JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 28

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: Which overlap can form a sigma bond along the x-axis?
(a) s of A with p\(_z\) of B
(b) p\(_x\) of A with p\(_y\) of B
(c) p\(_x\) of A with s of B
(d) p\(_z\) of A with p\(_x\) of B
Show Answer & Explanation

Answer: (c) p\(_x\) of A with s of B

Explanation:
1. Only overlaps with symmetry about the x-axis form sigma bonds: p\(_x\) with s is head-on. s with p\(_z\) or p\(_x\) with p\(_y\) overlaps cancel.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 2: The hybridisation of the central atom in \( \ce{AlF4-} \) is
(a) sp
(b) sp2
(c) sp3
(d) sp3d
Show Answer & Explanation

Answer: (c) sp3

Explanation:
1. \( \ce{AlF4-} \) has four bond pairs and no lone pair on Al, so it is tetrahedral sp3.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 3: The maximum bond energy among \( \ce{O2-} \), \( \ce{O2+} \), \( \ce{O2} \) and \( \ce{O2^2-} \) is in
(a) \( \ce{O2+} \)
(b) \( \ce{O2-} \)
(c) \( \ce{O2} \)
(d) \( \ce{O2^2-} \)
Show Answer & Explanation

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

Explanation:
1. Bond orders: \( \ce{O2+} \) 2.5, \( \ce{O2} \) 2, \( \ce{O2-} \) 1.5, \( \ce{O2^2-} \) 1
2. The highest bond order gives the highest bond energy.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 4: The H-N-H bond angle in ammonia is
(a) 104.5 degrees
(b) 109.5 degrees
(c) 120 degrees
(d) 106.7 degrees
Show Answer & Explanation

Answer: (d) 106.7 degrees

Explanation:
1. Ammonia is pyramidal with one lone pair that compresses the angle below the tetrahedral value.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 5: The maximum possible number of hydrogen bonds a water molecule can form is
(a) 2
(b) 4
(c) 3
(d) 1
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Each O atom can accept two H bonds through its lone pairs and each O-H can donate one, so 4.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 6: Which pair has the strongest hydrogen bonding between themselves?
(a) \( \ce{CH4} \) and \( \ce{CH3OH} \)
(b) \( \ce{SiH4} \) and \( \ce{SiF4} \)
(c) \( \ce{CH3COCH3} \) and \( \ce{CHCl3} \)
(d) \( \ce{HCOOH} \) and \( \ce{CH3COOH} \)
Show Answer & Explanation

Answer: (d) \( \ce{HCOOH} \) and \( \ce{CH3COOH} \)

Explanation:
1. Carboxylic acids contain both O-H donors and C=O acceptors, which gives strong H bonds (dimers).
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 7: \( \ce{SF6} \) has which geometry?
(a) octahedral
(b) tetrahedral
(c) pyramidal
(d) trigonal bipyramidal
Show Answer & Explanation

Answer: (a) octahedral

Explanation:
1. Six bond pairs and no lone pair on S give an octahedral shape.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 8: The pair of species \( \ce{CO2} \) and \( \ce{NO2+} \) is
(a) isosteric but not isoelectronic
(b) isoelectronic but not isosteric
(c) isosteric and isoelectronic
(d) neither
Show Answer & Explanation

Answer: (c) isosteric and isoelectronic

Explanation:
1. Each has 22 electrons and the same linear geometry.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 9: During the change of \( \ce{O2} \) to \( \ce{O2-} \) the added electron goes into the
(a) antibonding pi orbital (pi*2px or pi*2py)
(b) sigma*2pz orbital
(c) bonding pi orbital
(d) sigma2pz orbital
Show Answer & Explanation

Answer: (a) antibonding pi orbital (pi*2px or pi*2py)

Explanation:
1. The pi* orbitals are half-filled in \( \ce{O2} \) and take the next electron, giving \( \ce{O2-} \).
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 10: Which of the following has both ionic and covalent bonds?
(a) NaBr
(b) \( \ce{PCl3} \)
(c) \( \ce{CH3CH2OH} \)
(d) \( \ce{Ba(CN)2} \)
Show Answer & Explanation

Answer: (d) \( \ce{Ba(CN)2} \)

Explanation:
1. \( \ce{Ba(CN)2} \) has ionic bonds between \( \ce{Ba^2+} \) and \( \ce{CN-} \) and covalent bonds inside the cyanide ion.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 11: Which of the following molecules is non-polar?
(a) CO
(b) \( \ce{SO3} \)
(c) \( \ce{SO2} \)
(d) \( \ce{H2O} \)
Show Answer & Explanation

Answer: (b) \( \ce{SO3} \)

Explanation:
1. \( \ce{SO3} \) is trigonal planar and symmetrical, so the dipoles cancel
2. CO, \( \ce{SO2} \) and \( \ce{H2O} \) are polar.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 12: The overlap of atomic orbitals can form a covalent bond of the type
(a) sigma only
(b) sigma or pi
(c) pi only
(d) both sigma and pi in every case
Show Answer & Explanation

Answer: (b) sigma or pi

Explanation:
1. Head-on overlap gives sigma; sideways overlap gives pi, depending on the orbitals and the molecular axis.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 13: Which of the following is arranged in increasing order of covalent character?
(a) \( \ce{CCl4} \) < \( \ce{BeCl2} \) < \( \ce{BCl3} \) < \( \ce{LiCl} \)
(b) \( \ce{LiCl} \) < \( \ce{CCl4} \) < \( \ce{BeCl2} \) < \( \ce{BCl3} \)
(c) \( \ce{LiCl} \) < \( \ce{BeCl2} \) < \( \ce{CCl4} \) < \( \ce{BCl3} \)
(d) \( \ce{LiCl} \) < \( \ce{BeCl2} \) < \( \ce{BCl3} \) < \( \ce{CCl4} \)
Show Answer & Explanation

