JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 36

Chemistry Objective Questions and Answers: 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: Based on VSEPR theory, the number of 90 degree F-Br-F angles in \( \ce{BrF5} \) is
(a) 4
(b) 12
(c) 8
(d) 6
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. \( \ce{BrF5} \) is square pyramidal
2. The four basal F atoms each make a 90 degree angle with the apical F (4), and each pair of adjacent basal F atoms also makes 90 degrees (4)
3. The total is 8.

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 shape of the \( \ce{XeO2F2} \) molecule is
(a) trigonal bipyramidal
(b) square planar
(c) see-saw
(d) tetrahedral
Show Answer & Explanation

Answer: (c) see-saw

Explanation:
1. Xe has 4 bond atoms (two O and two F) and one lone pair, so 5 pairs
2. The lone pair sits equatorial, so the shape is see-saw.

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: Among the triatomic species \( \ce{BeCl2} \), \( \ce{N3^-} \), \( \ce{N2O} \), \( \ce{NO2^+} \), \( \ce{O3} \), \( \ce{SCl2} \), \( \ce{ICl2^-} \), \( \ce{I3^-} \) and \( \ce{XeF2} \), the number of linear species in which the hybridisation of the central atom does not involve d orbitals is
(a) 3
(b) 5
(c) 6
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. The linear species are \( \ce{BeCl2} \), \( \ce{N3^-} \), \( \ce{N2O} \), \( \ce{NO2^+} \), \( \ce{ICl2^-} \), \( \ce{I3^-} \) and \( \ce{XeF2} \)
2. \( \ce{ICl2^-} \), \( \ce{I3^-} \) and \( \ce{XeF2} \) are sp3d (involving d orbitals)
3. The other four are sp, so the answer is 4.

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 crystalline form of borax has (A) a tetranuclear \( \ce{[B4O5(OH)4]^2-} \) unit, (B) all boron atoms in the same plane, (C) equal numbers of sp2 and sp3 boron atoms, (D) one terminal hydroxide per boron atom. Which are correct?
(a) A and B only
(b) A, C and D
(c) B and C only
(d) A and D only
Show Answer & Explanation

Answer: (b) A, C and D

Explanation:
1. The anion has four B atoms: two sp2 (three-coordinate) and two sp3 (four-coordinate), so (A) and (C) are true
2. Each boron carries one terminal OH group (4 OH for 4 B), so (D) is true
3. The boron atoms are not all coplanar.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 5: The total number of compounds with at least one bridging oxo group among \( \ce{N2O3} \), \( \ce{N2O5} \), \( \ce{P4O6} \), \( \ce{P4O7} \), \( \ce{H4P2O5} \), \( \ce{H5P3O10} \), \( \ce{H2S2O3} \), \( \ce{H2S2O5} \) is
(a) 4
(b) 5
(c) 6
(d) 3
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. A bridging oxo group is an O joining two central atoms (X-O-X)
2. \( \ce{N2O5} \), \( \ce{P4O6} \), \( \ce{P4O7} \), \( \ce{H4P2O5} \) and \( \ce{H5P3O10} \) have one
3. \( \ce{N2O3} \) (O=N-NO2), \( \ce{H2S2O3} \) and \( \ce{H2S2O5} \) have the two central atoms joined directly
4. So 5.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 6: The hybridisation of the underlined atom changes in: (1) \( \ce{AlH3} \) to \( \ce{AlH4^-} \) (Al), (2) \( \ce{H2O} \) to \( \ce{H3O^+} \) (O), (3) \( \ce{NH3} \) to \( \ce{NH4^+} \) (N). Which is correct?
(a) in all cases
(b) in the second only
(c) in the first and second only
(d) only in the first
Show Answer & Explanation

Answer: (d) only in the first

Explanation:
1. Al changes from sp2 to sp3. In \( \ce{H3O^+} \) oxygen stays sp3 as in \( \ce{H2O} \), and nitrogen stays sp3 in \( \ce{NH4^+} \)
2. Only Al changes.

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: A bond angle of 109 degrees 28 minutes is found in
(a) \( \ce{NH4^+} \)
(b) \( \ce{NH3} \)
(c) \( \ce{H2O} \)
(d) \( \ce{CH3^+} \)
Show Answer & Explanation

Answer: (a) \( \ce{NH4^+} \)

Explanation:
1. \( \ce{NH4^+} \) is perfectly tetrahedral
2. \( \ce{NH3} \) (107) and \( \ce{H2O} \) (104.5) have lone pairs, and \( \ce{CH3^+} \) is trigonal planar with 120 degrees.

