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Chapter 4 Chemical Bonding and Molecular Structure Chemistry Worksheet for Class 11
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Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure Worksheet Pdf
Question. In PO43–ion, the formal charge on each oxygen atom and P—O bond order respectively are
(a) –0.75, 1.25
(b) –0.75, 1.0
(c) –0.75, 0.6
(d) –3, 1.25
Answer. A
Question. Among LiCl, BeCl2, BCl3and 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
Answer. C
Question. Among the following, which compound will show the highest lattice energy?
(a) KF
(b) NaF
(c) CsF
(d) RbF
Answer. D
Question. Which of the following, set of molecules will have zero dipole moment?
(a) Ammonia, beryllium difluoride, water,1, 4-dichlorobenzene
(b) Boron trifluoride, hydrogen fluoride, carbon dioxide, 1, 3-dichlorobenzene
(c) Nitrogen trifluoride, beryllium difluoride, water, 1, 3-dichlorobenzene
(d) Boron trifluoride, beryllium difluoride, carbon dioxide, 1, 4-dichlorobenzene
Answer. B
Question. Which of the following is the correct order of dipole moment ?
(a) NH3< BF3< NF3 < H2O
(b) BF3 < NF3 < NH3 < H2O
(c) BF3 < NH3 < NF3 < H2O
(d) H2O < NF3 < NH3 < BF3
Answer. D
Question. The species, having bond angles of 120° is
(a) ClF3
(b) NCl3
(c) BCl3
(d) PH3
Answer. B
Question. Consider the molecules CH4, NH3and H2O. Which of the given statements is false?
(a) The H — O — H bond angle in H2O is smaller than the H — N — H bond angle in NH3.
(b) The H — C — H bond angle in CH4is larger than the H — N — H bond angle in NH3.
(c) The H — C — H bond angle in CH4, the H — N — H bond angle in NH3, and the H — O — H bond angle in H2O are all greater than 90°.
(d) The H — O — H bond angle in H2O is larger than the H — C — H bond angle in CH4.
Answer. C
Question. Which of the following molecules has the maximum dipole moment?
(a) CO2
(b) CH4
(c) NH3
(d) NF3
Answer. D
Question. The correct order of increasing bond length of C – H, C – O, C – C and C C is
(a) C – H < C C < C – O < C – C
(b) C – C < C C < C – O < C – H
(c) C – O < C – H < C – C < C C
(d) C – H < C – O < C – C < C C
Answer. C
Question. The correct order of increasing bond angles in the following triatomic species is
(a) NO2+< NO2< NO2–
(b) NO2+ < NO2– < NO2
(c) NO2– < NO2+ < NO2
(d) NO2– < NO2 < NO2+
Answer. A
Question. The correct order of C – O bond length among CO,CO32–, CO2is
(a) CO < CO32– < CO2
(b) CO3 2– < CO2 < CO
(c) CO < CO2 < CO32–
(d) CO2 < CO3 2– < CO
Answer. D
Question. The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH3 (1.5 D) is larger than that of NF3 (0.2 D). This is because
(a) in NH3 the atomic dipole and bond dipole are in the opposite directions whereas in NF3 these are in the same direction
(b) in NH3 as well as in NF3 the atomic dipole and bond dipole are in the same direction
(c) in NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions
(d) in NH3as well as in NF3 the atomic dipole and bond dipole are in opposite directions.
Answer. C
Question. The correct order in which the O – O bond length increases in the following is
(a) O2 < H2O2 < O3
(b) O3 < H2O2 < O2
(c) H2O2 < O2 < O3
(d) O2 < O3 < H2O2
Answer. D
Question. The correct sequence of increasing covalent character is represented by
(a) LiCl < NaCl < BeCl2
(b) BeCl2< LiCl < NaCl
(c) NaCl < LiCl < BeCl2
(d) BeCl2 < NaCl < LiCl
Answer. C
Question. Which of the following would have a permanent dipole moment?
(a) SiF4
(b) SF4
(c) XeF4
(d) BF3
Answer. B
Question. H2O is dipolar, whereas BeF2 is not. It is because
(a) the electronegativity of F is greater than that of O
(b) H2O involves hydrogen bonding whereas BeF2 is a discrete molecule
(c) H2O is linear and BeF2 is angular
(d) H2O is angular and BeF2 is linear.
