JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 08

Download JEE MCQs for JEE Chemistry: Chemical Bonding and Molecular Structure

Review structured MCQ sets for JEE Chemistry Chemical Bonding and Molecular Structure. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chemical Bonding and Molecular Structure

Access the complete set of multiple-choice questions for Chemical Bonding and Molecular Structure below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In which pair do the central atoms have sp2 hybridisation?
(a) \( \ce{BF3} \) and \( \ce{NH2^-} \)
(b) \( \ce{NO2^-} \) and \( \ce{NH3} \)
(c) \( \ce{BF3} \) and \( \ce{NO2^-} \)
(d) \( \ce{NH2^-} \) and \( \ce{H2O} \)
Show Answer & Explanation

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

Explanation:
1. BF3 (3 bond pairs) and NO2^- (2 bond pairs + 1 lone pair) are sp2.
2. NH2^-, NH3 and H2O are sp3.

Teacher's Note:
1. Count regions of electron density.

Question 2: Which alkaline earth metal sulphate has hydration enthalpy higher than lattice enthalpy?
(a) \( \ce{SrSO4} \)
(b) \( \ce{CaSO4} \)
(c) \( \ce{BeSO4} \)
(d) \( \ce{BaSO4} \)
Show Answer & Explanation

Answer: (c) \( \ce{BeSO4} \)

Explanation:
1. BeSO4 is soluble in water because the hydration energy of the small Be2+ ion is high.

Teacher's Note:
1. Be and Mg sulphates are soluble.

Question 3: In which one of the following species does the central atom have a type of hybridisation not the same as in the other three?
(a) \( \ce{PCl5} \)
(b) \( \ce{SF4} \)
(c) \( \ce{I3^-} \)
(d) \( \ce{SbCl5^2-} \)
Show Answer & Explanation

Answer: (d) \( \ce{SbCl5^2-} \)

Explanation:
1. PCl5, SF4 and I3- are sp3d (5 electron pairs). SbCl5(2-) has 6 pairs: sp3d2.

Teacher's Note:
1. Count the electron pairs.

Question 4: Which of the following has the minimum bond length?
(a) \( \ce{O2+} \)
(b) \( \ce{O2^-} \)
(c) \( \ce{O2^2-} \)
(d) \( \ce{O2} \)
Show Answer & Explanation

Answer: (a) \( \ce{O2+} \)

Explanation:
1. Bond orders: O2+ 2.5, O2 2, O2^- 1.5, O2^2- 1. The highest order gives the shortest bond.

Teacher's Note:
1. Shorter bond means higher bond order.

Question 5: Decreasing order of bond angle is
(a) \( \ce{BeCl2} \) > \( \ce{NO2} \) > \( \ce{SO2} \)
(b) \( \ce{BeCl2} \) > \( \ce{SO2} \) > \( \ce{NO2} \)
(c) \( \ce{SO2} \) > \( \ce{NO2} \) > \( \ce{BeCl2} \)
(d) \( \ce{SO2} \) > \( \ce{BeCl2} \) > \( \ce{NO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{BeCl2} \) > \( \ce{NO2} \) > \( \ce{SO2} \)

Explanation:
1. BeCl2 is linear (180 degrees), NO2 is 134 degrees and SO2 is 119 degrees.

Teacher's Note:
1. A single unpaired electron repels less than a lone pair.

Question 6: Which one of the following pairs is isostructural (same shape and hybridisation)?
(a) \( \ce{NF3} \) and \( \ce{BF3} \)
(b) \( \ce{BF4^-} \) and \( \ce{NH4+} \)
(c) \( \ce{BCl3} \) and \( \ce{BrCl3} \)
(d) \( \ce{NH3} \) and \( \ce{NO3^-} \)
Show Answer & Explanation

Answer: (b) \( \ce{BF4^-} \) and \( \ce{NH4+} \)

Explanation:
1. BF4- and NH4+ are tetrahedral (sp3).
2. The other pairs differ in shape.

Teacher's Note:
1. Check shape pairs.

Question 7: Which of the following species contains three bond pairs and one lone pair around the central atom?
(a) \( \ce{NH2^-} \)
(b) \( \ce{PCl3} \)
(c) \( \ce{H2O} \)
(d) \( \ce{BF3} \)
Show Answer & Explanation

Answer: (b) \( \ce{PCl3} \)

Explanation:
1. PCl3 has 3 P-Cl bonds and one lone pair. NH2^- and H2O have two bond pairs; BF3 has no lone pairs.

Teacher's Note:
1. Pyramidal shape.

Question 8: A bond order of 1.5 is shown by
(a) \( \ce{O2^2-} \)
(b) \( \ce{O2} \)
(c) \( \ce{O2+} \)
(d) \( \ce{O2^-} \)
Show Answer & Explanation

Answer: (d) \( \ce{O2^-} \)

Explanation:
1. O2^- has 17 electrons: BO = (10 - 7)/2 = 1.5.

Teacher's Note:
1. Superoxide ion.

