JEE Chemistry - MCQ Practice Questions

Atomic radius increases down a group in the periodic table. As we go from Na to K to Rb to Cs, the number of shells increases, leading to increased atomic radius.

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.

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

In H₂S₂O₃ (thiosulfuric acid), two sulfur atoms have different oxidation states: one is +5 and one is -1. Average oxidation state = (+5-1)/2 = +2, but individual states are +5 and -1. Correct answer shows +3 as average consideration.

Fluorine forms HF which is the most stable hydride among halogens due to the strong H-F bond (high bond energy ~570 kJ/mol), resulting from fluorine's high electronegativity.

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.

Al₂O₃ is amphoteric, reacting with both strong acids (forming AlCl₃) and strong bases (forming aluminate ions). Na₂O is basic, SiO₂ is acidic, and P₂O₅ is acidic.

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.

Fluorine has the highest electronegativity value (4.0 on Pauling scale) among all elements due to its small size and high nuclear charge.

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.

N₂O₄ is colorless gas, not blue. NO is colorless, NO₂ is brown/reddish-brown, and N₂O is indeed called laughing gas.

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.

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.

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

BeCl₂ shows maximum covalent character among the options due to Be²⁺ ion's small size and high charge density, which polarizes the Cl⁻ ions significantly (Fajans rules).

At such low HCl concentration (10⁻⁸ M), contribution of H⁺ from water ionization (10⁻⁷ M) cannot be ignored. Total [H⁺] ≈ 10⁻⁷ + 10⁻⁸ ≈ 1.1 × 10⁻⁷, giving pH ≈ 6.98.

Noble gases have VERY HIGH ionization energies (not low) due to their stable electron configuration. All other options are correct properties of noble gases.

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.

Borax solution is basic because borate ions undergo hydrolysis: BO₃³⁻ + H₂O ⇌ HBO₃²⁻ + OH⁻, producing OH⁻ ions and making the solution basic.

In XeF₄, Xe has 8 valence electrons. 4 electrons form bonds with F atoms, leaving 2 lone pairs. With 4 bonding pairs and 2 lone pairs (octahedral electron geometry), the molecular geometry is square planar.