JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 05

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Chapter-wise Objective Questions: Chemical Bonding and Molecular Structure

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Multiple Choice Questions

Question 1: When the hybridisation state of carbon changes from sp3 to sp2 to sp, the angle between the hybridised orbitals
(a) Decreases considerably
(b) Increases progressively
(c) Decreases gradually
(d) All of these
Show Answer & Explanation

Answer: (b) Increases progressively

Explanation:
1. The angles are 109.5, 120 and 180 degrees, which increase progressively.

Teacher's Note:
1. More s-character means a larger angle.

Question 2: The hybridisation states of the central atoms in \( \ce{I3^-} \), \( \ce{ICl4^-} \) and \( \ce{ICl2^-} \) respectively are
(a) sp2, dsp2, sp3
(b) sp3d, sp3d2 and sp3d
(c) sp3d, sp3d, dsp2
(d) sp, sp, sp
Show Answer & Explanation

Answer: (b) sp3d, sp3d2 and sp3d

Explanation:
1. I3^-: 2 bond pairs + 3 lone pairs (5 pairs, sp3d). ICl4^-: 4 + 2 (6, sp3d2). ICl2^-: 2 + 3 (5, sp3d).

Teacher's Note:
1. Count electron pairs on the central atom.

Question 3: The molecular shapes of \( \ce{SF4} \), \( \ce{CF4} \) and \( \ce{XeF4} \) are
(a) The same, with 2, 0 and 1 lone pairs
(b) The same, with 1, 1 and 1 lone pairs
(c) Different, with 0, 1 and 2 lone pairs
(d) Different, with 1, 0 and 2 lone pairs
Show Answer & Explanation

Answer: (d) Different, with 1, 0 and 2 lone pairs

Explanation:
1. SF4 is see-saw (1 lone pair), CF4 is tetrahedral (0) and XeF4 is square planar (2).

Teacher's Note:
1. Different lone pairs mean different shapes.

Question 4: Which two are isostructural?
(a) \( \ce{XeF2} \), \( \ce{IF2^-} \)
(b) \( \ce{NH3} \), \( \ce{BF3} \)
(c) \( \ce{CO3^2-} \), \( \ce{SO3^2-} \)
(d) \( \ce{PCl5} \), \( \ce{ICl5} \)
Show Answer & Explanation

Answer: (a) \( \ce{XeF2} \), \( \ce{IF2^-} \)

Explanation:
1. Both XeF2 and IF2^- are sp3d with three lone pairs and are linear.
2. The other pairs differ in shape.

Teacher's Note:
1. Equal valence electron count with equal bonding gives the same shape.

Question 5: Which pair is correct for the molecular formula of the xenon compound and the hybridisation of xenon?
(a) \( \ce{XeF4} \), sp3
(b) \( \ce{XeF2} \), sp
(c) \( \ce{XeF2} \), sp3d
(d) \( \ce{XeF4} \), sp2
Show Answer & Explanation

Answer: (c) \( \ce{XeF2} \), sp3d

Explanation:
1. XeF2: 2 bond pairs + 3 lone pairs = 5 pairs, sp3d.
2. XeF4 is sp3d2.

Teacher's Note:
1. Lone pairs are counted in hybridisation.

Question 6: The xenon compounds isostructural with \( \ce{IBr2^-} \) and \( \ce{BrO3^-} \) respectively are
(a) Linear \( \ce{XeF2} \) and pyramidal \( \ce{XeO3} \)
(b) Bent \( \ce{XeF2} \) and pyramidal \( \ce{XeO3} \)
(c) Bent \( \ce{XeF2} \) and planar \( \ce{XeO3} \)
(d) Linear \( \ce{XeF2} \) and tetrahedral \( \ce{XeO3} \)
Show Answer & Explanation

Answer: (a) Linear \( \ce{XeF2} \) and pyramidal \( \ce{XeO3} \)

Explanation:
1. IBr2^- is linear (like XeF2), and BrO3^- is pyramidal (like XeO3).

Teacher's Note:
1. Same electron count means the same shape.

Question 7: Which statement is correct?
(a) All compounds having polar bonds have a dipole moment
(b) \( \ce{SO2} \) is non-polar
(c) \( \ce{H2O} \) molecule is non-polar, having polar bonds
(d) \( \ce{PH3} \) is a polar molecule having non-polar bonds
Show Answer & Explanation

Answer: (d) \( \ce{PH3} \) is a polar molecule having non-polar bonds

Explanation:
1. P-H bonds are almost non-polar (EN 2.1 each) but the pyramidal shape with a lone pair makes PH3 slightly polar.

Teacher's Note:
1. Symmetrical molecules cancel bond dipoles.

