Chemical Bonding MCQ For NEET Solved – PDF
Chemical Bonding MCQ For NEET
Que 1. Which of the following molecules is adequately represented by a single Lewis structure?
(a) O3
(b) NOCl
(c) SO2
(d) N2O
Ans 1. (b)
Reason: NOCl can be represented with only one Lewis diagram O::N:Cl. We cannot write other Lewis diagrams for NOCl but for other molecules, more than one Lewis diagram can be written.
Que 2. Among LiCl, BeCl2, BCl3, and CCl4, the covalent bond character follows the order
(a) BeCl2 > BCl3 > CCl4 < LiCl
(b) BeCl2 < BCl3 < CCl4 < LiCl
(c) LiCl < BeCl2 < BCl3 < CCl4
(d) LiCl > BeCl2 > BCl3 > CCl4
Ans 2. (c)
Reason: Along the period, as we move from Li → Be → B → C, the electronegativity increases, and hence the Electronegativity difference between the element and Cl decreases, and accordingly, the covalent character increases. Thus LiCl < BeCl2 < BCl3 < CCl4 is the correct order of covalent bond character.
Que 3. Which forms a crystal of NaCl
(a) NaCl molecules
(b) Na+ and Cl– ions
(c) Na and Cl atoms
(d) None of the above
Ans 3. (b)
Reason: NaCl is an ionic crystal so it is formed by Na+ and Cl– ions.
Que 4. Which of the following molecules has both covalent and ionic bond
(a) CH3Cl
(b) NH4Cl
(c) HCl
(d) BeCl2
Ans 4. (b)
Reason: Ammonium Chloride is an ionic compound as a bond is formed between ammonia (a polyatomic ion) and chlorine (a metal). The ammonium chloride molecule also consists of three covalent bonds, present between nitrogen and three hydrogen atoms, and a coordinate covalent bond between nitrogen and a hydrogen atom.
Que 5. According to Lewis and Kossel’s approach, which of the following molecule has a complete octet of the central atom?
(a) LiCl
(b) BeH2
(c) BCl3
(d) CO2
Ans 5. (d)
Reason: According to Lewis and Kossel’s approach, the CO2 molecule has a complete octet of the central atom LiCl, BeH2 and BCl3 are the molecules with an incomplete octet of the central atom. Lewis structure of CO2 can be represented as:
Chemical Bonding MCQ For NEET
Que 6. The compound with no dipole moment is
(a) Methyl chloride
(b) Carbon tetrachloride
(c) Methylene chloride
(d) Chloroform
Ans 6. (b)
Reason: Carbon tetrachloride is a nonpolar compound
Que 7. Sodium chloride is an ionic compound whereas hydrogen chloride is mainly covalent because
(a) Sodium is less reactive
(b) Hydrogen is non-metal
(c) Hydrogen chloride is a gas
(d) Electronegativity difference in the case of hydrogen and chlorine is less than 2.1.
Ans 7. (b)
Reason: A compound formed between two non-metal atoms is a covalent bond.
Que 8. Which of the following species contains an equal number of σ and π-bonds?
(a) HCO3–
(b) XeO4
(c) (CN2)
(d) CH2(CN)2
Ans 8.
Reason: XeO4 contains an equal number of s and p-bonds. the number of σ and π–bonds present in species are as follows:
Que 9. Some of the properties of the two species, NO3– and H3O+ are described below. Which one of them is correct?
(a) Similar in hybridization for the central atom with different structures.
(b) Dissimilar in hybridization for the central atom with different structures.
(c) Isostructural with the same hybridization for the central atom.
(d) Isostructural with different hybridization for the central atom.
Ans 9. (b)
Reason: In NO3–, nitrogen has sp2 hybridization, thus planar in shape. In H3O+, oxygen is in sp3 hybridization, thus tetrahedral geometry is expected but due to the presence of one lp of electrons on the central oxygen atom, it is pyramidal in shape.
Que 10. The following group does not contain a dative bond
(a) – NO2
(b) – N2Cl
(c) – NC
(d) All of these
Ans 10: (b)
Reason: A dative bond is formed between two atoms when the electrons of the shared pair are contributed solely by one of the two atoms and it is directed toward that atom that does not contribute the electrons. It is represented by an arrow on the bond which is directed towards the electron recipient atom. So, – N+ ≡ NCl– is the only group among the given options which does not contain a dative bond.
Que 11. An element (X) forms compounds of the formula XCl3, X2O5, and Ca3X2 but does not form XCl5. Which of the following is the element (X)?
(a) B
(b) Al
(c) N
(d) P
Ans 11. (c)
Reason: N can form NCl3, N2O5, and Ca3N2 but cannot form NCl5. Due to the absence of d–orbitals. N cannot expand its valency to 5.
Que 12. The formal charges on the three atoms in O3 molecule are
(a) 0, 0, 0
(b) 0, 0, –1
(c) 0, 0, +1
(d) 0, +1, –1
Ans 12. (d) 0, + 1, –1
Que 13. Among the following, which compound will show the highest lattice energy?
