CBSE · NCERT · Class 11 Chemistry · Chapter 4

NCERT Solutions: Class 11 Chemistry Chapter 4 - Chemical Bonding and Molecular Structure

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Chapter-wise NCERT intext questions and exercise answers for Chemical Bonding and Molecular Structure, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 36
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1Exercises36 questions
Q.4.1Explain the formation of a chemical bond.v
Solution

Atoms combine because the resulting species has lower energy than the separated atoms. Ionic bonds form by electron transfer followed by electrostatic attraction between ions. Covalent bonds form by sharing electron pairs so that atoms attain stable valence-shell configurations.

Answer:

A chemical bond forms when atoms combine to attain a lower-energy, more stable arrangement, usually by transfer or sharing of valence electrons.

Q.4.2Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.v
Solution

Lewis symbols show valence electrons. Mg is group 2, Na group 1, B group 13, O group 16, N group 15 and Br group 17. Therefore their Lewis symbols contain 2, 1, 3, 6, 5 and 7 dots respectively.

Answer:

Mg has 2 dots, Na has 1, B has 3, O has 6, N has 5 and Br has 7 valence-electron dots.

Q.4.3Write Lewis symbols for the following atoms and ions: S and S2-; Al and Al3+; H and H-v
Solution

Sulphur has 6 valence electrons and gains 2 electrons to form S2- with 8 valence electrons. Aluminium has 3 valence electrons and loses them to form Al3+. Hydrogen has 1 electron and H- has 2 electrons, the helium-like duplet.

Answer:

S has 6 dots and S2- has an octet; Al has 3 dots and Al3+ has no valence dots shown; H has 1 dot and H- has a duplet.

Q.4.5Define octet rule. Write its significance and limitations.v
Solution

The rule is significant because it relates bonding to stable noble-gas configurations and helps draw Lewis structures. Its limitations include electron-deficient molecules such as BeH2, BF3 and AlCl3; odd-electron molecules such as NO and NO2; and hypervalent molecules such as PF5, SF6 and H2SO4 where the central atom has more than eight valence electrons.

Answer:

Octet rule states that atoms tend to attain eight electrons in their valence shell by losing, gaining or sharing electrons. It explains many ionic and covalent bonds but fails for incomplete octets, odd-electron species and expanded octets.

Q.4.6Write the favourable factors for the formation of ionic bond.v
Solution

A metal should lose electrons easily, so low ionization enthalpy helps. A non-metal should accept electrons readily, so a large negative electron gain enthalpy helps. The oppositely charged ions should release substantial energy on forming the crystal lattice, so high lattice enthalpy stabilizes the ionic compound.

Answer:

Low ionization enthalpy of the metal, high negative electron gain enthalpy of the non-metal, and high lattice enthalpy of the ionic solid favour ionic bond formation.

Q.4.7Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3v
Solution

VSEPR arranges electron pairs to minimise repulsion. BeCl2 has two bond pairs and no lone pair on Be, so linear. BCl3 has three bond pairs, so trigonal planar. SiCl4 has four bond pairs, so tetrahedral. AsF5 has five bond pairs, so trigonal bipyramidal. H2S has two bond pairs and two lone pairs on S, so bent. PH3 has three bond pairs and one lone pair on P, so trigonal pyramidal.

Answer:

BeCl2 linear; BCl3 trigonal planar; SiCl4 tetrahedral; AsF5 trigonal bipyramidal; H2S bent; PH3 trigonal pyramidal.

Q.4.8Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.v
Solution

Electron pair repulsions follow lp-lp > lp-bp > bp-bp. NH3 has one lone pair and three bond pairs, giving a pyramidal molecule with bond angle about 107 degrees. H2O has two lone pairs and two bond pairs, so greater lone-pair repulsion reduces the H-O-H angle to about 104.5 degrees.

Answer:

H2O has two lone pairs on oxygen, while NH3 has one lone pair on nitrogen; stronger lone-pair repulsions in water compress its bond angle more.

Q.4.9How do you express the bond strength in terms of bond order ?v
Solution

Higher bond order means more bonding electron pairs or greater net bonding interaction between atoms. Therefore triple bonds are generally stronger than double bonds, which are stronger than single bonds; bond length usually decreases as bond order increases.

Answer:

Bond strength generally increases with bond order.

