526 questions on Chemical Bonding and Molecular Structure for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.
Free with a Google sign-in; NEET Daily needs no sign-in at all.
Answer: (C) Bent (V-shaped)
sp³, 2 bond pairs + 2 lone pairs → bent. Bond angle 104.5°.
Quick tip: H₂O=bent(104.5°), NH₃=trigonal pyramidal(107°), CH₄=tetrahedral(109.5°).
NCERT Class 11, Chapter 4, page 124
Answer: (C) LiF
Lattice enthalpy is proportional to (|z₊ · z₋|) / (r₊ + r₋). All four compounds are 1:1 salts. Li⁺ has the smallest cation radius and F⁻ has the smallest anion radius among the given pairs, making (r₊ + r₋) smallest for LiF, which imparts the highest lattice enthalpy.
Answer: (B) sp²
Counting bond pairs and lone pairs on the central atom of C₂H₄ gives sp² hybridisation.
Answer: (A) All statements A, B, and C are correct
Fe³⁺ has a +3 charge and smaller ionic radius than Fe²⁺ (+2). Hence, by Fajan's rules, Fe³⁺ has much greater polarizing power, conferring substantial covalent character to FeCl₃. As a result, FeCl₃ has a lower melting point than ionic FeCl₂ and readily dissolves in organic solvents like ether.
Answer: (A) bent
The bond pairs and lone pairs around the central atom of H₂O arrange to give a bent shape.
Answer: (B) +1
Formal charge = V - N - (B/2). For the central oxygen of ozone: V = 6, lone pair electrons N = 2, and bonding electrons B = 6 (one double bond and one single bond). Thus, Formal charge = 6 - 2 - (6/2) = +1.
Answer: (D) -644 kJ mol⁻¹
Using the Born-Haber cycle: Δ_fH° = Δ_subH°(Mg) + IE₁(Mg) + IE₂(Mg) + Δ_dissH°(Cl₂) + 2Δ_egH°(Cl) - Δ_latticeH°(MgCl₂) Δ_fH° = 148 + 738 + 1451 + 242 + 2(-349) - 2526 Δ_fH° = 2579 - 698 - 2526 = -645 ≈ -644 kJ mol⁻¹.
Answer: (D) BeCl₂
According to Fajan's rules, smaller cations have higher polarizing power, leading to greater covalent character. Be²⁺ is the smallest cation among the alkaline earth metals, making BeCl₂ predominantly covalent and thus most soluble in non-polar organic solvents like CCl₄.
Answer: (C) Statement I is incorrect but Statement II is correct
Canonical structures are hypothetical and do not possess real physical existence; thus, they cannot be isolated. The actual molecule is the resonance hybrid, which is more stable than any contributing canonical structure by an amount known as the resonance energy.
Answer: (A) Both Statement I and Statement II are correct
Statement I is correct because metallic bonds involve positive metal kernels submerged in a delocalized electron pool without fixed directional angles, whereas hydrogen bonding involves specific directional orientation between the polar covalent bond and the lone pair of an electronegative atom. Statement II is also correct as both fundamentally arise from electrostatic attractions (kernels to delocalized valence electrons in metals, and partially positive hydrogen to electronegative lone pairs in hydrogen bonding).
Answer: (B) Statement I is correct but Statement II is incorrect
Statement I is correct as electrovalent (ionic) bonding involves the transfer of electrons between electropositive and electronegative elements. Statement II is incorrect because the electrostatic field around an ion is non-directional, meaning ionic bonds are non-directional and do not possess directional molecular geometries like covalent molecules.
Answer: (C) SF₆
In SF₆, the central sulfur atom forms six single covalent bonds with six fluorine atoms, thereby sharing 12 electrons in its valence shell. This constitutes an expanded octet (hypervalent molecule). BCl₃ has an incomplete octet (6 electrons), while CCl₄ and NH₃ obey the octet rule.
Answer: (B) H₂Te has a significantly higher molecular mass and stronger London dispersion forces than H₂S
From H₂S to H₂Te, hydrogen bonding is virtually absent. The boiling points increase down the group from H₂S to H₂Te because of increasing molecular size and mass, which leads to stronger London dispersion forces.
Answer: (D) 76.8%
Theoretical dipole moment for 100% ionic character: μ_ionic = q × d = (1.602 × 10⁻¹⁹ C) × (1.596 × 10⁻¹⁰ m) = 2.557 × 10⁻²⁹ C m. % ionic character = (μ_experimental / μ_ionic) × 100 = (1.964 × 10⁻²⁹ / 2.557 × 10⁻²⁹) × 100 ≈ 76.8%.
Answer: (B) 5
The element with atomic number 15 is phosphorus (P). Its electronic configuration is [Ne] 3s² 3p³, which has 5 valence electrons. Hence, its Lewis symbol has 5 dots around it.