Question 8: Which of the following molecules has zero dipole moment?
(a) \( \ce{BF3} \)
(b) \( \ce{CH2Cl2} \)
(c) \( \ce{NF3} \)
(d) \( \ce{SO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{BF3} \)

Explanation:
1. BF3 is trigonal planar and symmetric: the bond dipoles cancel.

Teacher's Note:
1. CH2Cl2, NF3 and SO2 are polar.

Question 9: The dipole moment of \( \ce{NH3} \) is
(a) Less than the dipole moment of \( \ce{NCl3} \)
(b) Higher than the dipole moment of \( \ce{NCl3} \)
(c) Equal to the dipole moment of \( \ce{NCl3} \)
(d) None of these
Show Answer & Explanation

Answer: (b) Higher than the dipole moment of \( \ce{NCl3} \)

Explanation:
1. In NH3 the N-H bond dipoles and lone pair dipole point in the same direction. In NCl3 the polarity of N-Cl is small.
2. NH3 = 1.47 D, NCl3 about 0.6 D.

Teacher's Note:
1. Bond and lone-pair dipoles add or cancel.

Question 10: Which order of polarities is correct?
(a) HF > \( \ce{NH3} \) > \( \ce{PH3} \)
(b) \( \ce{CH4} \) > \( \ce{NH3} \) > \( \ce{H2O} \)
(c) \( \ce{CH3Cl} \) < \( \ce{CH2Cl2} \) < \( \ce{CHCl3} \)
(d) \( \ce{BF3} \) > \( \ce{BeF2} \) > \( \ce{F2} \)
Show Answer & Explanation

Answer: (a) HF > \( \ce{NH3} \) > \( \ce{PH3} \)

Explanation:
1. Dipole moments in D: HF 1.82, NH3 1.47, PH3 0.58.
2. The others are in the wrong order.

Teacher's Note:
1. CH4, BF3, BeF2 and F2 have zero dipole.

Question 11: Which set contains only polar molecules?
(a) \( \ce{XeF4} \), \( \ce{IF7} \), \( \ce{SO3} \)
(b) \( \ce{PCl5} \), \( \ce{C6H6} \), \( \ce{SF6} \)
(c) \( \ce{SnCl2} \), \( \ce{SO2} \), \( \ce{NO2} \)
(d) \( \ce{CO2} \), \( \ce{CS2} \), \( \ce{C2H4} \)
Show Answer & Explanation

Answer: (c) \( \ce{SnCl2} \), \( \ce{SO2} \), \( \ce{NO2} \)

Explanation:
1. SnCl2, SO2 and NO2 are bent and polar. The other sets contain symmetrical molecules.

Teacher's Note:
1. Check each for symmetry.

Question 12: Which of the following has a symmetrical structure?
(a) \( \ce{PCl3} \)
(b) \( \ce{CH2Cl2} \)
(c) \( \ce{CHCl3} \)
(d) \( \ce{CCl4} \)
Show Answer & Explanation

Answer: (d) \( \ce{CCl4} \)

Explanation:
1. CCl4 is a regular tetrahedron with four identical bonds.

Teacher's Note:
1. Symmetrical molecules have zero dipole.

Question 13: What conclusion can be drawn from the fact that \( \ce{BF3} \) has no dipole moment but \( \ce{PF3} \) does?
(a) \( \ce{BF3} \) is not symmetrical but \( \ce{PF3} \) is
(b) \( \ce{BF3} \) must be linear
(c) Atomic radius of P is larger than that of B
(d) \( \ce{BF3} \) must be planar triangular
Show Answer & Explanation

Answer: (d) \( \ce{BF3} \) must be planar triangular

Explanation:
1. A planar triangular geometry cancels the B-F bond dipoles; PF3 is pyramidal.

Teacher's Note:
1. Shape explains zero dipole.

Question 14: \( \ce{PCl5} \) is non-polar because
(a) The P-Cl bond is non-polar
(b) Its dipole moment is zero (symmetrical trigonal bipyramid)
(c) The P-Cl bond is polar
(d) P and Cl have equal electronegativity
Show Answer & Explanation

Answer: (b) Its dipole moment is zero (symmetrical trigonal bipyramid)

Explanation:
1. Bond dipoles cancel in the symmetrical trigonal bipyramidal shape.

Teacher's Note:
1. P-Cl bonds are polar individually.

Question 15: The dipole moment of \( \ce{CO2} \) is zero, which implies that
(a) Carbon and oxygen have equal electronegativities
(b) Carbon has no polar bond
(c) \( \ce{CO2} \) is a linear molecule
(d) Carbon has bond moments of zero value
Show Answer & Explanation

Answer: (c) \( \ce{CO2} \) is a linear molecule

Explanation:
1. The two C=O bond dipoles are equal and opposite in a linear molecule.

