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JEE Main MCQ questions — Mathematics, Physics, Chemistry for engineering entrance.

91 Q 3 Subjects 12th (PCM)
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Q.41 Hard JEE Chemistry Inorganic Chemistry
The reaction of concentrated H₂SO₄ with carbon produces primarily:
A CO₂ only
B CO only
C CO₂ and CO in 1:1 ratio
D CO and SO₂ in 2:1 ratio
Correct Answer:  D. CO and SO₂ in 2:1 ratio
EXPLANATION

C + 2H₂SO₄(conc) → CO₂ + 2SO₂ + 2H₂O at 25°C. At higher temperatures: C + 2H₂SO₄(conc) → CO + 2SO₂ + 2H₂O, with CO and SO₂ in 1:1 ratio. Overall primary products are CO and SO₂.

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Q.42 Hard JEE Chemistry Inorganic Chemistry
The bond order between atoms in the superoxide ion O₂⁻ is:
A 1
B 1.5
C 2
D 2.5
Correct Answer:  B. 1.5
EXPLANATION

O₂⁻ (superoxide) has configuration similar to O₂ with one additional electron in π* orbital. Bond order = (8-5)/2 = 1.5

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Q.43 Hard JEE Chemistry Inorganic Chemistry
In the preparation of potassium permanganate from pyrolusite (MnO₂), the ore is first fused with KOH. The product formed is:
A KMnO₄
B K₂MnO₄
C KMnO₃
D Mn₂O₃
Correct Answer:  B. K₂MnO₄
EXPLANATION

MnO₂ (Mn⁴⁺) is oxidized by air to MnO₄²⁻ (Mn⁶⁺) forming K₂MnO₄. This is then oxidized to KMnO₄ using oxidizing agents like Cl₂.

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Q.44 Hard JEE Chemistry Inorganic Chemistry
The stability of peroxides increases in the order:
A Li₂O₂ < Na₂O₂ < K₂O₂
B K₂O₂ < Na₂O₂ < Li₂O₂
C Na₂O₂ < Li₂O₂ < K₂O₂
D Li₂O₂ < K₂O₂ < Na₂O₂
Correct Answer:  A. Li₂O₂ < Na₂O₂ < K₂O₂
EXPLANATION

Larger cations better stabilize the larger O₂²⁻ ion through lattice energy considerations. K⁺ > Na⁺ > Li⁺, so K₂O₂ is most stable.

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Q.45 Hard JEE Chemistry Inorganic Chemistry
In the context of transition metals, what does the term 'lanthanide contraction' refer to?
A Decrease in size of lanthanides
B Unusual decrease in atomic radius across second and third transition series
C Contraction due to poor shielding by f-electrons
D Both B and C
Correct Answer:  D. Both B and C
EXPLANATION

Lanthanide contraction is the result of poor shielding by f-electrons, causing atomic radius to decrease unusually across the second and third transition series.

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Q.46 Hard JEE Chemistry Inorganic Chemistry
Which of the following complexes would show optical isomerism?
A [Zn(NH₃)₄]²⁺
B [Co(en)₃]³⁺
C [Ni(H₂O)₆]²⁺
D [CuCl₄]²⁻
Correct Answer:  B. [Co(en)₃]³⁺
EXPLANATION

[Co(en)₃]³⁺ is an octahedral complex with three bidentate ligands, forming non-superimposable mirror images (Δ and Λ isomers).

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Q.47 Hard JEE Chemistry Inorganic Chemistry
The structure of [PtCl₄]²⁻ is square planar because:
A Pt is in +2 oxidation state
B d⁸ configuration with strong field ligands favors square planar geometry
C Crystal field stabilization energy is maximum
D All of the above
Correct Answer:  D. All of the above
EXPLANATION

Pt²⁺ has d⁸ configuration. With strong field Cl⁻, square planar geometry provides maximum CFSE and is thermodynamically favored.

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Q.48 Hard JEE Chemistry Inorganic Chemistry
Which of the following chromium compounds would show maximum paramagnetic behavior?
A [Cr(H₂O)₆]³⁺
B [Cr(NH₃)₆]³⁺
C [Cr(CN)₆]³⁻
D CrO₄²⁻
Correct Answer:  A. [Cr(H₂O)₆]³⁺
EXPLANATION

[Cr(H₂O)₆]³⁺ is high spin with 3 unpaired electrons. CN⁻ is strong field ligand causing pairing. [Cr(NH₃)₆]³⁺ is intermediate.

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Q.49 Hard JEE Chemistry Inorganic Chemistry
In the extraction of iron from Fe₂O₃ using CO as reducing agent, the rate-determining step involves:
A Dissociation of CO
B Diffusion of CO through the solid layer of unreacted ore
C Formation of Fe
D Decomposition of Fe₂O₃
Correct Answer:  B. Diffusion of CO through the solid layer of unreacted ore
EXPLANATION

In solid-state reduction, diffusion through the ash layer is often the rate-determining step, creating a diffusion barrier.

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Q.50 Hard JEE Chemistry Inorganic Chemistry
The stability of peroxides and superoxides of alkali metals increases with increasing atomic number because:
A Larger cations can stabilize larger anions
B Smaller cations form stronger bonds
C Electronegativity increases
D Oxidation state increases
Correct Answer:  A. Larger cations can stabilize larger anions
EXPLANATION

Larger alkali metal cations (K⁺, Rb⁺) can stabilize larger peroxide and superoxide anions due to better lattice energy matching.

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