Both NH 3 and NF 3 have trigonal pyramidal shape, yet the dipole moment of NH 3 (1.47 D) is greater than that of NF 3 (0.23 D). Explain.

In both NH 3 and NF 3 , nitrogen carries a lone pair whose orbital dipole points upward away from the three bond pairs. In NH 3 , nitrogen is more electronegative than H, so the resultant of the three N–H bond dipoles points in the same direction as the lone-pair orbital dipole → vectors reinforce (μ = 1.47 D). In NF 3 , fluorine is more electronegative than N, so the resultant of the three N–F bond dipoles points opposite to the lone-pair orbital dipole → partial cancellation lowers μ to 0.23 D.

Explain why H 2 O has a smaller bond angle (104.5°) than NH 3 (107°), even though both have distorted tetrahedral electron pair arrangements.

According to VSEPR theory, the order of electron pair repulsion is: lp–lp > lp–bp > bp–bp . NH 3 (AB 3 E): Central N has 3 bond pairs and 1 lone pair ; lp–bp repulsion compresses the H–N–H angle from 109.5° to 107° . H 2 O (AB 2 E 2 ): Central O has 2 bond pairs and 2 lone pairs ; strong lp–lp repulsion compresses the H–O–H angle further down to 104.5° .

Use Molecular Orbital Theory to prove that He 2 molecule does not exist, whereas Li 2 exists in the vapour phase.

He 2 (4 e − ): (σ1s) 2 (σ*1s) 2 → N b = 2, N a = 2 → Bond Order = ½(2 − 2) = 0 . Since BO = 0, He 2 is unstable and does not exist. Li 2 (6 e − ): (σ1s) 2 (σ*1s) 2 (σ2s) 2 → N b = 4, N a = 2 → Bond Order = ½(4 − 2) = 1 (diamagnetic), so Li 2 is stable in the vapour phase.

Compare the relative stability, bond order, and magnetic properties of O 2 , O 2 + , O 2 − (superoxide), and O 2 2− (peroxide).

All four follow the >14 e − MO filling order; extra electrons enter antibonding π*2p orbitals: O 2 + (15 e − ): π*2p 1 → BO = 2.5 , Paramagnetic (1 unpaired e − ) O 2 (16 e − ): π*2p 2 → BO = 2.0 , Paramagnetic (2 unpaired e − ) O 2 − (17 e − ): π*2p 3 → BO = 1.5 , Paramagnetic (1 unpaired e − ) O 2 2− (18 e − ): π*2p 4 → BO = 1.0 , Diamagnetic (0 unpaired e − ) Relative Stability Order: O 2 + > O 2 > O 2 − > O 2 2−

Predict the shapes of (a) ClF 3 and (b) XeF 4 using VSEPR theory.

(a) ClF 3 (AB 3 E 2 ): Central Cl has 3 bond pairs + 2 lone pairs (5 total → trigonal bipyramidal arrangement). Both lone pairs occupy equatorial sites → T-shaped . (b) XeF 4 (AB 4 E 2 ): Central Xe has 4 bond pairs + 2 lone pairs (6 total → octahedral arrangement). Two lone pairs stay trans (180° opposite) → Square planar .

NCERT Plus One (Class 11) Chemistry • Chapter 4

Chemical Bonding and Molecular Structure

Q: Draw the molecular orbital configuration of O 2 molecule. Calculate its bond order and explain why O 2 is paramagnetic whereas N 2 is diamagnetic.

O 2 (16 e − ): (σ1s) 2 (σ*1s) 2 (σ2s) 2 (σ*2s) 2 (σ2p z ) 2 (π2p x 2 = π2p y 2 )(π*2p x 1 = π*2p y 1 ) Bond Order = ½(N b − N a ) = ½(10 − 6) = 2 . Presence of 2 unpaired electrons in degenerate π*2p orbitals makes O 2 paramagnetic . N 2 (14 e − ): KK(σ2s) 2 (σ*2s) 2 (π2p x 2 = π2p y 2 )(σ2p z ) 2 → Bond Order = ½(10 − 4) = 3 . All electrons are paired → diamagnetic .