Answer: (d) \( \ce{LiCl} \) < \( \ce{BeCl2} \) < \( \ce{BCl3} \) < \( \ce{CCl4} \)

Explanation:
1. The greater the charge and the smaller the cation, the greater its polarising power and the covalent character (Fajans' rules): Li+ < Be2+ < B3+ < C4+.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 14: The correct order of decreasing polarisability of the halide ions is
(a) \( \ce{F-} \) > \( \ce{Cl-} \) > \( \ce{Br-} \) > \( \ce{I-} \)
(b) \( \ce{Cl-} \) > \( \ce{Br-} \) > \( \ce{I-} \) > \( \ce{F-} \)
(c) \( \ce{I-} \) > \( \ce{Br-} \) > \( \ce{Cl-} \) > \( \ce{F-} \)
(d) \( \ce{F-} \) > \( \ce{I-} \) > \( \ce{Br-} \) > \( \ce{Cl-} \)
Show Answer & Explanation

Answer: (c) \( \ce{I-} \) > \( \ce{Br-} \) > \( \ce{Cl-} \) > \( \ce{F-} \)

Explanation:
1. Polarisability rises with the size of the anion, so iodide is the most polarisable and fluoride the least.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 15: \( \ce{SnCl4} \) is a covalent liquid because
(a) the Cl- clouds are weakly polarised
(b) of strong van der Waals forces between ions
(c) the electron clouds of the Cl- ions are strongly polarised by the highly charged Sn4+
(d) Sn shows inert pair effect
Show Answer & Explanation

Answer: (c) the electron clouds of the Cl- ions are strongly polarised by the highly charged Sn4+

Explanation:
1. The small, highly charged \( \ce{Sn^4+} \) polarises the large \( \ce{Cl-} \) ions strongly, giving a covalent molecular compound.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 16: Which of the following has a permanent dipole moment?
(a) \( \ce{PF3} \)
(b) \( \ce{SiF4} \)
(c) \( \ce{BF3} \)
(d) \( \ce{PF5} \)
Show Answer & Explanation

Answer: (a) \( \ce{PF3} \)

Explanation:
1. \( \ce{PF3} \) is pyramidal with a lone pair
2. \( \ce{SiF4} \), \( \ce{BF3} \) and \( \ce{PF5} \) are symmetric.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 17: The molecule with the least dipole moment is
(a) \( \ce{NF3} \)
(b) \( \ce{SO2} \)
(c) \( \ce{NH3} \)
(d) \( \ce{CO2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CO2} \)

Explanation:
1. \( \ce{CO2} \) is linear and non-polar, so its dipole moment is zero.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 18: The geometry of \( \ce{H2S} \) and its dipole moment are
(a) linear and zero
(b) angular and non-zero
(c) angular and zero
(d) linear and non-zero
Show Answer & Explanation

Answer: (b) angular and non-zero

Explanation:
1. \( \ce{H2S} \) has two bond pairs and two lone pairs and so is bent with a net dipole moment.
2. Only option (b) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 19: The decreasing order of dipole moment is
(a) \( \ce{CH3F} \) > \( \ce{CH3Cl} \) > \( \ce{CH3Br} \) > \( \ce{CH3I} \)
(b) \( \ce{CH3Cl} \) > \( \ce{CH3Br} \) > \( \ce{CH3I} \) > \( \ce{CH3F} \)
(c) \( \ce{CH3F} \) > \( \ce{CH3Cl} \) > \( \ce{CH3I} \) > \( \ce{CH3Br} \)
(d) \( \ce{CH3Cl} \) > \( \ce{CH3F} \) > \( \ce{CH3Br} \) > \( \ce{CH3I} \)
Show Answer & Explanation

Answer: (d) \( \ce{CH3Cl} \) > \( \ce{CH3F} \) > \( \ce{CH3Br} \) > \( \ce{CH3I} \)

Explanation:
1. Though C-F is the most polar bond, the short C-F length reduces the moment, so methyl chloride has the highest value (1.86 D) and the order is Cl > F > Br > I.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 20: The dipole moment of chlorobenzene is 1.73 D. The dipole moment of p-dichlorobenzene is expected to be
(a) 0 D
(b) 3.46 D
(c) 1.73 D
(d) 1.00 D
Show Answer & Explanation

Answer: (a) 0 D

Explanation:
1. The two C-Cl dipoles are equal and opposite in the para isomer, so they cancel.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 21: The strongest oxyacid among \( \ce{H2SO4} \), \( \ce{H3PO4} \), \( \ce{HClO4} \) and \( \ce{H2SiO3} \) is
(a) \( \ce{H2SO4} \)
(b) \( \ce{H3PO4} \)
(c) \( \ce{HClO4} \)
(d) \( \ce{H2SiO3} \)
Show Answer & Explanation

Answer: (c) \( \ce{HClO4} \)

Explanation:
1. Acidity of oxyacids of the same period increases with the oxidation state and electronegativity of the central atom, so \( \ce{HClO4} \) is the strongest.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 22: The anhydride of nitric acid is
(a) \( \ce{NO2} \)
(b) \( \ce{N2O3} \)
(c) \( \ce{N2O5} \)
(d) \( \ce{N2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{N2O5} \)

Explanation:
1. \( \ce{N2O5} \) + \( \ce{H2O} \) gives 2 \( \ce{HNO3} \).
2. Only option (c) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

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