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Question 151

JEE Main 2025• Chemistry Section A

A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is ___ .

[Given : molar mass of aluminium and chlorine are 27 g mol–1 and 35.5 g mol–1 respectively, Faraday constant = 96500 C mol–1].

  1. AOption (a)

    1.660 g

  2. BOption (b)

    1.007 g

  3. COption (c)

    0.336 g 

  4. DOption (d)

    0.441 g

View answer & explanation

Correct answer

Answer: (c)C.

0.336 g 

Why this answer is correct

CONCEPT:

Faraday's Laws of Electrolysis

  • Faraday's first law of electrolysis states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
  • The relationship can be expressed as:

m = (ItM)/(nF)

where m is the mass of the substance deposited, I is the current, t is the time, M is the molar mass of the substance, n is the number of electrons transferred per ion, and F is the Faraday constant.

EXPLANATION:

  • For the electrolysis of aluminium chloride (AlCl3), the reaction at the cathode is:

Al3+ + 3e- → Al

  • Given data:
  • Current (I) = 2A
  • Time (t) = 30 minutes = 30 × 60 = 1800 seconds
  • Molar mass of Aluminium (M) = 27 g mol-1
  • Faraday constant (F) = 96500 C mol-1
  • In the reaction, the number of electrons (n) transferred per ion is 3.
  • Using the formula:

m = (ItM)/(nF)

m = (2 × 1800 × 27)/(3 × 96500)

m = (97200)/(289500)

m = 0.336 g

Therefore, the amount of aluminium deposited at the cathode is 0.336 g.

Question 152

JEE Main 2025• Chemistry Section A

Which of the following statement is not true for radioactive decay ? 

  1. AOption (a)

    Amount of radioactive substance remained after three half lives is 18\frac{1}{8}th of original amount.

  2. BOption (b)

    Decay constant does not depend upon temperature.

  3. COption (c)

    Decay constant increases with increase in temperature. 

  4. DOption (d)

    Half life is ln 2 times of 1 rate constant \frac{1}{\text { rate constant }}

View answer & explanation

Correct answer

Answer: (c)C.

Decay constant increases with increase in temperature. 

Why this answer is correct

CONCEPT:

Radioactive Decay

  • Radioactive decay is a process by which the nucleus of an unstable atom loses energy by emitting radiation. The decay constant (λ) is a probability rate constant that quantifies the likelihood of decay of an atom per unit time.
  • Key characteristics of radioactive decay:
  • The decay constant (λ) is independent of temperature.
  • The half-life (t1/2) of a radioactive substance is the time required for half of the radioactive atoms to decay. The relationship between half-life and decay constant is given by:

t1/2 = ln(2) / λ

EXPLANATION:

  • Statement 1: Amount of radioactive substance remained after three half-lives is 1/8 of the original amount.
  • This is true. After one half-life, half of the original amount remains. After three half-lives, (1/2)3 = 1/8 of the original amount remains.
  • Statement 2: Decay constant does not depend upon temperature.
  • This is true. The decay constant is a fundamental property of the radioactive substance and is not affected by temperature.
  • Statement 3: Decay constant increases with increase in temperature.
  • This is not true. The decay constant is independent of temperature.
  • Statement 4: Half life is ln(2) times of 1/rate constant.
  • This is true. The half-life (t1/2) is given by t1/2 = ln(2) / λ, where λ is the decay constant.

Therefore, the statement that is not true for radioactive decay is: Decay constant increases with increase in temperature.

Question 153

JEE Main 2025• Chemistry Section A

How many different stereoisomers are possible for the given molecule ? 

  1. AOption (a)

    3

  2. BOption (b)

    1

  3. COption (c)

    2

  4. DOption (d)

    4

View answer & explanation

Correct answer

Answer: (d)D.

4

Why this answer is correct

CONCEPT:

Stereoisomers and Geometrical Isomerism

  • Stereoisomers are compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of atoms.
  • Geometrical isomerism occurs in compounds with restricted rotation around a double bond or ring, resulting in cis-trans isomers.
  • Each chiral center or double bond contributes to possible isomerism.

