Chemical Bonding and Molecular Structure Questions and Answers pdf
Chemical Bonding and Molecular Structure Questions and Answers pdf
Multiple Choice Questions – 1
Que 1. The lattice energy of an ionic compound depends upon
(a) Charge on the ion only (b) Size of the ion only
(c) Packing of ions only (d) Charge on the ion and size of the ion
Ans 1. (d) The value of lattice energy depends on the charges present on the two ions and the distance between them. It shell be high if charges are high and ionic radii are small.
Que 2. In the given bonds which one is most ionic
(a) Cs -Cl (b) Al -Cl
(c) C -Cl (d) H -Cl
Ans 2. (a) Cs is more electropositive
Que 3. Element x is strongly electropositive and y is strongly electronegative. Both elements are univalent, the compounds formed from their combination will be
(a) x+ y– (b) x– y+
(c) x – y (d) x → y
Ans 3. (a) X loses an electron, and Y gains it.
Que 4. In the formation of NaCl from Na and Cl
(a) Sodium and chlorine both give electrons
(b) Sodium and chlorine both accept electrons
(c) Sodium loses an electron and chlorine accepts an electron
(d) Sodium accepts an electron and chlorine loses an electron
Ans 4. (c) Formation of NaCl occurs by Na+ ion and Cl– ion.
Que 5. Which of the following is an electrovalent linkage
(a) CH4 (b) MgCl2
(c) SiCl4 (d) BF3
Ans 5. (b) MgCl2 has electrovalent linkage because magnesium is electropositive metal while chlorine is electronegative.
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 6. Electrovalent compounds do not have
(a) High M.P. and Low B.P. (b) High dielectric constant
(c) High M.P. and High B.P. (d) High polarity
Ans 6. (a) Electrovalent compounds generally have high m.pt and high b.pt due to stronger coulombic forces of attraction.
Que 7. Many ionic crystals dissolve in water because
(a) Water is an amphiprotic solvent
(b) Water is a high boiling liquid
(c) The process is accompanied by a positive heat of solution
(d) Water decreases the interionic attraction in the crystal lattice due to solvation
Ans 7. (d) Water is a polar solvent so it decreases the interionic attraction in the crystal lattice due to solvation.
Que 8. The electronic structure of four elements A, B, C, and D are
(A) 1s2 (B) 1s2, 2s2 2p2
(C) 1s2, 2s2 2p5 (D) 1s2, 2s2 2p6
The tendency to form an electrovalent bond is largest in
(a) A (b) B (c) C (d) D
Ans 8. (c) Element C has an electronic structure 1s2, 2s2, 2p5, it requires only one electron to complete its octet and it will form an anion so it will form an electrovalent bond.
Que 9. With which of the given pairs CO2 resembles
(a) HgCl2, C2H2 (b) HgCl2, SnCl4
(c) C2H2, NO2 (d) N2O and NO2
Ans 9. (a) All have a linear structure. O = C = O, Cl – Hg – Cl, HC ≡ CH
Que 10. The electron pair which forms a bond between two similar non-metallic atoms will be
(a) Dissimilar shared between the two
(b) By complete transfer from one atom to other
(c) In a similar spin condition
(d) Equally shared between the two
Ans 10. (d) Similar atoms form a covalent bond.
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 11. For the formation of a covalent bond, the difference in the value of electronegativities should be
(a) Equal to or less than 1.7 (b) More than 1.7
(c) 1.7 or more (d) None of these
Ans 11. (a) Covalent bond forms when the electronegativity difference of two atoms is equal to 1.7 or less than 1.7
Que 12. Which type of bond is formed between similar atoms
(a) Ionic (b) Covalent
(c) Coordinate (d) Metallic
Ans 12. (b) Similar atoms form a covalent bond
Que 13. Covalent compounds are generally …… in water
(a) Soluble (b) Insoluble
(c) Dissociated (d) Hydrolysed
Ans 13. (b) Water is a polar solvent while covalent compounds are nonpolar so they are usually insoluble in water.
Que 14. Which one is the electron-deficient compound
(a) ICl (b) NH3
(c) BCl3 (d) PCl3
Ans 14. (c) BCl3 is an electron-deficient compound because it has only ‘6’ electrons after forming the bond.
Que 15. Which among the following elements has the tendency to form covalent compounds
(a) Ba (b) Be
(c) Mg (d) Ca
Ans 15. (b) Due to its small size and 2 electrons in the s-orbital Be forms a covalent compound.
