Class 11 Chemistry Chapter 4 complete concise notes, Kössel-Lewis approach, octet rule, formal charge, VSEPR molecular shapes, Valence Bond Theory, Hybridisation ($sp$ to $sp^3d^2$), Molecular Orbital Theory (MOT), Hydrogen bonding, solved in-text problems (4.14.4), NCERT exercises (4.1 to 4.40), revision cheat-sheets, and 3-tier leveled tests.
Introduction: Why Do Atoms Form Chemical Bonds?
Matter is made up of elements. Under normal conditions, no element exists as an independent atom in nature, except the noble gases. All other atoms combine to form molecules.
Chemical Bond: The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species.
Every system in nature tends to attain maximum stability by achieving a state of minimum potential energy. Bonding is nature's mechanism of lowering the net potential energy of the system.
Major Theories of Chemical Bonding
Kössel-Lewis Approach (1916): Octet rule based on the inertness of noble gases.
VSEPR Theory (Sidgwick & Powell 1940; Gillespie & Nyholm 1957): Predicts 3D shapes based on electron-pair repulsions.
Valence Bond (VB) Theory (Heitler-London 1927; Pauling): Explains directional bonding, orbital overlap, and hybridisation.
Molecular Orbital (MO) Theory (Hund & Mulliken 1932): Linear combination of atomic orbitals (LCAO), bond orders, and magnetic properties.
4.1 Kössel-Lewis Approach to Chemical Bonding
In 1916, W. Kössel and G.N. Lewis independently provided the first logical electronic explanation of valence based on the inertness of noble gases.
Lewis's Model: Pictured the atom as a positively charged 'Kernel' (nucleus + inner core electrons) surrounded by an outer valence shell containing up to eight electrons located at the eight corners of a cube.
Lewis Symbols: Notations representing valence electrons as dots around the elemental symbol:
Group Valence: Equals the number of dots (valence electrons) OR $(8 - \text{number of dots})$.
Kössel's Key Postulates:
Highly electronegative halogens and electropositive alkali metals are separated by noble gases.
Formation of negative ions (electron gain) and positive ions (electron loss) leads to stable noble gas configurations ($ns^2np^6$, or $1s^2$ for He).
Oppositely charged ions are held by strong electrostatic (coulombic) attraction — forming an electrovalent (ionic) bond.
Lewis Cubical Atom & Electron Transfer in NaCl
4.1.1 Octet Rule
Electronic Theory of Chemical Bonding: Atoms combine either by transfer of valence electrons (forming ionic bonds) or by sharing of valence electrons (forming covalent bonds) in order to achieve an octet of electrons in their valence shell.
4.1.2 Covalent Bond & Lewis-Langmuir Theory
In 1919, Irving Langmuir abandoned the stationary cubical model and introduced the concept of the covalent bond:
Single Bond: Sharing of 1 electron pair (e.g., $Cl_2$, $H_2O$, $CH_4$, $CCl_4$).
Double Bond: Sharing of 2 electron pairs (e.g., $CO_2$ ($O=C=O$), $C_2H_4$ ($H_2C=CH_2$)).
Triple Bond: Sharing of 3 electron pairs (e.g., $N_2$ ($:N \equiv N:$), $C_2H_2$ ($H-C \equiv C-H$)).
4.1.3 Steps for Writing Lewis Dot Structures
Calculate Total Valence Electrons: Sum valence electrons of all combining atoms. Add $1\,e^-$ for each unit negative charge (anions); subtract $1\,e^-$ for each unit positive charge (cations).
Determine Central Atom: In general, the least electronegative atom occupies the central position (e.g., in $NF_3$ and $CO_3^{2-}$, N and C are central; H and F always occupy terminal positions).
Draw Skeletal Framework: Connect atoms with single bonds (1 shared pair = 2 electrons each).
Complete Terminal Octets: Distribute remaining electrons to complete octets of terminal atoms (or duplet for H).
Form Multiple Bonds if Needed: If the central atom lacks an octet, convert lone pairs on terminal atoms into bonding shared pairs (double or triple bonds).
4.1
Write the Lewis dot structure of the carbon monoxide (CO) molecule.
Step 1: Total valence electrons = C($2s^2 2p^2$) + O($2s^2 2p^4$) = $4 + 6 = 10$ electrons. Step 2: Skeletal structure: C — O (single bond uses 2 electrons; 8 remaining). Step 3: Completing oxygen's octet leaves only 2 electrons as a lone pair on C ($:C - :O:$), leaving C with only 4 electrons. Therefore, two electron pairs from oxygen must be shared as multiple bonds.
Lewis Structure: $:C \equiv O:$ (Triple bond with 1 lone pair on C and 1 lone pair on O).
4.2
Write the Lewis structure of the nitrite ion ($NO_2^-$).
Step 1: Valence electrons = N($5$) + 2×O($2 \times 6 = 12$) + 1 (negative charge) = 18 electrons. Step 2: Skeletal structure: O — N — O (central atom N is less electronegative). Two single bonds use 4 electrons, leaving 14 electrons. Step 3: Completing the octets on terminal oxygen atoms uses 12 electrons; placing the remaining 2 electrons on N leaves it with only 6 electrons. Step 4: Shift one lone pair from one oxygen to make a double bond with N:
Structure: $[:\ddot{O} = \ddot{N} - \ddot{O}:]^-$
4.1.4 Formal Charge
In polyatomic ions, the charge resides on the ion as a whole. However, assigning a formal charge to each individual atom helps track valence electrons and select the lowest-energy Lewis structure (the most stable structure has the smallest formal charges):
$$\text{Formal Charge} = V - L - \frac{1}{2} S$$
Where $V$ = Total valence electrons in free atom, $L$ = Total non-bonding electrons (lone pair electrons), $S$ = Total bonding (shared) electrons.
Formal Charge in Ozone ($O_3$):
Central $O(1)$: $6 - 2 - \frac{1}{2}(6) = \mathbf{+1}$
Noble gases (Xe, Kr) form compounds ($XeF_2, XeF_4, XeOF_2$); rule does not explain molecular shape or thermodynamic stability.
4.2 Ionic or Electrovalent Bond & Lattice Enthalpy
An ionic bond is formed by the complete transfer of one or more electrons from an electropositive atom to an electronegative atom.
Favourable Factors for Ionic Bond Formation
Low Ionization Enthalpy ($\Delta_i H$): Cation formation from metal ($M \rightarrow M^+ + e^-$) is always endothermic; low $\Delta_i H$ facilitates easy electron loss.
High Negative Electron Gain Enthalpy ($\Delta_{eg} H$): Non-metal readily gains electron ($X + e^- \rightarrow X^-$) with large energy release.
High Lattice Enthalpy ($\Delta_{\text{lattice}} H$): Energy released when gaseous ions pack into the 3D crystal lattice.
Lattice Enthalpy Definition: The energy required to completely separate one mole of a solid ionic compound into its constituent gaseous ions to an infinite distance.
$$\text{NaCl(s)} \rightarrow \text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)}; \quad \Delta_{\text{lattice}}H = +788\text{ kJ mol}^{-1}$$
Even though $\Delta_i H(Na) + \Delta_{eg} H(Cl) = 495.8 - 348.7 = +147.1\text{ kJ mol}^{-1}$ (positive/unfavourable), the huge lattice enthalpy release ($-788\text{ kJ mol}^{-1}$) drives the net reaction spontaneously!
Rock Salt (NaCl) 3D Lattice Enthalpy Model
4.3 Bond Parameters
Parameter
Definition & Key Concept
Important Relations & Examples
Bond Length
Equilibrium distance between the nuclei of two bonded atoms in a molecule.
Single bond > Double bond > Triple bond ($C-C\text{ 154 pm} > C=C\text{ 133 pm} > C\equiv C\text{ 120 pm}$).