Answer. A
Question. Which of the following molecules does not possess a permanent dipole moment?
(a) CS2
(b) SO3
(c) H2S
(d) SO2
Answer. D
Question. The table shown below gives the bond dissociation energies (Ediss) for single covalent bonds of carbon (C) atoms with element A, B, C and D. Which element has the smallest atoms?
Bond Ediss(kJ mol–1)
C-A 240
C-B 328
C-C 276
C-D 485
(a) C
(b) D
(c) A
(d) B
Answer. D
Question. Strongest bond is in between
(a) CsF
(b) NaCl
(c) both (a) and (b)
(d) none of the above.
Answer. D
Question. Which of the following bonds will be most polar?
(a) N – Cl
(b) O – F
(c) N – F
(d) N – N
Answer. C
Question. In the structure of ClF3, the number of lone pairs of electrons on central atom ‘Cl’ is
(a) one
(b) two
(c) four
(d) three
Answer. C
Question. Predict the correct order among the following :
(a) bond pair - bond pair > lone pair - bond pair > lone pair - lone pair
(b) lone pair - bond pair > bond pair - bond pair > lone pair - lone pair
(c) lone pair - lone pair > lone pair - bond pair > bond pair - bond pair
(d) lone pair - lone pair > bond pair - bond pair > lone pair - bond pair
Answer. B
Question. Which of the following species contains three bond pairs and one lone pair around the central atom?
(a) H2O
(b) BF3
(c) NH2–
(d) PCl3
Answer. A
Question. Which of the following is not a correct statement?
(a) Multiple bonds are always shorter than corresponding single bonds.
(b) The electron-deficient molecules can act as Lewis acids.
(c) The canonical structures have no real existence.
(d) Every AB5 molecule does in fact have square pyramid structure.
Answer. B
Question. Which of the following is not isostructural with SiCl4?
(a) NH4 +
(b) SCl4
(c) SO4 2–
(d) PO4 3–
Answer. C
Question. In which of the following molecules all the bonds are not equal?
(a) NF3
(b) ClF3
(c) BF3
(d) AlF3
Answer. C
Question.Which of the following molecules has trigonal planar geometry?
(a) BF3
(b) NH3
(c) PCl3
(d) IF3
Answer. C
Question. In a regular octahedral molecule, MX6 the number of X – M – X bonds at 180° is
(a) three
(b) two
(c) six
(d) four.
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. A
Question. In NO3– ion, number of bond pair and lone pair of electrons on nitrogen atom are
(a) 2, 2
(b) 3, 1
(c) 1, 3
(d) 4, 0.
Answer. C
Question. In which of the following bond angle is maximum?
(a) NH3
(b) NH4+
(c) PCl3
(d) SCl2
Answer. D
Question. The BCl3 is a planar molecule whereas NCl3 is pyramidal because
(a) nitrogen atom is smaller than boron atom
(b) BCl3 has no lone pair but NC3l has a lone pair of electrons
(c) B—Cl bond is more polar than N—Cl bond
(d) N—Cl bond is more covalent than B—Cl bond.
Answer. C
Question. In compound X, all the bond angles are exactly 109°28′, X is
(a) chloromethane
(b) carbon tetrachloride
(c) iodoform
(d) chloroform.
Answer. D
Question. Which of the following species contains equal number of s and p-bonds?
(a) (CN)2
(b) (CH)2(CN)2
(c) HCO3–
(d) XeO4,
Answer. D
Question. Which one of the following molecules contains no p bond?
(a) SO2
(b) NO2
(c) CO2
(d) H2O
Answer. D
Question. Which one of the following statements is not correct for sigma- and pi- bonds formed between two carbon atoms?
(a) Sigma-bond is stronger than a pi-bond.
(b) Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively.
(c) Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond.
(d) Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard.
Answer. B
Question. Main axis of a diatomic molecule is z, molecular orbital px and py overlap to form which of the following orbitals?
(a) p molecular orbital
(b) s molecular orbital
(c) d molecular orbital
(d) No bond will form.
Answer. A
Question. Which statement is not correct?
(a) A sigma bond is weaker than a pi bond.
