Please refer to the MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure with Answers. The following Chemical Bonding and Molecular Structure Class 11 Chemistry MCQ Questions have been designed based on the latest syllabus and examination pattern for Class 11. Our experts have designed MCQ Questions for Class 11 Chemistry with Answers for all chapters in your NCERT Class 11 Chemistry book.
Chemical Bonding and Molecular Structure Class 11 MCQ Questions with Answers
See below Chemical Bonding and Molecular Structure Class 11 Chemistry MCQ Questions, solve the questions and compare your answers with the solutions provided below.
Question. According to VSEPR theory the geometry of a covalent molecules 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
Answer
D
Question. The geometry of ClO3– ion according to Valence Shell ElectronPair Repulsion (VSEPR) theory will be
(a) planar triangular
(b) pyramidal
(c) tetrahedral
(d) square planar
Answer
B
Question. 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
Answer
C
Question. 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
Answer
A
Question. The covalent bond length is the shortest in which one of the following bonds?
(a) C — O
(b) C — C
(c) C ≡ N
(d) O — H
Answer
D
Question. Hydrogen chloride molecule contains
(a) polar covalent bond
(b) double bond
(c) co-ordinate bond
(d) electrovalent bond
Answer
A
Question. The number of lone pair and bond pair of electrons on the sulphur atom in sulphur dioxide molecule are respectively
(a) 1 and 3
(b) 4 and 1
(c) 3 and 1
(d) 1 and 4
Answer
D
Question. Among the following molecules : SO2, SF4, CIF3, BrF5 and XeF4, which of the following shapes does not describe any of the molecules mentioned?
(a) Bent
(b) Trigonal bipyramidal
(c) See-saw
(d) T-shape
Answer
B
Question. A σ-bonded molecule MX3 is T-shaped. The number of non-bonding pairs of electron is
(a) 0
(b) 2
(c) 1
(d) can be predicted only if atomic number of M is known.
Answer
B
Question. Shape of methane molecule is
(a) tetrahedral
(b) pyramidal
(c) octahedral
(d) square planar
Answer
A
Question. The shape of stannous chloride molecule is
(a) see-saw
(b) square planar
(c) trigonal pyramidal
(d) bent
Answer
D
Question. Which of the following statements is false ?
(a) H2 molecule has one sigma bond
(b) HCl molecule has one sigma bond
(c) Water molecule has two sigma bonds and two lone pairs
(d) Acetylene molecule has three pi bonds and three sigma bonds
Answer
D
Question. The number of sigma (σ) and pi (π) bonds present in 1,3,5,7 octatetraene respectively are
(a) 14 and 3
(b) 17 and 4
(c) 16 and 5
(d) 15 and 4
Answer
B
Question. 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
Answer
A
Question. Using VSEPR theory, predict the species which has square pyramidal shape
(a) SnCl2
(b) CCl4
(c) SO3
(d) BrF5
Answer
D
Question. Allyl cyanide molecule contains
(a) 9 sigma bonds, 4 pi bonds and no lone pair
(b) 9 sigma bonds, 3 pi bonds and one lone pair
(c) 8 sigma bonds, 5 pi bonds and one lone pair
(d) 8 sigma bonds, 3 pi bonds and two lone pairs
Answer
B
Question. The molecule not having π-bond is
(a) Cl2
(b) O2
(c) N2
(d) CO2
Answer
A
Question. The enolic form of a acetone contains
(a) 9 sigma bonds, 1 pi bond and 2 lone pairs
(b) 8 sigma bonds, 2 pi bonds and 2 lone pairs
(c) 10 sigma bonds, 1 pi bond and 1 lone pair
(d) 9 sigma bonds, 2 pi bonds and 1 lone pair
Answer
A
Question. Linear combination of two hybridized orbitals belonging to two atoms and each having one electron leads to a
(a) sigma bond
(b) double bond
(c) co-ordinate covalent bond
(d) pi bond.