Que 16. Silicon has 4 electrons in the outermost orbit. In forming the bonds
(a) It gains electrons (b) It loses electrons
(c) It shares electrons (d) None of these
Ans 16. (a) It gains electrons
Que 17. According to VSEPR theory, the geometry of a covalent molecule depends upon
(a) the number of bond pairs of electrons
(b) the number of lone pairs of electrons
(c) the number of electron pairs present in the outer shell of the central atom
(d) All the above
Ans 17. (d)
Que 18. The geometry of – ClO– ion according to Valence Shell Electron Pair Repulsion (VSEPR) theory will be
(a) planar triangular (b) pyramidal
(c) tetrahedral (d) square planar
Ans 18. (b) Hybridisation is sp3 and shape pyramidal.
Que 19. In BrF3 molecule, the lone pairs occupy equatorial positions to minimize
(a) lone pair-bond pair repulsion only
(b) bond pair-bond pair repulsion only
(c) lone pair – lone pair repulsion and lone pair-bond pair repulsion
(d) lone pair – lone pair repulsion only
Ans 19. (c)In BrF3, both bond pairs, as well as lone pairs of electrons, are present. Due to the presence of lone pairs of electrons (lp) in the valence shell, the bond angle is contracted and the molecule takes a T shape. This is due to greater repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pairs.
Que 20. Which of the correct increasing order of lone pair of electrons on the central atom?
(a) IF7 < IF5 < CIF3 < XeF2
(b) IF7 < XeF2 < CIF2 < IF5
(c) IF7 < CIF3 < XeF2 < IF5
(d) IF7 < XeF2 < IF5 < CIF3
Ans 20. (a) The number of lone pairs of electrons on the central atom in various given species is 0, 1, 2, and 3 respectively
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 21. The number of lone pair and bond pair of electrons on the sulphur atom in sulphur dioxide molecule is respectively
(a) 1 and 3 (b) 4 and 1
(c) 3 and 1 (d) 1 and 4
Ans 21. (d)
Que 22. A molecule has two lone pairs and two bond pairs around the central atom. The molecule shape is expected to be
(a) V-shaped (b) triangular
(c) linear (d) tetrahedral
Ans 22. (a) V-shaped H2O like structure
Que 23. Using VSEPR theory, predict the species which has a square pyramidal shape
(a) SnCl2 (b) CCl4
(c) SO3 (d) BrF5
Ans 23. d) BrF5 has square pyramidal geometry.
Que 24. The ground state electronic configuration of valence shell electrons in nitrogen molecule (N2) is written as KK σ2s, σ*2s, л2px, л2py, σ2pz Bond order in nitrogen molecule is
(a) 0 (b) 1 (c) 2 (d) 3
Ans 24. (d) Reason: In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding molecular orbital. So that Nb = 8 and Na = 2
Bond order = ½ [8 – 2] = 3
Que 25. Bond order in benzene is
(a) 1 (b) 2 (c) 1.5 (d) None of these
Ans 25. (c) Benzene has the following resonance structures–
Hence, its bond order is
no of possible resonating structures = 1.5.
2
Que 26. In O2–, O2, and O2– molecular species, the total number of antibonding electrons respectively are
(a) 7, 6, 8 (b) 1, 0, 2 (c) 6, 6, 6 (d) 8, 6, 8
Ans 26. (a) Hence the number of antibonding electrons is 7, 6, and 8 respectively.
Que 27. Hybridization and structure of I3– are
(a) sp2 and trigonal planar (b) sp3d2 and linear
(c) sp3d and linear (d) sp3 and T-shape
Ans 27. I3– has sp3d hybridization and has a linear structure
Que 28. In which of the following species, all three types of hybrid carbons are present?
(a) CH2 = C = CH2
(b) CH3 – CH = CH – CH2+
(c) CH3 – C ≡ C – CH2+
(d) CH3 – CH = CH – CH2–
Ans 28. (c)
(a) CH2 = C= CH2
sp2 sp sp2
(b) CH — CH = CH— CH
sp3 sp2 sp2 sp2
(c) CH3 – C ≡ C – CH2+
sp3 sp sp sp2
(d) CH3 – CH = CH – CH2–
sp3 sp2 sp2 sp3
Que 29. Which of the following molecules has trigonal planar geometry?
(a) BF3 (b) NH3 (c) PCl3 (d) IF3
Ans 29: (a)
Reason: BF3 is sp2 hybridized. So, it is a trigonal planner. NH3, PCl3 has sp3 hybridization hence has trigonal bipyramidal shape, IF3, has sp3d hybridization and has a linear shape.
Que 30. Which of the following molecules is planar?
(a) SF4 (b) XeF4 (c) NF3 (d) SiF4
Ans 30. (b)
Reason: XeF4 hybridisation is = ½ [ V +X – C + A]
V → Valance electrons = 8
X → no. of monovalent atom = 4
C → charge on cation = 0
A → charge on anion = 0
= ½ [ 8 + 4 + 0 + 0] = 6
Therefore XeF4 = Square Planer
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 31. Hybridization present in ClF3 is
(a) sp2 (b) sp3 (c) dsp2 (d) sp3d
Ans 31. (d)
Reason: Hybridisation present in a molecule can find out by the following formula.
