JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 44

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Chapter-wise Objective Questions: Chemical Bonding and Molecular Structure

View or download the dedicated Chemical Bonding and Molecular Structure MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: Which of the following compounds forms a non-conducting solution in water?
(a) Blue vitriol
(b) Chile saltpetre
(c) Indian saltpetre
(d) Toluene
Show Answer & Explanation

Answer: (d) Toluene

Explanation:
1. Toluene is a non-polar liquid and does not dissolve in water. The other three are ionic salts that give conducting solutions.

Teacher's Note:
1. Only ions conduct in solution.

Question 2: The molecule that contains sigma bonds, pi bonds and lone pairs in a 1 : 1 : 1 ratio is:
(a) \( \ce{C2N2} \) (cyanogen)
(b) \( \ce{C6H6} \) (benzene)
(c) \( \ce{C(CN)4} \)
(d) \( \ce{C3O2} \) (carbon suboxide)
Show Answer & Explanation

Answer: (d) \( \ce{C3O2} \) (carbon suboxide)

Explanation:
1. O=C=C=C=O has 4 sigma bonds, 4 pi bonds and 4 lone pairs (two on each O). C2N2 gives 3 : 4 : 2, benzene 12 : 3 : 0 and C(CN)4 gives 8 : 8 : 4.

Teacher's Note:
1. Count all three types explicitly for each structure.

Question 3: The maximum number of identical bond lengths is present in:
(a) \( \ce{IF7} \)
(b) \( \ce{PCl5} \)
(c) \( \ce{SF6} \)
(d) \( \ce{SO4^{2-}} \)
Show Answer & Explanation

Answer: (c) \( \ce{SF6} \)

Explanation:
1. SF6 has six equivalent S-F bonds. IF7 has two axial and five equatorial bonds (two lengths), PCl5 has axial and equatorial bonds, and sulphate has only four.

Teacher's Note:
1. Count the equivalent bonds.

Question 4: Which orbital is not involved in the formation of the PCl5 molecule?
(a) \( d_{x^2-y^2} \)
(b) s
(c) \( d_{z^2} \)
(d) \( p_z \)
Show Answer & Explanation

Answer: (a) \( d_{x^2-y^2} \)

Explanation:
1. PCl5 is sp3d with s, px, py, pz and d(z2) orbitals of P involved. d(x2-y2) is not used.

Teacher's Note:
1. Trigonal bipyramid uses d(z2).

Question 5: The salt having the least solubility in water is:
(a) \( \ce{BaCl2} \)
(b) \( \ce{BaSO4} \)
(c) \( \ce{Ba(NO3)2} \)
(d) \( \ce{MgSO4} \)
Show Answer & Explanation

Answer: (b) \( \ce{BaSO4} \)

Explanation:
1. BaSO4 has a very high lattice energy and a low hydration energy, so it is practically insoluble (Ksp about 1e-10).

Teacher's Note:
1. Large cation with large anion gives low solubility.

Question 6: Which of the following orders is not correct?
(a) Covalent character: CrO < \( \ce{Cr2O3} \) < \( \ce{CrO3} \)
(b) Ionic character: ZnO > ZnS
(c) Ionic character: \( \ce{BeCl2} \) > \( \ce{MgCl2} \) > \( \ce{CaCl2} \) > \( \ce{SrCl2} \) > \( \ce{BaCl2} \)
(d) Covalent character: KCl < AgCl
Show Answer & Explanation

Answer: (c) Ionic character: \( \ce{BeCl2} \) > \( \ce{MgCl2} \) > \( \ce{CaCl2} \) > \( \ce{SrCl2} \) > \( \ce{BaCl2} \)

Explanation:
1. Ionic character increases down group 2, so BeCl2 is the least ionic and the reverse order is wrong.

Teacher's Note:
1. Be compounds are the most covalent in group 2.

Question 7: Of CHCl3, CH4 and SF4, the molecules not having a regular geometry are:
(a) CHCl3 only
(b) CHCl3 and SF4
(c) CH4 only
(d) CH4 and SF4
Show Answer & Explanation

Answer: (b) CHCl3 and SF4

Explanation:
1. CH4 is regular tetrahedral. CHCl3 is a distorted tetrahedron and SF4 is see-saw.

Teacher's Note:
1. Regular means all bond angles and lengths are equal.

Question 8: Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is:
(a) \( \ce{Be2} \)
(b) \( \ce{B2} \)
(c) \( \ce{N2} \)
(d) \( \ce{C2} \)
Show Answer & Explanation

Answer: (d) \( \ce{C2} \)

Explanation:
1. Without s-p mixing the order has sigma2pz below pi2p. C2 then has its last two electrons in the degenerate pi2p orbitals, unpaired. B2 fills sigma2pz and is diamagnetic, and N2 and Be2 are diamagnetic.

Teacher's Note:
1. Degenerate orbitals fill singly first.

Question 9: Which of the following species are hypervalent? (I) ClO4-, (II) BF3, (III) SO4 2-, (IV) CO3 2-
(a) I, II and III
(b) III and IV
(c) I and III
(d) I, III and IV
Show Answer & Explanation

Answer: (c) I and III

Explanation:
1. Hypervalent species have more than 8 electrons around the central atom in the Lewis structure: ClO4- and SO4 2- (with S=O and Cl=O bonds). BF3 has 6 and CO3 2- has 8.