(b) A sigma bond is stronger than a pi bond.
(c) A double bond is stronger than a single bond.
(d) A double bond is shorter than a single bond.
Answer. A
Question. Linear combination of two hybridized orbitals belonging to two atoms and each having one electron leads to the formation of
(a) sigma bond
(b) double bond
(c) co-ordinate covalent bond
(d) pi bond.
Answer. A
Question. Which of the following does not apply to metallic bond?
(a) Overlapping valence orbitals
(b) Mobile valence electrons
(c) Delocalized electrons
(d) Highly directed bonds
Answer. D
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. Which of the following pairs of compounds is isoelectronic and isostructural?
(a) TeI2, XeF2
(b) IBr2–, XeF2
(c) IF3, XeF2
(d) BeCl2, XeF2
Answer. None
Question. The hybridizations of atomic orbitals of nitrogen in NO+2, NO–3 and NH4+ respectively are
(a) sp, sp3 and sp2
(b) sp2, sp3 and sp
(c) sp, sp2 and sp3
(d) sp2, sp and sp3
Answer. C
Question. Which of the following pairs of ions is isoelectronic and isostructural?
(a) CO32–, NO3–
(b) ClO3–, CO32–
(c) SO32–, NO3–
(d) ClO3–, SO32–
Answer. A,D
Question. The correct geometry and hybridization for XeF4 are
(a) octahedral, sp3d2
(b) trigonal bipyramidal, sp3d
(c) planar triangle, sp3d3
(d) square planar, sp3d2.
Answer. A
Question. Among the following, which one is a wrong statement?
(a) PH5 and BiCl5 do not exist.
(b) pp-dp bonds are present in SO2.
(c) SeF4 and CH4 have same shape.
(d) I3 + has bent geometry.
Answer. C
Question. In which of the following pairs, both the species are not isostructural?
(a) Diamond, Silicon carbide
(b) NH3, PH3
(c) XeF4, XeO4
(d) SiCl4, PCl4 +
Answer. C
Question. Maximum bond angle at nitrogen is present in which of the following?
(a) NO2+
(b) NO3–
(c) NO2
(d) NO2–
Answer. A
Question. Which one of the following species has planar triangular shape?
(a) N3
(b) NO3–
(c) NO2–
(d) CO2
Answer. B
Question. XeF2 is isostructural with
(a) SbCl3
(b) BaCl2
(c) TeF2
(d) ICl2–
Answer. D
Question. Which of the following is a polar molecule?
(a) SiF4
(b) XeF4
(c) BF3
(d) SF4
Answer. D
Question. The outer orbitals of C in ethene molecule can be considered to be hybridized to give three equivalent sp2 orbitals. The total number of sigma (s) and pi (p) bonds in ethene molecule is
(a) 3 sigma (s) and 2 pi (p) bonds
(b) 4 sigma (s) and 1 pi (p) bonds
(c) 5 sigma (s) and 1 pi (p) bonds
(d) 1 sigma (s) and 2 pi (p) bonds.
Answer. C
Question. Identify a molecule which does not exist.
(a) He2
(b) Li2
(c) C2
(d) O2
Answer. A
Question. Which of the following diatomic molecular species has only p bonds according to Molecular Orbital Theory?
(a) Be2
(b) O2
(c) N2
(d) C2
Answer. D
Question. Which of the following is paramagnetic?
(a) N2
(b) H2
(c) Li2
(d) O2
Answer. D
Question. Decreasing order of stability of O2, O2–, O2+ andO22– is
(a) O22– > O2– > O2 > O2+
(b) O2 > O2+ > O22– > O2–
(c) O2– > O22– > O2+ > O2
(d) O2+ > O2 > O2– > O22–
Answer. D
Question. The correct bond order in the following species is
(a) O2+ < O2– < O22+
(b) O2– < O2+ < O22+
(c) O22+ < O2+ < O2–
(d) O22+ < O2– < O2+
Answer. B
Question. Bond order of 1.5 is shown by
(a) O2+
(b) O2–
(c) O22–
(d) O2
Answer. B
Question. Which of the following has the minimum bond length?
(a) O2+
(b) O2–
(c) O22–
(d) O2
Answer. A
Question. The pairs of species of oxygen and their magnetic behaviour are noted below. Which of the following presents the correct description?