Bond Angle
Angle between orbitals containing bonding electron pairs around the central atom.
$CH_4$ ($109.5^\circ$), $NH_3$ ($107^\circ$), $H_2O$ ($104.5^\circ$). Determines 3D shape.
Bond Enthalpy
Energy required to break 1 mole of bonds of a particular type between two atoms in gaseous state.
$H_2$ ($435.8\text{ kJ/mol}$), $O=O$ ($498$), $N\equiv N$ ($946.0\text{ kJ/mol}$). Polyatomics use average bond enthalpy.
Bond Order
Number of bonds between two atoms in a molecule.
$H_2=1, O_2=2, N_2=3, CO=3$. Isoelectronic species have identical bond orders ($N_2, CO, NO^+ = 3$).
Golden Correlation of Bond Parameters
$$\text{Bond Order} \uparrow \implies \text{Bond Enthalpy (Strength)} \uparrow \implies \text{Bond Length} \downarrow$$
Higher bond order = tightly held atoms = shorter bond = maximum dissociation energy!
4.3.5 Resonance & Canonical Structures
When a single Lewis structure cannot accurately depict the observed properties of a molecule (e.g., identical bond lengths), the actual structure is a resonance hybrid of two or more contributing canonical forms.
Ozone ($O_3$): Has two canonical structures with one single bond ($148\text{ pm}$) and one double bond ($121\text{ pm}$). The resonance hybrid exhibits two identical $O-O$ bond lengths of $128\text{ pm}$ (intermediate between single and double bond).
Carbonate Ion ($CO_3^{2-}$): Hybrid of 3 canonical forms; all three $C-O$ bonds are identical with bond order $\frac{4}{3} = 1.33$.
Carbon Dioxide ($CO_2$): Experimental $C-O$ distance is $115\text{ pm}$ (intermediate between $C=O$ $121\text{ pm}$ and $C\equiv O$ $110\text{ pm}$).
Key Misconceptions Dispelled: Canonical forms have no real existence; molecules do not alternate between them; resonance hybrid has lower energy than any canonical form (stabilization energy = resonance energy).
Resonance in Ozone (O₃) Molecule
4.3.6 Polarity of Bonds & Dipole Moment ($\mu$)
In a heteronuclear bond (e.g., $H-F$), the shared pair shifts towards the more electronegative atom, generating partial charges ($H^{\delta+} - F^{\delta-}$).
Dipole Moment ($\mu$): Product of the magnitude of charge ($Q$) and distance of separation ($r$):
$$\mu = Q \times r \qquad \text{Units: Debye (D), where } 1\text{ D} = 3.33564 \times 10^{-30}\text{ C m}$$
Vector Nature: In chemistry, represented by a crossed arrow $\mu$ with the cross on the positive end and arrow pointing towards the negative centre ($\overset{+}{\longrightarrow}$).
Molecular Dipole Moment: Vector sum of individual bond dipoles:
$H_2O$ vs $BeF_2$: $H_2O$ has bent geometry ($104.5^\circ$) with $\mu = 1.85\text{ D}$, whereas $BeF_2$ is linear with $\mu = 0$.
$NH_3$ vs $NF_3$: Both are pyramidal with 1 lone pair. However, $\mu(NH_3) = 1.47\text{ D} \gg \mu(NF_3) = 0.24\text{ D}$! In $NH_3$, the lone pair dipole points in the same direction as the resultant of the three $N-H$ bond dipoles, whereas in $NF_3$, fluorine is more electronegative than nitrogen, so the three $N-F$ bond dipoles point downward, opposing the lone pair dipole!
Dipole Moment Comparison: NH₃ (1.47 D) vs NF₃ (0.24 D)
4.3.7 Fajans' Rules (Partial Covalent Character of Ionic Bonds)
Just as covalent bonds possess partial ionic character (electronegativity difference), ionic bonds possess partial covalent character due to polarisation of the anion by the cation:
Small Cation Size: Smaller cation $\implies$ high charge density $\implies$ higher polarising power ($LiCl > NaCl > KCl$ covalent character).
Large Anion Size: Larger anion $\implies$ outer electrons loosely held $\implies$ highly polarisable ($LiI > LiBr > LiCl > LiF$).
High Charge on Ions: Greater charge on cation or anion $\implies$ higher polarising power ($AlCl_3 > MgCl_2 > NaCl$).
Pseudo-Noble Gas Configuration: Cations with $(n-1)d^{10}ns^0$ configuration have higher polarising power than $ns^2np^6$ cations of similar size and charge ($CuCl > NaCl$).
4.4 Valence Shell Electron Pair Repulsion (VSEPR) Theory
Proposed by Sidgwick and Powell (1940) and refined by Nyholm and Gillespie (1957) to predict the 3D geometry of molecules based on minimising electron-pair repulsions.
Order of Repulsive Interactions:
$$\text{Lone Pair (lp) Lone Pair (lp)} > \text{Lone Pair (lp) Bond Pair (bp)} > \text{Bond Pair (bp) Bond Pair (bp)}$$
Why? Lone pairs are localised on the central atom under the influence of only one nucleus, so they occupy greater spatial volume. Bond pairs are shared between two nuclei and are held more tightly.
Summary of Molecular Geometries (VSEPR Matrix)
Type
Bond Pairs
Lone Pairs
Electron Geometry
Molecular Shape
Bond Angle
Examples
$AB_2$
2
0
Linear
Linear
$180^\circ$
$BeCl_2, HgCl_2, CO_2$
$AB_3$
3
0
Trigonal planar
Trigonal planar
$120^\circ$
$BF_3, BCl_3, AlCl_3$
$AB_2E$
2
1
Trigonal planar
Bent / V-shaped
$119.5^\circ$
$SO_2, O_3, NO_2^-$
$AB_4$
4
0
Tetrahedral
Tetrahedral
$109.5^\circ$
$CH_4, CCl_4, SiF_4, NH_4^+$
$AB_3E$
3
1
Tetrahedral
Trigonal pyramidal
$107^\circ$
$NH_3, NF_3, PCl_3$
$AB_2E_2$
2
2
Tetrahedral
Bent / Angular
$104.5^\circ$
$H_2O, H_2S, OF_2, SCl_2$
$AB_5$
5
0
Trigonal bipyramidal
Trigonal bipyramidal
$120^\circ, 90^\circ$
$PCl_5, PF_5$
$AB_4E$
4
1
Trigonal bipyramidal
See-saw
$117^\circ, 89^\circ$
$SF_4$ (lp at equatorial)
$AB_3E_2$
3
2
Trigonal bipyramidal
T-shaped
$87.5^\circ$
$ClF_3, BrF_3$ (lps at equatorial)
$AB_2E_3$
2
3
Trigonal bipyramidal
Linear
$180^\circ$
$XeF_2, I_3^-$ (3 lps equatorial)
$AB_6$
6
0
Octahedral
Octahedral
$90^\circ$
$SF_6$
$AB_5E$
5
1
Octahedral
Square pyramidal
$<90^\circ$
$BrF_5, IF_5$
$AB_4E_2$
4
2
Octahedral
Square planar
$90^\circ$
$XeF_4$ (lps trans/axial)
VSEPR 3D Geometries at a Glance
4.5 Valence Bond (VB) Theory
Introduced by Heitler and London (1927) and developed by Pauling. Explains bond formation based on quantum mechanics, orbital overlap, and energetics.
Formation of $H_2$ Molecule: As two H atoms approach, attractive forces (nucleus of one attracting electron of other) and repulsive forces (electron-electron and nucleus-nucleus) operate. Experimentally, attractive forces exceed repulsive forces, causing the potential energy to decrease to a minimum ($-435.8\text{ kJ mol}^{-1}$ at equilibrium bond distance of $74\text{ pm}$).
Orbital Overlap Concept: A covalent bond forms by partial merging of two half-filled atomic orbitals with paired electrons of opposite spins. Greater overlap $\implies$ stronger bond.
