NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry

Question 1.1:

Explain the formation of a chemical bond.

Solution:

A chemical bond is defined as the attractive force that holds constituent atoms, ions, or molecules together in a chemical species.

Driving Forces for Chemical Bond Formation:

• Tendency to attain stability (Octet Rule): Atoms combine to achieve a noble gas electronic configuration with 8 valence electrons.
• Tendency to lower potential energy: As two atoms approach each other, attractive forces between their nuclei and electrons operate alongside repulsive forces. When attractive forces dominate, potential energy decreases, resulting in a stable chemical bond.

Question 1.2:

Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N and Br.

Solution:

Lewis dot symbols show valence shell electrons around the element symbol:

Question 1.3:

Write Lewis symbols for S and S²⁻, Al and Al³⁺, H and H⁻.

Solution:

Question 1.4:

Draw the Lewis structures of H₂S, SiCl₄, BeF₂, CO₃²⁻ and HCOOH.

Solution:

 

Question 1.5:

Define the octet rule. Write its significance and limitations.

Solution:

Octet Rule: During the formation of a chemical bond, atoms tend to gain, lose, or share electrons to attain 8 valence electrons (a stable octet).

Significance: It successfully explains the structure, bonding, and chemical behavior of most organic compounds and main group elements.

Limitations:

• Incomplete Octet: Central atoms in LiCl, BeH₂, BCl₃ have fewer than 8 valence electrons.
• Odd-electron Molecules: Species like NO and NO₂ have an odd number of electrons, so octet rule is violated for all atoms.
• Expanded Octet: Elements from period 3 onwards (PF₅, SF₆, H₂SO₄) have more than 8 valence electrons around the central atom.
• Other flaws: It assumes noble gas compounds do not form, yet Xe and Kr form compounds like XeF₂, XeF₄. It also fails to predict molecular geometry and bond energy.

Question 1.6:

Write the favourable factors for the formation of an ionic bond.

Solution:

The favourable conditions for ionic bond formation are:

• Low Ionization Enthalpy of the metal to easily lose electron(s) and form a cation.
• High Negative Electron Gain Enthalpy of the non-metal to readily accept electron(s) and form an anion.
• High Lattice Enthalpy of the resulting ionic compound to stabilize the crystal structure.

Question 1.7:

Discuss the shapes of the molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S and PH₃.

Solution:

Question 1.8:

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

Solution:

Both NH₃ and H₂O undergo sp³ hybridisation with tetrahedral electron-pair geometry:

• In NH₃: Nitrogen has 3 bond pairs and 1 lone pair. Lone pair-bond pair repulsion reduces the bond angle to 107°.
• In H₂O: Oxygen has 2 bond pairs and 2 lone pairs. According to VSEPR theory, repulsive forces follow: Lone Pair–Lone Pair > Lone Pair–Bond Pair > Bond Pair–Bond Pair.

The presence of two lone pairs in water produces greater repulsion on bond pairs than a single lone pair in ammonia, compressing the H-O-H bond angle further down to 104.5°.

Question 1.9:

How do you express the bond strength in terms of bond order?

Solution:

Bond strength is directly proportional to bond order. A higher bond order signifies a greater number of shared electron pairs between two atoms, resulting in stronger attraction, higher bond dissociation enthalpy, and shorter bond length.

Example: Bond order of N₂ is 3 (Bond Enthalpy = 946 kJ/mol) whereas O₂ is 2 (Bond Enthalpy = 498 kJ/mol). Thus, N₂ bond strength > O₂ bond strength.

Question 1.10:

Define bond length.

Solution:

Bond Length:

Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is commonly expressed in picometres (pm) or angstroms (Å), where 1 Å = 10⁻¹⁰ m and 1 pm = 10⁻¹² m.

For an ionic compound, the internuclear distance is approximately the sum of the ionic radii: d = r⁺ + r⁻. For a covalent bond between atoms A and B, it is approximately the sum of their covalent radii: d = rA + rB.

Question 1.11:

Explain the important aspects of resonance with reference to the CO₃²⁻ ion.

Solution:

Resonance applies to species that cannot be accurately described by a single Lewis structure:

• All three C–O bonds are experimentally identical in length (129 pm), intermediate between a single bond (143 pm) and a double bond (122 pm).
• The actual structure is a resonance hybrid that lowers overall energy and stabilizes the ion by evenly delocalizing the -2 charge (-2/3 per Oxygen).

Question 1.12:

H₃PO₃ can be represented by structures 1 and 2 shown below. Can the two commonly shown structures of H₃PO₃ be treated as resonance forms? Give a reason.

Solution:

No, these two structures cannot be treated as resonance forms.

Reason: Canonical (resonance) structures must have identical positions of atomic nuclei and differ only in the arrangement of electrons. In Structure 1, Hydrogen is bonded directly to Phosphorus (P–H), whereas in Structure 2, that Hydrogen is attached to Oxygen (O–H). Since atomic positions change, they are structural tautomers, not resonance structures.