(a) KF
(b) NaF
(c) CsF
(d) RbF
Ans 13. (b)
Reason: For compounds containing ions of the same charge, lattice energy increases as the size of ions decreases. Thus, NaF has the highest lattice energy.
Que 14. The electronic configuration of four elements L, P, Q, and R are given in brackets
The formulae of ionic compounds that can be formed between these elements are
(a) L (1s2, 2s2, 2p4)
(b) Q (1s2, 2s2, 2p6, 3s2, 3p5)
(c) P (1s2, 2s2, 2p6, 3s1)
(d) R (1s2, 2s2, 2p6, 3s2)
The formulae of ionic compounds that can be formed between these elements are
(a) L2P, RL, PQ, and R2Q
(b) LP, RL, PQ and RQ
(c) P2L, RL, PQ, and RQ2
(d) LP, R2L, P2Q and RQ
Ans 14. (c)
Reason: Valences of L, Q, P, and R is – 2, – 1, + 1, and + 2 respectively so they will form, P2L, RL, PQ, and RQ2 .
Que 15. What is the maximum number of water molecules that can attach to one water molecule through intermolecular hydrogen bonds?
(a) 2
(b) 3
(c) 4
(d) 1
Ans 15. (c) 4
Reason: The maximum number of molecules that one water molecule can hold through hydrogen bonding is 4 because the two hydrogen atoms of water can form two hydrogen bonds and the two lone pairs of oxygen can also form two hydrogen bonds.
Chemical Bonding MCQ For NEET
Que 16. The correct order of increasing bond length of CH, C–O, C–C and C=C is
(a) C – H < C – O < C – C < C = C
(b) C – H < C = C < C – O < C – C
(c) C – C < C = C < C – O < C – H
(d) C – O < C – H < C – C < C = C
Ans 16.
Reason: The C–H is the shortest due to the small size of both atoms and the large electronegativity difference. Double bonds are shorter than single bonds. Hence, the C = C bond is shorter than C – O, and C – C Due to the electronegativity difference between C and O, the C – O bond length is shorter than the C – C bond. Hence, the correct order is C — H < C = C < C — O < C — C.
Que 17. Carbon tetrachloride has no net dipole moment because of
(a) Its planar structure
(b) Its regular tetrahedral structure
(c) Similar sizes of carbon and chlorine atoms
(d) Similar electron affinities of carbon and chlorine
Ans 17. (b)
Reason: In CCl4, each C – Cl bond is polar but due to the very symmetrical tetrahedral arrangement they exactly cancel each other, and hence CCl4 has zero dipole moment
Que 18. Which one of the following molecules will have unequal M – F bond lengths?
(a) NF3
(b) BF3
(c) PF5
(d) SF4
Ans 18. (c)
Reason: In PF5, P has sp3d− hybridization, and thus has a trigonal bipyramidal structure. In PF5, the axial P−F bond length is not the same as the equatorial P−F bond.
Que 19. In PO43- ion the formal charge on the oxygen atom of P – O bond is
(a) + 1 (b) -1 (c) -0 75. (d) + 0 75
Ans 19. (c)
Reason:
Que 20. Which one of the following is the correct order of interactions?
(a) covalent < hydrogen bonding < van der Waals < dipole-dipole
(b) van der Waals < hydrogen bonding < dipole-dipole < covalent
(c) van der Waals < dipole-dipole < hydrogen bonding < covalent
(d) dipole-dipole < van der Waals < hydrogen bonding < covalent
Ans 20. (c)
Que 21. Which of the following statement(s) is not true for the given species? N2, CO, CN– and NO+
(a) All species have a linear shape
(b) All species have some dipole moments
(c) All species are isoelectronic
(d) All species have identical bond order and they are diamagnetic in nature.
Ans 21. (d)
Reason: All diatomic species are linear. The dipole moment of N2 is zero but the remaining species have some dipole moments. All species have 14 e–, bond order is 3, and are diamagnetic in nature.
Que 22. Which of the following order is incorrect?
(a) Ionic character: MCl < MCl2 < MCl3
(b) Polarizability: F– < Cl– < Br– < I–
(c) Polarising power: Na+ < Ca+2 < Mg+2 < Al+3
(d) Covalent character: LiF < LiCl < LiBr < LiI
Ans 22. (a)
Reason: As the positive oxidation state increases or, the size of the cation decreases, polarising power of the cation increases hence, the correct order of ionic character is: MCl < MCl2 < MCl3
Que 23. Select the wrong statement:
(a) A transition metal ion has more polarising power than S-block ions of comparable size and charge.
(b) Order of solubility in water is AgF > AgCl > AgBr > AgI
(c) LiCl is soluble in organic solvents
(d) The hydration of ions involves absorption of heat.
Ans 23. (d)
Reason: Out of cations having a comparable size and charge the one having Noble gas configuration has less polarizing power. LiCl is a covalent compound hence, it is soluble in organic solvents. The order of solubility of heavy metal halides depends on order of hydration energy.
(AgF > AgCl > AgBr > AgI).
The hydration of ions involves the evolution of heat.