Types of Overlapping:
Sigma ($\sigma$) Bond: Formed by end-to-end (head-on / axial) overlap along the internuclear axis ($s-s$, $s-p$, $p-p$). Symmetrical electron cloud around internuclear axis; large extent of overlap $\implies$ strong bond.
Pi ($\pi$) Bond: Formed by lateral (sidewise) overlap of atomic orbitals perpendicular to the internuclear axis. Overlapping consists of two saucer-shaped charge clouds above and below the plane $\implies$ weaker than $\sigma$ bond.
Orbital Overlap: Sigma (σ) Head-on vs Pi (π) Lateral Overlap
4.6 Hybridisation of Atomic Orbitals
Introduced by Pauling to explain the equivalent bond lengths and tetrahedral geometry of $CH_4$ (pure $p$ orbitals at $90^\circ$ cannot explain $109.5^\circ$ angles).
Hybridisation: The phenomenon of intermixing of atomic orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of equivalent orbitals having identical energy, shape, and directional symmetry.
Types of Hybridisation
$sp$ Hybridisation (Diagonal): Mixing of one $s$ + one $p$ orbital $\implies$ two linear $sp$ hybrids ($180^\circ$, 50% $s$, 50% $p$). Example: $BeCl_2$, $C_2H_2$ (ethyne has $1\sigma + 2\pi$ between carbons).
$sp^2$ Hybridisation (Trigonal): Mixing of one $s$ + two $p$ orbitals $\implies$ three trigonal planar hybrids ($120^\circ$, 33.3% $s$, 66.7% $p$). Example: $BCl_3$, $C_2H_4$ (ethene has $1\sigma + 1\pi$ between carbons).
$sp^3$ Hybridisation (Tetrahedral): Mixing of one $s$ + three $p$ orbitals $\implies$ four tetrahedral hybrids ($109.5^\circ$, 25% $s$, 75% $p$). Example: $CH_4$ ($109.5^\circ$), $NH_3$ ($107^\circ$, 1 lp), $H_2O$ ($104.5^\circ$, 2 lps).
Why are axial bonds longer than equatorial bonds in $PCl_5$?
The 3 equatorial $P-Cl$ bonds lie in a plane at $120^\circ$. The 2 axial bonds lie above and below the plane at $90^\circ$ to the equatorial bonds, suffering repulsion from three equatorial pairs at $90^\circ$. To minimise this repulsion, axial bonds are longer ($219\text{ pm}$) and weaker than equatorial bonds ($204\text{ pm}$) $\implies PCl_5$ easily dissociates into $PCl_3 + Cl_2$ upon heating!
$sp^3d^2$ Hybridisation (Octahedral): Mixing of $1s + 3p + 2d$ ($d_{x^2-y^2}, d_z^2$) orbitals $\implies$ 6 octahedral hybrids directed at $90^\circ$. Example: $SF_6$. All 6 $S-F$ bonds are equivalent ($158\text{ pm}$).
Multiple Bonding: σ and π Framework in Ethene (C₂H₄) and Ethyne (C₂H₂)
4.7 Molecular Orbital (MO) Theory
Developed by F. Hund and R.S. Mulliken (1932). Electrons in a molecule occupy molecular orbitals that spread over the entire molecule (polycentric), formed by the Linear Combination of Atomic Orbitals (LCAO).
Linear Combination of Atomic Orbitals (LCAO)
Bonding Molecular Orbital ($\sigma$): Formed by constructive interference (addition of wave functions: $\psi_{\text{MO}} = \psi_A + \psi_B$). Electron density is concentrated between the nuclei, shielding nuclear repulsion $\implies$ lower energy & higher stability.
Antibonding Molecular Orbital ($\sigma^*$): Formed by destructive interference (subtraction: $\psi_{\text{MO}}^* = \psi_A - \psi_B$). Has a nodal plane between nuclei $\implies$ higher energy & destabilising.
Conditions for Combination of Atomic Orbitals
Combining atomic orbitals must have comparable energies (e.g., $1s$ combines with $1s$, but not with $2s$).
Must have the same symmetry about the molecular axis ($z$-axis by convention: $2p_z$ combines with $2p_z$ to form $\sigma$ orbitals; $2p_x$ and $2p_y$ form $\pi$ orbitals).
Must have maximum overlap.
MO Energy Level Sequences for Diatomic Molecules
The Crucial Energy Sequence Switch
1. For $O_2$ and $F_2$ ($> 14$ electrons — No $s-p$ mixing):
$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \mathbf{\sigma 2p_z} < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$
2. For $B_2, C_2, N_2$ ($\le 14$ electrons — Due to significant $2s-2p_z$ mixing):
$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \mathbf{(\pi 2p_x = \pi 2p_y)} < \mathbf{\sigma 2p_z} < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$
Notice that for $\le 14\,e^-$, $\pi 2p_{x,y}$ is lower in energy than $\sigma 2p_z$!
Molecular Orbital Energy Diagram for O₂ (Showing Paramagnetism)
4.7.5 Stability, Bond Order & Magnetic Properties
Bond Order: $\text{B.O.} = \frac{1}{2}(N_b - N_a)$.
Diamagnetic: All electrons paired (repelled by magnetic field), e.g., $H_2, Li_2, C_2, N_2, F_2$.
Paramagnetic: Contains one or more unpaired electrons (attracted by magnetic field), e.g., $B_2$ (2 unpaired in $\pi 2p$), $O_2$ (2 unpaired in $\pi^* 2p$). MOT triumphantly explains the paramagnetism of $O_2$ where Lewis theory completely failed!
Special Case of $C_2$: Double bond in $C_2$ consists of both $\pi$ bonds (4 electrons in $\pi 2p_x^2 = \pi 2p_y^2$). In most other molecules, double bonds consist of one $\sigma$ and one $\pi$.
4.8 Hydrogen Bonding
When hydrogen is covalently bonded to a highly electronegative small atom (F, O, N), electron density shifts away from H, giving it a high fractional positive charge ($\delta^+$). This partially positive H forms an electrostatic attraction with a lone pair on another electronegative atom:
$$\cdots \text{H}^{\delta+} - \text{F}^{\delta-} \cdots \text{H}^{\delta+} - \text{F}^{\delta-} \cdots \text{H}^{\delta+} - \text{F}^{\delta-} \cdots$$
Dotted line ($\cdots$) represents the hydrogen bond; solid line (—) represents the covalent bond. Strength: $10\text{ to }40\text{ kJ mol}^{-1}$ (weaker than covalent, but much stronger than van der Waals).
Types of Hydrogen Bonding
Type
Definition & Characteristics
Consequences & Examples
Intermolecular H-Bonding
Occurs between different molecules of the same or different compounds.
Causes association, abnormally high boiling points ($H_2O$ is liquid at room temperature whereas $H_2S$ is gas; $HF$ boiling point > $HCl$). High solubility of alcohols/sugars in water.
Intramolecular H-Bonding
Occurs within the same single molecule between H and an electronegative atom nearby (forming a 5- or 6-membered chelate ring).
Prevents intermolecular association $\implies$ lower boiling point, higher steam volatility. Classic example: o-nitrophenol (intramolecular, steam-volatile) vs p-nitrophenol (intermolecular, higher b.p.).
Hydrogen Bonding: Intermolecular (Water) vs Intramolecular (o-Nitrophenol)
Chapter Summary (Unit 4)
Kössel-Lewis: Octet rule governs chemical bonding; ionic bonds form by electron transfer; covalent bonds form by electron sharing.
Formal Charge: $V - L - \frac{1}{2}S$; lowest energy structures minimise formal charges.
Valence Bond Theory: Head-on overlap forms strong $\sigma$ bonds; lateral overlap forms weaker $\pi$ bonds.