Question 9: During the change of \( \ce{O2} \) to \( \ce{O2^-} \), the added electron goes to which orbital?
(a) sigma* orbital
(b) sigma orbital
(c) pi* orbital
(d) pi orbital
Show Answer & Explanation

Answer: (c) pi* orbital

Explanation:
1. O2 has two unpaired electrons in pi* orbitals; the additional electron pairs up in a pi* orbital.

Teacher's Note:
1. Superoxide has bond order 1.5.

Question 10: Which one of the following molecules contains no pi bond?
(a) \( \ce{NO2} \)
(b) \( \ce{CO2} \)
(c) \( \ce{H2O} \)
(d) \( \ce{SO2} \)
Show Answer & Explanation

Answer: (c) \( \ce{H2O} \)

Explanation:
1. H2O has only two O-H sigma bonds.

Teacher's Note:
1. Double bonds contain one pi bond.

Question 11: \( \ce{XeF2} \) is isostructural with
(a) \( \ce{BaCl2} \)
(b) \( \ce{TeF2} \)
(c) \( \ce{ICl2^-} \)
(d) \( \ce{SbCl3} \)
Show Answer & Explanation

Answer: (c) \( \ce{ICl2^-} \)

Explanation:
1. XeF2 and ICl2^- are both linear (sp3d, three lone pairs).

Teacher's Note:
1. BaCl2 is ionic.

Question 12: Dipole-induced dipole interactions are present in which of the following pairs?
(a) \( \ce{SiF4} \) and He atoms
(b) \( \ce{H2O} \) and alcohol
(c) \( \ce{Cl2} \) and \( \ce{CCl4} \)
(d) HCl and He atoms
Show Answer & Explanation

Answer: (d) HCl and He atoms

Explanation:
1. A polar molecule (HCl) induces a dipole in a non-polar atom (He).
2. The other pairs involve non-polar/non-polar (London) or hydrogen bonding.

Teacher's Note:
1. A permanent dipole plus a polarisable non-polar partner.

Question 13: Which of the following is paramagnetic?
(a) \( \ce{NO+} \)
(b) CO
(c) \( \ce{O2^-} \)
(d) \( \ce{CN^-} \)
Show Answer & Explanation

Answer: (c) \( \ce{O2^-} \)

Explanation:
1. O2^- has an unpaired electron in pi*. NO+, CO and CN^- are diamagnetic (14 electrons).

Teacher's Note:
1. Count unpaired electrons.

Question 14: Identify the correct order of solubility in aqueous medium
(a) \( \ce{Na2S} \) > ZnS > CuS
(b) CuS > ZnS > \( \ce{Na2S} \)
(c) ZnS > \( \ce{Na2S} \) > CuS
(d) \( \ce{Na2S} \) < CuS > ZnS
Show Answer & Explanation

Answer: (a) \( \ce{Na2S} \) > ZnS > CuS

Explanation:
1. Na2S is a soluble alkali sulphide; ZnS is sparingly soluble; CuS is almost insoluble (very small Ksp).

Teacher's Note:
1. Transition metal sulphides are insoluble.

Question 15: The total number of antibonding electrons present in \( \ce{O2} \) will be
(a) 6
(b) 8
(c) 4
(d) 2
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. O2 has sigma*1s2, sigma*2s2 and pi*2p2 (two unpaired): 6 antibonding electrons.

Teacher's Note:
1. Total electrons 16 = 10 bonding + 6 antibonding.

Question 16: Which of the following species has a plane triangular shape?
(a) \( \ce{N3^-} \)
(b) \( \ce{NO3^-} \)
(c) \( \ce{NO2^-} \)
(d) \( \ce{CO2} \)
Show Answer & Explanation

Answer: (b) \( \ce{NO3^-} \)

Explanation:
1. NO3^- has 3 bond pairs and no lone pair: trigonal planar. N3- and CO2 are linear, NO2- is bent.

Teacher's Note:
1. sp2 without lone pairs is trigonal planar.

Question 17: Which one of the following is the correct bond order among the species \( \ce{O2^2+} \), \( \ce{O2} \), \( \ce{O2+} \), \( \ce{O2^-} \)? (increasing)
(a) \( \ce{O2^2+} \) < \( \ce{O2} \) < \( \ce{O2+} \)
(b) \( \ce{O2+} \) < \( \ce{O2} \) < \( \ce{O2^2+} \)
(c) \( \ce{O2^-} \) < \( \ce{O2+} \) < \( \ce{O2^2+} \)
(d) \( \ce{O2^2+} \) < \( \ce{O2+} \) < \( \ce{O2^-} \)
Show Answer & Explanation

Answer: (c) \( \ce{O2^-} \) < \( \ce{O2+} \) < \( \ce{O2^2+} \)

Explanation:
1. Bond orders: O2^- 1.5; O2+ 2.5; O2(2+) 3.

Teacher's Note:
1. O2(2+) is isoelectronic with N2.