Q.4.10Define the bond length.v
Solution

At the equilibrium bond distance, attractive and repulsive forces balance and the molecule has minimum potential energy. This internuclear distance is called bond length.

Answer:

Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Q.4.11Explain the important aspects of resonance with reference to the CO3^2- ion.v
Solution

In each canonical form of carbonate, one C-O bond is double and two C-O bonds are single with negative charge on oxygen atoms. Since any oxygen can form the double bond, three equivalent structures are possible. The real ion is not any one structure; it is a hybrid in which all C-O bonds are identical with bond order about 1.33 and the charge is delocalised over the three oxygen atoms.

Answer:

CO3^2- is represented by three equivalent resonance structures; the actual ion is a resonance hybrid with three equal C-O bonds and delocalised negative charge.

Q.4.12H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same.v
Solution

Canonical resonance forms must have the same skeletal arrangement of atoms and differ only in the distribution of electrons. In the two shown structures of H3PO3, hydrogen is attached differently, changing atom connectivity. Therefore they are not resonance forms of the same molecule.

Answer:

No. They cannot be canonical forms because the positions/connectivity of atoms differ; resonance structures must differ only in electron arrangement.

Q.4.14Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.v
Solution

Potassium loses one electron, while sulphur gains two, so two K atoms transfer one electron each to S. Calcium loses two electrons and oxygen gains two. Aluminium loses three electrons and nitrogen gains three. The resulting oppositely charged ions are held by ionic bonds.

Answer:

(a) 2K + S -> K2S, giving 2K+ and S2-. (b) Ca + O -> CaO, giving Ca2+ and O2-. (c) Al + N -> AlN, giving Al3+ and N3-.

Q.4.15Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.v
Solution

In CO2, the C=O dipoles are equal and opposite along a straight line. In H2O, lone pairs on oxygen make the molecule bent; the two O-H dipoles are at an angle and their vector sum is non-zero. This difference in dipole moment is consistent with linear CO2 and bent H2O.

Answer:

CO2 is linear and its two equal bond dipoles cancel, so its dipole moment is zero. H2O is bent, so its O-H bond dipoles do not cancel and it has a net dipole moment.

Q.4.16Write the significance/applications of dipole moment.v
Solution

A non-zero dipole moment indicates a polar molecule. Dipole moment values help distinguish linear and bent or symmetrical and unsymmetrical structures, estimate percentage ionic character of bonds, and compare polarity in related molecules.

Answer:

Dipole moment helps determine molecular polarity, distinguish molecular shapes, compare ionic character and identify bond/molecular symmetry.

Q.4.17Define electronegativity. How does it differ from electron gain enthalpy ?v
Solution

Electronegativity is a relative, dimensionless property of bonded atoms and depends on bonding environment. Electron gain enthalpy is a thermodynamic quantity for isolated gaseous atoms, measured in kJ mol^-1.

Answer:

Electronegativity is the tendency of an atom in a molecule to attract shared electrons. Electron gain enthalpy is the enthalpy change when an isolated gaseous atom gains an electron.

Q.4.18Explain with the help of suitable example polar covalent bond.v
Solution

In HCl, chlorine is more electronegative than hydrogen, so the shared electron pair is displaced toward chlorine. Chlorine becomes partially negative and hydrogen partially positive, making the H-Cl bond polar covalent.

Answer:

A polar covalent bond is a covalent bond in which the shared electron pair is drawn more toward the more electronegative atom, giving partial charges. Example: H-Cl.

Q.4.20The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.v
Solution

Acetic acid contains a methyl group bonded to a carboxyl carbon. The carboxyl carbon forms one double bond to oxygen and one single bond to the hydroxyl oxygen; the hydroxyl oxygen is bonded to hydrogen. Each oxygen has two lone pairs in the Lewis structure.

Answer:

The correct Lewis structure is H3C-C(=O)-O-H, with the carbonyl oxygen double-bonded to carbon and the hydroxyl oxygen single-bonded to carbon and hydrogen.

Q.4.21Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ?v
Solution

Methane has four bond pairs and no lone pair around carbon. VSEPR predicts a tetrahedral arrangement for minimum bond-pair repulsion. Valence bond theory also gives sp3 hybridisation with orbitals directed toward the corners of a tetrahedron. Square planar geometry would not give equivalent maximum-overlap sp3 bonds for carbon.