Hybridisation: Intermixing of valence orbitals produces equivalent directional hybrids ($sp$ linear, $sp^2$ planar, $sp^3$ tetrahedral, $sp^3d$ trigonal bipyramidal with long axial bonds, $sp^3d^2$ octahedral).
Molecular Orbital Theory: Linear Combination of Atomic Orbitals ($\psi_{\text{MO}} = \psi_A \pm \psi_B$); Bonding MOs are lower in energy, antibonding MOs have nodal planes. Successfully explains $O_2$ paramagnetism.
Hydrogen Bonding: Attractive force binding H bonded to F, O, N with another electronegative atom. Strong influence on melting/boiling points and solubility.
Practice Quiz (25 Conceptual Questions)
Test your understanding of Lewis structures, octet exceptions, lattice enthalpy, dipole moments, VSEPR shapes, hybridisation, MOT, and hydrogen bonding.
Q01 In Lewis's electronic cubical model of the atom, the 'Kernel' represents:
Incorrect: Valence electrons occupy the corners of the cube.
Incorrect: Kernel is positively charged.
Correct Answer: Lewis defined the Kernel as the positively charged inner core consisting of the nucleus and inner electrons.
Incorrect: It includes the inner closed-shell electrons as well.
Q02 Which of the following molecules has an incomplete octet on the central atom?
Incorrect: Carbon has an 8-electron octet in CCl₄.
Correct Answer: Boron in BCl₃ has only 3 bonded pairs (6 electrons), forming an incomplete octet.
Incorrect: SF₆ has an expanded octet (12 electrons).
Incorrect: Oxygen in H₂O has an octet (2 bp + 2 lp).
Q03 Which of the following is an odd-electron molecule that violates the octet rule?
Incorrect: CO₂ has 16 valence electrons (even).
Incorrect: N₂O has 16 valence electrons (even).
Incorrect: SO₂ has 18 valence electrons (even).
Correct Answer: NO₂ has 5 + 2(6) = 23 valence electrons (odd electron molecule, paramagnetic).
Q04 In the ozone (O₃) molecule, what is the formal charge on the central oxygen atom?
Incorrect: The terminal double-bonded oxygen has a formal charge of 0.
Incorrect: The singly-bonded terminal oxygen has a formal charge of −1.
Incorrect: Formal charge is +1.
Q05 Why can sulfur and phosphorus form expanded octets (e.g., SF₆, PF₅), while oxygen and nitrogen cannot?
Incorrect: Not the structural reason for expanding valence shell.
Correct Answer: Period 3 elements have vacant 3d orbitals that can participate in bonding, whereas Period 2 elements (N, O) have only 2s and 2p orbitals (max covalency = 4).
Incorrect: P and S are non-metals.
Incorrect: P and S are less electronegative than N and O.
Q06 Which set of conditions most strongly favours the formation of a stable ionic bond?
Incorrect: High ionization energy opposes cation formation.
Incorrect: High lattice enthalpy provides stability.
Correct Answer: Low energy to remove electron + high energy release upon electron gain + large lattice energy = most stable ionic solid.
Incorrect: That leads to covalent bonding.
Q07 What is the correct order of bond lengths for carbon-carbon bonds?
Incorrect: Triple bonds are the shortest.
Incorrect: Single bonds are the longest.
Incorrect: Double bond length is intermediate.
Correct Answer: Increasing bond order pulls nuclei closer, decreasing bond length.
Q08 The experimentally determined O−O bond length in ozone (O₃) is:
Correct Answer: Resonance averages both bonds to 128 pm (intermediate between single 148 pm and double 121 pm).
Incorrect: Those are the theoretical values in individual canonical forms, not observed in the real molecule.
Incorrect: Too short (typical of triple bonds).
Incorrect: That is a C−C single bond length.
Q09 Why is the net dipole moment of BeF₂ and BF₃ zero?
Incorrect: The individual bonds are highly polar.
Correct Answer: BeF₂ is linear (180°) and BF₃ is trigonal planar (120°); vector sum of bond dipoles is zero.
Incorrect: Neither Be in BeF₂ nor B in BF₃ has a lone pair.
Incorrect: BeF₂ and BF₃ are covalent molecules in gaseous state.
Q10 Although fluorine is more electronegative than nitrogen, NH₃ (1.47 D) has a higher dipole moment than NF₃ (0.24 D) because:
Incorrect: Both are pyramidal.
Incorrect: Both have 1 lone pair on nitrogen.
Correct Answer: In NF₃, F pulls electron density downward, opposing the upward lone pair orbital dipole.
Incorrect: F is the most electronegative element.
Q11 According to Fajans' rules, covalent character in an ionic compound increases with:
Correct Answer: Small cation has high polarising power; large anion is easily distorted.
Incorrect: That favours purely ionic character (e.g., CsF).
Incorrect: Low charge decreases polarising power.
Incorrect: Pseudo-noble gas core (d¹⁰) increases covalent character.
Q12 What is the correct order of repulsive interaction between electron pairs according to VSEPR theory?
Incorrect: Lone pairs repel most strongly.
Incorrect: lplp repulsion is the strongest.
Correct Answer: Because lone pairs are under the influence of only one nucleus, they spread out and exert maximum repulsion.
Incorrect: Repulsions are not equal.
Q13 Why does the bond angle decrease from CH₄ (109.5°) → NH₃ (107°) → H₂O (104.5°)?
Incorrect: Electronegativity actually increases from C to O.
Incorrect: All three have sp³ hybridisation.
Incorrect: Atomic radius decreases from C to O.
Correct Answer: As lone pairs increase, lpbp repulsion pushes bond pairs closer together, compressing the bond angle.
Q14 In PCl₅ (trigonal bipyramidal), why are the axial P−Cl bonds longer than the equatorial bonds?
Incorrect: Axial bonds involve pz and dz² orbitals.
Correct Answer: Axial bonds experience three 90° repulsions from equatorial pairs, pushing them farther out to minimise repulsion.
Incorrect: All chlorine atoms are identical.
Incorrect: All five bonds are sigma bonds.
Q15 What is the molecular shape of SF₄ (AB₄E type) according to VSEPR theory?
Correct Answer: The 1 lone pair occupies an equatorial position in a trigonal bipyramid to minimise 90° repulsions, giving a see-saw shape.
Incorrect: Square planar is AB₄E₂ (e.g., XeF₄).
Incorrect: SF₄ has 5 electron pairs (4 bp + 1 lp).
Incorrect: That is AB₃E (e.g., NH₃).
Q16 How many sigma (σ) and pi (π) bonds are present in an ethyne (C₂H₂) molecule?
Incorrect: Two C−H bonds and one C−C bond make 3 sigma bonds.
Incorrect: That is ethene (C₂H₄).
Correct Answer: Ethyne (H−C≡C−H) has two C−H σ bonds, one C−C σ bond, and two C−C π bonds.
Incorrect: Forgets the two C−H bonds.
Q17 Which of the following atomic orbital overlaps is NOT possible to form a covalent bond?
Incorrect: Forms a σ bond.
Incorrect: Forms a σ bond.
Incorrect: Forms a π bond.
Correct Answer: S-orbital overlaps equally with positive and negative lobes of py, causing net overlap to be zero.
Q18 What is the hybridisation of the central boron atom in BCl₃ and central phosphorus atom in PCl₅?
Incorrect: BCl₃ has only 3 bonds.
Correct Answer: BCl₃ is trigonal planar (sp²); PCl₅ is trigonal bipyramidal (sp³d).
Incorrect: Does not match steric numbers 3 and 5.
Incorrect: PCl₅ uses outer 3d orbitals (sp³d).
Q19 According to Molecular Orbital Theory, which of the following diatomic species CANNOT exist?
Correct Answer: Both have Bond Order = ½(Nb − Na) = ½(2−2) or ½(4−4) = 0, so they are unstable and cannot exist.