EXPLANATION:

It has 4 stereoisomers [R cis R trans  ScisStrans]\left[\begin{array}{ll} \mathrm{R} \text { cis } & \mathrm{R} \text { trans } \ \mathrm{Scis} & \mathrm{Strans} \end{array}\right]

  • Two factors contribute to stereoisomerism here:
  • The double bond between the third and fourth carbon introduces cis and trans isomers.
  • The chiral carbon (second carbon with -CH3, -CH=CH-CH3, -OH, and -H as substituents) introduces R and S configurations.
  • Combining these, the molecule can exhibit the following configurations:
  • Cis with R configuration
  • Cis with S configuration
  • Trans with R configuration
  • Trans with S configuration
  • Hence, the molecule has 4 possible stereoisomers.

Therefore, the total number of stereoisomers for the given molecule is 4.

Question 154

JEE Main 2025• Chemistry Section A

Which of the following electronegativity order is incorrect?

  1. AOption (a)

    Al < Mg < B < N 

  2. BOption (b)

    Al < Si < C < N

  3. COption (c)

    Mg < Be < B < N

  4. DOption (d)

    S < Cl < O < F

View answer & explanation

Correct answer

Answer: (a)A.

Al < Mg < B < N 

Why this answer is correct

EXPLANATION:​.

Electronegativity

  • Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used scale for electronegativity.
  • The electronegativity values (on the Pauling scale) for some elements are as follows:
  • | | Li | Be | B | C | N | O | F | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = On pauling scale | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4.0 |

For elements in the second period and third period:

  • | | Na | Mg | Al | Si | P | S | Cl | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = | 0.9 | 1.2 | 1.5 | 1.8 | 2.1 | 2.5 | 3.0
    |

 

Therefore, the electronegativity order that is incorrect is: Al < Mg < B < N

Question 155

JEE Main 2025• Chemistry Section A

Lanthanoid ions with 4f7 configuration are :

(A) Eu2+

(B) Gd3+

(C) Eu3+

(D) Tb3+

(E) Sm2+

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) and (B) only

  2. BOption (b)

    (A) and (D) only

  3. COption (c)

    (B) and (E) only

  4. DOption (d)

    (B) and (C) only 

View answer & explanation

Correct answer

Answer: (a)A.

(A) and (B) only

Why this answer is correct

CONCEPT:

Electronic Configuration of Lanthanoids

  • Lanthanoids are a series of 15 elements from lanthanum (La) to lutetium (Lu) in the periodic table, characterized by the filling of the 4f orbitals.
  • The general electron configuration of lanthanoids is [Xe] 4fn 5dx 6s2.
  • The common oxidation states of lanthanoids include +2, +3, and sometimes +4.
  • Key electron configurations:
  • Europium (Eu): [Xe] 4f7 6s2
  • Gadolinium (Gd): [Xe] 4f7 5d1 6s2
  • Terbium (Tb): [Xe] 4f9 6s2
  • Samarium (Sm): [Xe] 4f6 6s2

EXPLANATION:

  • To determine which ions have a 4f7 configuration, we must consider the electron loss when forming the given ions:
  • Europium (Eu):
  • Eu2+: [Xe] 4f7 (Loses 2 electrons from 6s2)
  • Eu3+: [Xe] 4f6 (Loses 2 electrons from 6s2 and 1 electron from 4f7)
  • Gadolinium (Gd):
  • Gd3+: [Xe] 4f7 (Loses 2 electrons from 6s2 and 1 electron from 5d1)
  • Terbium (Tb):
  • Tb3+: [Xe] 4f8 (Loses 2 electrons from 6s2 and 1 electron from 4f9)
  • Samarium (Sm):
  • Sm2+: [Xe] 4f6 (Loses 2 electrons from 6s2)

Therefore, the lanthanoid ions with a 4f7 configuration are Eu2+ and Gd3+.