Hybridisation = ½ [ V +X – C + A]
V → Valance electrons = 7
X → no. of monovalent atom = 3
C → charge on cation = 0
A → charge on anion = 0
= ½ [ 7 + 3 + 0 + 0] = 5
So, we have sp3d Hybridisation ClF3.
Que 32. Match Column-I (molecule) with Column-II (type of hybridization) and choose the correct option from the codes given below.
Column-I Column-II
(Molecule) (Type of hybridisation)
(A) SF6 (p) sp3d
(B) PF5 (q) sp3
(C) BCl3 (r) sp3d2
(D) C2H6 (s) sp2
(a) A – (r), B – (p), C – (s), D – (q)
(b) A – (r), B – (p), C – (q), D – (s)
(c) A – (p), B – (r), C – (q), D – (s)
(d) A – (p), B – (r), C – (s), D – (q)
Ans 32. (a)
SF6 → sp3d2
PF5 → sp3d
BCl3 → sp2
C2H6 → sp3
Que 33. Match Column-I with Column-II and Column-III and choose the correct option from the given codes.
Column-I Column-II Column-III
Molecule (No. of lone pairs and bond pairs) (Shape of the molecule)
(A) NH3 (i) 1, 2 (p) Bent
(B) SO2 (ii) 1, 4 (q) Trigonal pyramidal
(C) SF4 (iii) 2, 3 (r) T-shape
(D) ClF3 (iv) 1, 3 (s) See-Saw
(a) A – (iv, q); B – (ii, p); C – (i, r); D – (iii, s)
(b) A – (iv, q); B – (i, p); C – (ii, s); D – (iii, r)
(c) A – (i, p); B – (iii, s); C – (iv, r); D – (ii, q)
(d) A – (iv, p); B – (i, r); C – (iii, q); D – (ii, s)
Ans 33. (b)
NH3 → 1lp, 3bp → Trigonal pyramidal
SO2 → 1lp, 2bp → Bent
SF4 → 1lp, 4bp → See-saw
ClF3 → 2lp, 3bp → T-shape
Que 34. Match the columns
Column-I Column-II
(A) HCl (p) Covalent compound with directional bond
(B) CO2 (q) Ionic compound with non-directional bonds
(C) NaCl (r) Polar molecule
(D) CCl4 (s) Non-polar molecule
(a) A – (p, q, r), B – (q, r), C – (p, q), D – (r)
(b) A – (q), B – (r), C – (p), D – (s)
(c) A – (p, r), B – (p, s), C – (q), D – (p, s)
(d) A – (q), B – (r), C – (p, q), D – (s)
Ans 34. (c)
Que 35. Assertion: The lesser the lattice enthalpy more stable is the ionic compound.
Reason: The lattice enthalpy is greater, for ions of the highest charge and smaller radii.
Ans 35. (d) The assertion is false but the reason is true.
Reason: The greater the lattice enthalpy, the more stable the ionic compound.
Que 36. Assertion: Sulphur compounds like SF6 and H2SO4 have 12 valence electrons around the S atom.
Reason: All sulphur compounds do not follow the octet rule
Ans 36. (c) The assertion is correct, but the reason is incorrect
Reason: Sulphur forms many compounds in which the octet rule is obeyed. For example, SCl2 has an octet of electrons around it
Que 37. Assertion: The shape of the NH3 molecule is tetrahedral.
Reason: In NH3 nitrogen is sp3 hybridized.
Ans 37. (d) The assertion is false but the reason is true.
NH3 molecule is pyramidal in shape, because out of four electron pairs, three are bonding pairs and one is lone pair
Que 38. Assertion: pi bonds are weaker than V bonds.
Reason: pi bonds are formed by the overlapping of p-p orbitals along their axes.
Ans 38. (a) Both assertion and reason are true and the reason is the correct explanation of assertion.
Reason: pi bonds are formed by the overlapping of p-p orbitals perpendicular to their axis i.e., sidewise overlap
Que 39. Read the following statements and choose the correct sequence of T and F from the given codes. Here T represents true and F represents false statement.
(i) The number of dots in the Lewis symbol represents the number of valence electrons.
(ii) Number of valence electrons helps to calculate the group valence of elements.
(iii) Group valence is given as 8 minus the number of inner-shell electrons.
(a) T T T (b) T F F (c) T T F (d) F F F
Ans 39. (c)
Reason: The group valence of the elements is generally either equal to the number of dots in Lewis symbols or 8 minus the number of dots or valence electrons.
Que 40. Read the following statements and choose the correct option. Here T stands for True and F stands for False statement.
(i) The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
(ii) The smaller the charge on the cation, the greater the covalent character of the ionic bond.