Incorrect: Li₂ exists in vapour phase with Bond Order = 1.
Incorrect: C₂ exists with Bond Order = 2.
Incorrect: H₂⁺ has Bond Order = 0.5 and exists.
Q20 Why is the oxygen molecule (O₂) paramagnetic according to Molecular Orbital Theory?
Incorrect: O₂ has 16 electrons (an even number).
Incorrect: Lewis octet rule falsely predicted all electrons paired.
Correct Answer: By Hund's rule, the two highest-energy electrons occupy separate π* orbitals singly, making O₂ paramagnetic.
Incorrect: Bond order of O₂ is 2.
Q21 In the C₂ molecule, the double bond consists of:
Incorrect: That is true for ethene and O₂, but not C₂.
Correct Answer: In C₂ (12 e⁻), the last 4 valence electrons occupy π 2px² = π 2py² orbitals, so both bonds are pi bonds.
Incorrect: Two atoms cannot be bonded by two sigma bonds.
Incorrect: Homonuclear diatomic molecule has no coordinate bond.
Q22 What is the bond order and relative stability order of O₂, O₂⁺, O₂⁻, and O₂²⁻?
Incorrect: Adding electrons to antibonding orbitals decreases stability.
Incorrect: O₂⁺ has a higher bond order (2.5) than O₂ (2.0).
Incorrect: Bond orders are 2.5, 2.0, 1.5, and 1.0 respectively.
Correct Answer: Higher bond order corresponds to shorter bond length and higher thermodynamic stability.
Q23 Which elements form strong hydrogen bonds?
Correct Answer: High electronegativity and small atomic radii are prerequisites for significant hydrogen bonding.
Incorrect: Too large in size to create high enough electrostatic charge density.
Incorrect: Electronegativities are too low.
Incorrect: These are electropositive metals.
Q24 Why is water (H₂O) a liquid at room temperature while hydrogen sulfide (H₂S) is a gas?
Incorrect: H₂S (34 g/mol) is heavier than H₂O (18 g/mol).
Incorrect: H₂S does not form hydrogen bonds due to large S atom size.
Correct Answer: Oxygen's high electronegativity leads to 3D H-bonded networks in water, drastically raising its boiling point.
Incorrect: Water is a covalent molecule.
Q25 o-Nitrophenol is steam-volatile and has a lower boiling point than p-nitrophenol because:
Incorrect: p-Nitrophenol has intermolecular H-bonding.
Correct Answer: The −OH and −NO₂ groups in ortho position form a 6-membered chelate ring, preventing association and making it steam-volatile.
Incorrect: Both are polar.
Incorrect: Both are structural isomers with identical molecular weights.
NCERT Textbook Exercises (4.1 to 4.40 Solved)
Complete step-by-step solutions for all 40 end-of-chapter questions from CBSE / NCERT Class 11 Chemistry Chapter 4.
4.1
Explain the formation of a chemical bond.
A chemical bond is formed when two atoms approach each other and their potential energy decreases, resulting in greater stability:
Energetics: When atoms approach, attractive forces (nucleus-electron) and repulsive forces (nucleus-nucleus, electron-electron) operate. When attractive forces exceed repulsive forces, energy is released and potential energy reaches a minimum at an equilibrium internuclear distance (bond length).
Electronic Configuration: Atoms tend to acquire a stable noble gas configuration ($ns^2np^6$, octet rule) by transfer of electrons (ionic bond) or sharing of electrons (covalent bond).
A chemical bond is the attractive force holding atoms/ions together, formed to lower the system's potential energy and achieve noble gas stability.
4.2
Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Lewis symbols represent the valence shell electrons as dots surrounding the elemental symbol:
Na ($[Ne]3s^1$): $1$ valence electron $\implies$ •Na
Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻.
S ($3s^2 3p^4$): 6 dots $\implies$ :S̈: | S²⁻: gained 2 electrons $\implies$ [:S̈: ]²⁻ (8 dots with 2− charge).
Al ($3s^2 3p^1$): 3 dots $\implies$ •Ȧl• | Al³⁺: lost 3 valence electrons $\implies$ [Al]³⁺ (no valence dots, $[Ne]$ core).
H ($1s^1$): 1 dot $\implies$ H• | H⁻: gained 1 electron $\implies$ [H:]⁻ (2 dots, noble gas He configuration).
Anions display full octet/duplet with brackets and charge; simple metal cations show no dots with positive charge.
4.4
Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH.
H₂S: Central S has 6 valence e⁻ + 2 from H = 8 e⁻. Two single bonds and 2 lone pairs on S: $H - \ddot{S} - H$.
SiCl₄: Central Si has 4 valence e⁻ + 4×7 from Cl = 32 e⁻. Four single bonds with Si-Cl and 3 lone pairs on each Cl.
BeF₂: Central Be has 2 valence e⁻ + 2×7 from F = 16 e⁻. Linear single bonds: $:F̈ - Be - F̈:$ (incomplete octet on Be).
CO₃²⁻: Total 24 e⁻ (C=4, 3×O=18, charge=2). Resonance hybrid of one $C=O$ double bond and two $C-O^-$ single bonds.
HCOOH (Formic Acid): C is bonded to H, double-bonded to carbonyl O, and single-bonded to OH group: $H - C(=O) - \ddot{O} - H$.
All atoms attain octet (or duplet for H), except Be in BeF₂ (incomplete octet, 4 e⁻).
4.5
Define octet rule. Write its significance and limitations.
Definition: Atoms combine by gaining, losing, or sharing valence electrons in order to attain a stable outer-shell configuration of eight electrons (an octet), resembling the nearest noble gas. Significance: Successfully explains the chemical reactivity and formulas of the vast majority of organic compounds and representative covalent molecules of Period 2. Limitations:
Incomplete octet (e.g., $LiCl, BeH_2, BCl_3$).
Odd-electron molecules (e.g., $NO, NO_2$).
Expanded octet in Period 3 and beyond (e.g., $PF_5, SF_6, H_2SO_4$).
Compounds of noble gases (e.g., $XeF_2, XeF_4$).
Does not explain molecular shapes or thermodynamic bond energies.
Useful empirical rule for Period 2, but has major exceptions with electron-deficient, odd-electron, and expanded-octet species.
4.6
Write the favourable factors for the formation of ionic bond.
Low Ionization Enthalpy of the Metal ($\Delta_i H$): Cation formation requires less input energy.
High Negative Electron Gain Enthalpy of Non-metal ($\Delta_{eg} H$): Large energy release upon gaining electrons.
High Lattice Enthalpy of Crystal ($\Delta_{\text{lattice}} H$): Strong electrostatic stabilization during packing of opposite ions into the 3D solid lattice.
Low ΔᵢH(metal) + High negative Δ_egH(non-metal) + High lattice enthalpy.
4.7
Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.
BeCl₂: 2 bp, 0 lp $\implies$ $AB_2$ $\implies$ Linear ($180^\circ$).
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Both molecules have $sp^3$ hybridised central atoms:
In NH₃, nitrogen has 3 bond pairs and 1 lone pair. The repulsion is between $1\text{ lp} - 3\text{ bp}$, compressing the bond angle from $109.5^\circ$ to $107^\circ$.
In H₂O, oxygen has 2 bond pairs and 2 lone pairs. According to VSEPR theory, $lp-lp\text{ repulsion} > lp-bp\text{ repulsion} > bp-bp\text{ repulsion}$. The strong repulsion between the two lone pairs pushes the two $O-H$ bonding pairs even closer together, further reducing the bond angle to $104.5^\circ$.
Due to greater repulsion from two lone pairs in H₂O (lplp repulsion) compared to only one lone pair in NH₃.
4.9
How do you express the bond strength in terms of bond order?