Answer:

CH4 is not square planar because carbon uses four equivalent sp3 hybrid orbitals directed tetrahedrally, which minimises repulsions and gives four equivalent C-H bonds.

Q.4.22Explain why BeH2 molecule has a zero dipole moment although the Be-H bonds are polar.v
Solution

The Be atom in BeH2 is sp hybridised and the molecule has a linear H-Be-H geometry. Although each Be-H bond is polar, the two bond dipoles act in opposite directions along the same line, giving zero net dipole moment.

Answer:

BeH2 is linear, so the two equal and opposite Be-H bond dipoles cancel.

Q.4.23Which out of NH3 and NF3 has higher dipole moment and why ?v
Solution

Both molecules are trigonal pyramidal. In NH3, the N-H bond dipoles and the lone-pair contribution act in the same general direction, increasing the net dipole. In NF3, the N-F bond dipoles oppose the lone-pair contribution, so the net dipole moment is smaller.

Answer:

NH3 has the higher dipole moment.

Q.4.24What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.v
Solution

In sp hybridisation, one s and one p orbital mix to form two equivalent orbitals oriented 180 degrees apart. In sp2 hybridisation, one s and two p orbitals form three equivalent orbitals in one plane at 120 degrees. In sp3 hybridisation, one s and three p orbitals form four equivalent orbitals directed tetrahedrally at about 109.5 degrees.

Answer:

Hybridisation is mixing of atomic orbitals of comparable energy on the same atom to form equivalent hybrid orbitals. sp is linear, sp2 trigonal planar and sp3 tetrahedral.

Q.4.25Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + Cl- -> AlCl4-v
Solution

In AlCl3, aluminium forms three sigma bonds and is trigonal planar, corresponding to sp2 hybridisation. On accepting a lone pair from Cl-, Al forms four sigma bonds in AlCl4-, which is tetrahedral and uses sp3 hybrid orbitals.

Answer:

Al changes from sp2 in AlCl3 to sp3 in AlCl4-.

Q.4.26Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF3 + NH3 -> F3B.NH3v
Solution

In BF3, boron is trigonal planar and sp2 hybridised. In the adduct F3B.NH3, nitrogen donates a lone pair to boron and boron becomes tetra-coordinate, so B becomes sp3 hybridised. Nitrogen in NH3 already has three bonds and one lone pair, so it remains approximately sp3.

Answer:

B changes from sp2 to sp3; N remains sp3.

Q.4.28What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4v
Solution

In C2H2, H-C≡C-H has two C-H sigma bonds and one C-C sigma bond, plus two pi bonds in the triple bond. In C2H4, four C-H bonds and one C-C bond are sigma bonds, while the C=C double bond has one pi bond.

Answer:

(a) C2H2 has 3 sigma bonds and 2 pi bonds. (b) C2H4 has 5 sigma bonds and 1 pi bond.

Q.4.29Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s.v
Solution

A sigma bond requires head-on overlap along the internuclear axis. With the x-axis as internuclear axis, 2py orbitals are perpendicular to the axis and overlap sideways, forming a pi-type interaction rather than a sigma bond. The other listed overlaps can occur along the internuclear axis.

Answer:

(c) 2py and 2py will not form a sigma bond.

Q.4.30Which hybrid orbitals are used by carbon atoms in the following molecules? CH3-CH3; (b) CH3-CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOHv
Solution

Carbon with four single bonds is sp3 hybridised. Carbon in a C=C double bond or C=O carbonyl is sp2 hybridised. Applying these rules gives the listed hybridisations.

Answer:

CH3-CH3: both C sp3. CH3-CH=CH2: CH3 carbon sp3, double-bond carbons sp2. CH3-CH2-OH: both C sp3. CH3-CHO: CH3 carbon sp3, carbonyl carbon sp2. CH3COOH: methyl carbon sp3, carboxyl carbon sp2.

Q.4.31What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.v
Solution

Bond pairs form covalent bonds and lie between bonded atoms. Lone pairs remain on one atom and influence shape through repulsion. In ammonia, nitrogen has three N-H bond pairs and one lone pair.

Answer:

A bond pair is an electron pair shared between two atoms; a lone pair is a valence electron pair not involved in bonding. Example: the shared pair in H-H is a bond pair, while the lone pair on N in NH3 is a lone pair.

