The ionization energy of nitrogen is higher than oxygen because:
Answer: B
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.
Q.102Medium
Which compound shows the highest ionic character?
Answer: B
Ionic character depends on electronegativity difference and charge density. MgO has the highest charge density (smallest cation with 2+ charge) and highest electronegativity difference.
Q.103Medium
The solubility of alkaline earth metal hydroxides increases down the group with the exception of:
Answer: D
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.
Q.104Medium
The correct order of lattice energies is:
Answer: A
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.
Q.105Medium
In the reaction 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O, the reducing agent is:
Answer: B
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.
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Q.106Medium
The thermal stability of Group 16 hydrides decreases in the order:
Answer: A
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.
Q.107Medium
Which of the following transition metals can form stable complexes with CO?
Answer: C
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.
Q.108Medium
The correct order of reducing power of Group 17 hydrides is:
Answer: B
Reducing power increases with increasing size of halogen (decreasing electronegativity). HI is the strongest reducing agent due to the weak H-I bond.
Q.109Medium
In the extraction of copper from chalcopyrite (CuFeS₂), the role of silica (SiO₂) is:
Answer: A
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.
Q.110Medium
Which of the following statements about borax (Na₂B₄O₅(OH)₄·8H₂O) is correct?
Answer: B
Borax solution is basic because borate ions undergo hydrolysis: BO₃³⁻ + H₂O ⇌ HBO₃²⁻ + OH⁻, producing OH⁻ ions and making the solution basic.
Q.111Medium
Which statement about diagonal relationship in periodic table is correct?
Answer: C
Be and Al show diagonal relationship due to similar charge-to-radius ratio. Both form covalent compounds and amphoteric oxides.
Q.112Medium
The most stable oxidation state of chromium is:
Answer: B
+3 is the most stable oxidation state for Cr with d³ configuration. Cr³⁺ complexes are highly stable.
Q.113Medium
In the reaction of P₄ with excess oxygen, the product is:
The orange color arises from charge transfer between oxygen and chromium, not from d-d transition within Cr³⁺.
Q.115Medium
Which of the following metals can displace hydrogen from dilute acids?
Answer: C
Zn is more reactive than hydrogen and can displace it from dilute acids. Cu, Ag, and Au are less reactive than hydrogen.
Q.116Medium
The aqueous solution of which salt would be basic?
Answer: C
Na₂CO₃ solution is basic because CO₃²⁻ hydrolyzes to produce OH⁻ ions. It's the salt of weak acid and strong base.
Q.117Medium
Which complex ion shows maximum paramagnetism?
Answer: C
[CoF₆]³⁻ is high spin with 4 unpaired electrons (d⁶). [Fe(CN)₆]⁴⁻ is low spin with 0 unpaired electrons. [Fe(H₂O)₆]²⁺ has 4 unpaired electrons but [CoF₆]³⁻ has higher magnetic moment.
Q.118Medium
The lattice energy of NaCl is higher than that of NaI because:
Answer: B
Lattice energy is inversely proportional to interionic distance. Cl⁻ is smaller than I⁻, so NaCl has higher lattice energy.
Q.119Medium
Which of the following elements forms the most stable +6 oxidation state compound?
Answer: B
Molybdenum forms very stable +6 oxidation state compounds like MoO₃ and MoO₄²⁻. This is due to d-block stability and orbital availability.
Q.120Medium
Which halide of boron exists as a dimer in the gaseous state?
Answer: B
BCl₃ forms a dimer (B₂Cl₆) in gaseous state due to bridging chlorine atoms. BF₃ remains monomeric due to strong B-F bonds and p-p repulsion.