Bond strength is directly proportional to bond order:
$$\text{Bond Strength (Bond Enthalpy)} \propto \text{Bond Order} \propto \frac{1}{\text{Bond Length}}$$
As the bond order increases (single $\rightarrow$ double $\rightarrow$ triple), the number of shared electron pairs between the atoms increases, pulling the nuclei closer together and requiring more energy to dissociate the bond.
Higher bond order implies greater bond dissociation enthalpy and shorter, stronger bonds (e.g., N≡N with BO=3 has Δ_aH = 946 kJ/mol).
4.10
Define the bond length.
Definition: Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
It is typically determined experimentally using spectroscopic techniques, X-ray diffraction, or electron diffraction, and is measured in picometres (pm) or Ångströms (Å).
Equilibrium internuclear distance between two bonded atoms (e.g., HH = 74 pm, ClCl = 198 pm).
4.11
Explain the important aspects of resonance with reference to the CO₃²⁻ ion.
A single Lewis structure depicts one $C=O$ double bond and two $C-O^-$ single bonds, predicting different bond lengths ($121\text{ pm}$ and $143\text{ pm}$).
However, experimental evidence shows that all three carbon-to-oxygen bonds in $CO_3^{2-}$ are completely identical with equal bond length ($129\text{ pm}$) and equal bond order of $\frac{4}{3} = 1.33$.
Therefore, $CO_3^{2-}$ is best represented as a resonance hybrid of three equivalent canonical structures where the double bond is delocalised over all three oxygen atoms.
Resonance stabilizes the ion and makes all three CO bonds identical in length, energy, and character.
4.12
H₃PO₃ can be represented by two structures with different positions of H atoms. Can these be taken as canonical forms of a resonance hybrid? If not, give reasons.
No. In canonical structures of a resonance hybrid, the relative positions of all atomic nuclei must remain strictly identical; only the distribution of electrons (bonding and lone pairs) can differ.
Since hydrogen atoms change their bonding positions from P to O, these structures represent tautomers (distinct chemical isomers in equilibrium), not resonance structures!
No. The atomic positions differ. Resonance involves only delocalization of electrons with stationary nuclei.
4.13
Write the resonance structures for SO₃, NO₂ and NO₃⁻.
SO₃: Can be represented as a hybrid of canonical forms having $S=O$ double bonds delocalised over three planar positions (all three $S-O$ bond lengths are identical).
NO₂: Odd-electron molecule with 2 canonical structures showing the double bond and unpaired electron alternating between the two oxygen atoms ($[:\ddot{O} = \dot{N} - \ddot{O}:] \longleftrightarrow [:\ddot{O} - \dot{N} = \ddot{O}:]$).
NO₃⁻: Planar nitrate ion has 3 equivalent canonical forms where the $N=O$ double bond shifts among the three oxygens; each $N-O$ bond order is $\frac{4}{3} \approx 1.33$.
Each species delocalises multiple bond character across terminal oxygen atoms to attain equivalence and resonance stability.
4.14
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.
(a) K and S: Two K atoms ($4s^1$) each lose $1\,e^-$ to one S atom ($3s^2 3p^4$):
$$2\text{K}\cdot + :\ddot{S}: \longrightarrow 2\text{K}^+ + [:\ddot{S}:]^{2-} \implies \text{K}_2\text{S}$$
(b) Ca and O: One Ca atom ($4s^2$) loses $2\,e^-$ to one O atom ($2s^2 2p^4$):
$$\text{Ca}: + :\ddot{O}: \longrightarrow \text{Ca}^{2+} + [:\ddot{O}:]^{2-} \implies \text{CaO}$$
(c) Al and N: One Al atom ($3s^2 3p^1$) loses $3\,e^-$ to one N atom ($2s^2 2p^3$):
$$\cdot\dot{\text{Al}}\cdot + :\dot{\text{N}}\cdot \longrightarrow \text{Al}^{3+} + [:\ddot{\text{N}}:]^{3-} \implies \text{AlN}$$
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.
Experimental dipole moment measurements reveal:
CO₂ ($\mu = 0\text{ D}$): The individual $C=O$ bonds are polar. The net dipole moment is zero only if the two equal and opposite bond dipoles cancel out completely $\implies$ CO₂ must be strictly linear ($180^\circ$).
H₂O ($\mu = 1.85\text{ D}$): The net dipole moment is large and non-zero. If H₂O were linear, the two $O-H$ dipoles would cancel to zero. Because it has a net resultant dipole of $1.85\text{ D}$, the two bonds must be oriented at an angle $\implies$ H₂O has an angular / bent shape ($104.5^\circ$).
Linear CO₂ cancels bond dipoles (μ = 0); angular H₂O yields a non-zero vector resultant (μ = 1.85 D).
4.16
Write the significance / applications of dipole moment.
Predicting Bond Polarity: Non-polar if $\mu = 0$; polar if $\mu > 0$.
Determining Molecular Geometry: Symmetrical molecules have $\mu = 0$ ($BeF_2$ linear, $BF_3$ trigonal planar, $CH_4$ tetrahedral). Non-zero $\mu$ indicates bent or unsymmetrical geometry ($H_2O, NH_3$).
Distinguishing Cis and Trans Isomers: Cis isomers generally have higher dipole moments than symmetrical trans isomers ($\mu_{\text{trans}} \approx 0$).
Define electronegativity. How does it differ from electron gain enthalpy?
Feature
Electronegativity
Electron Gain Enthalpy (Δ_egH)
Definition
Tendency of an atom in a molecule to attract the shared pair of electrons towards itself.
Energy change when an electron is added to an isolated neutral gaseous atom.
State
Applies to a bonded atom in a chemical compound.
Applies to an isolated gaseous atom in ground state.
Nature
Relative, dimensionless qualitative value (e.g. Pauling scale).
Thermodynamic, measurable quantity with units (kJ mol⁻¹).
Electronegativity is relative attraction in a bonded molecule; Δ_egH is quantitative energy released upon electron addition to a free gaseous atom.
4.18
Explain with the help of a suitable example polar covalent bond.
When a covalent bond is formed between two atoms having different electronegativities (e.g., $HF$), the shared electron pair is not equally shared.
Fluorine has a much higher electronegativity ($4.0$) than hydrogen ($2.1$), pulling electron density toward itself.
This induces a partial negative charge ($\delta^-$) on fluorine and an equal partial positive charge ($\delta^+$) on hydrogen:
$$H^{\delta+} - F^{\delta-}$$
Such a bond having charge separation is called a polar covalent bond.
Bond formed between atoms of differing electronegativity with fractional positive and negative charges (e.g., HF, HCl, OH).
4.19
Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.
Ionic character depends on the electronegativity difference ($\Delta\chi$) between the bonded atoms:
$ClF_3$: Cl (3.0) and F (4.0) $\implies \Delta\chi = 1.0$.
$SO_2$: S (2.5) and O (3.5) $\implies \Delta\chi = 1.0$ (polar covalent).
$K_2O$: K (0.8) and O (3.5) $\implies \Delta\chi = 2.7$ (predominantly ionic).
$LiF$: Li (1.0) and F (4.0) $\implies \Delta\chi = 3.0$ (highest ionic character).
N₂ < ClF₃ < SO₂ < K₂O < LiF
4.20
The skeletal structure of CH₃COOH as usually shown has some incorrect bonds. Write the correct Lewis structure for acetic acid.
In acetic acid ($CH_3COOH$):
Methyl carbon ($C_1$) forms three $C-H$ single bonds and one $C-C$ single bond (4 bonds = 8 e⁻).
Carboxyl carbon ($C_2$) forms a double bond with carbonyl oxygen ($C=O$), a single bond with hydroxyl oxygen ($C-O$), and a single bond with $C_1$ (4 bonds = 8 e⁻).
Carbonyl oxygen has 2 lone pairs; hydroxyl oxygen has 2 lone pairs and forms a single bond with H.
H₃C — C(=Ö:) — Ö̈ — H (All atoms satisfy the octet, H satisfies duplet).
