Download JEE MCQs for JEE Chemistry: Chemical Bonding and Molecular Structure
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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
(a) much less than in \( \ce{NH3} \)
(b) equal to that in \( \ce{NH3} \)
(c) much greater than in \( \ce{NH3} \)
(d) slightly more than in \( \ce{NH3} \)
Show Answer & Explanation
Answer: (a) much less than in \( \ce{NH3} \)
Explanation:
1. \( \ce{PH3} \) has an angle of about 93.5 degrees against 107 degrees in \( \ce{NH3} \).
2. P is less electronegative and bigger, so the bond pairs lie farther from P and repel less, and the lone pair takes more space.
Teacher's Note:
1. Down a group the bond angle in hydrides falls.
2. Phosphorus uses almost pure p orbitals.
(a) The polarising power of a cation is directly proportional to its size.
(b) The polarisability of an anion is directly proportional to its size.
(c) Ionic interactions are directional.
(d) An element with low ionisation potential most readily forms a covalent bond with an element of high electron affinity.
Show Answer & Explanation
Answer: (b) The polarisability of an anion is directly proportional to its size.
Explanation:
1. A larger anion has a loosely held electron cloud, so it is more easily polarised.
2. Polarising power of a cation rises with charge and falls with size, so (a) is wrong.
3. Ionic bonds are non-directional, and low ionisation energy with high electron affinity gives an ionic bond.
Teacher's Note:
1. Fajans' rules: small, highly charged cation and large, highly charged anion favour covalency.
2. Ionic interactions act equally in all directions.
(a) the cation is large and the anion is small
(b) the cation is small and highly charged, and the anion is large
(c) the cation is large and highly charged, and the anion is small
(d) both ions are small and have low charge
Show Answer & Explanation
Answer: (b) the cation is small and highly charged, and the anion is large
Explanation:
1. A small, highly charged cation has a high polarising power.
2. A large anion is easily polarised.
3. The distortion of the anion cloud gives partial sharing of electrons, i.e. covalent character.
Teacher's Note:
1. Remember: small cation, large anion, high charges.
2. Example: LiI is more covalent than LiF.
(a) All diatomic molecules are non-polar.
(b) SO2 is polar whereas CO2 is non-polar.
(c) NH3 is less polar than NF3.
(d) All molecules having polar bonds are polar.
Show Answer & Explanation
Answer: (b) SO2 is polar whereas CO2 is non-polar.
Explanation:
1. SO2 is bent, so the bond dipoles add up.
2. CO2 is linear, so they cancel.
3. HCl is a polar diatomic molecule, NH3 (1.47 D) is more polar than NF3 (0.23 D), and symmetrical molecules with polar bonds can be non-polar.
Teacher's Note:
1. Shape decides whether bond dipoles cancel.
2. Heteronuclear diatomics are always polar.
(a) NaCl is ionic and gives \( \ce{Cl-} \) ions, while \( \ce{CCl4} \) is covalent and gives none
(b) \( \ce{CCl4} \) is polar
(c) NaCl is covalent
(d) \( \ce{CCl4} \) is soluble in water
Show Answer & Explanation
Answer: (a) NaCl is ionic and gives \( \ce{Cl-} \) ions, while \( \ce{CCl4} \) is covalent and gives none
Explanation:
1. NaCl gives free \( \ce{Cl-} \) ions in solution, which precipitate as AgCl.
2. \( \ce{CCl4} \) is a non-polar covalent liquid that does not ionise, so no chloride ions are available.
Teacher's Note:
1. Precipitation needs free ions.
2. Covalent halides like CCl4 do not give the test directly.
(a) \( sp^3 \) and bent (V-shaped)
(b) sp and linear
(c) \( sp^2 \) and bent
(d) \( sp^3d \) and T-shaped
Show Answer & Explanation
Answer: (a) \( sp^3 \) and bent (V-shaped)
Explanation:
1. Oxygen has two bond pairs and two lone pairs, i.e. four electron pairs.
2. This gives \( sp^3 \) hybridisation and a bent shape.
3. Oxidation numbers are +2 for O and -1 for F.
Teacher's Note:
1. Two lone pairs push the bond angle below 109.5 degrees.
2. The molecule is like water but with F in place of H.
(a) linear, square planar, trigonal bipyramidal
(b) bent, tetrahedral, square pyramidal
(c) linear, tetrahedral, octahedral
(d) bent, square planar, trigonal bipyramidal
Show Answer & Explanation
Answer: (a) linear, square planar, trigonal bipyramidal
Explanation:
1. \( \ce{XeF2} \) has 2 bond pairs and 3 lone pairs, so it is linear.
