Practice MCQs for JEE Chemistry Chemical Bonding and Molecular Structure
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Multiple Choice Questions
(a) Solubility in water: \( \ce{BeC2O4} \) < \( \ce{BaC2O4} \)
(b) Solubility in water: \( \ce{AgCl} \) < \( \ce{AgF} \)
(c) Solubility in acetone: KCl < KI
(d) Solubility in water: \( \ce{CaCrO4} \) > \( \ce{BaCrO4} \)
Show Answer & Explanation
Answer: (a) Solubility in water: \( \ce{BeC2O4} \) < \( \ce{BaC2O4} \)
Explanation:
1. BeC2O4 is soluble because of the high hydration enthalpy of the small Be2+ ion, while BaC2O4 is sparingly soluble. The stated order is reversed.
Teacher's Note:
1. For small cations the hydration enthalpy can outweigh the lattice energy.
(a) IF7: planar non-polar; SO2: planar polar; SF4: non-planar polar; CS2: non-planar non-polar
(b) IF7: planar polar; SO2: non-planar polar; SF4: non-planar non-polar; CS2: planar polar
(c) IF7: non-planar non-polar; SO2: planar polar; SF4: non-planar polar; CS2: planar non-polar
(d) IF7: non-planar polar; SO2: planar non-polar; SF4: planar polar; CS2: non-planar polar
Show Answer & Explanation
Answer: (c) IF7: non-planar non-polar; SO2: planar polar; SF4: non-planar polar; CS2: planar non-polar
Explanation:
1. IF7 is a pentagonal bipyramid and symmetric (non-planar, non-polar). SO2 is bent (planar, polar). SF4 is see-saw (non-planar, polar). CS2 is linear (planar, non-polar).
Teacher's Note:
1. Combine shape and symmetry for polarity.
(a) Bonding molecular orbitals have low electron density between the two nuclei
(b) Bonding molecular orbitals have less energy than the atomic orbitals from which they are formed
(c) Every electron in a bonding MO contributes to the attraction between atoms
(d) They are formed when the lobes of the combining atomic orbitals have the same sign
Show Answer & Explanation
Answer: (a) Bonding molecular orbitals have low electron density between the two nuclei
Explanation:
1. Bonding MOs are formed by constructive overlap, so electron density is concentrated (high) between the nuclei.
Teacher's Note:
1. Bonding means high density between nuclei.
(a) Both carbonate and nitrate ions
(b) \( \ce{CO3^{2-}} \) only
(c) \( \ce{NO3-} \) only
(d) \( \ce{SiO2} \)
Show Answer & Explanation
Answer: (a) Both carbonate and nitrate ions
Explanation:
1. In carbonate and nitrate the central C or N (second period) forms a p-pi p-pi bond with oxygen. SiO2 has only single bonds.
Teacher's Note:
1. Second-period elements form p-pi p-pi bonds.
(a) CsF < CsCl < CsBr < CsI
(b) \( \ce{MgF2} \) < \( \ce{MgCl2} \) < \( \ce{MgBr2} \) < \( \ce{MgI2} \)
(c) \( \ce{Be(OH)2} \) < \( \ce{Mg(OH)2} \) < \( \ce{Ca(OH)2} \) < \( \ce{Sr(OH)2} \)
(d) \( \ce{Li2O} \) < \( \ce{Na2O} \) < \( \ce{K2O} \) < \( \ce{Rb2O} \) < \( \ce{Cs2O} \)
Show Answer & Explanation
Answer: (a) CsF < CsCl < CsBr < CsI
Explanation:
1. CsF is the most soluble of the caesium halides because hydration enthalpy of F- compensates its high lattice energy, so the order is reversed.
Teacher's Note:
1. Large cation, small anion: hydration energy wins.
(a) \( \ce{SF4} \) and \( \ce{CCl4} \)
(b) \( \ce{BF3} \) and \( \ce{NF3} \)
(c) \( \ce{CO2} \) and \( \ce{SO2} \)
(d) \( \ce{ClO4-} \) and \( \ce{ClO2-} \)
Show Answer & Explanation
Answer: (d) \( \ce{ClO4-} \) and \( \ce{ClO2-} \)
Explanation:
1. In ClO4- Cl has 4 electron pairs and in ClO2- also 4 (2 bp + 2 lp): both sp3.
Teacher's Note:
1. Count electron pairs for each central atom.
(a) II < IV < III < I < V
(b) III < II < I < V < IV
(c) II < IV < III < V < I
(d) II < III < IV < I < V
Show Answer & Explanation
Answer: (d) II < III < IV < I < V
Explanation:
1. s-character: XeF4 sp3d2 (1/6 = 16.7%), I3- sp3d (20%), NCl3 sp3 (25%), CO3 2- sp2 (33.3%), BeCl2 sp (50%).