4.21
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar. Explain why CH₄ is tetrahedral and not square planar.
According to VSEPR theory, electron pairs around the central atom arrange themselves to maximise distance and minimise repulsive interactions:
In a square planar geometry, the four $C-H$ bond pairs would be confined to a 2D plane with bond angles of only $90^\circ$.
In a tetrahedral geometry, the bond pairs spread out in 3D space with bond angles of $109.5^\circ$.
Since $109.5^\circ > 90^\circ$, repulsions between electron pairs are far lower in the tetrahedral arrangement, conferring greater stability.
Tetrahedral arrangement provides maximum angular separation (109.5° vs 90°) and minimum repulsion.
4.22
Explain why BeH₂ molecule has a zero dipole moment although the BeH bonds are polar.
Beryllium in $BeH_2$ is $sp$ hybridised, giving the molecule a strictly linear geometry ($180^\circ$):
$$H \xleftarrow{\quad} Be \xrightarrow{\quad} H$$
The two $Be-H$ bond dipoles are equal in magnitude but point in exactly opposite directions ($180^\circ$).
Consequently, the two bond dipole vectors cancel each other out completely:
$$\mu_{\text{net}} = \mu_1 - \mu_2 = 0\text{ D}$$
Linear geometry causes equal and opposite bond dipole vectors to cancel completely (μ = 0).
4.23
Which out of NH₃ and NF₃ has higher dipole moment and why?
NH₃ has a much higher dipole moment ($1.47\text{ D}$) than NF₃ ($0.24\text{ D}$):
Both molecules have pyramidal shape with 1 lone pair on nitrogen.
In NH₃, nitrogen is more electronegative than hydrogen, so the three $N-H$ bond moments point upward toward N. The lone pair orbital dipole also points upward $\implies$ they reinforce each other.
In NF₃, fluorine is more electronegative than nitrogen, so the three $N-F$ bond moments point downward toward F. The lone pair dipole points upward $\implies$ they oppose and cancel each other.
NH₃ (1.47 D) > NF₃ (0.24 D). Lone pair dipole reinforces bond dipoles in NH₃, but opposes them in NF₃.
4.24
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
Hybridisation: The phenomenon of intermixing of atomic orbitals of slightly different energies of an isolated atom to produce a new set of equivalent orbitals of identical shape, energy, and directional orientation.
$sp$ Hybridisation: 1 s + 1 p orbital $\implies$ 2 linear hybrid orbitals oriented at $180^\circ$ (e.g., $BeCl_2$).
$sp^2$ Hybridisation: 1 s + 2 p orbitals $\implies$ 3 trigonal planar hybrid orbitals oriented at $120^\circ$ (e.g., $BCl_3$).
$sp^3$ Hybridisation: 1 s + 3 p orbitals $\implies$ 4 tetrahedral hybrid orbitals directed toward corners of a tetrahedron at $109.5^\circ$ (e.g., $CH_4$).
Describe the change in hybridisation (if any) of the Al atom in the reaction: AlCl₃ + Cl⁻ → AlCl₄⁻.
In AlCl₃, Al has 3 valence electrons forming 3 single bonds with no lone pairs $\implies$ steric number = 3 $\implies$ $sp^2$ hybridised (trigonal planar).
In AlCl₄⁻, the incoming $Cl^-$ donates an electron pair into the vacant $3p$ orbital of Al, forming a fourth coordinate covalent bond $\implies$ steric number = 4 $\implies$ $sp^3$ hybridised (tetrahedral).
Hybridisation changes from sp² (trigonal planar) to sp³ (tetrahedral).
4.26
Is there any change in the hybridisation of B and N atoms as a result of the reaction: BF₃ + NH₃ → F₃B·NH₃?
Boron (B): In $BF_3$, B has 3 bond pairs and 0 lone pairs $\implies$ $sp^2$ hybridised. In the adduct $F_3B \leftarrow NH_3$, B accepts a lone pair into its vacant $2p$ orbital, forming a fourth bond $\implies$ changes to $sp^3$ hybridised.
Nitrogen (N): In $NH_3$, N has 3 bond pairs and 1 lone pair $\implies$ $sp^3$ hybridised. In the adduct, the lone pair forms a coordinate bond, still maintaining 4 electron pairs $\implies$ remains $sp^3$ hybridised.
B changes from sp² to sp³; N remains sp³ (geometry around B becomes tetrahedral).
4.27
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Ethene (C₂H₄): Each carbon is $sp^2$ hybridised. One $sp^2$ orbital from each carbon overlaps axially to form a $C-C$ $\sigma$ bond; two $sp^2$ orbitals overlap with $1s$ orbitals of H atoms to form four $C-H$ $\sigma$ bonds. The unhybridised $2p_z$ orbitals overlap laterally to form one $\pi$ bond.
Ethyne (C₂H₂): Each carbon is $sp$ hybridised. One $sp$ orbital forms a $C-C$ $\sigma$ bond; one forms a $C-H$ $\sigma$ bond. The two mutually perpendicular unhybridised orbitals ($2p_x, 2p_y$) overlap laterally to form two $\pi$ bonds.
C₂H₄ contains 1 CC σ + 1 CC π bond; C₂H₂ contains 1 CC σ + 2 CC π bonds.
4.28
What is the total number of sigma and pi bonds in the following molecules? (a) C₂H₂ (b) C₂H₄
(a) C₂H₂: 3 σ and 2 π bonds; (b) C₂H₄: 5 σ and 1 π bond.
4.29
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 2px (c) 2py and 2py (d) 1s and 2s.
Internuclear axis is the $x$-axis:
(a) 1s and 1s: Overlap along $x$-axis is end-to-end $\implies$ forms a $\sigma$ bond.
(b) 1s and 2px: $2p_x$ lies along the $x$-axis, so axial overlap with $1s$ forms a $\sigma$ bond.
(c) 2py and 2py: $2p_y$ orbitals are perpendicular to the $x$-axis. When approaching along the $x$-axis, they overlap sidewise (laterally) to form a $\pi$ bond, NOT a $\sigma$ bond!
(d) 1s and 2s: Spherical orbitals overlap axially $\implies$ forms a $\sigma$ bond.
(c) 2py and 2py will not form a sigma bond; they undergo lateral overlap to form a pi (π) bond.
4.30
Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH₃CH₃ (b) CH₃CH=CH₂ (c) CH₃CH₂OH (d) CH₃CHO (e) CH₃COOH
Count the number of $\sigma$ bonds formed by each carbon:
(a) CH₃CH₃: Both carbons have 4 $\sigma$ bonds $\implies$ $sp^3$ and $sp^3$.
(b) CH₃CH=CH₂: $C_1$ (methyl) has 4 $\sigma$ $\implies$ $sp^3$; $C_2$ ($=CH-$) has 3 $\sigma$ $\implies$ $sp^2$; $C_3$ ($=CH_2$) has 3 $\sigma$ $\implies$ $sp^2$.
(c) CH₃CH₂OH: Both carbons have 4 $\sigma$ bonds $\implies$ $sp^3$ and $sp^3$.
(d) CH₃CHO: Methyl carbon has 4 $\sigma$ $\implies$ $sp^3$; aldehyde carbon ($-\text{CH}=\text{O}$) has 3 $\sigma$ $\implies$ $sp^2$.
(e) CH₃COOH: Methyl carbon has 4 $\sigma$ $\implies$ $sp^3$; carboxylic carbon ($-\text{COOH}$) has 3 $\sigma$ $\implies$ $sp^2$.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Bond Pair (bp): A shared pair of valence electrons involved in forming a covalent bond between two atoms (shared between two nuclei). Example: In $CH_4$, all 4 pairs are bond pairs.
Lone Pair (lp) / Non-bonding pair: A pair of valence electrons that belongs exclusively to one atom and does not participate in bonding (localised on one nucleus). Example: In $H_2O$, oxygen has 2 lone pairs.