Answer
A
Question. Which of the following statements is not correct ?
(a) Double bond is shorter than a single bond
(b) Sigma bond is weaker than a π(pi) bond
(c) Double bond is stronger than a single bond
(d) Covalent bond is stronger than hydrogen bond
Answer
B
Question. As the s-character of hybridised orbital increases, the bond angle
(a) increase
(b) decrease
(c) becomes zero
(d) does not change
Answer
A
Question. In hexa-1, 3-diene-5-yne the number of C —C δ, C — C π and C — H σ bonds, respectively are
(a) 5, 4 and 6
(b) 6, 3 and 5
(c) 5, 3 and 6
(d) 6, 4 and 5
Answer
A
Question. The angle between the overlapping of one s-orbital and one p-orbital is
(a) 180°
(b) 120°
(c) 109°28′
(d) 120° 60′
Answer
A
Question. The nature of hybridisation in the ammonia molecule is
(a) sp2
(b) dp2
(c) sp
(d) sp3
Answer
D
Question. Which of the following will have sp3 d3 hybridisation?
(a) BrF5
(b) PCl5
(c) XeF6
(d) SF6
Answer
C
Question. The shape of CO2 molecule is
(a) linear
(b) tetrahedral
(b) planar
(d) pyramidal
Answer
A
Question. The hybridisation state of carbon in fullerene is
(a) sp
(b) sp2
(c) sp3
(d) sp3d
Answer
B
Question. Which of the following statements is true for an ion having sp3 hybridisation?
(a) all bonds are ionic
(b) H-bonds are situated at the corners of a square
(c) all bonds are co-ordinate covalent
(d) H-atoms are situated at the corners of tetrahedron
Answer
D
Question. The shape of sulphate ion is
(a) square planar
(b) triagonal
(c) trigonal planar
(d) tetrahedral
Answer
D
Question. The strength of bonds formed by s–s and p–p, s–p overlap in the order of
(a) s–p > s–s > p–p
(b) p–p > s–s > s–p
(c) s–s > p–p > s–p
(d) s–s > s–p > p–p
Answer
D
Question. Which of the following molecule does not have a linear arrangement of atoms ?
(a) H2S
(b) C2H2
(c) BeH2
(d) CO2
Answer
A
Question. In which one of the following molecules the central atom said to adopt sp2 hybridization?
(a) BeF2
(b) BF3
(c) C2H2
(d) NH3
Answer
B
Question. The trigonal bipyramidal geometry is obtained from the hybridisation
(a) dsp3 or sp3d
(b) dsp2 or sp2d
(c) d 2sp3 or sp3d2
(d) None of these
Answer
A
Question. A sp3-hybrid orbital contains
(a) 25% s-character
(b) 75% s-character
(c) 50% s-character
(d) 25% p-character
Answer
A
Question. The types of hybridisation of the five carbon atoms from left to right in the molecule CH3 — CH == C == CH — CH3 are
(a) sp3, sp2, sp2, sp2, sp3
(b) sp3, sp, sp2, sp2, sp3
(c) sp3, sp2, sp, sp2, sp3
(d) sp3, sp2, sp2, sp, sp3
Answer
C
Question. Pick out the incorrect statement from the following
(a) sp hybrid orbitals are equivalent and are at an angle of 180° with each other
(b) sp2 hybrid orbitals are equivalent and bond angle between any two of them is 120°
(c) sp3d2 hybrid orbitals are equivalent and are oriented towards corners of a regular octahedron
(d) sp3d3 hybrid orbitals are not equivalent
Answer
D
Question. All carbon atoms are sp2 hybridised in
(a) 1, 3-butadiene
(b) CH2 = C = CH2
(c) cyclohexane
(d) 2-butene
Answer
A
Question. Which one of the following is not correct in respect of hybridization of orbitals?
