JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 09

Multiple Choice Questions (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.

Practice Chemical Bonding and Molecular Structure MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: Which statement is false for \( \ce{CH4} \), \( \ce{NH3} \) and \( \ce{H2O} \)?
(a) All bond angles are greater than 90 degrees
(b) The H-O-H angle in \( \ce{H2O} \) is larger than the H-C-H angle in \( \ce{CH4} \)
(c) The H-O-H angle in \( \ce{H2O} \) is smaller than the H-N-H angle in \( \ce{NH3} \)
(d) The H-C-H angle in \( \ce{CH4} \) is larger than the H-N-H angle in \( \ce{NH3} \)
Show Answer & Explanation

Answer: (b) The H-O-H angle in \( \ce{H2O} \) is larger than the H-C-H angle in \( \ce{CH4} \)

Explanation:
1. Bond angles: CH4 109.5, NH3 107, H2O 104.5. So the H-O-H angle is smaller than the H-C-H angle, making statement 2 false.

Teacher's Note:
1. More lone pairs reduce the angle.

Question 2: The correct order of repulsion between electron pairs is
(a) lp-lp > lp-bp > bp-bp
(b) lp-lp > bp-bp > lp-bp
(c) bp-bp > lp-bp > lp-lp
(d) lp-bp > lp-lp > bp-bp
Show Answer & Explanation

Answer: (a) lp-lp > lp-bp > bp-bp

Explanation:
1. Lone pairs occupy more space, so lp-lp repulsion is the strongest.

Teacher's Note:
1. This explains why the angle of H2O is less than that of NH3.

Question 3: Match the compounds in column I with the shape in column II: (a) \( \ce{XeF6} \) (b) \( \ce{XeO3} \) (c) \( \ce{XeOF4} \) (d) \( \ce{XeF4} \); (i) distorted octahedral, (ii) square planar, (iii) pyramidal, (iv) square pyramidal
(a) a-i, b-iii, c-iv, d-ii
(b) a-i, b-ii, c-iv, d-iii
(c) a-iv, b-iii, c-i, d-ii
(d) a-iv, b-i, c-ii, d-iii
Show Answer & Explanation

Answer: (a) a-i, b-iii, c-iv, d-ii

Explanation:
1. XeF6 distorted octahedral, XeO3 pyramidal, XeOF4 square pyramidal and XeF4 square planar.

Teacher's Note:
1. Use VSEPR with lone pairs on Xe.

Question 4: The correct geometry and hybridisation for \( \ce{XeF4} \) are
(a) Planar triangle, sp3d3
(b) Square planar, sp3d2
(c) Octahedral, sp3d2
(d) Trigonal bipyramidal, sp3d
Show Answer & Explanation

Answer: (b) Square planar, sp3d2

Explanation:
1. XeF4 has 4 bond pairs and 2 lone pairs (6 pairs): sp3d2, square planar.

Teacher's Note:
1. Lone pairs are trans.

Question 5: Which one is a WRONG statement?
(a) \( \ce{SeF4} \) and \( \ce{CH4} \) have the same shape
(b) \( \ce{I3+} \) has a bent geometry
(c) \( \ce{PH5} \) and \( \ce{BiCl5} \) do not exist
(d) pi(p)-pi(d) bonds are present in \( \ce{SO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{SeF4} \) and \( \ce{CH4} \) have the same shape

Explanation:
1. SeF4 is see-saw (sp3d) while CH4 is tetrahedral.

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

Question 6: Which fluoro-compound is most likely to behave as a Lewis base?
(a) \( \ce{CF4} \)
(b) \( \ce{SiF4} \)
(c) \( \ce{BF3} \)
(d) \( \ce{PF3} \)
Show Answer & Explanation

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

Explanation:
1. PF3 has a lone pair on P. CF4 is saturated; SiF4 and BF3 are Lewis acids.

Teacher's Note:
1. A lone pair is needed for Lewis basicity.

Question 7: Which of the following pairs of ions is isoelectronic and isostructural?
(a) \( \ce{SO3^2-} \), \( \ce{NO3^-} \)
(b) \( \ce{ClO3^-} \), \( \ce{SO3^2-} \)
(c) \( \ce{CO3^2-} \), \( \ce{NO3^-} \)
(d) \( \ce{ClO3^-} \), \( \ce{CO3^2-} \)
Show Answer & Explanation

Answer: (c) \( \ce{CO3^2-} \), \( \ce{NO3^-} \)

Explanation:
1. CO3(2-) and NO3- both have 32 electrons and are trigonal planar.

Teacher's Note:
1. The pair ClO3-/SO3(2-) is also isoelectronic and isostructural; here the option (c) is the planar pair.