Q: What is hydrogen bonding? Distinguish between intermolecular and intramolecular hydrogen bonding with one example each.

Hydrogen bond: The attractive force binding a hydrogen atom (covalently attached to a strongly electronegative atom F, O, or N) with another electronegative atom (F, O, or N). Intermolecular H-bond: Formed between two different molecules of the same or different substances (e.g., HF, H 2 O, alcohols, p -nitrophenol). Intramolecular H-bond: Formed when H lies between two electronegative atoms within the same molecule (e.g., o -nitrophenol).

Kossel-Lewis Octet, Formal Charge, Ionic & Lattice Enthalpy, VSEPR Shapes, Hybridization, MOT & Hydrogen Bonding

📄 Official NCERT Textbook PDF

Topic 4.1: Kössel–Lewis Octet Rule, Covalent Bonds, Formal Charge & Octet Limitations

Picture atoms arriving at a cosmic celebration. The noble gases (with 8 outer electrons, or 2 for Helium) are already wearing complete matching outfits—calm and unreactive. Sodium walks in holding one extra valence electron it desperately wants to hand off, while Chlorine is just one electron short of a full eight-electron set! They trade outright to become Na⁺ and Cl⁻. Meanwhile, two Chlorine atoms meet each other: neither wants to give up an electron, so they clasp hands and share a pair right down the middle. Every bond is simply nature's clever way of lowering energy to reach stability!

The attractive force holding constituent atoms or ions together in different chemical species is called a chemical bond. In 1916, Kössel and G.N. Lewis proposed the Electronic Theory of Chemical Bonding based on noble gas inertness.

Lewis Symbols: Dots around an element's symbol denote its valence electrons. Group valence equals either the number of dots or 8 minus the number of dots.
Octet Rule: Atoms combine either by transferring electrons (electrovalent/ionic bond) or by sharing valence electron pairs (covalent bond, named by Langmuir in 1919) to attain an outer octet (ns²np⁶; duplet for H and Li).
Multiple Covalent Bonds: Sharing 1 pair = single bond (Cl₂, H₂O, CCl₄); 2 pairs = double bond (O₂, CO₂, C₂H₄); 3 pairs = triple bond (N₂, CO, C₂H₂).

Formal Charge (FC)

For polyatomic molecules or ions, formal charge tracks valence electrons assuming pure equal covalent sharing:

Formal Charge (FC) = [Total valence e⁻ in free atom (V)] − [Non-bonding lone-pair e⁻ (L)] − ½ [Bonding shared e⁻ (S)]

NCERT Example — Ozone (O₃): In the Lewis structure O(2)=O(1)−O(3):
• Central O(1) [1 lone pair, 3 bonds = 6 shared e⁻]: FC = 6 − 2 − ½(6) = +1
• Double-bonded O(2) [2 lone pairs, 2 bonds = 4 shared e⁻]: FC = 6 − 4 − ½(4) = 0
• Single-bonded O(3) [3 lone pairs, 1 bond = 2 shared e⁻]: FC = 6 − 6 − ½(2) = −1
Rule: Among competing Lewis structures, the lowest-energy (most stable) structure has the smallest formal charges.

Three Key Limitations of the Octet Rule

Exception Category	Explanation	NCERT Examples
1. Incomplete Octet	Central atom has fewer than 8 electrons (elements with <4 valence e⁻).	LiCl (2), BeH₂ (4), BCl₃ / BF₃ / AlCl₃ (6 e⁻)
2. Odd-Electron Molecules	Total valence e⁻ is odd; all atoms cannot achieve an octet.	NO (11 valence e⁻), NO₂ (17 valence e⁻)
3. Expanded Octet	Period 3+ elements use empty 3d orbitals alongside 3s and 3p to hold >8 e⁻.	PF₅ (10 e⁻), SF₆ (12 e⁻), H₂SO₄ (12 e⁻)

Other drawbacks: Noble gases like Xe and Kr actually form compounds (XeF₂, KrF₂, XeOF₂); the octet rule cannot predict molecular shapes or explain relative bond energies.