(a) The orbitals present in the valence shell only are hybridized
(b) The orbitals undergoing hybridization have almost equal energy
(c) Promotion of electron is not essential condition for hybridization
(d) Pure atomic orbitals are more effective in forming stable bonds than hybrid orbitals
Answer
D
Question. Molecular orbital theory was given by
(a) Kossel
(b) Mosley
(c) Mulliken
(d) Werner
Answer
C
Question. With increasing bond order, stability of bond
(a) Remain unaltered
(b) Decreases
(c) Increases
(d) None of these
Answer
C
Question. Which of the following corresponds unstable molecule? Here Nb is number of bonding electrons and Na is number of antibonding electrons.
(a) Nb > Na
(b) Nb < Na
(c) Na = Nb
(d) Both (b) and (c)
Answer
D
Question. If Nx is the number of bonding orbitals of an atom and Ny is the number of antibonding orbitals, then the molecule/atom will be stable if
(a) Nx > Ny
(b) Nx = Ny
(c) Nx < Ny
(d) Nx ≤ Ny
Answer
A
Question. In the molecular orbital diagram for O2 + ion, the highest occupied orbital is
(a) σ MO orbital
(b) π MO orbital
(c) π* MO orbital
(d) σ* MO orbital
Answer
C
Question. The theory capable of explaining paramagnetic behaviour of oxygen is
(a) resonance theory
(b) V.S.E.P.R. theory
(c) molecular orbital theory
(d) valence bond energy
Answer
C
Question. In an anti-bonding molecular orbital, electron density is minimum
(a) around one atom of the molecule
(b) between the two nuclei of the molecule
(c) at the region away from the nuclei of the molecule
(d) at no place
Answer
B
Question. When two atomic orbitals combine, they form
(a) one molecular orbital
(b) two molecular orbital
(c) three molecular orbital
(d) four molecular orbital
Answer
B
Question. Paramagnetism is exhibited by molecules
(a) not attracted into a magnetic field
(b) containing only paired electrons
(c) carrying a positive charge
(d) containing unpaired electrons
Answer
D
Question. The difference in energy between the molecular orbital formed and the combining atomic orbitals is called
(a) bond energy
(b) activation energy
(c) stabilization energy
(d) destabilization energy
Answer
C
Question. The bond order in N2+ is
(a) 1.5
(b) 3.0
(c) 2.5
(d) 2.0
Answer
C
Question. Which molecule has the highest bond order?
(a) N2
(b) Li2
(c) He2
(d) O2
Answer
A
Question. Which of the following order of bond angles is correct?
(a) BF3 < NF3 < PF3 < ClF3
(b) BF3 < NF3 < PF3 > ClF3
(c) ClF3 < PF3 < NF3 < BF3
(d) BF3 < NF3 < PF3 < ClF3
Answer
C
Question. The percentage of p -character of the hybrid orbitals in graphite and diamond are respecitvely,
(a) 50 and 75
(b) 67 and 75
(c) 33 and 75
(d) 33 and 25
Answer
B
Question. The state of hybridisation of the central atom and the number of lone pairs over the central atom in POCI3 are
(a) sp ,0
(b) sp2, 0
(c) sp3, 0
(d) dsp2, 1
Answer
C
Question. The hybridisation of orbitals ofN atom in NO3– ,No2+ and NH4+arerespectively
(a) sp, sp2, sp3
(b) sp2, sp, sp3
(c) sp, sp3, sp2
(d) sp2, sp3, sp
Answer
B
Question. The structure of IF7 is
(a) square pyramid
(b) trigonal bipyramid
(c) octahedral
(d) pentagonal bipyramid
Answer
D
Question. Which of the following has maximum number oflone pairs associated with Xe ?