Question 156

JEE Main 2025• Chemistry Section A

Match List-I with List-II 

List-IList-II
(A)Al3+ < Mg2+ < Na+ < F–(I)Ionisation Enthalpy
(B)B < C < O < N(II)Metallic character
(C)B < Al < Mg < K(III)Electronegativity
(D)Si < P < S < Cl(IV)Ionic radii
  1. AOption (a)

    A-IV, B-I, C-III, D-II 

  2. BOption (b)

    A-II, B-III, C-IV, D-I

  3. COption (c)

    A-IV, B-I, C-II, D-III

  4. DOption (d)

    A-III, B-IV, C-II, D-I

View answer & explanation

Correct answer

Answer: (c)C.

A-IV, B-I, C-II, D-III

Why this answer is correct

EXPLANATION:

Trends in the Periodic Table

  • Ionic Radii: The size of the ionic radii decreases as the positive charge of the ion increases.
  • Order: Al3+ < Mg2+ < Na+ < F-
  • Ionisation Enthalpy: The energy required to remove an electron from an atom in its gaseous state. It generally increases across a period from left to right.
  • Order: B < C < O < N
  • Metallic Character: The degree to which an element exhibits metallic properties, which decreases across a period and increases down a group.
  • Order: B < Al < Mg < K
  • Electronegativity: The tendency of an atom to attract electrons in a chemical bond. It generally increases across a period from left to right.
  • Order: Si < P < S < Cl
List-IList-II
(A) Al3+ < Mg2+ < Na+ < F-(IV) Ionic Radii
(B) B < C < O < N(I) Ionisation Enthalpy
(C) B < Al < Mg < K(II) Metallic Character
(D) Si < P < S < Cl(III) Electronegativity

The correct matching is  A-IV, B-I, C-II, D-III

Question 157

JEE Main 2025• Chemistry Section A

Which of the following acids is a vitamin ? 

  1. AOption (a)

    Adipic acid

  2. BOption (b)

    Aspartic acid

  3. COption (c)

    Ascorbic acid

  4. DOption (d)

    Saccharic acid

View answer & explanation

Correct answer

Answer: (c)C.

Ascorbic acid

Why this answer is correct

CONCEPT:

Vitamins and their Chemical Names

  • Vitamins are essential nutrients required by the body to function properly. Each vitamin has a specific chemical name.
  • Ascorbic acid is the chemical name for Vitamin C.

EXPLANATION:

  • Adipic acid: This is not a vitamin, but a dicarboxylic acid commonly used in the production of nylon.
  • Aspartic acid: This is not a vitamin, but an amino acid involved in the synthesis of proteins.
  • Ascorbic acid: This is Vitamin C, an essential nutrient for the repair of tissues and the enzymatic production of certain neurotransmitters.
  • Saccharic acid: This is not a vitamin, but an oxidized derivative of glucose.

Therefore, the acid that is a vitamin is ascorbic acid (Vitamin C).

The correct answer is: (3) Ascorbic acid

Question 158

JEE Main 2025• Chemistry Section A

A liquid when kept inside a thermally insulated closed vessel at 25°C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ? 

  1. AOption (a)

    ΔU > 0, q = 0, w > 0

  2. BOption (b)

    ΔU = 0, q = 0, w = 0

  3. COption (c)

    ΔU < 0, q = 0, w > 0

  4. DOption (d)

    ΔU = 0, q < 0, w > 0

View answer & explanation

Correct answer

Answer: (a)A.

ΔU > 0, q = 0, w > 0

Why this answer is correct

CONCEPT:

First Law of Thermodynamics

  • The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (q) plus the work done on the system (w):
  • ΔU = q + w
  • In a thermally insulated (adiabatic) system, there is no heat exchange with the surroundings, so:
  • q = 0
  • When work is done on the system (for example, by mechanically stirring the liquid), the work done (w) is positive, which increases the internal energy (ΔU) of the system.