(a) O2–, O22– - Both diamagnetic
(b) O+, O22– - Both paramagnetic
(c) O2+, O2 - Both paramagnetic
(d) O, O22– - Both paramagnetic
Answer. C
Question. Which one of the following species does not exist under normal conditions?
(a) Be2+
(b) Be2
(c) B2
(d) Li2
Answer. B
Question. According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order?
(a) N22– < N2– < N2
(b) N2 < N22– < N2–
(c) N2– < N22– < N2
(d) N2 – < N2 < N2 2–
Answer. A
Question. Right order of dissociation energy N2 and N2+ is
(a) N2> N2+
(b) N2= N2+
(c) N2+ > N2
(d) none.
Answer. A
Question. N2 and O2 are converted into monocations, N2+ andO2+ respectively. Which is wrong?
(a) In O+2 paramagnetism decreases.
(b) N2+ becomes diamagnetic.
(c) In N2+, the N–N bond weakens.
(d) In O2+, the O–O bond order increases.
Answer. B
Question. N2 and O2 are converted into monoanions N2– andO2– respectively, which of the following statements iswrong?
(a) In O2–, bond length increases.
(b) N2– becomes diamagnetic.
(c) In N2–, N–N bond weakens.
(d) In O2–, the O–O bond order decreases.
Answer. B
Question. The ground state electronic configuration of valence shell electrons in nitrogen molecule (N2) is written as KK, s2s2, s*2s2, p2px2 = p2py 2s2pz2. Hence the bond order in nitrogen molecule is
(a) 2
(b) 3
(c) 0
(d) 1
Answer. B
Question. Which of the following molecules has the highest bond order?
(a) O2–
(b) O2
(c) O2+
(d) O22–
Answer. C
Question. Which one of the following compounds shows the presence of intramolecular hydrogen bond?
(a) H2O2
(b) HCN
(c) Cellulose
(d) Concentrated acetic acid
Answer. C
Question. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?
(a) Dipole-dipole interaction
(b) Covalent bonds
(c) London dispersion force
(d) Hydrogen bonding
Answer. D
Question. In X – H Y, X and Y both are electronegative elements. Then
(a) electron density on X will increase and on H will decrease
(b) in both electron density will increase
(c) in both electron density will decrease
(d) on X electron density will decrease and on H increases.
Answer. A
Question. Strongest hydrogen bond is shown by
(a) water
(b) ammonia
(c) hydrogen fluoride
(d) hydrogen sulphide.
Answer. C
Question. Which one shows maximum hydrogen bonding?
(a) H2O
(b) H2Se
(c) H2S
(d) HF
Answer. A
Short Answer
Question. What is octet rule?
Answer. According to the octet rule, the atoms can combine either by the transfer of the valence electrons from one atom to another by gaining or losing or by sharing of the valence electrons in order to have an octet in their valence shells.
Question. What is the significance of Lewis symbols?
Answer. The number of dots around the symbol represents the number of the valence electrons. A Lewis Symbol is constructed by placing dots representing electrons in the outer energy around the symbol for the element. The Lewis symbols and Lewis structures help visualize the valence electrons of atoms and molecules, whether they exist as lone pairs or within bonds.
Question. What do you mean by the term covalent bond?
Answer. Covalent bonds are chemical bonds between two non-metal atoms. An example is water, where hydrogen and oxygen bond together to make. Covalent bonds are formed by atoms sharing valence electrons. When two atoms share one electron pair they are said to be joined by a single covalent bonds. If two atoms share two pairs of the electrons the covalent bond between them is called a double bond.
Question. What are the conditions on which the formations of the ionic compounds are depend?
Answer. The conditions on which the formations of the ionic compounds are depend on:
1. The ease of formation of the positive and negative ions from the respective neutral atoms.
2. The arrangement of the positive and negative ions in the solid i.e., the lattice of the crystalline compound.
Question. What is the formal charge?
Answer. The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of the valance electrons of that atom in an isolated or free state and the number of the electrons assigned to that atom. When multiple Lewis structures can represent the same compound, the different Lewis formulas are called resonance structures.
Long Answer
Question. Explain the steps for the Lewis dot structure?