Teacher's Note:
1. Hypervalence is only possible beyond period 2.

Question 10: Intramolecular hydrogen bonding is not present in:
(a) Chloral hydrate
(b) Salicylaldehyde
(c) o-Fluorophenol
(d) Urea
Show Answer & Explanation

Answer: (d) Urea

Explanation:
1. Urea forms only intermolecular hydrogen bonds. The others have an H bond within the same molecule.

Teacher's Note:
1. Intramolecular H-bonding needs a favourable ring.

Question 11: Which of the following statements is incorrect?
(a) In \( \ce{ICl4-} \) the Cl-I-Cl angle is 109 degrees
(b) \( \ce{ClF3} \) has a T shape
(c) In \( \ce{SF4} \) the equatorial F-S-F angle is less than 120 degrees
(d) The shape of \( \ce{I3-} \) is linear
Show Answer & Explanation

Answer: (a) In \( \ce{ICl4-} \) the Cl-I-Cl angle is 109 degrees

Explanation:
1. ICl4- is square planar, so the Cl-I-Cl angle is 90 degrees.

Teacher's Note:
1. AX4E2 = square planar.

Question 12: Which of the following compounds has a number of p-pi p-pi bonds equal to the number of p-pi d-pi bonds?
(a) \( \ce{SO3} \)
(b) \( \ce{SOCl2} \)
(c) \( \ce{SO2Cl2} \)
(d) \( \ce{SO2} \)
Show Answer & Explanation

Answer: (d) \( \ce{SO2} \)

Explanation:
1. In SO2 one S=O bond is p-pi p-pi and the other is p-pi d-pi. In SO3 the ratio is 1 : 2.

Teacher's Note:
1. Count the pi bonds of each type.

Question 13: Which of the following statements is incorrect?
(a) The boiling point of H2O2 is greater than that of H2O
(b) Ethylene glycol is less viscous than glycerol
(c) o-Nitrophenol can be separated from its meta and para isomers by steam distillation
(d) In ice each O atom is surrounded tetrahedrally by four H atoms at equal distance
Show Answer & Explanation

Answer: (d) In ice each O atom is surrounded tetrahedrally by four H atoms at equal distance

Explanation:
1. In ice each oxygen has two short covalent O-H bonds and two longer hydrogen bonds, so the four H atoms are not at equal distances.

Teacher's Note:
1. Hydrogen bonds are longer than covalent O-H bonds.

Question 14: In which of the following sets are the two species not both sp3d hybridised?
(a) \( \ce{ICl2+} \) and \( \ce{SF4} \)
(b) \( \ce{PF4-} \) and \( \ce{BrF3} \)
(c) \( \ce{XeF2} \) and \( \ce{I3-} \)
(d) \( \ce{AsF4-} \) and \( \ce{SCl4} \)
Show Answer & Explanation

Answer: (a) \( \ce{ICl2+} \) and \( \ce{SF4} \)

Explanation:
1. ICl2+ has 2 bp + 2 lp (sp3) while SF4 is sp3d. The other pairs are both sp3d (5 electron pairs).

Teacher's Note:
1. Count bond pairs plus lone pairs.

Question 15: How many of CH3-, CH3+, NH3, CO3 2-, BCl3 and PCl5 have a bond angle of 120 degrees?
(a) 3
(b) 5
(c) 6
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. CH3+ (sp2), CO3 2-, BCl3 and PCl5 (equatorial angles) have 120 degrees. CH3- and NH3 are pyramidal.

Teacher's Note:
1. Trigonal planar species have 120 degrees.

Question 16: Arrange in order of increasing dipole moment: (I) 1,3,5-trichlorobenzene, (II) 1,2,4-trichlorobenzene, (III) 1,2,3,4-tetrachlorobenzene, (IV) p-dichlorobenzene.
(a) IV < II < I < III
(b) I = IV < II < III
(c) IV = I < III < II
(d) IV < I < II < III
Show Answer & Explanation

Answer: (b) I = IV < II < III

Explanation:
1. I and IV are symmetric (zero dipole). 1,2,4-trichloro has a net dipole of about 1.4 D, and 1,2,3,4-tetrachloro has about 2.5 D (equal to o-dichlorobenzene, since C6Cl6 is non-polar).

Teacher's Note:
1. A polychlorobenzene's dipole equals that of the missing groups reversed.

Question 17: The solubility of inert gases in water is due to:
(a) Keesom attraction
(b) London force alone
(c) Ion-dipole attraction
(d) Debye (dipole-induced dipole) attraction
Show Answer & Explanation

Answer: (d) Debye (dipole-induced dipole) attraction

Explanation:
1. Water is polar, so it induces a dipole in the non-polar inert gas atom.

Teacher's Note:
1. Polar plus non-polar gives dipole-induced dipole.

