JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 07

Practice MCQs for JEE Chemistry Chemical Bonding and Molecular Structure

Access targeted multiple-choice questions for Chemical Bonding and Molecular Structure designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: The correct order of covalent character of alkaline earth metal chlorides is
(a) \( \ce{BeCl2} \) < \( \ce{MgCl2} \) < \( \ce{CaCl2} \) < \( \ce{SrCl2} \)
(b) \( \ce{BeCl2} \) < \( \ce{CaCl2} \) < \( \ce{SrCl2} \) < \( \ce{MgCl2} \)
(c) \( \ce{BeCl2} \) > \( \ce{MgCl2} \) > \( \ce{CaCl2} \) > \( \ce{SrCl2} \)
(d) \( \ce{SrCl2} \) > \( \ce{BeCl2} \) > \( \ce{CaCl2} \) > \( \ce{MgCl2} \)
Show Answer & Explanation

Answer: (c) \( \ce{BeCl2} \) > \( \ce{MgCl2} \) > \( \ce{CaCl2} \) > \( \ce{SrCl2} \)

Explanation:
1. The smaller the cation, the higher the polarising power, so the covalent character falls down the group.

Teacher's Note:
1. Be2+ is the smallest.

Question 2: Which of the silver halides is the least soluble in water?
(a) AgF
(b) AgCl
(c) AgBr
(d) AgI
Show Answer & Explanation

Answer: (d) AgI

Explanation:
1. AgI is the most covalent and least soluble. AgF is quite soluble.

Teacher's Note:
1. Solubility: AgF > AgCl > AgBr > AgI.

Question 3: The melting point of \( \ce{SnCl4} \) is less than that of \( \ce{SnCl2} \) because
(a) There is more charge on \( \ce{Sn^4+} \)
(b) The size of \( \ce{Sn^4+} \) is small
(c) The ionic potential of \( \ce{Sn^4+} \) is high
(d) The shape of \( \ce{SnCl4} \) is tetrahedral
Show Answer & Explanation

Answer: (a) There is more charge on \( \ce{Sn^4+} \)

Explanation:
1. Sn4+ has a high charge and strongly polarises Cl-, making SnCl4 covalent (molecular) with a low melting point.

Teacher's Note:
1. Fajans' rules.

Question 4: Ionic conductances of hydrated alkali metal ions are in the order
(a) \( \ce{Li+} \) > \( \ce{Na+} \) > \( \ce{K+} \) > \( \ce{Rb+} \) > \( \ce{Cs+} \)
(b) \( \ce{Li+} \) < \( \ce{Na+} \) < \( \ce{K+} \) < \( \ce{Rb+} \) < \( \ce{Cs+} \)
(c) \( \ce{Li+} \) > \( \ce{Na+} \) > \( \ce{K+} \) < \( \ce{Rb+} \)
(d) \( \ce{Li+} \) < \( \ce{Na+} \) < \( \ce{K+} \) > \( \ce{Rb+} \)
Show Answer & Explanation

Answer: (b) \( \ce{Li+} \) < \( \ce{Na+} \) < \( \ce{K+} \) < \( \ce{Rb+} \) < \( \ce{Cs+} \)

Explanation:
1. Li+ is the most heavily hydrated, so its hydrated size is the largest and its mobility the lowest.
2. Conductance rises from Li+ to Cs+.

Teacher's Note:
1. The hydrated radius, not the bare radius, determines mobility.

Question 5: Which of the following halides has the highest melting point?
(a) NaCl
(b) KCl
(c) NaBr
(d) NaF
Show Answer & Explanation

Answer: (d) NaF

Explanation:
1. NaF has the smallest ions and the highest lattice energy (m.p. 993 C).

Teacher's Note:
1. Smaller ions mean a larger lattice energy.

Question 6: Which of the following does NOT give an oxide on heating?
(a) \( \ce{MgCO3} \)
(b) \( \ce{Li2CO3} \)
(c) \( \ce{ZnCO3} \)
(d) \( \ce{K2CO3} \)
Show Answer & Explanation

Answer: (d) \( \ce{K2CO3} \)

Explanation:
1. K2CO3 is thermally stable up to its melting point; the others decompose to oxide and CO2.