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: Which compound has the smallest bond angle in its molecule?
(a) \( \ce{SO2} \)
(b) \( \ce{H2O} \)
(c) \( \ce{H2S} \)
(d) \( \ce{NH3} \)
Show Answer & Explanation

Answer: (c) \( \ce{H2S} \)

Explanation:
1. The angles are \( \ce{H2S} \) about 92, \( \ce{NH3} \) 107, \( \ce{H2O} \) 104.5 and \( \ce{SO2} \) 119 degrees
2. \( \ce{H2S} \) has the smallest because S uses nearly pure p orbitals.

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: The maximum number of 90 degree bond pair-bond pair angles is observed with which hybridisation?
(a) sp3d2
(b) sp3d
(c) dsp2
(d) sp3
Show Answer & Explanation

Answer: (a) sp3d2

Explanation:
1. An octahedral arrangement (sp3d2) has 12 angles of 90 degrees
2. A trigonal bipyramid has 6 and a square planar arrangement has 4.

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: The correct order of bond angles (smallest first) in \( \ce{H2S} \), \( \ce{NH3} \), \( \ce{BF3} \) and \( \ce{SiH4} \) is
(a) \( \ce{H2S} \) < \( \ce{SiH4} \) < \( \ce{NH3} \) < \( \ce{BF3} \)
(b) \( \ce{NH3} \) < \( \ce{H2S} \) < \( \ce{SiH4} \) < \( \ce{BF3} \)
(c) \( \ce{H2S} \) < \( \ce{NH3} \) < \( \ce{BF3} \) < \( \ce{SiH4} \)
(d) \( \ce{H2S} \) < \( \ce{NH3} \) < \( \ce{SiH4} \) < \( \ce{BF3} \)
Show Answer & Explanation

Answer: (d) \( \ce{H2S} \) < \( \ce{NH3} \) < \( \ce{SiH4} \) < \( \ce{BF3} \)

Explanation:
1. The angles are \( \ce{H2S} \) about 92, \( \ce{NH3} \) 107, \( \ce{SiH4} \) 109.5 and \( \ce{BF3} \) 120 degrees.
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: The molecular shapes of \( \ce{SF4} \), \( \ce{CF4} \) and \( \ce{XeF4} \) are
(a) the same, with 2, 0 and 1 lone pairs respectively
(b) different, with 1, 0 and 2 lone pairs on the central atom respectively
(c) the same, with 1, 1 and 1 lone pair respectively
(d) different, with 0, 1 and 2 lone pairs respectively
Show Answer & Explanation

Answer: (b) different, with 1, 0 and 2 lone pairs on the central atom respectively

Explanation:
1. \( \ce{SF4} \) is see-saw (1 lone pair), \( \ce{CF4} \) tetrahedral (0) and \( \ce{XeF4} \) square planar (2)
2. The shapes are all different.

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 hybridisations of the N atom in \( \ce{NO3^-} \), \( \ce{NO2^+} \) and \( \ce{NH4^+} \) respectively are
(a) sp, sp2, sp3
(b) sp2, sp, sp3
(c) sp, sp3, sp2
(d) sp2, sp3, sp
Show Answer & Explanation

Answer: (b) sp2, sp, sp3

Explanation:
1. \( \ce{NO3^-} \) has 3 sigma bonds (sp2), \( \ce{NO2^+} \) has 2 sigma bonds (sp) and \( \ce{NH4^+} \) has 4 sigma bonds (sp3).
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: The structure of \( \ce{IF7} \) is
(a) square pyramidal
(b) trigonal bipyramidal
(c) octahedral
(d) pentagonal bipyramidal
Show Answer & Explanation

Answer: (d) pentagonal bipyramidal

Explanation:
1. Iodine has 7 bond pairs and no lone pair, so it is sp3d3 with a pentagonal bipyramidal shape.
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 14: Which of \( \ce{NCl3} \), \( \ce{AsCl3} \), \( \ce{SbCl3} \) and \( \ce{PCl3} \) has the smallest bond angle?
(a) \( \ce{SbCl3} \)
(b) \( \ce{NCl3} \)
(c) \( \ce{AsCl3} \)
(d) \( \ce{PCl3} \)
Show Answer & Explanation

Answer: (a) \( \ce{SbCl3} \)