2. \( \ce{XeF4} \) has 4 bond pairs and 2 lone pairs, so it is square planar.
3. \( \ce{PBr5} \) has 5 bond pairs and no lone pair, so it is trigonal bipyramidal.
Teacher's Note:
1. Count bond pairs and lone pairs.
2. Lone pairs go to equatorial positions in a trigonal bipyramid.
(a) 1.24 A
(b) 0.62 A
(c) 2.49 A
(d) 1.08 A
Show Answer & Explanation
Answer: (a) 1.24 A
Explanation:
1. Dipole moment for 100 percent ionic character = 1.08 / 0.1808 = 5.97 D.
2. Also this equals e x d, i.e. d = 5.97 x 10^-18 esu cm / 4.8 x 10^-10 esu = 1.24 x 10^-8 cm.
3. So the distance is 1.24 A.
Teacher's Note:
1. 1 D = 10^-18 esu cm and e = 4.8 x 10^-10 esu.
2. Convert cm to angstrom (1 A = 10^-8 cm).
(a) the electronegativity of the central atom decreases down the group, so bond pairs move away from it and repel less
(b) the size of H increases
(c) the number of lone pairs increases
(d) the hybridisation changes to sp2
Show Answer & Explanation
Answer: (a) the electronegativity of the central atom decreases down the group, so bond pairs move away from it and repel less
Explanation:
1. Oxygen is the most electronegative, so the bonding pairs are close to it and repel strongly, widening the angle.
2. Down the group the central atom is less electronegative and bigger, so the bond pairs lie farther away and repel less.
3. The lone pairs then dominate and the angle shrinks towards 90 degrees.
Teacher's Note:
1. Lower electronegativity of the central atom gives smaller bond angle.
2. Heavier hydrides use nearly pure p orbitals.
(a) 2.52 A
(b) 1.54 A
(c) 3.08 A
(d) 1.78 A
Show Answer & Explanation
Answer: (a) 2.52 A
Explanation:
1. The C-C-C angle is 109.5 degrees.
2. Distance = 2 x 1.54 x sin(109.5 / 2) = 3.08 x sin 54.75 = 3.08 x 0.8165.
3. This is 2.52 A.
Teacher's Note:
1. Use the cosine rule or the half-angle formula for an isosceles triangle.
2. The tetrahedral angle is 109.5 degrees.
(a) 1.41 A
(b) 0.71 A
(c) 2.82 A
(d) 0.28 A
Show Answer & Explanation
Answer: (a) 1.41 A
Explanation:
1. Dipole moment for 100 percent ionic character = 2.6 x 10^-30 / 0.115 = 2.26 x 10^-29 C m.
2. d = 2.26 x 10^-29 / 1.602 x 10^-19 = 1.41 x 10^-10 m.
3. So the spacing is 1.41 A.
Teacher's Note:
1. Divide by the ionic fraction first.
2. Use e = 1.602 x 10^-19 C.
(a) 84 percent
(b) 66 percent
(c) 100 percent
(d) 42 percent
Show Answer & Explanation
Answer: (a) 84 percent
Explanation:
1. Dipole moment for 100 percent ionic character = 4.8 x 1.56 = 7.49 D (using e = 4.8 x 10^-10 esu and d = 1.56 A).
2. Percentage ionic = 6.32 / 7.49 x 100 = 84.4 percent.
Teacher's Note:
1. Be careful with the unit of distance: 156 pm = 1.56 A.
2. Ionic character shows LiF is largely ionic.
(a) 25
(b) 12.5
(c) 50
(d) 100
Show Answer & Explanation
Answer: (a) 25
Explanation:
1. Charge q = mu / d = 1.2 x 10^-18 esu cm / 1.0 x 10^-8 cm = 1.2 x 10^-10 esu.
2. Fraction of electron charge = 1.2 x 10^-10 / 4.8 x 10^-10 = 0.25.
3. So 25 percent of an electronic charge is on each atom.
Teacher's Note:
1. mu = q x d.