Question 1.13:

Write the resonance structures for SO₃, NO₂ and NO₃⁻.

Solution:

Question 1.14:

Use 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.

Solution:Question 1.15:

Although 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.

Solution:

• CO₂ (μ = 0 D): CO₂ has zero dipole moment. The two polar C=O bond dipoles are equal in magnitude and point in opposite directions (180°), cancelling out vectorially. Hence, CO₂ is **linear** (O = C = O).
• H₂O (μ = 1.85 D): H₂O has a net dipole moment because Oxygen contains 2 lone pairs. The O–H bond dipoles add up vectorially, giving a **bent / V-shaped** structure.

Question 1.16:

Write the significance / applications of dipole moment.

Solution:

• Distinguishing polar and non-polar molecules: μ = 0 indicates non-polar molecules (e.g., CCl₄, CO₂), while μ > 0 indicates polar molecules (e.g., HCl, H₂O).
• Determining molecular geometry: Helps predict if a triatomic molecule is linear (μ = 0) or bent (μ > 0).
• Calculating percentage ionic character: % Ionic character = (μ_experimental / μ_theoretical) × 100.
• Differentiating cis and trans isomers: Cis-isomers typically possess higher dipole moments than symmetrical trans-isomers (often μ = 0).

Question 1.17:

Define electronegativity. How does it differ from electron gain enthalpy?

Solution:

Electronegativity: The relative tendency of a bonded atom to attract shared electron pairs towards itself.

Question 1.18:

Explain with the help of suitable example polar covalent bond.

Solution:

A polar covalent bond is a covalent bond in which the shared pair of electrons is unequally shared between two atoms due to a difference in their electronegativity.

As a result:

• The more electronegative atom acquires a partial negative charge (δ⁻).
• The less electronegative atom acquires a partial positive charge (δ⁺).
• This creates a dipole in the molecule.

Example: In H^δ⁺–Cl^δ⁻, Chlorine is more electronegative than Hydrogen. The shared electron pair shifts towards Cl, creating a permanent dipole moment.

Question 1.19:

Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.

Solution:

Ionic character increases with electronegativity difference (ΔEN):

N₂ < ClF₃ < SO₂ < LiF < K₂O

Question 1.20:

The 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.

Solution:

In acetic acid (CH₃COOH), the carbonyl Carbon is double-bonded to Oxygen and single-bonded to the hydroxyl Oxygen atom:

Correct Lewis Structure:

Question 1.21:

Apart 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?

Solution:

Electronic configuration of carbon atom:

₆C: 1s²2s²2p²

In the excited state, the orbital diagram of carbon can be represented as:

Hence, carbon atom undergoes sp³-hybridisation in CH₄ molecule and has a tetrahedral shape.

For a square planar shape, the hybridisation of the central atom should be dsp² hybridised. But C atom do not have energetically accessible d-orbitals, it cannot undergo dsp²-hybridisation. So, the structure of CH₄ cannot be square planar.

For square planer geometry the bond angle between the atoms would be 90°, the stability of CH₄ will be very less because of the repulsion between the bond pairs. VSEPR theory also supports a tetrahedral structure for CH₄.

Question 1.22:

Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.

Solution:

The Lewis structure for BeH₂ is as follows:

The central atom (Be) has no lone pair of electrons and there are two bond pairs. Hence, BeH₂ is of the type AB₂. It has a linear structure.

Dipole moments of each H–Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH2 molecule has zero dipole moment.

Question 1.23:

Which out of NH₃ and NF₃ has higher dipole moment and why?

Solution:

NH₃ has a higher dipole moment (1.47 D) than NF₃ (0.24 D).

Reason:

In NH₃, N is more electronegative than H; the lone pair orbital dipole and the three N–H bond dipoles act in the **same direction**, reinforcing each other. In NF₃, F is more electronegative than N; the three N–F bond dipoles point away from N, **opposing** the lone pair orbital dipole and reducing net dipole moment.

Question 1.24:

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

Solution:

Hybridisation is the process of intermixing of atomic orbitals of slightly different energies of the same atom to form an equal number of new equivalent orbitals, called hybrid orbitals.

Characteristics of Hybrid Orbitals

• They have equal energy (degenerate).
• They have identical shape and size.
• They are more directional than pure atomic orbitals.
• They are oriented in space to minimize electron pair repulsion.
• The number of hybrid orbitals formed is equal to the number of atomic orbitals mixed.

• sp: 1 s + 1 p → 2 sp orbitals. Shape: Linear (180°).
• sp²: 1 s + 2 p → 3 sp² orbitals. Shape: Trigonal Planar (120°).
• sp³: 1 s + 3 p → 4 sp³ orbitals. Shape: Tetrahedral (109.5°).
  

Question 1.25:

Describe the change in hybridisation (if any) of the Al atom in the reaction: AlCl₃ + Cl⁻ → AlCl₄⁻

Solution:

Electronic configuration of 13Al = 1s2 2s2 2p6 3s1 3px13py1
(excited state)

Hence, hybridisation will be SP2

In AlCl–4, the empty 3pz orbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral.