Explanation:
1. With the same halogen, the angle falls as the central atom gets larger and less electronegative: \( \ce{NCl3} \) (107) > \( \ce{PCl3} \) (100) > \( \ce{AsCl3} \) (98.5) > \( \ce{SbCl3} \) (97).
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 15: Which one does NOT have a pyramidal shape?
(a) \( \ce{(CH3)3N} \)
(b) \( \ce{P(CH3)3} \)
(c) \( \ce{(SiH3)3N} \)
(d) \( \ce{P(SiH3)3} \)
Show Answer & Explanation

Answer: (c) \( \ce{(SiH3)3N} \)

Explanation:
1. The nitrogen lone pair in \( \ce{(SiH3)3N} \) is delocalised into empty d orbitals of silicon, so N becomes sp2 and the NSi3 skeleton is planar
2. The others are pyramidal.

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 16: According to VSEPR theory, the geometry of \( \ce{XeOF4} \) is
(a) square pyramidal
(b) pentagonal planar
(c) octahedral
(d) trigonal bipyramidal
Show Answer & Explanation

Answer: (a) square pyramidal

Explanation:
1. Xe has 4 F, 1 O and 1 lone pair: 6 pairs and an octahedral arrangement
2. The molecule itself, counting only atoms, is square pyramidal with the lone pair opposite O.

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 17: Which of the following compounds has a P-P bond?
(a) \( \ce{H4P2O5} \)
(b) \( \ce{(HPO3)3} \)
(c) \( \ce{H4P2O7} \)
(d) \( \ce{H4P2O6} \)
Show Answer & Explanation

Answer: (d) \( \ce{H4P2O6} \)

Explanation:
1. Hypophosphoric acid \( \ce{H4P2O6} \) has two phosphorus atoms joined directly (P-P)
2. The others, \( \ce{H4P2O5} \), \( \ce{(HPO3)3} \) and \( \ce{H4P2O7} \), have P-O-P links.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 18: Which is the incorrect formula for an element X forming compounds with X in the +3 state?
(a) \( \ce{X2O3} \)
(b) \( \ce{X2Cl3} \)
(c) \( \ce{X2(SO4)3} \)
(d) \( \ce{XPO4} \)
Show Answer & Explanation

Answer: (b) \( \ce{X2Cl3} \)

Explanation:
1. With X as +3, the formulae are \( \ce{XCl3} \), \( \ce{X2O3} \), \( \ce{X2(SO4)3} \) and \( \ce{XPO4} \)
2. \( \ce{X2Cl3} \) would need X to be +1.5, so it is the incorrect one.

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 19: The bond angle H-X-H is the greatest in
(a) \( \ce{NH3} \)
(b) \( \ce{CH4} \)
(c) \( \ce{PH3} \)
(d) \( \ce{H2O} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH4} \)

Explanation:
1. \( \ce{CH4} \) has the tetrahedral angle of 109.5 degrees; \( \ce{NH3} \) 107, \( \ce{H2O} \) 104.5 and \( \ce{PH3} \) 93.5 degrees.
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 20: Identify the pair in which the geometries are T-shaped and square pyramidal respectively.
(a) \( \ce{ICl2^-} \) and \( \ce{ICl5} \)
(b) \( \ce{IO3^-} \) and \( \ce{IO2F2^-} \)
(c) \( \ce{ClF3} \) and \( \ce{IO4^-} \)
(d) \( \ce{XeOF2} \) and \( \ce{XeOF4} \)
Show Answer & Explanation

Answer: (d) \( \ce{XeOF2} \) and \( \ce{XeOF4} \)

Explanation:
1. \( \ce{XeOF2} \) has 5 pairs with 2 lone pairs (T-shaped) and \( \ce{XeOF4} \) has 6 pairs with 1 lone pair (square pyramidal)
2. \( \ce{ICl2^-} \) is linear, \( \ce{IO3^-} \) pyramidal and \( \ce{IO4^-} \) tetrahedral.

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 21: In graphite and diamond, the percentage of p-character in the hybrid orbitals is respectively
(a) 67 and 75
(b) 33 and 25
(c) 50 and 75
(d) 33 and 75
Show Answer & Explanation

Answer: (a) 67 and 75

Explanation:
1. Graphite carbon is sp2 (33 per cent s, so 67 per cent p) and diamond carbon is sp3 (25 per cent s, so 75 per cent p).
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.

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