 55×10^-7

4s³ = 55×10^-7

s³ = 5500 / 4 × 10^-9

\(\begin{array}{l}s = \left ( \frac{2250}{2} \right )^{\frac{1}{3}}\times 10^{-3}\end{array} \)

s = (1125)^⅓× 10^-3

s = 1.11 × 10^-2

SECTION B

Question 1. If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ______molal.

(Rounded-off to the nearest integer)

[K_b=0.52 K kg mol^-1]

Answer: 3

AB → A⁺ + B^–

1 – α α α

α = 3 / 4

N = 2

i = [1+(2-1)α]
2.5 = [1+(2-1)3/4] × 0.52 × m

\(\begin{array}{l}m = \frac{2.5}{0.52\times \frac{7}{4}} = \frac{10}{3.64} = 2.747\end{array} \)

m= 2.747≈ 3 mol/kg

Question 2. The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ____BM.

Answer: 2

₂₉Cu⁺² → [Ar]¹⁸3d⁹

No. of unpaired e^– = 1

Magnetic moment = μ = √n(n + 2)

μ = √(1)(1 + 2) = √3B.M.

= 1.73 B.M

Question 3. The number of compound/s given below which contain/s —COOH group is ______

a. Sulphanilic acid
b. Picric acid
c. Aspirin
d. Ascorbic acid

Answer: (a)

Question 4. The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m³is ______.

[Molar mass of Cu : 63.54 g; Avogadro number = 6.022×10²³]

Answer: 9077

a = 3.596 Å

\(\begin{array}{l}d = \frac{Z\times GMM}{N_{A}\times a^{3}}\end{array} \)

\(\begin{array}{l}d = \frac{Z4\times63.54\times 10^{-3}}{6.022\times 10^{23}\times (3.956\times 10^{-10})^{3}}\end{array} \)

d = 0.9076 × 104 = 9076.2 kg/m3

Question 5. Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.

(i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL

If the volume of oxalic acid taken was 10.0 ml. then the molarity of the NaOH solution is ____M. (Rounded-off to the nearest integer)

Answer: 6

Eq. of NaOH = Eq. of oxalic acid
[NaOH] × 1 × 4.4 = 5/4 × 2 ×10
[NaOH] = 100 / 4 × 4.4 = 25 / 4.4 = 5.68

Nearest integer = 6 M

Question 6. Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol^—1 is____. (Rounded off to the nearest integer)

[h=6.63×10^-34Js, c = 3.00×10⁸ms^-1, NA=6.02×1023 mol^-1]

Answer: 180

Energy required to ionize an atom of metal ‘A’ = hc / λ = hc / 663 nm for 1 mole atoms of ‘A’

Total energy required = N_A × hc / λ

=
\(\begin{array}{l}\frac{6.023\times10^{23} \times6.63 \times10^{-34} \times3 \times10^{8}}{663\times 10^{-9}}\end{array} \)

= 6.023 × 3 × 10^23-34+8+7

= 18.04 × 10⁴ J/mol

= 180.4 kJ/mol

Nearest Integer = 180 kJ/Mol.

Question 7. The rate constant of a reaction increases by five times on increasing temperature from 27⁰ C to 52⁰C. The value of activation energy in kJ mol^-1 is ______.

(Rounded off to the nearest integer)
[R=8.314 J K^-1 mol^-1]

Answer: 52

E_a = 51524.96 J/mol

E_a = 51.524 kJ/mol

52 Ans.

Question 8. Copper reduces into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed [Cu²⁺] and P_NO=P_NO2), the HNO₃ concentration at which the thermodynamic tendency for reduction of into NO and NO₂ by copper is same is 10^x M.

The value of 2x is ______. (Rounded-off to the nearest integer)
[Given:
\(\begin{array}{l}E_{Cu^{2+} / Cu}^{0} = 0.34V, E_{No_{3}^{-} / NO}^{0} = 0.96V, E_{No_{3}^{-} / NO_{2}}^{0} = 0.79V\end{array} \)
and at 298 K,
\(\begin{array}{l}\frac{RT}{F}(2.303)= 0.059\end{array} \)
]

Answer: 4

Anode

Cu(s) → Cu+2 + 2e-

Cathode (1)

log (HNO₃) = 2.16
[HNO₃] = 10^2.16 = 10^x

x = 2.16 ⇒ 2x = 4.32 ≈ 4

Question 9. Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against a constant external 4.3 MPa. The heat transferred in this process is ____kJ mol^-1. (Rounded-off of the nearest integer)

[Use R = 8.314 J mol^-1 K^-1]

Answer: 15

Moles (n) = 5

T = 293 K

Process = Iso T. → Irreversible

Pⁱⁿⁱ = 2.1 M Pa

P^t = 1.3 M Pa

P^ext = 4.3 mPa

Work = – P^ext Δv

\(\begin{array}{l}= -4.3\times \left ( \frac{5\times 293R }{1.3}-\frac{5\times293 }{2.1} \right )\end{array} \)

\(\begin{array}{l}= -5\times293 \times8.314 \times43 \left ( \frac{1}{13}-\frac{1}{21} \right )\end{array} \)

\(\begin{array}{l}= \frac{5\times293 \times8.314 \times43\times 8 }{21 \times 13}\end{array} \)

= -15347.7049 J

= – 15.34 kJ

Isothermal process, so ΔU = 0

w = – Q

Q = 15.34 kJ / mol

So, answer is 15.

Question 10. Among the following, the number of metal/s which can be used as electrodes in the photoelectric cell is _____(Integer answer).

a. Li

b. Na

c. Rb

d. Cs

Answer: (a)

Cs is used in photoelectric cells due to its very low ionization potential.

JEE Main 2021 February 25 – Shift 2 Chemistry Question Paper with Solutions

JEE Main 2021 February 25 – Shift 2 Chemistry Question Paper with Solutions

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