Q.4.32Distinguish between a sigma and a pi bond.v
Solution

Sigma electron density is concentrated along the line joining the nuclei and can arise from s-s, s-p or p-p head-on overlap. Pi electron density lies above and below the internuclear axis from lateral p-p overlap. A double bond has one sigma and one pi bond; a triple bond has one sigma and two pi bonds.

Answer:

A sigma bond is formed by head-on overlap along the internuclear axis; a pi bond is formed by sideways overlap of parallel orbitals. Sigma bonds are generally stronger and allow freer rotation than pi bonds.

Q.4.33Explain the formation of H2 molecule on the basis of valence bond theory.v
Solution

Each hydrogen atom has one electron in a 1s orbital. As two H atoms approach, their 1s orbitals overlap and electron density increases between the two nuclei. If the electron spins are opposite, attraction overcomes repulsion and the system reaches minimum energy at the H-H bond length. This stable overlap forms the H-H sigma bond.

Answer:

H2 forms by overlap of two half-filled hydrogen 1s orbitals containing electrons with opposite spins, producing a sigma bond and lowering the energy.

Q.4.34Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.v
Solution

In LCAO, combining atomic orbitals should be close in energy so neither dominates completely. They must have the same symmetry with respect to the molecular axis so constructive/destructive combination is possible. Greater overlap gives more effective molecular orbital formation.

Answer:

Atomic orbitals must have comparable energies, proper symmetry about the internuclear axis, and appreciable overlap.

Q.4.35Use molecular orbital theory to explain why the Be2 molecule does not exist.v
Solution

Each Be atom has configuration 1s2 2s2, so Be2 has eight electrons. Its MO configuration fills sigma1s, sigma*1s, sigma2s and sigma*2s equally. The number of bonding and antibonding electrons is equal, so bond order = (Nb - Na)/2 = 0. Therefore no net bond forms.

Answer:

Be2 has bond order zero, so it is not stable.

Q.4.36Compare the relative stability of the following species and indicate their magnetic properties; O2, O2+, O2- (superoxide), O2^2- (peroxide)v
Solution

Using MO theory, bond orders are: O2+ = 2.5, O2 = 2, O2- = 1.5 and O2^2- = 1. Higher bond order means greater stability, giving the listed order. O2+ and O2- have one unpaired electron each, O2 has two unpaired electrons, and peroxide O2^2- has all electrons paired.

Answer:

Stability order: O2+ > O2 > O2- > O2^2-. O2+, O2 and O2- are paramagnetic; O2^2- is diamagnetic.

Q.4.37Write the significance of a plus and a minus sign shown in representing the orbitals.v
Solution

Atomic orbitals are wave functions and can have positive or negative phase. When orbitals of the same phase overlap, constructive interference gives bonding molecular orbitals. Opposite phases overlap destructively and give antibonding molecular orbitals.

Answer:

The plus and minus signs indicate the phase/sign of the orbital wave function, not electric charge.

Q.4.38Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?v
Solution

Phosphorus forms five sigma bonds in PCl5 using sp3d hybrid orbitals arranged as a trigonal bipyramid. Three equatorial bonds lie in one plane at 120 degrees, while two axial bonds are perpendicular to that plane. Each axial bond has three 90-degree interactions with equatorial bonds, so it experiences greater repulsion and becomes longer and weaker than equatorial bonds.

Answer:

P in PCl5 is sp3d hybridised with trigonal bipyramidal geometry. Axial P-Cl bonds are longer because they experience greater repulsion from three equatorial bonds.

Q.4.39Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?v
Solution

Hydrogen bonding occurs when H is covalently bonded to F, O or N and becomes strongly partially positive. It can then interact with a lone pair on another electronegative atom. This interaction is directional and stronger than ordinary van der Waals forces.

Answer:

A hydrogen bond is the attractive interaction between H bonded to a highly electronegative atom and another electronegative atom such as F, O or N. It is stronger than van der Waals forces but weaker than covalent bonds.

Q.4.40What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2-.v
Solution

In MO theory, bond order is (Nb - Na)/2. For N2, the net bonding count gives bond order 3. For O2, bond order is (10 - 6)/2 = 2. Removing one electron from antibonding O2 gives O2+ with bond order 2.5. Adding one electron to antibonding O2 gives O2- with bond order 1.5.

Answer:

Bond order = (number of bonding electrons - number of antibonding electrons)/2. N2 = 3, O2 = 2, O2+ = 2.5 and O2- = 1.5.