EXPLANATION:

  • Since the vessel is thermally insulated, there is no heat exchange with the surroundings:
  • q = 0
  • The work is done on the liquid by mechanically stirring it, so:
  • w > 0
  • According to the first law of thermodynamics:
  • ΔU = q + w
  • Since q = 0, we have ΔU = w
  • When w > 0, ΔU > 0

Therefore, the correct option for the thermodynamic parameters is  ΔU > 0, q = 0, w > 0​

Question 159

JEE Main 2025• Chemistry Section A

Radius of the first excited state of Helium ion is given as :

a0 → radius of first stationary state of hydrogen atom.

  1. AOption (a)

    r=a02r=\frac{a_{0}}{2}

  2. BOption (b)

    r=a04r=\frac{a_{0}}{4}

  3. COption (c)

    r = 4a0

  4. DOption (d)

    r = 2a0

View answer & explanation

Correct answer

Answer: (d)D.

r = 2a0

Why this answer is correct

CONCEPT:

Bohr's Model of the Atom

  • In Bohr's model, the radius of the ( n )th orbit of an atom is given by:
  • rn=n2a0Zr_n = \frac{n^2 a_0}{Z}
  • where:
  • ( rn ) = radius of the ( n )th orbit
  • ( a0 ) = Bohr radius (radius of the first stationary state of the hydrogen atom)
  • ( n ) = principal quantum number
  • ( Z ) = atomic number of the element
  • For the first excited state, ( n = 2 ).

APPLICATION TO HELIUM ION (He+):

  • For the Helium ion (He+), the atomic number ( Z ) is 2.
  • For the first excited state, ( n = 2 ).
  • Substituting these values into the formula for the radius:
  • r=n2a0Z =(2)2a02 =4a02 =2a0r = \frac{n^2 a_0}{Z} \ = \frac{(2)^2 a_0}{2} \ = \frac{4 a_0}{2} \ = 2 a_0

Therefore, the radius of the first excited state of the Helium ion is r = 2a0

Question 160

JEE Main 2025• Chemistry Section A

Given below are two statements :

Statement I : CH3 – O – CH2 – Cl will undergo SN1 reaction though it is a primary halide. 

Statement II : 

 will not undergo SN2 reaction very easily though it is a primary halide.

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

  1. AOption (a)

    Statement I is incorrect but Statement II is correct. 

  2. BOption (b)

    Both Statement I and Statement II are incorrect

  3. COption (c)

    Statement I is correct but Statement II is incorrect

  4. DOption (d)

    Both Statement I and Statement II are correct. 

View answer & explanation

Correct answer

Answer: (d)D.

Both Statement I and Statement II are correct. 

Why this answer is correct

CONCEPT:

SN1 and SN2 Reactions

  • SN1 Mechanism: This is a two-step mechanism where the rate-determining step involves the formation of a carbocation intermediate. It is favored by:
  • Stable carbocations (tertiary > secondary > primary).
  • Polar protic solvents.
  • SN2 Mechanism: This is a single-step mechanism involving a backside attack by the nucleophile. It is favored by:
  • Less steric hindrance (methyl > primary > secondary; tertiary is unfavorable).
  • Polar aprotic solvents.

EXPLANATION:

CH3 – O – CH2 – Cl will undergo SN1 mechanism because \(\mathrm{CH}{3}-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H}{2}\)  is highly stable. 

  • Statement I: CH3–O–CH2–Cl undergoes the SN1 mechanism even though it is a primary halide because the intermediate carbocation (CH3–O–CH2+) formed is stabilized by resonance with the oxygen atom.
  • Statement II: CH3–C(CH3)2–CH2–Cl (neopentyl chloride) does not undergo the SN2 mechanism easily due to significant steric hindrance around the reactive center, which obstructs the nucleophilic backside attack.
  • Both statements are correct based on the mechanisms and the structural properties of the molecules.

Therefore, both Statement I and Statement II are correct.