Formal charge is NOT a real physical charge! It is purely an electron-accounting tool assuming 100% equal sharing. Also, Period 2 elements (C, N, O, F) have no 2d orbitals and can NEVER expand their octet beyond 8 electrons—only Period 3 and heavier elements (P, S, Cl, Xe) can form expanded octets!

Topic 4.2: Ionic (Electrovalent) Bond, Crystal Packing & Lattice Enthalpy

Wait—here is a puzzle! Removing an electron from Sodium gas costs +495.8 kJ/mol, while Chlorine giving that electron a home only pays back −348.7 kJ/mol. That leaves us +147.1 kJ/mol in debt! So why on Earth does table salt (NaCl) form spontaneously and sit rock-solid in your kitchen? Because once Na⁺ and Cl⁻ ions snap together into a giant 3D checkerboard crystal, nature pays a massive energy jackpot of −788 kJ/mol—the Lattice Formation Enthalpy!

An ionic (electrovalent) bond forms by complete electron transfer from a metal atom to a non-metal atom, producing oppositely charged ions held together by strong electrostatic attraction.

Favourable Conditions for Ionic Bond Formation

Low Ionization Enthalpy (Δ_iH) of the electropositive metal (easier cation formation; ionization is always endothermic).
High Negative Electron Gain Enthalpy (Δ_egH) of the electronegative non-metal (easier anion formation).
High Lattice Enthalpy (stronger 3D crystal stabilization; note NH₄⁺ is an exception of a cation formed by two non-metals).

Why NaCl(s) is Stable (NCERT Energetics):
• Na(g) → Na⁺(g) + e⁻   (Δ_iH = +495.8 kJ mol⁻¹)
• Cl(g) + e⁻ → Cl⁻(g)   (Δ_egH = −348.7 kJ mol⁻¹)
• Gas-phase net = +495.8 − 348.7 = +147.1 kJ mol⁻¹ (still unfavourable!)
• Na⁺(g) + Cl⁻(g) → NaCl(s)   (Enthalpy of lattice formation = −788 kJ mol⁻¹) → Net exothermic & rock-solid!

4.2.1 Lattice Enthalpy

Lattice Enthalpy: Energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.
NaCl(s) → Na⁺(g) + Cl⁻(g) (Lattice Enthalpy = +788 kJ mol⁻¹)

Because crystals are 3D, lattice enthalpy depends on ion charges, ionic sizes (smaller ions pack closer → higher lattice enthalpy), and crystal geometry (such as the rock-salt structure of NaCl).

Do NOT confuse Lattice Enthalpy (+788 kJ mol⁻¹, energy required to break 1 mol solid into gaseous ions) with Enthalpy of Lattice Formation (−788 kJ mol⁻¹, energy released when gaseous ions form 1 mol crystal). Also remember: ionic compounds owe their stability to lattice enthalpy, not merely to gaseous ions attaining octets!

Topic 4.3.1 – 4.3.5: Bond Parameters (Length, Angle, Enthalpy, Order) & Resonance

Look at Ozone (O₃). Draw it on paper and you are forced to draw one single O–O bond (148 pm) and one double O=O bond (121 pm). Yet when scientists measured real ozone with spectroscopy, they got a shock: BOTH bonds measured an identical 128 pm! Neither single drawing tells the truth—just like a rhinoceros isn't a unicorn flipping back and forth into a dragon, ozone is a single blended Resonance Hybrid!

