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JEE Chemistry
Inorganic Chemistry

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

49 Q 5 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 41–49 of 49
Topics in JEE Chemistry
In the extraction of copper from chalcopyrite (CuFeS₂), the role of silica (SiO₂) is:
A To act as a flux and form slag with iron oxide
B To reduce copper oxide
C To increase the melting point
D To oxidize iron sulfide
Correct Answer:  A. To act as a flux and form slag with iron oxide
EXPLANATION

SiO₂ acts as a flux. It combines with Fe₂O₃ (formed from iron oxidation) to form iron silicate slag (FeSiO₃), which is removed from molten copper.

Test
The correct order of reducing power of Group 17 hydrides is:
A HF > HCl > HBr > HI
B HI > HBr > HCl > HF
C HCl > HF > HI > HBr
D HBr > HI > HF > HCl
Correct Answer:  B. HI > HBr > HCl > HF
EXPLANATION

Reducing power increases with increasing size of halogen (decreasing electronegativity). HI is the strongest reducing agent due to the weak H-I bond.

Test
Which of the following transition metals can form stable complexes with CO?
A Only 3d metals
B Only 4d and 5d metals
C Both 3d, 4d, and 5d metals
D Only Fe and Co
Correct Answer:  C. Both 3d, 4d, and 5d metals
EXPLANATION

Transition metals from 3d, 4d, and 5d series can form stable carbonyl complexes. Examples include Fe(CO)₅, Ni(CO)₄, Mo(CO)₆, and W(CO)₆.

Test
The thermal stability of Group 16 hydrides decreases in the order:
A H₂O > H₂S > H₂Se > H₂Te
B H₂Te > H₂Se > H₂S > H₂O
C H₂S > H₂O > H₂Se > H₂Te
D H₂O > H₂Te > H₂Se > H₂S
Correct Answer:  A. H₂O > H₂S > H₂Se > H₂Te
EXPLANATION

Thermal stability of hydrides decreases down the group due to decreasing bond strength (H-O > H-S > H-Se > H-Te). The H-O bond is the strongest.

Test
In the reaction 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O, the reducing agent is:
A KMnO₄
B HCl
C MnCl₂
D Cl₂
Correct Answer:  B. HCl
EXPLANATION

In this reaction, HCl acts as a reducing agent because Cl⁻ (oxidation state -1) is oxidized to Cl₂ (oxidation state 0). Mn in KMnO₄ is reduced from +7 to +2.

Test
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.

Test
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.

Test
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.

Test
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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