Teacher's Note:
1. Alkali carbonates (except Li) are stable.

Question 7: Which of the following forms a metal oxide on heating?
(a) \( \ce{Na2CO3} \)
(b) \( \ce{Li2CO3} \)
(c) \( \ce{K2SO4} \)
(d) \( \ce{NaHCO3} \)
Show Answer & Explanation

Answer: (b) \( \ce{Li2CO3} \)

Explanation:
1. \( \ce{Li2CO3 -> Li2O + CO2} \). NaHCO3 gives Na2CO3 and not an oxide.

Teacher's Note:
1. Lithium is anomalous.

Question 8: The increasing order of stability of I. \( \ce{K2CO3} \), II. \( \ce{MgCO3} \), III. \( \ce{Na2CO3} \) is
(a) I < II < III
(b) II < III < I
(c) II < I < III
(d) I < III < II
Show Answer & Explanation

Answer: (b) II < III < I

Explanation:
1. Thermal stability increases with the electropositivity of the metal: MgCO3 < Na2CO3 < K2CO3.

Teacher's Note:
1. Group 1 carbonates are more stable than group 2.

Question 9: Which carbonate will NOT decompose on heating?
(a) \( \ce{BaCO3} \)
(b) \( \ce{ZnCO3} \)
(c) \( \ce{Na2CO3} \)
(d) \( \ce{Li2CO3} \)
Show Answer & Explanation

Answer: (c) \( \ce{Na2CO3} \)

Explanation:
1. Na2CO3 melts without decomposition.

Teacher's Note:
1. BaCO3 decomposes only at very high temperature.

Question 10: The decomposition temperature is maximum for
(a) \( \ce{BeCO3} \)
(b) \( \ce{CaCO3} \)
(c) \( \ce{K2CO3} \)
(d) \( \ce{Li2CO3} \)
Show Answer & Explanation

Answer: (c) \( \ce{K2CO3} \)

Explanation:
1. K2CO3 is stable to a very high temperature.

Teacher's Note:
1. The less polarising the cation, the more stable the carbonate.

Question 11: Which of the following species has a linear shape?
(a) \( \ce{O3} \)
(b) \( \ce{NO2^-} \)
(c) \( \ce{SO2} \)
(d) \( \ce{NO2+} \)
Show Answer & Explanation

Answer: (d) \( \ce{NO2+} \)

Explanation:
1. NO2+ has 2 bond pairs and no lone pair (sp): linear. The others are bent.

Teacher's Note:
1. 16-valence-electron triatomics are linear.

Question 12: Which of the following is NOT isostructural with \( \ce{SiCl4} \)?
(a) \( \ce{NH4+} \)
(b) \( \ce{SCl4} \)
(c) \( \ce{SO4^2-} \)
(d) \( \ce{PO4^3-} \)
Show Answer & Explanation

Answer: (b) \( \ce{SCl4} \)

Explanation:
1. SCl4 is see-saw (sp3d). The others are tetrahedral.

Teacher's Note:
1. SiCl4 is tetrahedral.

Question 13: The electronegativity difference between N and F is greater than that between N and H, yet the dipole moment of \( \ce{NH3} \) (1.5 D) is larger than that of \( \ce{NF3} \) (0.2 D). This is because
(a) In NH3 the atomic dipole and bond dipole are in opposite directions, whereas in NF3 they are in the same direction
(b) In NH3 as well as in NF3 the atomic dipole and bond dipole are in the same direction
(c) In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions
(d) In NH3 as well as NF3 the atomic dipole and bond dipole are in opposite directions
Show Answer & Explanation

Answer: (c) In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions

Explanation:
1. In NH3, N is negative with the lone pair pointing the same way as the N-H bond dipoles (adding).
2. In NF3, the N-F dipoles point towards F and oppose the lone pair dipole.

Teacher's Note:
1. Vector addition decides the net dipole.