Bond Parameter	Definition & Unit	Key NCERT Values & Trends
Bond Length	Equilibrium distance between nuclei of two bonded atoms (pm). R = r_A + r_B.	In Cl₂: covalent radius r_c = 99 pm (198/2), van der Waals radius r_vdw = 180 pm (360/2). C–C (154 pm) > C=C (133/134 pm) > C≡C (120 pm).
Bond Angle	Angle between orbitals containing bonding electron pairs around central atom (°).	Determines molecular geometry; e.g., H–O–H = 104.5°, H–N–H = 107°, H–C–H = 109.5°.
Bond Enthalpy	Energy to break 1 mol of gaseous bonds (kJ mol⁻¹). Larger value = stronger bond.	H₂: 435.8; HCl: 431.0; O₂: 498; N₂: 946.0 kJ mol⁻¹. In H₂O, O–H bonds break at 502 and 427 → mean bond enthalpy = 464.5 kJ mol⁻¹.
Bond Order (BO)	Number of covalent bonds between two atoms.	H₂ = 1, O₂ = 2, N₂ = 3, CO = 3. Isoelectronic species have equal BO: (F₂, O₂²⁻ → BO 1) and (N₂, CO, NO⁺ → BO 3).

Master Stability Correlation: ↑ Bond Order ⇒ ↑ Bond Enthalpy (Stronger) ⇒ ↓ Bond Length (Shorter)

4.3.5 Resonance Structures

Whenever a single Lewis structure cannot match experimental properties, multiple contributing canonical structures (with identical nuclear positions and similar energies, linked by ↔) combine into one real, lower-energy resonance hybrid.

Classic NCERT Examples:
• Ozone (O₃): 2 canonical forms (O=O–O ↔ O–O=O) blend into equal 128 pm bonds (between 121 and 148 pm).
• Carbonate Ion (CO₃²⁻, Problem 4.3): 3 equivalent canonical forms make all three C–O bonds identical (BO = 1.33).
• Carbon Dioxide (CO₂, Problem 4.4): Experimental C–O length is 115 pm, between C=O (121 pm) and C≡O (110 pm).

Biggest Resonance Misconception: Canonical forms do NOT exist in reality, and the molecule does NOT flip rapidly from one form to another (unlike keto-enol tautomerism!). Only the single resonance hybrid exists—and its energy is ALWAYS lower than any individual canonical structure.

Topic 4.3.6: Polarity of Bonds, Dipole Moment (BF₃ vs NH₃ vs NF₃) & Fajans' Rules

Here is one of NCERT Chemistry's greatest magic tricks: Fluorine is the most electronegative element on the periodic table—far stronger at tugging electrons than Hydrogen. Yet when Nitrogen bonds with three Fluorines (NF₃), its net dipole moment is a tiny 0.23 D, while Ammonia (NH₃) boasts a huge 1.47 D! Why does the stronger electron-puller lose the tug-of-war? Watch the lone pair's orbital dipole!

In homonuclear molecules (H₂, Cl₂), electrons sit halfway → nonpolar covalent bond. In heteronuclear molecules (HF), the more electronegative atom pulls electron density toward itself → polar covalent bond.

Dipole Moment (μ) = Charge (Q) × Distance of Separation (r) | 1 Debye (D) = 3.33564 × 10⁻³⁰ C m

In chemistry, dipole moment is drawn with a crossed arrow (+→) pointing from the positive centre toward the negative electron shift.

Polyatomic Vector Sums & Classic Dipole Comparisons

Molecule	Geometry	Dipole Moment μ (D)	Why?
HF / HCl / HBr / HI	Linear	1.78 / 1.07 / 0.79 / 0.38 D	Decreases as electronegativity difference drops.
BeF₂ / CO₂	Linear (180°)	0 D	Two equal opposing bond dipoles cancel completely.
H₂O	Bent (104.5°)	1.85 D (6.17 × 10⁻³⁰ C m)	Bent shape prevents cancellation; O–H dipoles + lone pairs add up.
BF₃	Trigonal planar (120°)	0 D	Resultant of any two B–F dipoles is equal & opposite to the third.
NH₃	Trigonal pyramidal	1.47 D (4.90 × 10⁻³⁰ C m)	Lone-pair orbital dipole and resultant of 3 N–H dipoles point in the SAME direction (add up!).
NF₃	Trigonal pyramidal	0.23 D (0.80 × 10⁻³⁰ C m)	Lone-pair orbital dipole points UP while resultant of 3 N–F dipoles points DOWN (OPPOSE & cancel partially!).
CCl₄ / CHCl₃	Tetrahedral	CCl₄: 0 D \| CHCl₃: 1.04 D	Symmetric CCl₄ cancels completely; CHCl₃ is asymmetric.