Answer. The steps for the Lewis dot structure are:
1. Total number of the electrons required for the writing the structures are obtained by the adding the valence electrons of the combining atoms.
2. For atoms each negative charge would mean addition of one electron, for cations each positive charge would result in subtraction of one electron from the total number of the valence electrons.
3. Knowing the chemical symbols of the combining atoms and having knowledge of the skeletal structure of the compound.
4. It is easy to distribute the total number of the electrons as bonding shared pairs between the atoms in proportion to the total bonds.
5. The least electronegative atom occupies the central position in the molecule or ion.
6. After the accounting for the shared pairs of the electrons for the single bonds, the remaining electron pairs are either utilized for the multiple bonding or remain as the lone pairs.
Question. What are the important Kossel factors in relation to chemical bonding?
Answer. The important Kossel factors in relation to chemical bonding are:
1. In the periodic table the highly electronegative halogens and the highly electropositive alkali metals are separated by the noble gases.
2. The formation of a negative ion from a halogen atom and a positive ion from an alkali metal atom is associated with the gain and loss of an electron by the respective atoms.
3. The negative and positive ions so the formed attain the stable noble gas electronic configurations.
4. The noble gases have a particularly stable outer shell configuration of eight i.e., octet electrons.
5. The negative and positive ions are stabilized by electrostatic attraction.
Question. State the difference between single covalent bond and double covalent bond?
Answer. 1. Single covalent bond is a form of covalent bond that forms between two atoms which can share one electron pair between them whereas double covalent bond is a form of covalent bond that forms between two atoms which can share two electrons pairs between them.
2. Single covalent bond has one sigma bond whereas double covalent bond has one sigma bond and one pi bond.
3. Single covalent bond has one electron pair whereas double covalent bond has two electrons pairs.
4. Single covalent bond is longer than the double bond whereas double covalent bond is shorter than the single bond.
5. Single covalent bond disassociation energy is relatively low whereas double covalent bond disassociation energy is relatively high.
Question. Explain the limitations of the octet rule?
Answer. Limitations of the octet rule:
1. The incomplete octet of the central atom: An incomplete octet occurs when the central element in a Lewis Dot Structure is surrounded by less than 8 valence electrons. When this occurs, the atom of the element within the molecule is said to contain an incomplete octet, for example LiCl, Bcl3 and BeH3.
2. Odd-electron molecules: There are a number of molecules whose total number of valence electrons is an odd number. It is not possible for all of the atoms in such a molecule to satisfy the octet rule, for example NO2 nitrogen di oxide.
3. The expanded octet: Expanded octet refers to the Lewis structures where the central atom ends up with more than an octet, for example Phosphorus pentachloride (PCl5) and sulphur hexafluoride (SF6).
4. This theory does not account for the shape of the molecules.
5. It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.
Question. Write short note on electrovalent bond?
Answer.
An electrovalent bond is formed when a metal atom transfers one or more electrons to a non-metal atom. Electrovalent bond is also known as ionic bond. The atom that loses the electrons becomes a positively charged ion or cation, while the one that gains them becomes a negatively charged ion or anion. Electrovalent bonds compounds are generally soluble in water and insoluble in solvents such as kerosene, petrol, etc. They have high melting and boiling points. This is because a considerable amount of energy is required to break the strong inter-ionic attraction between the molecules, for example, the bond between the sodium and chlorine atoms in sodium chloride (NaCl) is formed by the transfer of an electron from sodium to chlorine, creating Na + and Cl –ions. The electrostatic attraction between these ions provides the bonding in NaCl. Electrovalent compounds are formed due to the reaction between highly electropositive and highly electronegative atoms.
Due to the presence of a strong force of attraction between cations and anions in ionic bonded molecules, the following properties are:
1. The electrovalent bonds are the strongest of all the bonds.
2. The electrovalent bonds have a charge separation. So, they are the most reactive of all the bonds in the proper medium.
3. The electrovalent bonded molecules have high melting and boiling point.
4. The electrovalent bonded molecules in their aqueous solutions or in the molten state are good conductors of electricity. This is due to the presence of ions which acts as charge carriers.
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Chapter 4 Chemical Bonding and Molecular Structure CBSE Class 11 Chemistry Worksheet
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