Teacher's Note:
1. Zero overall dipole does not mean zero bond dipoles.

Question 16: Which of the following molecules does NOT possess a permanent dipole moment?
(a) \( \ce{H2S} \)
(b) \( \ce{SO2} \)
(c) \( \ce{SO3^2-} \)
(d) \( \ce{CS2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CS2} \)

Explanation:
1. CS2 is linear and symmetrical (S=C=S) with zero dipole.

Teacher's Note:
1. H2S and SO2 are bent.

Question 17: Which of the following has the highest dipole moment?
(a) HCl
(b) HF
(c) HI
(d) HBr
Show Answer & Explanation

Answer: (b) HF

Explanation:
1. HF has the greatest electronegativity difference: 1.82 D.

Teacher's Note:
1. Dipole decreases HF > HCl > HBr > HI.

Question 18: The ion isoelectronic with CO and having the same bond order is
(a) \( \ce{CN^-} \)
(b) \( \ce{O2+} \)
(c) \( \ce{O2^-} \)
(d) \( \ce{N2+} \)
Show Answer & Explanation

Answer: (a) \( \ce{CN^-} \)

Explanation:
1. CO and CN^- each have 14 electrons and bond order 3.

Teacher's Note:
1. N2 and NO+ are also in this family.

Question 19: Which of the following is paramagnetic?
(a) \( \ce{O2^-} \)
(b) \( \ce{CN^-} \)
(c) CO
(d) \( \ce{NO+} \)
Show Answer & Explanation

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

Explanation:
1. O2^- has 17 electrons, with one unpaired electron in pi*2p. The others are 14-electron diamagnetic species.

Teacher's Note:
1. Odd-electron species are paramagnetic.

Question 20: The bond order of CO on the basis of molecular orbital theory is
(a) Zero
(b) 2
(c) 3
(d) 1
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. CO has 10 valence electrons: 8 bonding and 2 antibonding. BO = (8 - 2)/2 = 3.

Teacher's Note:
1. Isoelectronic with N2.

Question 21: The energy of the sigma 2s orbital is greater than sigma 1s because
(a) The sigma 2s orbital is bigger than sigma 1s*
(b) The sigma 2s orbital is bonding whereas sigma 1s* is antibonding
(c) The sigma 2s orbital has a greater value of n than the sigma 1s* orbital
(d) None
Show Answer & Explanation

Answer: (c) The sigma 2s orbital has a greater value of n than the sigma 1s* orbital

Explanation:
1. The 2s level lies in the second shell (n = 2), higher in energy than 1s-derived orbitals.

Teacher's Note:
1. Energy increases with principal quantum number.

Question 22: Which species has the minimum bond energy?
(a) \( \ce{N2} \)
(b) \( \ce{N2^-} \)
(c) \( \ce{N2+} \)
(d) \( \ce{N2^2-} \)
Show Answer & Explanation

Answer: (d) \( \ce{N2^2-} \)

Explanation:
1. Bond orders: N2 = 3, N2^- = 2.5, N2+ = 2.5, N2^2- = 2. The lowest BO has the lowest bond energy.

Teacher's Note:
1. Bond energy increases with bond order.

Question 23: Which of the following ions does NOT have bond order 2.5?
(a) \( \ce{O2^-} \)
(b) \( \ce{O2+} \)
(c) \( \ce{N2+} \)
(d) \( \ce{N2^-} \)
Show Answer & Explanation

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

Explanation:
1. O2^- has bond order 1.5. O2+, N2+ and N2^- are all 2.5.

Teacher's Note:
1. Count electrons in bonding and antibonding MOs.

Question 24: In a coordinate bond, the acceptor atom must essentially contain in its valency shell
(a) Paired electrons
(b) A single electron
(c) No electron (a vacant orbital)
(d) Three electrons
Show Answer & Explanation

Answer: (c) No electron (a vacant orbital)

Explanation:
1. The acceptor provides an empty orbital to accommodate the donated lone pair.

Teacher's Note:
1. H+, BF3 and AlCl3 are good acceptors.

Question 25: The bonds present in \( \ce{N2O5} \) are
(a) Only ionic
(b) Covalent and coordinate
(c) Only covalent
(d) Covalent and ionic
Show Answer & Explanation

Answer: (b) Covalent and coordinate

Explanation:
1. N2O5 contains sigma and pi covalent bonds plus coordinate bonds between N and O.

Teacher's Note:
1. The coordinate bonds are N -> O.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Bonding and Molecular Structure

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