Bond pairs are shared between two atoms; lone pairs remain unshared on the central atom (e.g. NH₃ has 3 bp and 1 lp).
4.32
Distinguish between a sigma and a pi bond.
Property
Sigma (σ) Bond
Pi (π) Bond
Mode of Overlap
End-to-end (axial / head-on) along internuclear axis.
Sidewise (lateral / parallel) perpendicular to internuclear axis.
Extent of Overlap
Large extent of overlap $\implies$ strong bond.
Small extent of overlap $\implies$ comparatively weaker bond.
Electron Cloud
Cylindrically symmetrical around the bond axis.
Two saucer-shaped lobes above and below the internuclear plane.
Rotation
Free rotation about a $\sigma$ bond is possible.
Rotation about a $\pi$ bond is restricted (hindered).
Existence
Can exist independently (single bond).
Always formed in addition to a $\sigma$ bond (multiple bonds).
Sigma is formed by axial overlap (strong, free rotation); Pi is formed by lateral overlap (weaker, restricted rotation).
4.33
Explain the formation of H₂ molecule on the basis of valence bond theory.
Consider two hydrogen atoms $H_A$ and $H_B$ with electrons $e_A$ and $e_B$:
At infinite distance, there is no interaction ($PE = 0$).
As they approach, new attractive forces ($H_A - e_B, H_B - e_A$) and repulsive forces ($e_A - e_B, H_A - H_B$) develop.
Attractive forces are stronger than repulsive forces, causing the system's potential energy to decrease.
At an equilibrium distance of $74\text{ pm}$ (bond length), attractive and repulsive forces balance, achieving a potential energy minimum of $-435.8\text{ kJ mol}^{-1}$ (bond enthalpy). The $1s$ orbitals overlap, pairing electrons with opposite spins to form a stable $H_2$ molecule.
Attractive forces exceed repulsions, lowering potential energy to a minimum of −435.8 kJ/mol at internuclear distance 74 pm.
4.34
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Comparable Energies: Combining atomic orbitals must have the same or nearly the same energy (e.g., $1s$ combines with $1s$, but not with $2s$).
Identical Symmetry: Combining orbitals must have the same symmetry about the molecular axis ($z$-axis). For instance, $2p_z$ combines with $2p_z$, but cannot combine with $2p_x$ or $2p_y$.
Maximum Overlap: The combining orbitals must overlap to the maximum extent to ensure high electron density between the nuclei.
1. Similar energies; 2. Same symmetry along molecular axis; 3. Maximum extent of overlap.
4.35
Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Beryllium ($Z = 4$) has electronic configuration $1s^2 2s^2$. In a hypothetical $Be_2$ molecule, there are $8$ electrons:
$$\text{MO Configuration: } (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2$$
Number of bonding electrons ($N_b$) = 4; Number of antibonding electrons ($N_a$) = 4.
$$\text{Bond Order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(4 - 4) = 0$$
Since the bond order is zero, no chemical bond is formed and the $Be_2$ molecule is unstable and cannot exist.
Bond order = ½(4 − 4) = 0. Zero bond order implies no net binding force, so Be₂ does not exist.
4.36
Compare the relative stability of the following species and indicate their magnetic properties: O₂, O₂⁺, O₂⁻ (superoxide), O₂²⁻ (peroxide).
Species
Total e⁻
Bond Order = ½(Nb − Na)
Magnetic Property
O₂⁺
15
½(10 − 5) = 2.5
Paramagnetic (1 unpaired e⁻ in π* 2p)
O₂
16
½(10 − 6) = 2.0
Paramagnetic (2 unpaired e⁻ in π* 2p)
O₂⁻ (superoxide)
17
½(10 − 7) = 1.5
Paramagnetic (1 unpaired e⁻ in π* 2p)
O₂²⁻ (peroxide)
18
½(10 − 8) = 1.0
Diamagnetic (all paired)
Stability Order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻. O₂⁺, O₂, O₂⁻ are paramagnetic; O₂²⁻ is diamagnetic.
4.37
Write the significance of a plus and a minus sign shown in representing the orbitals.
The plus ($+$) and minus ($-$) signs represent the mathematical sign (phase) of the orbital wave function ($\psi$) in that region of space.
They do not indicate electrical charge!
Overlap of lobes with the same sign ($+ / +$ or $- / -$) represents constructive interference, leading to a bonding molecular orbital with increased electron density.
Overlap of lobes with opposite signs ($+ / -$) represents destructive interference, producing an antibonding orbital with a nodal plane.
Signs indicate phase of the wave function (ψ), not electrical charge; same-phase overlap forms bonding orbitals.
4.38
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
In $PCl_5$, the central phosphorus atom in its excited state ($3s^1 3p^3 3d^1$) undergoes $sp^3d$ hybridisation, forming five hybrid orbitals with a trigonal bipyramidal geometry:
Three equatorial bonds lie in one plane at angles of $120^\circ$.
Two axial bonds lie perpendicular ($90^\circ$) to the equatorial plane (one above, one below).
The axial bond pairs suffer repulsion from three equatorial pairs at $90^\circ$, whereas equatorial pairs suffer repulsion from only two axial pairs at $90^\circ$.
To minimise this greater repulsive interaction, axial bonds stretch and become longer ($219\text{ pm}$) and weaker than equatorial bonds ($204\text{ pm}$).
sp³d hybridisation. Axial bonds suffer greater 90° repulsion from 3 equatorial pairs, making them longer (219 pm vs 204 pm).
4.39
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Definition: Hydrogen bond is the attractive electrostatic force which binds a hydrogen atom (already covalently linked to a highly electronegative atom like F, O, or N) with another electronegative atom of the same or another molecule.
Strength Comparison:
$$\text{Covalent Bond (200400 kJ/mol)} \gg \text{Hydrogen Bond (1040 kJ/mol)} \gg \text{van der Waals Forces (< 8 kJ/mol)}$$
Therefore, the hydrogen bond is much stronger than van der Waals forces, although weaker than a true covalent bond.
Attractive force between bonded H (with F, O, N) and another electronegative atom. It is much stronger than van der Waals forces.
4.40
What is meant by the term bond order? Calculate the bond order of: N₂, O₂, O₂⁺ and O₂⁻.
Bond Order: In molecular orbital theory, bond order is defined as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$):
$$\text{Bond Order} = \frac{1}{2}(N_b - N_a)$$
Calculations:
Correct Answer: Smaller cation polarises anion more strongly: Li⁺ (smallest) > Na⁺ > K⁺. Hence LiCl has the highest covalent character.
L3.3
In the adduct formed by BF₃ and NH₃ (F₃B·NH₃), what is the hybridisation of Boron and Nitrogen respectively?
Incorrect: That was the state before adduct formation.
Correct Answer: Boron accepts a coordinate bond, changing its steric number from 3 to 4 (sp³). Nitrogen retains 4 electron pairs (sp³).
Incorrect: Both have 4 single bonds in the adduct.
Incorrect: Boron has no d orbitals in Period 2.
L3.4
Which of the following molecules has a non-zero dipole moment?
Incorrect: XeF₄ is square planar (symmetrical, μ = 0).
Incorrect: SF₆ is octahedral (symmetrical, μ = 0).
Correct Answer: SF₄ has a see-saw shape with 1 equatorial lone pair; bond dipoles do not cancel (μ > 0).
Incorrect: CCl₄ is tetrahedral (symmetrical, μ = 0).
L3.5
In solid sodium chloride (NaCl), the sum of ΔᵢH(Na) and Δ_egH(Cl) is +147.1 kJ/mol. Why is NaCl formation highly exothermic and spontaneous?
Correct Answer: The huge electrostatic coulombic lattice energy release (−788 kJ/mol) more than compensates for the positive +147.1 kJ/mol, giving a net highly exothermic process.