Question 8: The species with minimum number of lone pairs on the central atom is
(a) \( \ce{XeO3} \)
(b) \( \ce{XeF4} \)
(c) \( \ce{ClF3} \)
(d) HOCl
Show Answer & Explanation

Answer: (a) \( \ce{XeO3} \)

Explanation:
1. XeO3 has 1 lone pair on Xe; XeF4 has 2; ClF3 has 2; HOCl has 2 on Cl.

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

Question 9: In which molecule is the dipole moment more than zero but less than one debye?
(a) \( \ce{NH3} \)
(b) \( \ce{NF3} \)
(c) \( \ce{BeF2} \)
(d) \( \ce{BF3} \)
Show Answer & Explanation

Answer: (b) \( \ce{NF3} \)

Explanation:
1. NF3 = 0.2 D; NH3 = 1.47 D; BeF2 and BF3 are zero.

Teacher's Note:
1. In NF3 the bond dipoles oppose the lone pair dipole.

Question 10: Select the incorrect statement:
(a) Hybrid orbitals are maximally repelled from each other
(b) Only half-filled atomic orbitals participate in hybridisation
(c) The bond energy of hybrid orbitals is more than unhybrid atomic orbitals
(d) Without hybridisation, the geometry of molecules is uncertain
Show Answer & Explanation

Answer: (b) Only half-filled atomic orbitals participate in hybridisation

Explanation:
1. Even filled orbitals (as in NH3 and H2O with lone pairs) take part in hybridisation.

Teacher's Note:
1. Hybridisation involves orbitals, not just half-filled electrons.

Question 11: Which of the following has a tetrahedral structure?
(a) \( \ce{XeF4} \)
(b) \( \ce{H3PO4} \)
(c) \( \ce{SF4} \)
(d) \( \ce{ClF3} \)
Show Answer & Explanation

Answer: (b) \( \ce{H3PO4} \)

Explanation:
1. In H3PO4, P is surrounded by four O atoms (sp3), tetrahedral.

Teacher's Note:
1. XeF4 is square planar; SF4 is see-saw; ClF3 is T-shaped.

Question 12: Which of the following molecules has more than one lone pair on the central atom?
(a) \( \ce{SO2} \)
(b) \( \ce{XeF2} \)
(c) \( \ce{PCl5} \)
(d) \( \ce{IF7} \)
Show Answer & Explanation

Answer: (b) \( \ce{XeF2} \)

Explanation:
1. XeF2 has three lone pairs on Xe. SO2 has one; PCl5 and IF7 have none.

Teacher's Note:
1. Linear XeF2 arises from three equatorial lone pairs.

Question 13: Match the interhalogen compounds in column I with their geometry in column II: (a) XX' (b) XX'3 (c) XX'5 (d) XX'7; (i) T-shape, (ii) pentagonal bipyramidal, (iii) linear, (iv) square pyramidal, (v) tetrahedral
(a) a-iii, b-i, c-iv, d-ii
(b) a-v, b-iv, c-iii, d-ii
(c) a-iv, b-iii, c-ii, d-i
(d) a-iii, b-iv, c-i, d-ii
Show Answer & Explanation

Answer: (a) a-iii, b-i, c-iv, d-ii

Explanation:
1. XX' is linear; XX'3 is T-shaped; XX'5 is square pyramidal; XX'7 is pentagonal bipyramidal.

Teacher's Note:
1. Count lone pairs on X.

Question 14: Which pair of compounds is isoelectronic and isostructural?
(a) \( \ce{TeI2} \), \( \ce{XeF2} \)
(b) \( \ce{IBr2^-} \), \( \ce{XeF2} \)
(c) \( \ce{IF3} \), \( \ce{XeF2} \)
(d) \( \ce{BeCl2} \), \( \ce{XeF2} \)
Show Answer & Explanation

Answer: (b) \( \ce{IBr2^-} \), \( \ce{XeF2} \)

Explanation:
1. IBr2^- and XeF2 both have 22 valence electrons and are linear.

Teacher's Note:
1. Count valence electrons.

Question 15: The species having bond angles of 120 degrees is
(a) \( \ce{ClF3} \)
(b) \( \ce{NCl3} \)
(c) \( \ce{BCl3} \)
(d) \( \ce{PH3} \)
Show Answer & Explanation

Answer: (c) \( \ce{BCl3} \)

Explanation:
1. BCl3 is trigonal planar (sp2): 120 degrees.

Teacher's Note:
1. NCl3 and PH3 are pyramidal, ClF3 T-shaped.

Question 16: Which pair of species has the same bond order?
(a) \( \ce{O2} \), \( \ce{NO+} \)
(b) \( \ce{CN^-} \), CO
(c) \( \ce{N2} \), \( \ce{O2^-} \)
(d) CO, NO
Show Answer & Explanation

Answer: (b) \( \ce{CN^-} \), CO

Explanation:
1. CN^- and CO both have bond order 3.

Teacher's Note:
1. Both are isoelectronic with N2.