M+ (g) + H2O → M+ (aq); ΔH = – Ve
Que 24. Two elements X and Y have following electronic configuration: X: 1s2 2s2 2p6 3s2 3p6 4s2, Y: 1s2 2s2 2p6 3s2 3p5. The expected compound formed by the combination of X and Y Will be expressed as-
(a) XY2
(b) X5Y2
(c) X2Y5
(d) XY5
Ans 24. (a)
Reason: The valency of element X is 2 (2 electrons in the outermost shell) while that of element Y is 1(1 electron in the outermost shell). So the formula of the compound between X and Y is XY2.
Que 25. The species in which the N-atom is in a state of sp hybridization is:
(a) NO2–
(b) NO3–
(c) NO2
(d) NO2+
Ans 25. (d)
Chemical Bonding MCQ For NEET
Que 26. The intermolecular interaction that is dependent on the inverse cube of the distance between the molecules is:
(a) Ion-ion interaction
(b) Ion-dipole interaction
(c) London force
(d) Hydrogen bond
Ans 26. (d)
Que 27. Which one of the following molecules is expected to exhibit diamagnetic behavior?
(a) S2
(b) C2
(c) N2
(d) O2
Ans 27. (b), (c)
Reason: In C2 and N2, all molecular orbitals are paired.
Que 28. Stability of the species Li2, Li2– and Li2+ increases in the order of:
(a) Li2– < Li2 < Li2+
(b) Li2 < Li2+ < Li2–
(c) Li2– < Li2+ < Li2
(d) Li2 < Li2– < Li2+
Ans 28. (c)
Reason: Both Li2 + and Li2– have similar bond order but Li2– has a greater number of antibonding e– than Li2+ hence, Li2+, is more stable than Li2–.
Que 30. The molecule having smallest bond angle is:
(a) NCl3
(b) AsCl3
(c) SbCl3
(d) PCl3
Ans 30. (c)
Reason: Order of bond angle: NCl3 > PCl3 > AsCl3 > SbCl3 (Order depends on electronegativity of central atom)
Que 31. The hybridization of orbitals of N atom in NO3–, NO2+ and NH4+ are respectively:
(a) sp, sp2, sp3
(b) sp2, sp, sp3
(c) sp, sp3, sp2
(d) sp2, sp3, sp
Ans 31. (b)
Que 32. The correct order of the thermal stability of hydrogen halides (H-X) is:
(a) HI > HBr > HCl > HF
(b) HF > HCl > HBr > HI
(c) HCl < HF < HBr < HI
(d) Hl < HCl < HF < HBr
Ans 32. (b)
Reason: As bond length increases, bond breaking becomes easier.
Que 33. The molecular shapes of SF4, CF4, and XeF4 are:
(a) Different with 1, 0, and 2 lone pairs of electrons on the central atoms, respectively
(b) Different with 0, 1, and 2 lone pairs of electrons on the central atoms, respectively
(c) Different with 1, 1, and 1 lone pairs of electrons on the central atoms, respectively
(d) Different with 2, 0, and 1 lone pairs of electrons on the central atoms, respectively
Ans 33. (a)
Que 34. The decreasing values of bond angles from NH3(106°) to SbH3 (101°) down group-15 of the periodic table is due to:
(a) Increasing bp-bp repulsion
(b) Increasing p-orbital character in sp3
(c) Decreasing lp-bp repulsion
(d) Decreasing electronegativity
Ans 34. (d)
Reason: Order of bond angles: NH3 > PH3 > AsH3 > SbH3
Que 35. The maximum number of 90° angles between bond pair-bond pair of electrons is observed in,
(a) dsp3 hybridization
(b) sp3d hybridization
(c) dsp2 hybridization
(d) sp3d2 hybridization
Ans 35. (d)
Reason: In sp3d2 hybridization (octahedral geometry), 12, 90° angles are observed between bp – bp of electrons.
Chemical Bonding MCQ For NEET
Que 36. The shape of XeO2F2 molecule is:
(a) Trigonal bipyramidal
(b) Square planar
(c) Tetrahedral
(d) See-saw
Ans 36. (d)
Que 37. The species having pyramidal shape is:
(a) SO3
(b) BrF3
(c) SiO32-
(d) OSF2
Ans 37. (d)
Que 38. Which species has the maximum number of lone pairs of electrons on the central atom?
(a) ClO3–
(b) XeF4
(c) SF4
(d) I3–
Ans 38. (d)
Que 39. Which of the following are isoelectronic and isostructural? NO3–, CO32-, ClO3–, SO3
(a) NO3–, CO32-
(b) SO3, NO3–
(c) ClO3–, CO32-
(d) CO32-, SO3
Ans 39. (a)
Reason: Both NO3 – and CO32– have 32 e– and they have trigonal planar shape.
Que 40. Which of the following molecular species has unpaired electrons?
(a) N2
(b) F2
(c) O2–
(d) O22-
Ans 40. (c)
Reason: O2– is an odd e– species. It has 1 unpaired e–.
Chemical Bonding MCQ For NEET