Teacher's Note:
1. s-character = 1 / number of hybridising orbitals.
(a) \( \ce{BH4-} \) and \( \ce{AlH4-} \)
(b) \( \ce{BCl4-} \) and \( \ce{ICl4-} \)
(c) \( \ce{NH4+} \) and \( \ce{PH4+} \)
(d) \( \ce{PCl6-} \) and \( \ce{SiF6^{2-}} \)
Show Answer & Explanation
Answer: (b) \( \ce{BCl4-} \) and \( \ce{ICl4-} \)
Explanation:
1. BCl4- is tetrahedral (4 bp, 0 lp) whereas ICl4- is square planar (4 bp, 2 lp).
Teacher's Note:
1. Same number of bond pairs does not mean the same shape.
(a) \( \ce{SiF4} \) < \( \ce{SiCl4} \)
(b) \( \ce{CH4} \) < \( \ce{SiH4} \) < \( \ce{GeH4} \)
(c) \( \ce{CH4} \) < \( \ce{CD4} \)
(d) He > H2
Show Answer & Explanation
Answer: (d) He > H2
Explanation:
1. Boiling points: H2 20 K, He 4.2 K, so He < H2. The other orders agree with mass and surface area.
Teacher's Note:
1. London forces grow with size and mass.
(a) \( \ce{SF4} \)
(b) \( \ce{B2H6} \)
(c) \( \ce{PCl5} \)
(d) \( \ce{SiF4} \)
Show Answer & Explanation
Answer: (d) \( \ce{SiF4} \)
Explanation:
1. SiF4 is regular tetrahedral. SF4 and PCl5 have axial and equatorial bonds of different lengths, and B2H6 has terminal and bridging B-H bonds.
Teacher's Note:
1. Regular geometries give equal bond lengths.
(a) \( \ce{XeF2} \)
(b) \( \ce{SO2} \)
(c) \( \ce{SCl2} \)
(d) \( \ce{XeO2F2} \)
Show Answer & Explanation
Answer: (a) \( \ce{XeF2} \)
Explanation:
1. XeF2 has 3 lone pairs on Xe. SCl2 has 2, SO2 and XeO2F2 have 1 each.
Teacher's Note:
1. XeF2 is AX2E3.
(a) 5
(b) 3
(c) 4
(d) 6
Show Answer & Explanation
Answer: (a) 5
Explanation:
1. No d orbitals: PCl4+, ICl2+, SF2, AsF4+ and SiF4 (all sp3). The rest use sp3d or sp3d2.
Teacher's Note:
1. Count electron pairs: 4 pairs means sp3.
(a) \( \ce{B2} \)
(b) \( \ce{Be2} \)
(c) \( \ce{Li2} \)
(d) \( \ce{H2+} \)
Show Answer & Explanation
Answer: (b) \( \ce{Be2} \)
Explanation:
1. Be2 has bond order zero, so it does not form a stable molecule. B2 (bond order 1), Li2 (1) and H2+ (0.5) exist.
Teacher's Note:
1. Zero bond order means no stable molecule.
(a) the bond strength increases
(b) the bond length decreases
(c) the bond angle decreases
(d) the size of the orbital decreases
Show Answer & Explanation
Answer: (c) the bond angle decreases
Explanation:
1. Less s-character means the orbital is longer and more diffuse, giving weaker, longer bonds, bigger orbitals and smaller bond angles (sp 180, sp2 120, sp3 109.5).
Teacher's Note:
1. s-character is held more tightly by the nucleus.
(a) \( \ce{XeO2F2} \)
(b) \( \ce{SO2Cl2} \)
(c) \( \ce{SO3} \)
(d) \( \ce{XeO3} \)
Show Answer & Explanation
Answer: (c) \( \ce{SO3} \)
Explanation:
1. In SO3 one S=O bond is p-pi p-pi and two are p-pi d-pi. The other molecules contain only p-pi d-pi bonds (from the third period or higher).
Teacher's Note:
1. Sulphur can form both kinds in the same molecule.
(a) The solution will consist of Al3+ and Cl- (unhydrated)
(b) It will remain covalent in aqueous solution
(c) None of these
(d) The solution will consist of hydrated Al3+ and Cl-
Show Answer & Explanation
Answer: (d) The solution will consist of hydrated Al3+ and Cl-
Explanation:
1. Total hydration enthalpy = 4665 + 3 x 381 = 5808 kJ per mol, which exceeds the lattice energy 5137, so the dissolution is exothermic and the ions are hydrated.