Question 17: Which of the following has hydrogen-bonded water?
(a) \( \ce{BaCl2.2H2O} \)
(b) \( \ce{CrCl3.5H2O} \)
(c) \( \ce{CuSO4.5H2O} \)
(d) \( \ce{CaCl2.6H2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{CuSO4.5H2O} \)

Explanation:
1. In CuSO4.5H2O one of the water molecules is hydrogen-bonded to the sulphate (lattice water).

Teacher's Note:
1. The fifth water molecule is H-bonded.

Question 18: Which of the following is diamagnetic?
(a) NO
(b) \( \ce{Cl2+} \)
(c) \( \ce{S2^2-} \)
(d) ClO
Show Answer & Explanation

Answer: (c) \( \ce{S2^2-} \)

Explanation:
1. S2(2-) is like O2(2-): all electrons paired. NO, Cl2+ and ClO have odd electrons.

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

Question 19: Among \( \ce{CaH2} \), \( \ce{BeH2} \) and \( \ce{BaH2} \), the order of ionic character is
(a) \( \ce{BeH2} \) < \( \ce{CaH2} \) < \( \ce{BaH2} \)
(b) \( \ce{CaH2} \) < \( \ce{BeH2} \) < \( \ce{BaH2} \)
(c) \( \ce{BeH2} \) < \( \ce{BaH2} \) < \( \ce{CaH2} \)
(d) \( \ce{BaH2} \) < \( \ce{BeH2} \) < \( \ce{CaH2} \)
Show Answer & Explanation

Answer: (a) \( \ce{BeH2} \) < \( \ce{CaH2} \) < \( \ce{BaH2} \)

Explanation:
1. Ionic character increases with the size of the metal ion, so BeH2 (covalent) < CaH2 < BaH2.

Teacher's Note:
1. Fajans' rules.

Question 20: Among \( \ce{CN+} \), \( \ce{CN^-} \), NO and CN, which has the highest bond order?
(a) NO
(b) \( \ce{CN^-} \)
(c) \( \ce{CN+} \)
(d) CN
Show Answer & Explanation

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

Explanation:
1. Bond orders: CN- 3; NO 2.5; CN 2.5; CN+ 2.

Teacher's Note:
1. CN- is isoelectronic with N2.

Question 21: In the structure of \( \ce{ClF3} \), the number of lone pairs of electrons on the central atom Cl is
(a) One
(b) Two
(c) Four
(d) Three
Show Answer & Explanation

Answer: (b) Two

Explanation:
1. Cl has 7 valence electrons; 3 form bonds with F, leaving 4 electrons = 2 lone pairs.

Teacher's Note:
1. T-shape.

Question 22: Which statement is correct regarding \( \ce{N2O5} \)?
(a) It is the anhydride of nitrous acid
(b) It is a brown gas
(c) It is paramagnetic
(d) It exists as \( \ce{NO2+} \) and \( \ce{NO3^-} \) in the solid state
Show Answer & Explanation

Answer: (d) It exists as \( \ce{NO2+} \) and \( \ce{NO3^-} \) in the solid state

Explanation:
1. Solid N2O5 is ionic: nitronium nitrate (NO2+ NO3-).

Teacher's Note:
1. It is the anhydride of nitric acid.

Question 23: In s-p overlapping, which statement is wrong?
(a) A nodal plane is formed in ABMO
(b) Some orbital may have higher energy than combining orbitals
(c) Combining orbitals should have the same symmetry
(d) A nodal plane is found in BMO
Show Answer & Explanation

Answer: (d) A nodal plane is found in BMO

Explanation:
1. The bonding MO has no nodal plane between the nuclei; the antibonding MO does.

Teacher's Note:
1. Antibonding orbitals have more nodes.

Question 24: In which of the following is at least one sigma bond of the np (sigma-np) type present?
(a) \( \ce{B2} \)
(b) \( \ce{O2} \)
(c) \( \ce{C2} \)
(d) \( \ce{Li2} \)
Show Answer & Explanation

Answer: (b) \( \ce{O2} \)

Explanation:
1. O2 has a sigma2pz bond; B2 and C2 have only pi bonds (net) and Li2 uses s orbitals.

Teacher's Note:
1. The order of MO energy changes after N2.

Question 25: Which is correct for dipole moment?
(a) \( \ce{NH3} \) < \( \ce{NF3} \)
(b) \( \ce{SO2} \) < \( \ce{SO3} \)
(c) \( \ce{NF3} \) > \( \ce{BF3} \)
(d) \( \ce{SO2} \) > \( \ce{H2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{NF3} \) > \( \ce{BF3} \)

Explanation:
1. NF3 (0.2 D) > BF3 (0). NH3 > NF3, SO3 is zero, and H2O (1.85 D) > SO2 (1.6 D).

Teacher's Note:
1. Planar symmetrical BF3 has zero dipole.

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

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