Question 18: Which of the following groups of species have the same geometry?
(a) \( \ce{SbF5^{2-}} \), \( \ce{XeF5-} \), \( \ce{IF5} \)
(b) \( \ce{SF4} \), \( \ce{XeOF4} \), \( \ce{PCl4+} \)
(c) \( \ce{COCl2} \), \( \ce{SOCl2} \), \( \ce{XeOF2} \)
(d) \( \ce{SF6} \), \( \ce{PF6-} \), \( \ce{IOF5} \)
Show Answer & Explanation

Answer: (d) \( \ce{SF6} \), \( \ce{PF6-} \), \( \ce{IOF5} \)

Explanation:
1. SF6, PF6- and IOF5 are octahedral (IOF5 has no lone pair on I). The other sets mix different geometries.

Teacher's Note:
1. Check each species' lone pairs.

Question 19: The species in which a 2 pi bond and a half sigma bond are present is:
(a) \( \ce{O2-} \)
(b) \( \ce{N2+} \)
(c) \( \ce{O2+} \)
(d) \( \ce{N2-} \)
Show Answer & Explanation

Answer: (b) \( \ce{N2+} \)

Explanation:
1. N2+ has configuration sigma2s2 sigma*2s2 pi2p4 sigma2p1: two pi bonds (4 electrons) and a half sigma bond (1 electron), bond order 2.5. O2- and O2+ have fewer pi bonds, and N2- has full sigma and 1.5 pi.

Teacher's Note:
1. Bond order 2.5 can be split into pi and sigma contributions from the MO diagram.

Question 20: Which of the following compounds gives a paramagnetic gas on heating?
(a) All of these
(b) \( \ce{LiNO3} \) only
(c) \( \ce{NaNO3} \) only
(d) \( \ce{KNO3} \) only
Show Answer & Explanation

Answer: (a) All of these

Explanation:
1. LiNO3 gives Li2O + NO2 + O2 and NaNO3 and KNO3 give nitrite + O2. NO2 and O2 are both paramagnetic.

Teacher's Note:
1. Both NO2 (odd electron) and O2 (two unpaired) are paramagnetic.

Question 21: The correct order of bond angle is:
(a) \( \ce{NH4+} \) > \( \ce{NH3} \) > \( \ce{NH2-} \)
(b) \( \ce{NO2+} \) < \( \ce{NO2} \) < \( \ce{NO2-} \)
(c) \( \ce{BeCl2} \) > \( \ce{BF3} \) < \( \ce{CF4} \)
(d) \( \ce{OF2} \) > \( \ce{OCl2} \) > \( \ce{OBr2} \)
Show Answer & Explanation

Answer: (a) \( \ce{NH4+} \) > \( \ce{NH3} \) > \( \ce{NH2-} \)

Explanation:
1. NH4+ is 109.5, NH3 107 and NH2- about 104 degrees (more lone pairs mean smaller angle). The others are wrong: NO2+ (180) > NO2 (134) > NO2- (115); BeCl2 (180) > BF3 (120) > CF4 (109.5); OCl2 (111) > OF2 (103).

Teacher's Note:
1. Lone pair repulsion reduces angle.

Question 22: In which of the following compounds are all the bond angles equal?
(a) \( \ce{SF4} \)
(b) \( \ce{CHCl3} \)
(c) \( \ce{XeF6} \)
(d) \( \ce{CCl4} \)
Show Answer & Explanation

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

Explanation:
1. CCl4 is perfectly tetrahedral, with all angles 109.5 degrees. CHCl3 is distorted by the unlike atom, SF4 is see-saw and XeF6 is distorted octahedral.

Teacher's Note:
1. Only symmetric molecules with identical substituents have equal angles.

Question 23: Which of the following species is non-polar although it has polar bonds and lone pairs of electrons on the central atom?
(a) \( \ce{XeF4} \)
(b) \( \ce{CO2} \)
(c) \( \ce{SF4} \)
(d) \( \ce{CF4} \)
Show Answer & Explanation

Answer: (a) \( \ce{XeF4} \)

Explanation:
1. XeF4 is square planar with two lone pairs and symmetrical bond dipoles that cancel. CO2 and CF4 have no lone pair on the central atom and SF4 is polar.

Teacher's Note:
1. Symmetric arrangement cancels dipoles.

Question 24: Which statement is not correct from the point of view of molecular orbital theory?
(a) \( \ce{Be2} \) is not a stable molecule
(b) \( \ce{He2} \) is not stable but \( \ce{He2+} \) is expected to exist
(c) The bond strength of N2 is the maximum among homonuclear diatomic molecules
(d) In the F2 molecule the order of MO energies is \( \pi_{2p_x}=\pi_{2p_y}<\sigma_{2p_z} \)
Show Answer & Explanation

Answer: (d) In the F2 molecule the order of MO energies is \( \pi_{2p_x}=\pi_{2p_y}<\sigma_{2p_z} \)

Explanation:
1. For O2 and F2 the sigma 2pz orbital lies below the pi 2p orbitals (no s-p mixing), so the order stated for F2 is wrong. Be2 has bond order zero, He2+ has bond order 0.5 and N2 has bond order 3.

Teacher's Note:
1. Mixing reverses the order only for B2, C2 and N2.

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

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