JEE Chemistry — MCQ Practice Questions & Answers

Question 1

The ionization energy of nitrogen is higher than oxygen because:

Accepted Answer

Correct answer: B. Nitrogen has a half-filled 2p orbital configuration which is more stable — 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.

Answer

A. Nitrogen has a larger atomic radius

Answer

C. Oxygen has more electrons

Answer

D. Nitrogen has more protons

Question 2

Which compound shows the highest ionic character?

Accepted Answer

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

Answer

A. NaCl

Answer

C. CaO

Answer

D. KCl

Question 3

The solubility of alkaline earth metal hydroxides increases down the group with the exception of:

Accepted Answer

Correct answer: D. Be(OH)₂ is least soluble — 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.

Answer

A. Mg(OH)₂ → Ca(OH)₂

Answer

B. Ca(OH)₂ → Sr(OH)₂

Answer

C. Ba(OH)₂ is more soluble than Sr(OH)₂

Question 4

The correct order of lattice energies is:

Accepted Answer

Correct answer: A. NaF > NaCl > NaBr > NaI — 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.

Answer

B. NaI > NaBr > NaCl > NaF

Answer

C. NaCl > NaF > NaBr > NaI

Answer

D. NaBr > NaCl > NaF > NaI

Question 5

In the reaction 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O, the reducing agent is:

Accepted Answer

Correct answer: B. HCl — 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.

Answer

A. KMnO₄

Answer

C. MnCl₂

Answer

D. Cl₂

Question 6

The thermal stability of Group 16 hydrides decreases in the order:

Accepted Answer

Correct answer: A. H₂O > H₂S > H₂Se > H₂Te — 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.

Answer

B. H₂Te > H₂Se > H₂S > H₂O

Answer

C. H₂S > H₂O > H₂Se > H₂Te

Answer

D. H₂O > H₂Te > H₂Se > H₂S

Question 7

Which of the following transition metals can form stable complexes with CO?

Accepted Answer

Correct answer: C. Both 3d, 4d, and 5d metals — 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.

Answer

A. Only 3d metals

Answer

B. Only 4d and 5d metals

Answer

D. Only Fe and Co

Question 8

The correct order of reducing power of Group 17 hydrides is:

Accepted Answer

Correct answer: B. HI > HBr > HCl > HF — Reducing power increases with increasing size of halogen (decreasing electronegativity). HI is the strongest reducing agent due to the weak H-I bond.

Answer

A. HF > HCl > HBr > HI

Answer

C. HCl > HF > HI > HBr

Answer

D. HBr > HI > HF > HCl

JEE Chemistry - MCQ Practice Questions