Question 1.26:

Is there any change in the hybridisation of B and N atoms as a result of the reaction: BF₃ + NH₃ → F₃B·NH₃?

Solution:

• Boron (B): Changes from sp² in BF₃ to sp³ in F₃B·NH₃ adduct.
• Nitrogen (N): Is sp³ in NH₃ and remains sp³ in the adduct after donating its lone pair.

Question 1.27:

Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.

Solution:

Question 1.28:

What is the total number of sigma and pi bonds in the following molecules? (a) C₂H₂ (b) C₂H₄

Solution:

• (a) C₂H₂ (H–C≡C–H): 3 σ bonds (2 C–H, 1 C–C) and 2 π bonds.
• (b) C₂H₄ (H₂C=CH₂): 5 σ bonds (4 C–H, 1 C–C) and 1 π bond.

Question 1.29:

Considering 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 2p_x (c) 2p_y and 2p_y (d) 1s and 2s.

Solution:

(c) 2p_y and 2p_y will NOT form a sigma bond.

Reason: Since the x-axis is the internuclear axis, 2p_y orbitals lie perpendicular to the axis and undergo lateral (sideways) overlap to form a pi (π) bond instead.

Question 1.30:

Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃–CH₃   (b) CH₃–CH=CH₂   (c) CH₃-CH₂-OH   (d) CH₃-CHO   (e) CH₃COOH

Solution:

• (a) CH₃–CH₃: C1 = sp³, C2 = sp³

• (b) CH₃–CH=CH₂: C1 = sp³, C2 = sp², C3 = sp²

• (c) CH₃-CH₂-OH: C1 = sp³, C2 = sp³

• (d) CH₃-CHO: C1 = sp³, C2 = sp²

• (e) CH₃COOH: C1 = sp³, C2 = sp²

Question 1.31:

What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Solution:

• Bond Pair: A pair of valence electrons shared between two atoms to form a covalent bond. Example: In H₂O, there are 2 H–O bond pairs.
• Lone Pair: A valence electron pair that is not shared with another atom. Example: In H₂O, Oxygen carries 2 lone pairs.

Question 1.32:

Distinguish between sigma and pi bonds.

Solution:

Question 1.33:

Explain the formation of H₂ molecule on the basis of valence bond theory.

Solution:

Let us assume that two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (eA and eB) are taken to undergo a reaction to form a hydrogen molecule.

When A and B are at a large distance, there is no interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating.

Attractive force arises between:

(a) Nucleus of one atom and its own electron i.e., NA – eA and NB – eB.

(b) Nucleus of one atom and electron of another atom i.e., NA – eB and NB – eA.

Repulsive force arises between:

(a) Electrons of two atoms i.e., eA – eB.

(b) Nuclei of two atoms i.e., NA – NB.

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

Question 1.34:

Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Solution:

Conditions for LCAO (Linear Combination of Atomic Orbitals):

• Same or nearly same energy: e.g., 1s combines with 1s, not with 2s.
• Same symmetry about molecular axis: 2p_z overlaps with 2p_z along z-axis.
• Maximum overlap: Atomic orbitals must overlap significantly to produce strong bonding.

Question 1.35:

Use molecular orbital theory to explain why the Be₂ molecule does not exist.

Solution:

Beryllium (Z = 4) has configuration 1s² 2s². Total electrons in Be₂ = 8.

MO Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)²

Bond Order = ½ (N_b – N_a) = ½ (4 – 4) = 0.

Since bond order is zero, Be₂ molecule does not exist.

Question 1.36:

Compare the relative stability of the following species and indicate their magnetic properties: O₂, O₂⁺, O₂⁻ (superoxide), O₂²⁻ (peroxide).

Solution:

Stability increases with higher bond order:
O₂⁺ > O₂ > O₂⁻ > O₂²⁻

Question 1.37:

Write the significance of a plus and a minus sign shown in representing the orbitals.

Solution:

The (+) and (-) signs represent the **phase of the electron wave function ($\Psi$)**, not electrical charges.

• Same signs overlapping (+ with + or – with -): Constructive interference occurs, producing a **bonding molecular orbital**.
• Opposite signs overlapping (+ with -): Destructive interference occurs, producing an **antibonding molecular orbital**.

Question 1.38:

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

Solution:

The ground state and excited state electronic configurations of phosphorus (₁₅P)

Ground state electronic configuration:

Phosphorus atom is sp³d-hybridised in the excited state. Five chlorine atoms share their outer valence electrons with the sp³d –hybrid orbitals of P.

The five sp³d-hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl₅ can be represented as:

In PCl₅ molecule, five P-Cl sigma bonds are present, 3 of them lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.

The remaining two P–Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds. The axial bonds are slightly longer than equatorial bonds as the axial bond pairs suffer more repulsion than the equatorial bond pairs.

Question 1.39:

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

Solution:

Hydrogen Bond: An attractive force that binds a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) to another electronegative atom of the same or another molecule.

A hydrogen bond is stronger than van der Waals forces, though weaker than a covalent bond.