Question 14: In which of the following molecules are all the bonds NOT equal?
(a) \( \ce{NF3} \)
(b) \( \ce{ClF3} \)
(c) \( \ce{BF3} \)
(d) \( \ce{AlF3} \)
Show Answer & Explanation

Answer: (b) \( \ce{ClF3} \)

Explanation:
1. ClF3 is T-shaped with two axial bonds longer than the equatorial one.

Teacher's Note:
1. The others have identical bonds.

Question 15: In which pair are the two species isostructural?
(a) \( \ce{SO3^2-} \) and \( \ce{NO3^-} \)
(b) \( \ce{BF3} \) and \( \ce{NF3} \)
(c) \( \ce{BrO3^-} \) and \( \ce{XeO3} \)
(d) \( \ce{SF4} \) and \( \ce{XeF4} \)
Show Answer & Explanation

Answer: (c) \( \ce{BrO3^-} \) and \( \ce{XeO3} \)

Explanation:
1. BrO3^- and XeO3 are pyramidal (sp3, 1 lone pair).
2. SO3(2-) is pyramidal but NO3- is planar; BF3 planar, NF3 pyramidal; SF4 see-saw, XeF4 square planar.

Teacher's Note:
1. Compare the number of lone pairs.

Question 16: The correct order of the C-O bond length among CO, \( \ce{CO3^2-} \) and \( \ce{CO2} \) is
(a) CO < \( \ce{CO3^2-} \) < \( \ce{CO2} \)
(b) \( \ce{CO3^2-} \) < \( \ce{CO2} \) < CO
(c) CO < \( \ce{CO2} \) < \( \ce{CO3^2-} \)
(d) \( \ce{CO2} \) < CO < \( \ce{CO3^2-} \)
Show Answer & Explanation

Answer: (c) CO < \( \ce{CO2} \) < \( \ce{CO3^2-} \)

Explanation:
1. Bond orders: CO 3, CO2 2, CO3(2-) 1.33. The longer the bond, the lower the order: CO < CO2 < CO3(2-).

Teacher's Note:
1. Bond length is inversely related to bond order.

Question 17: In which of the following is the hydration energy higher than the lattice energy?
(a) \( \ce{MgSO4} \)
(b) \( \ce{CaSO4} \)
(c) \( \ce{BaSO4} \)
(d) \( \ce{SrSO4} \)
Show Answer & Explanation

Answer: (a) \( \ce{MgSO4} \)

Explanation:
1. Small Mg2+ has a very high hydration energy, which exceeds the lattice energy, so MgSO4 is soluble.

Teacher's Note:
1. Solubility of group 2 sulphates falls down the group.

Question 18: Four diatomic species are listed. Which presents the correct order of increasing bond order?
(a) \( \ce{C2^2-} \) < \( \ce{He2+} \) < NO < \( \ce{O2^-} \)
(b) \( \ce{He2+} \) < \( \ce{O2^-} \) < NO < \( \ce{C2^2-} \)
(c) \( \ce{O2^-} \) < NO < \( \ce{C2^2-} \) < \( \ce{He2+} \)
(d) NO < \( \ce{C2^2-} \) < \( \ce{O2^-} \) < \( \ce{He2+} \)
Show Answer & Explanation

Answer: (b) \( \ce{He2+} \) < \( \ce{O2^-} \) < NO < \( \ce{C2^2-} \)

Explanation:
1. Bond orders: He2+ 0.5, O2- 1.5, NO 2.5, C2(2-) 3.

Teacher's Note:
1. Compute (Nb - Na)/2.

Question 19: The angular shape of the ozone molecule consists of
(a) 1 sigma and 1 pi bond
(b) 2 sigma and 1 pi bond
(c) 1 sigma and 2 pi bonds
(d) 2 sigma and 2 pi bonds
Show Answer & Explanation

Answer: (b) 2 sigma and 1 pi bond

Explanation:
1. In O3, the central O forms two sigma bonds and there is one pi bond delocalised over the two O-O bonds.