2. Compare q with the charge of one electron.
(a) 2, 3, 1
(b) 1, 2, 3
(c) 4, 1, 2
(d) 3, 2, 1
Show Answer & Explanation
Answer: (d) 3, 2, 1
Explanation:
1. Xe has 8 valence electrons.
2. \( \ce{XeF2} \) uses 2, leaving 6 electrons, i.e. 3 lone pairs.
3. \( \ce{XeF4} \) uses 4, leaving 2 lone pairs, and \( \ce{XeF6} \) uses 6, leaving 1 lone pair.
Teacher's Note:
1. Lone pairs on Xe = (8 - number of F) / 2.
2. This works for all xenon fluorides.
(a) \( \ce{NH3} \) and \( \ce{BF4-} \)
(b) \( \ce{NH4+} \) and \( \ce{BF4-} \)
(c) \( \ce{NH3} \) and \( \ce{BF3} \)
(d) \( \ce{NH2-} \) and \( \ce{BF3} \)
Show Answer & Explanation
Answer: (b) \( \ce{NH4+} \) and \( \ce{BF4-} \)
Explanation:
1. A bond angle of 109 degrees 28 minutes belongs to a perfect tetrahedron with no lone pairs.
2. \( \ce{NH4+} \) and \( \ce{BF4-} \) are both regular tetrahedra.
3. \( \ce{NH3} \) and \( \ce{NH2-} \) have lone pairs, and \( \ce{BF3} \) is planar (120 degrees).
Teacher's Note:
1. Only AX4 species have the exact tetrahedral angle.
2. Lone pairs reduce the angle.
(a) \( \ce{O2^2-} < \ce{O2-} < \ce{O2} < \ce{O2+} \)
(b) \( \ce{O2-} < \ce{O2^2-} < \ce{O2} < \ce{O2+} \)
(c) \( \ce{O2^2-} < \ce{O2-} < \ce{O2+} < \ce{O2} \)
(d) \( \ce{O2+} < \ce{O2} < \ce{O2-} < \ce{O2^2-} \)
Show Answer & Explanation
Answer: (a) \( \ce{O2^2-} < \ce{O2-} < \ce{O2} < \ce{O2+} \)
Explanation:
1. Bond orders are \( \ce{O2^2-} \) 1, \( \ce{O2-} \) 1.5, \( \ce{O2} \) 2 and \( \ce{O2+} \) 2.5.
2. Bond strength rises with bond order.
Teacher's Note:
1. Add electrons to antibonding orbitals and bond order falls.
2. Remove them and bond order rises.
(a) \( \ce{AlH3} \) changes to \( \ce{AlH4-} \) (Al)
(b) \( \ce{H2O} \) changes to \( \ce{H3O+} \) (O)
(c) \( \ce{NH3} \) changes to \( \ce{NH4+} \) (N)
(d) all of these
Show Answer & Explanation
Answer: (a) \( \ce{AlH3} \) changes to \( \ce{AlH4-} \) (Al)
Explanation:
1. Al in \( \ce{AlH3} \) is sp2 (three bond pairs) and in \( \ce{AlH4-} \) it is sp3.
2. O in \( \ce{H2O} \) and \( \ce{H3O+} \) is sp3 in both, and N in \( \ce{NH3} \) and \( \ce{NH4+} \) is sp3 in both.
Teacher's Note:
1. Compare the steric number before and after.
2. Adding a proton to a lone pair does not change the steric number.
(a) \( \ce{CH3-COOH} \) (carboxyl carbon)
(b) \( \ce{CH3-CH2-OH} \) (the CH2 carbon)
(c) \( \ce{CH3-CO-CH3} \) (carbonyl carbon)
(d) \( \ce{CH2=CH-CH3} \) (the CH= carbon)
Show Answer & Explanation
Answer: (b) \( \ce{CH3-CH2-OH} \) (the CH2 carbon)
Explanation:
1. The CH2 carbon of ethanol has four sigma bonds, so it is sp3.
2. The carboxyl carbon, the carbonyl carbon and the alkene carbon each have a double bond and are sp2.
Teacher's Note:
1. Four sigma bonds means sp3.
2. A double bond carbon is sp2.
(a) dipolar character of ether
(b) alcohols having resonance structures
(c) intermolecular hydrogen bonding in ether
(d) intermolecular hydrogen bonding in alcohols
Show Answer & Explanation
Answer: (d) intermolecular hydrogen bonding in alcohols
Explanation:
1. Alcohols have O-H bonds and form intermolecular hydrogen bonds.
2. This raises their boiling point and reduces volatility.
3. Ethers have no O-H and cannot hydrogen bond to each other.
Teacher's Note:
1. Isomeric ethers boil lower than alcohols.
2. Volatility goes down as intermolecular attraction goes up.
(a) electronic orbitals of carbon atom are hybridised
(b) the C=O bond is weaker than the C-O bond
(c) the anion \( \ce{HCOO-} \) has two resonating structures
(d) the anion is obtained by removal of a proton from the acid molecule
Show Answer & Explanation
Answer: (c) the anion \( \ce{HCOO-} \) has two resonating structures
Explanation:
1. The negative charge is delocalised over both oxygen atoms.
2. Two equivalent resonance structures exist, with the double bond on either oxygen.
3. So each C-O bond has partial double bond character and the two bonds are equal in length.
Teacher's Note:
1. Equal bond lengths in an ion are a sign of resonance.
2. The same applies to the carbonate ion.
(a) \( \ce{H2O} \)
(b) \( \ce{SH2} \)
(c) \( \ce{NH3} \)
(d) \( \ce{SO2} \)
Show Answer & Explanation
Answer: (b) \( \ce{SH2} \)
Explanation:
1. The angles are about 104.5 degrees in \( \ce{H2O} \), 92 in \( \ce{H2S} \), 107 in \( \ce{NH3} \) and 119 in \( \ce{SO2} \).
2. Sulphur is less electronegative than oxygen and nearly pure p orbitals are used, so \( \ce{H2S} \) has the smallest angle.
Teacher's Note:
1. Hydrides of the third period have angles near 90 degrees.
2. Drago's rule explains this.
(a) dipole-dipole interaction
(b) hydrogen bonding
(c) electrostatic attractions
(d) van der Waals forces
Show Answer & Explanation
Answer: (b) hydrogen bonding
Explanation:
1. The two strands are held together by hydrogen bonds between complementary base pairs (A-T and G-C).
2. This gives the double helix its structure.
Teacher's Note:
1. Hydrogen bonding is weak but very specific.
2. The pairing makes the replication of DNA possible.
(a) \( \ce{NO2} \) and \( \ce{CO2} \)
(b) \( \ce{NO2} \) and \( \ce{O3} \)
(c) \( \ce{SiF4} \) and \( \ce{CO2} \)
(d) \( \ce{SiF4} \) and \( \ce{NO2} \)
Show Answer & Explanation
Answer: (b) \( \ce{NO2} \) and \( \ce{O3} \)
Explanation:
1. \( \ce{NO2} \) and \( \ce{O3} \) are both bent and have a net dipole moment.
2. \( \ce{CO2} \) is linear and \( \ce{SiF4} \) is tetrahedral, and both are non-polar.
Teacher's Note:
1. Bent molecules are polar.
2. Symmetrical linear and tetrahedral molecules are not.
(a) \( \ce{XeF2} \), \( \ce{CO2} \)
(b) \( \ce{BF3} \), \( \ce{PCl3} \)
(c) \( \ce{PF5} \), \( \ce{IF5} \)
(d) \( \ce{CF4} \), \( \ce{SF4} \)
Show Answer & Explanation
Answer: (a) \( \ce{XeF2} \), \( \ce{CO2} \)
Explanation:
1. \( \ce{XeF2} \) and \( \ce{CO2} \) are both linear.
2. \( \ce{BF3} \) is planar while \( \ce{PCl3} \) is pyramidal, \( \ce{PF5} \) is trigonal bipyramidal while \( \ce{IF5} \) is square pyramidal, and \( \ce{CF4} \) is tetrahedral while \( \ce{SF4} \) is see-saw.
Teacher's Note:
1. Compare lone pairs and bond pairs.
2. The same shape needs the same pattern.
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FAQs
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