(iii) For cations of the same size and charge, the one, with electronic configuration (n – 1)dn ns0, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.
(a) T T T (b) T T F (c) T F T (d) F T T
Ans 40. (c)
Reason: The greater the charge on the cation, the greater the covalent character of the ionic bond
Que 41. Which of the following statements is/are not correct for the combination of atomic orbitals?
(i) The combining atomic orbitals must have the same or nearly the same energy.
(ii) Greater the extent of overlap, the greater will be the electron density between the nuclei of a molecular orbital.
(iii) 2pz orbital of one atom can combine with either of 2px, 2py, or 2pz orbital of other atoms as these orbitals have the same energy.
(a) (i) and (ii) (b) (iii) only
(c) (i) only (d) (ii) and (iii)
Ans 41. (b)
Reason: Atomic orbitals having the same or nearly the same energy will not combine if they do not have the same symmetry. 2pz orbital of one atom cannot combine with 2px or 2py orbital of another atom because of their different symmetries
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 42. Intramolecular hydrogen bond exists in
(a) ortho nitrophenol (b) ethyl alcohol
(c) water (d) diethyl ether
Ans 42. (a)
Que 43. The boiling point of p-nitrophenol is higher than that of o-nitrophenol because
(a) NO2 group at p-position behaves in a different way from that at o-position.
(b) intramolecular hydrogen bonding exists in p-nitrophenol
(c) there is intermolecular hydrogen bonding in p-nitrophenol
(d) p-nitrophenol has a higher molecular weight than o-nitrophenol.
Ans 43. (c)
Reason: The b.p. of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-bonding but in o-nitrophenol it is intramolecular H-bonding.
Que 44. Which one of the following is the correct order of interactions?
(a) Covalent < hydrogen bonding < vander Waals < dipole-dipole
(b) vander Waals < hydrogen bonding < dipole < covalent
(c) vander Waals < dipole-dipole < hydrogen bonding < covalent
(d) Dipole-dipole < vander Waals < hydrogen bonding < covalent.
Ans 44. (b)
Que 45. The strongest hydrogen bond is shown by
(a) water (b) ammonia
(c) hydrogen fluoride (d) hydrogen sulphide
Ans 45. (c)
Reason: H–F shows the strongest H-bonds due to the high electronegativity of the F atom
Que 46. The low density of ice compared to water is due to
(a) induced dipole-induced dipole interactions
(b) dipole-induced dipole interactions
(c) hydrogen bonding interactions
(d) dipole-dipole interactions
Ans 46. (c)
Reason: Ice has many hydrogen bonds which give rise to the cage-like structure of water molecules. This structure possesses a larger volume and thus makes the density of ice low.
Chemical Bonding and Molecular Structure Questions and Answers pdf
Que 47. Complete the following statements. With ————-(A) in bond order, ———-(B) increases and ———–(C) decreases.
(a) A = increase, B = bond length, C = bond enthalpy
(b) A = decrease, B = bond enthalpy, C = bond length
(c) A = increase, B = bond enthalpy, C = bond length
(d) A = increase, B = bond angle, C = bond enthalpy
Ans 47. (c) A = increase, B = bond enthalpy, C = bond length
Que 48. The correct order of dipole moments of HF, H2S, and H2O is
(a) HF ˂ H2S ˂ H2O (b) HF ˂ H2S ˃ H2O
(c) HF ˃ H2S ˃ H2O (d) HF ˃ H2O ˂ H2S
Ans 48. (a)
Reason: The correct order of dipole moments of HF, H2S, and H2O.
Que 49. The most polar bond is
(a) C – F (b) C – O (c) C – Br (d) C – S
Ans 49. (a)
Reason: The difference between the electronegativity of carbon and fluorine is the highest.
Que 50. Which of the following is the electron-deficient molecule?
(a) C2H6 (b) B2H6
(c) SiH4 (d) PH3
Ans 50. (b)
Reason: The compounds in which the octet of the central atom is incomplete are known as electron deficient compounds. Hence B2H6 is an electron-deficient compound.
Que 51. Which of the following compounds does not follow the octet rule for electron distribution?
(a) PCl5 (b) PCl3 (c) H2O (d) PH3
Ans 51. (a)
Reason: PCl5 does not follow the octet rule, it has 10 electrons in its valence shell.
Chemical Bonding and Molecular Structure Questions and Answers pdf
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For further studies click the links below:
MOLECULAR ORBITAL THEORY AND HYDROGEN BONDING
CONCEPT OF HYBRIDISATION WITH EXAMPLES
VALANCE ELECTRON THEORIES AND ORBITAL OVERLAP
RESONANCE DIPOLE MOMENT AND BOND PARAMETERS
PROPERTIES OF IONIC BOND AND OCTET RULE
