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JEE Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

225 Q 5 Topics Take Test
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Difficulty: All Easy Medium Hard 121–130 of 225
Topics in JEE Chemistry
Q.121 Medium Inorganic Chemistry
The correct order of lattice energies is:
A NaF > NaCl > NaBr > NaI
B NaI > NaBr > NaCl > NaF
C NaCl > NaF > NaBr > NaI
D NaBr > NaCl > NaF > NaI
Correct Answer:  A. NaF > NaCl > NaBr > NaI
EXPLANATION

Lattice energy is inversely proportional to the size of ions. As we go from F⁻ to I⁻, ionic size increases, so lattice energy decreases. NaF has the smallest anion, hence highest lattice energy.

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Q.122 Medium Inorganic Chemistry
The solubility of alkaline earth metal hydroxides increases down the group with the exception of:
A Mg(OH)₂ → Ca(OH)₂
B Ca(OH)₂ → Sr(OH)₂
C Ba(OH)₂ is more soluble than Sr(OH)₂
D Be(OH)₂ is least soluble
Correct Answer:  D. Be(OH)₂ is least soluble
EXPLANATION

Be(OH)₂ is amphoteric and has low solubility. Generally, solubility increases from Mg to Ba, but Be(OH)₂ doesn't follow this trend due to its amphoteric nature.

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Q.123 Medium Inorganic Chemistry
Which compound shows the highest ionic character?
A NaCl
B MgO
C CaO
D KCl
Correct Answer:  B. MgO
EXPLANATION

Ionic character depends on electronegativity difference and charge density. MgO has the highest charge density (smallest cation with 2+ charge) and highest electronegativity difference.

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Q.124 Medium Inorganic Chemistry
The ionization energy of nitrogen is higher than oxygen because:
A Nitrogen has a larger atomic radius
B Nitrogen has a half-filled 2p orbital configuration which is more stable
C Oxygen has more electrons
D Nitrogen has more protons
Correct Answer:  B. Nitrogen has a half-filled 2p orbital configuration which is more stable
EXPLANATION

Nitrogen (1s² 2s² 2p³) has a half-filled 2p orbital which provides extra stability. Removing an electron from this configuration requires more energy than from oxygen's configuration.

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Q.125 Medium Organic Chemistry
In the nitration of benzene with HNO₃/H₂SO₄, the rate-determining step involves attack by:
A NO₂⁺ (nitronium ion)
B NO₂• (nitrogen dioxide radical)
C HNO₃ (nitric acid molecule)
D NO₃⁻ (nitrate ion)
Correct Answer:  A. NO₂⁺ (nitronium ion)
EXPLANATION

The HNO₃/H₂SO₄ mixture generates NO₂⁺ (nitronium ion), which is the electrophile attacking the benzene ring in the rate-determining step. This is a classic electrophilic aromatic substitution.

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Q.126 Medium Organic Chemistry
The addition of HBr to propene in the presence of peroxides follows:
A Markovnikov's rule to give 2-bromopropane
B Anti-Markovnikov's rule to give 1-bromopropane
C Free radical mechanism to give equal amounts of both isomers
D Ionic mechanism to give 2-bromopropane
Correct Answer:  B. Anti-Markovnikov's rule to give 1-bromopropane
EXPLANATION

Peroxides initiate free radical mechanism (Kharasch effect). In free radical addition, HBr adds anti-Markovnikov to propene, giving 1-bromopropane as the major product. The Br radical adds first to the terminal carbon.

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Q.127 Medium Organic Chemistry
In the dehydration of 2-methylbutan-2-ol, the major product is:
A 2-methylbut-1-ene
B 2-methylbut-2-ene
C But-1-ene
D Pent-2-ene
Correct Answer:  B. 2-methylbut-2-ene
EXPLANATION

Dehydration of 2-methylbutan-2-ol follows Zaitsev's rule, producing the most stable (most substituted) alkene. 2-methylbut-2-ene is a trisubstituted alkene and is the major product.

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Q.128 Medium Organic Chemistry
The reaction of phenol with excess bromine in water produces:
A 2,4,6-tribromophenol
B 4-bromophenol
C 2,4-dibromophenol
D 2,6-dibromophenol
Correct Answer:  A. 2,4,6-tribromophenol
EXPLANATION

Phenol is highly activated towards electrophilic aromatic substitution due to the electron-donating -OH group. Bromine can add at all three ortho and para positions with excess bromine, giving 2,4,6-tribromophenol.

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Q.129 Medium Organic Chemistry
Which of the following alkenes will undergo hydroboration-oxidation to give a secondary alcohol as the major product?
A 1-methylcyclohexene
B 2-methylbut-2-ene
C pent-1-ene
D 2,3-dimethylbut-2-ene
Correct Answer:  A. 1-methylcyclohexene
EXPLANATION

Hydroboration-oxidation follows anti-Markovnikov's rule and gives Markovnikov's hydration product after oxidation. 1-methylcyclohexene gives secondary alcohol, while pent-1-ene gives primary and 2-methylbut-2-ene gives tertiary alcohol.

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Q.130 Medium Organic Chemistry
The compound that will show maximum hydrogen bonding is:
A CH3-CH2-OH
B CH3-CO-NH2
C CH3-CH2-NH2
D CH3-CHO
Correct Answer:  B. CH3-CO-NH2
EXPLANATION

Amides have both N-H (hydrogen bond donor) and C=O (hydrogen bond acceptor), allowing formation of strong intermolecular hydrogen bonds.

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