Teacher's Note:
1. Resonance gives equal O-O bonds of order 1.5.

Question 20: The correct order of increasing bond angles in the triatomic species \( \ce{NO2+} \), \( \ce{NO2} \) and \( \ce{NO2^-} \) is
(a) \( \ce{NO2+} \) < \( \ce{NO2} \) < \( \ce{NO2^-} \)
(b) \( \ce{NO2+} \) < \( \ce{NO2^-} \) < \( \ce{NO2} \)
(c) \( \ce{NO2^-} \) < \( \ce{NO2+} \) < \( \ce{NO2} \)
(d) \( \ce{NO2^-} \) < \( \ce{NO2} \) < \( \ce{NO2+} \)
Show Answer & Explanation

Answer: (d) \( \ce{NO2^-} \) < \( \ce{NO2} \) < \( \ce{NO2+} \)

Explanation:
1. NO2- (115 degrees, lone pair repels more) < NO2 (134, single electron) < NO2+ (180, linear).

Teacher's Note:
1. Fewer non-bonding electrons give a larger angle.

Question 21: In which of \( \ce{BF3} \), \( \ce{NO2^-} \), \( \ce{NH2^-} \) and \( \ce{H2O} \) is the central atom sp2 hybridised?
(a) \( \ce{BF3} \) and \( \ce{NO2^-} \)
(b) \( \ce{NO2^-} \) and \( \ce{NH2^-} \)
(c) \( \ce{NH2^-} \) and \( \ce{H2O} \)
(d) \( \ce{NO2^-} \) and \( \ce{H2O} \)
Show Answer & Explanation

Answer: (a) \( \ce{BF3} \) and \( \ce{NO2^-} \)

Explanation:
1. BF3: 3 bond pairs (sp2). NO2^-: 2 bond pairs + 1 lone pair (sp2). NH2^- and H2O are sp3.

Teacher's Note:
1. Count sigma bonds plus lone pairs.

Question 22: In the case of alkali metal halides, the covalent character decreases in the order
(a) MI > MBr > MCl > MF
(b) MCl > MI > MBr > MF
(c) MF > MCl > MBr > MI
(d) MF > MCl > MI > MBr
Show Answer & Explanation

Answer: (a) MI > MBr > MCl > MF

Explanation:
1. Larger anions are more polarisable, so the covalent character increases from F to I.
2. The order is MI > MBr > MCl > MF.

Teacher's Note:
1. Fajans: large anion, small cation, high charge raise covalent character.

Question 23: The dominant intermolecular force that must be overcome in converting liquid \( \ce{CH3OH} \) to a gas is
(a) London dispersion force
(b) Hydrogen bonding
(c) Dipole-dipole interaction
(d) Covalent bonds
Show Answer & Explanation

Answer: (b) Hydrogen bonding

Explanation:
1. Methanol molecules are held together mainly by O-H...O hydrogen bonds.

Teacher's Note:
1. Hydrogen bonding is the strongest intermolecular force in alcohols.

Question 24: Considering \( \ce{NO3^-} \) and \( \ce{H3O+} \), which statement is correct?
(a) Isostructural with the same hybridisation for the central atom
(b) Isostructural with different hybridisation
(c) Similar hybridisation for the central atom with different structures
(d) Dissimilar hybridisation for the central atom with different structures
Show Answer & Explanation

Answer: (d) Dissimilar hybridisation for the central atom with different structures

Explanation:
1. NO3^- is trigonal planar (N sp2) and H3O+ is pyramidal (O sp3).

Teacher's Note:
1. Different hybridisation leads to different shapes.

Question 25: In which of the following molecules does the central atom NOT have sp3 hybridisation?
(a) \( \ce{SF4} \)
(b) \( \ce{BF4^-} \)
(c) \( \ce{NH4+} \)
(d) \( \ce{CH4} \)
Show Answer & Explanation

Answer: (a) \( \ce{SF4} \)

Explanation:
1. SF4 is sp3d (5 electron pairs).

Teacher's Note:
1. BF4-, NH4+ and CH4 have 4 electron pairs.

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