(a) XeO3
(b) XeF4
(c) XeF6
(d) XeF2
Answer
D
Question. The number of types of bonds between two carbon atoms in calcium carbide is
(a) one sigma(σ) two pi(π)
(b) one sigma(σ)one pi(π)
(c) two sigma(σ)one pi(π)
(d) two sigma(σ)two pi(π)
Answer
A
Question. Number of non-bonding electron pair on Xe in XeF6 , XeF4 and XeF2 respectively will be
(a) 6,4,2
(b) l,2,3
(c) 3,2, 1
(d) 0,3,2
Answer
B
Question. The number of sigma (σ) and pi (n) covalent bonds respectively in benzene nitrile are
(a) 5, 13
(b) 15, 3
(c) 13, 5
(d) 16, 2
Answer
C
Question. Hybridisation shown by carbon and oxygen of—OH group in phenol are respectively
(a) sp2, sp2
(b) sp3, sp3
(c) sp, sp2
(d) sp2, sp3
Answer
D
Question. Based on VSEPR theory, the number of 90 degree F—Br—F angles in BrF5 is
(a) 0
(b) 1
(c) 2
(d) 3
Answer
A
Question. The species having pyramidal shape is
(a) SO3
(b) BrF3
(c) SiO32–
(d) OSF2
Answer
D
Question. Which one of the following conversions involve change in both hybridisation and shape?
(a) CH4 → C2H6
(b) NH3 → NH4+
(c) BF3 → BF4
(d) H2O → H3O+
Answer
C
Question. Which of the following has sp2 hybridisation?
(a) C2H6
(b) C2H4
(c) BeCl2
(d) C2H2
Answer
B
Question. The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are
(a) dx2 -y2, dz2, s, Px, Py
(b) dx2, s, Px, Pz
(c) S,Px,Py ,Pz,dz2
(d) dx2 -y2, S, px, py
Answer
C
Question. The enolic fom1 of acetone contains
(a) 9 sigma bonds, 1 pi bond and 2 lone pairs
(b) 8 sigma bonds, 2 pi bond and 2 lone pairs
(c) 10 sigma bonds, l pi bond and l Ione pair
(d) 9 sigma bonds, 2 pi bond and l lone pair
Answer
A
Question. A a-bonded molecule MX3 is T-shaped. The number non-bonding pairs of electron is
(a) 0
(b) 2
(c) 1
(d) can be predicted only if atomic number of Mis known
Answer
B
Question. In which of the following species, all the 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–
Answer
C
Question. In XeF6, oxidation state and state of hybridisation of Xe and shape of the molecule are, respectively
(a) + 6, sp3d3, distorted octahedral
(b) +4, sp3d2, square planar
(c) + 6, sp3, pyramidal
(d) + 6, sp3d2, square pyramidal
Answer
A
Question. Which of the following species is non-linear?
(a) ICI2–
(b) I3–
(c) N3–
(d) CIO2–
Answer
D
Question. The state of hybridisation ofS in SF4 is
(a) sp3 and has a lone pair of electron
(b) sp2 and has tetrahedral structure
(c) sp3 d and has a trigonal bipyramidal structure
(d) sp3d2 and has an octahedral structure
Answer
C
Question. The bond angle and percentage of d-character in SF6 are
(a) 120°, 20%
(b) 90°, 33%
(c) 109°, 25%
(d) 90°, 25%
Answer
B
Question. The compound in which underlined carbon uses only its sp3 hybrid orbitals for bond formation is
(a) CH3COOH
(b) CH3CONH2
(c) CH3CH2OH
(d) CH2CH=CH2
Answer
C
Question. In which of the following pair both molecules do not possess same type ofhybtidisation?
(a) CH4 and H2O
(c) SF6 and XeF4
(b) PCI5 and SF4
(d) BCl3 and NCl3
Answer
D
Question. I. H—C—H angle in CH4
II. CI—B—CI angle in BCl3
III. F—I—F angle in IF7 in a plane
IV. I—I—I angle in I3–
Increasing order of above bond angles is
(a) I < II < III < IV
(c) III < I < II < IV
(b) II < I < III < IV
(d) IV < II < I < III
Answer
C