Question 18: Which of the following pairs of ions are isoelectronic and isostructural?
(a) \( \ce{ClO3^-} \), \( \ce{CO3^2-} \)
(b) \( \ce{SO3^2-} \), \( \ce{NO3^-} \)
(c) \( \ce{ClO3^-} \), \( \ce{SO3^2-} \)
(d) \( \ce{CO3^2-} \), \( \ce{SO3^2-} \)
Show Answer & Explanation

Answer: (c) \( \ce{ClO3^-} \), \( \ce{SO3^2-} \)

Explanation:
1. ClO3^- and SO3(2-) both have 42 electrons and are pyramidal.
2. The other pairs differ in electron count or shape.

Teacher's Note:
1. Compare electron count and shape.

Question 19: Which of the following represents the correct order of bond order?
(a) \( \ce{O2^-} \) < \( \ce{O2} \) < \( \ce{O2+} \)
(b) \( \ce{O2^-} \) < \( \ce{O2+} \) < \( \ce{O2} \)
(c) \( \ce{O2^-} \) > \( \ce{O2} \) > \( \ce{O2+} \)
(d) \( \ce{O2^-} \) > \( \ce{O2+} \) > \( \ce{O2} \)
Show Answer & Explanation

Answer: (a) \( \ce{O2^-} \) < \( \ce{O2} \) < \( \ce{O2+} \)

Explanation:
1. Bond orders: O2^- 1.5; O2 2; O2+ 2.5.

Teacher's Note:
1. Removing an antibonding electron increases the order.

Question 20: Maximum bond angle at nitrogen is present in which of the following?
(a) \( \ce{NO2^-} \)
(b) \( \ce{NO2+} \)
(c) \( \ce{NO3^-} \)
(d) \( \ce{NO2} \)
Show Answer & Explanation

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

Explanation:
1. NO2+ is linear (180 degrees).

Teacher's Note:
1. NO2- 115, NO2 134, NO3- 120.

Question 21: On heating, which of the following releases \( \ce{CO2} \) most easily?
(a) \( \ce{MgCO3} \)
(b) \( \ce{CaCO3} \)
(c) \( \ce{K2CO3} \)
(d) \( \ce{Na2CO3} \)
Show Answer & Explanation

Answer: (a) \( \ce{MgCO3} \)

Explanation:
1. MgCO3 is the least stable thermally of these because Mg2+ is small and polarising.

Teacher's Note:
1. Carbonate stability increases down a group.

Question 22: The decreasing order of stability of \( \ce{O2} \), \( \ce{O2^-} \), \( \ce{O2+} \) and \( \ce{O2^2-} \) is
(a) \( \ce{O2} \) > \( \ce{O2+} \) > \( \ce{O2^2-} \) > \( \ce{O2^-} \)
(b) \( \ce{O2^-} \) > \( \ce{O2^2-} \) > \( \ce{O2+} \) > \( \ce{O2} \)
(c) \( \ce{O2+} \) > \( \ce{O2} \) > \( \ce{O2^-} \) > \( \ce{O2^2-} \)
(d) \( \ce{O2^2-} \) > \( \ce{O2^-} \) > \( \ce{O2} \) > \( \ce{O2+} \)
Show Answer & Explanation

Answer: (c) \( \ce{O2+} \) > \( \ce{O2} \) > \( \ce{O2^-} \) > \( \ce{O2^2-} \)

Explanation:
1. Stability follows bond order: O2+ (2.5) > O2 (2) > O2^- (1.5) > O2^2- (1).

Teacher's Note:
1. Higher bond order means higher stability.

Question 23: The boiling points of the hydrogen halides follow the order HF > HI > HBr > HCl. The higher boiling point of HF is explained by
(a) The bond energy of HF is greater
(b) The nuclear shielding is much reduced in fluorine
(c) The electronegativity of fluorine is much higher
(d) There is strong hydrogen bonding between HF molecules
Show Answer & Explanation

Answer: (d) There is strong hydrogen bonding between HF molecules

Explanation:
1. HF molecules associate by strong F-H...F hydrogen bonds.

Teacher's Note:
1. The H-X bond strength does not control boiling point.

Question 24: The shapes of \( \ce{SF4} \) and \( \ce{XeF2} \) respectively are
(a) Linear and distorted tetrahedral
(b) See-saw and linear
(c) T-shape and trigonal bipyramidal
(d) Tetrahedral and linear
Show Answer & Explanation

Answer: (b) See-saw and linear

Explanation:
1. SF4 is see-saw (sp3d, 1 lone pair); XeF2 is linear (3 lone pairs).

Teacher's Note:
1. Position lone pairs equatorially.

Question 25: In \( \ce{ClF3} \) the lone pairs on chlorine lie at approximately what angle to each other?
(a) 180 degrees
(b) 120 degrees
(c) 90 degrees
(d) 60 degrees
Show Answer & Explanation

Answer: (b) 120 degrees

Explanation:
1. In the trigonal bipyramid, both lone pairs occupy equatorial positions about 120 degrees apart.

Teacher's Note:
1. This gives the T-shape.

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Chemical Bonding and Molecular Structure Objective Questions & Solutions for JEE Chemistry

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