Teacher's Note:
1. Compare total hydration energy with lattice energy.
(a) 6
(b) 8
(c) 4
(d) 12
Show Answer & Explanation
Answer: (b) 8
Explanation:
1. The two lone pairs sit above and below the plane (180 degrees apart). Each lone pair is at 90 degrees to all four bond pairs, so there are 2 x 4 = 8 lp-bp repulsions at 90 degrees.
Teacher's Note:
1. Square planar AX4E2 has lone pairs on the axial positions.
(a) \( \ce{Ag2S} \) is much less soluble than \( \ce{Ag2O} \)
(b) \( \ce{BeCO3} \) is thermally less stable than \( \ce{BaCO3} \)
(c) \( \ce{BaCO3} \) is less soluble than \( \ce{MgCO3} \)
(d) The melting point of \( \ce{AlCl3} \) is much lower than that of NaCl
Show Answer & Explanation
Answer: (c) \( \ce{BaCO3} \) is less soluble than \( \ce{MgCO3} \)
Explanation:
1. The solubility trend of the alkaline earth carbonates depends on lattice versus hydration enthalpy, not polarisation. The other three are classic Fajans examples.
Teacher's Note:
1. Solubility of s-block salts is decided by lattice and hydration energies.
(a) C-C bond: diamond < graphite
(b) C-O bond: \( \ce{CO3^{2-}} \) < \( \ce{HCO3-} \)
(c) S-O bond: \( \ce{SO2} \) > \( \ce{SO3^{2-}} \) > \( \ce{SO4^{2-}} \)
(d) O-O bond: \( \ce{O2-} \) < \( \ce{O2} \) < \( \ce{O2+} \)
Show Answer & Explanation
Answer: (c) S-O bond: \( \ce{SO2} \) > \( \ce{SO3^{2-}} \) > \( \ce{SO4^{2-}} \)
Explanation:
1. Bond orders of S-O: SO2 = 2, SO4 2- = 1.5 and SO3 2- = 1.33, so SO4 2- is stronger than SO3 2-; the stated order is wrong. Graphite has partial double bond character so its C-C bond is stronger than diamond's.
Teacher's Note:
1. Higher bond order means a stronger and shorter bond.
(a) 4
(b) 5
(c) 6
(d) 3
Show Answer & Explanation
Answer: (b) 5
Explanation:
1. Conductors: HCl(aq), AlF3(molten, ionic), graphite, Na(s) and AlCl3(aq): 5. Molten HCl and molten AlCl3 are covalent (molecular).
Teacher's Note:
1. Conduction requires free ions or electrons.
(a) \( \ce{O2} \) and \( \ce{O2^{2-}} \)
(b) \( \ce{O2} \) and \( \ce{O2+} \)
(c) \( \ce{O2-} \) and \( \ce{O2^{2-}} \)
(d) \( \ce{O2+} \) and \( \ce{O2^{2-}} \)
Show Answer & Explanation
Answer: (a) \( \ce{O2} \) and \( \ce{O2^{2-}} \)
Explanation:
1. Unpaired electrons: O2 has 2 (2.83 BM), O2+ 1 (1.73 BM), O2- 1 and O2 2- 0. The largest difference is between O2 (2.83 BM) and O2 2- (0).
Teacher's Note:
1. Count unpaired electrons from the MO diagram.
(a) Acidic nature
(b) Viscosity
(c) Basic strength
(d) Oxidation state
Show Answer & Explanation
Answer: (b) Viscosity
Explanation:
1. Viscosity depends on the extent of hydrogen bonding, which is greatest for H3PO4 (three OH groups) and least for HCl. Acidic strength follows the opposite order for several of these.
Teacher's Note:
1. Count OH groups for viscosity.
(a) Polarisability: \( \ce{F-} \) < \( \ce{Cl-} \) < \( \ce{Br-} \) < \( \ce{I-} \)
(b) Polarising power: \( \ce{Na+} \) < \( \ce{Ca^{2+}} \) < \( \ce{Mg^{2+}} \) < \( \ce{Al^{3+}} \)
(c) Covalent character: LiF < LiCl < LiBr < LiI
(d) Ionic character: MCl < MCl2 < MCl3
Show Answer & Explanation
Answer: (d) Ionic character: MCl < MCl2 < MCl3
Explanation:
1. Higher oxidation state of the metal increases polarisation, so ionic character decreases: MCl > MCl2 > MCl3.
Teacher's Note:
1. Fajans' rule: higher cation charge, more covalent character.
(a) 5
(b) 3
(c) 2
(d) 4
Show Answer & Explanation
Answer: (d) 4
Explanation:
1. S in SF4 uses sp3d hybridisation, using one of the five 3d orbitals. So 4 vacant 3d orbitals remain.
Teacher's Note:
1. Hybridisation uses only some of the d orbitals.
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Practice MCQs for JEE Chemistry Chemical Bonding and Molecular Structure
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