Fajans' Rules — Partial Covalent Character of Ionic Bonds

Just as covalent bonds have partial ionic character, ionic bonds have partial covalent character when a cation distorts (polarises) the electron cloud of an anion:

Smaller cation & larger anion → greater covalent character (e.g., LiI is more covalent than KI or LiF).
Greater charge on cation → stronger polarising power → greater covalent character.
Pseudo-noble-gas configuration (n−1)dⁿns⁰ (transition metal cations like Cu⁺) polarises anions MORE than noble-gas ns²np⁶ cations (Na⁺) of similar size & charge!

Exam Goldmine: Students often think NF₃ > NH₃ because F is more electronegative than H! Always check vector directions: in NH₃ the three N←H bond vectors point UP toward N, reinforcing the upward lone pair (1.47 D); in NF₃ the three N→F vectors point DOWN away from N, fighting the upward lone pair (0.23 D)!

Topic 4.4: VSEPR Theory: Predicting 3D Shapes & Lone Pair Repulsion

Imagine tying four identical balloons together at their knots: they naturally push away from one another to point toward the four corners of a tetrahedron (109.5°). Now imagine replacing one balloon with a huge invisible pufferfish—a lone pair! Because a lone pair belongs only to the central atom instead of being stretched between two nuclei, it hogs extra space and squeezes the remaining bond balloons closer together: 109.5° in CH₄ shrinks to 107° in NH₃, and two pufferfish in H₂O squeeze it further to 104.5°!

Proposed by Sidgwick and Powell (1940) and refined by Nyholm and Gillespie (1957), Valence Shell Electron Pair Repulsion (VSEPR) theory predicts 3D molecular shapes simply by minimizing repulsions among valence electron pairs around the central atom.

Order of Repulsive Interaction: Lone Pair–Lone Pair (lp–lp) > Lone Pair–Bond Pair (lp–bp) > Bond Pair–Bond Pair (bp–bp)
Note: A multiple bond (double or triple) is treated as a single "super pair" for geometry prediction.

Complete NCERT VSEPR Geometry Reference (Tables 4.6 – 4.8)

Type (BP + LP)	Electron Pair Geometry	Molecular Shape	Bond Angle	Examples
AB₂ (2 + 0)	Linear	Linear	180°	BeCl₂, HgCl₂
AB₃ (3 + 0)	Trigonal planar	Trigonal planar	120°	BF₃
AB₂E (2 + 1)	Trigonal planar	Bent / V-shape	119.5°	SO₂, O₃
AB₄ (4 + 0)	Tetrahedral	Tetrahedral	109.5°	CH₄, NH₄⁺
AB₃E (3 + 1)	Tetrahedral	Trigonal pyramidal	107°	NH₃
AB₂E₂ (2 + 2)	Tetrahedral	Bent / Angular	104.5°	H₂O
AB₅ (5 + 0)	Trigonal bipyramidal	Trigonal bipyramidal	120° (eq), 90° (ax)	PCl₅, AsF₅
AB₄E (4 + 1)	Trigonal bipyramidal	See-saw (LP equatorial)	~102°, ~173°	SF₄
AB₃E₂ (3 + 2)	Trigonal bipyramidal	T-shape (2 LPs eq)	~87.5°	ClF₃
AB₆ (6 + 0)	Octahedral	Octahedral	90°	SF₆
AB₅E (5 + 1)	Octahedral	Square pyramidal	<90°	BrF₅
AB₄E₂ (4 + 2)	Octahedral	Square planar (2 LPs trans)	90°	XeF₄

Exam Classic — SF₄ & ClF₃ Lone Pair Placement: In a trigonal bipyramid (5 electron pairs), lone pairs ALWAYS go into equatorial positions, NEVER axial! Why? An equatorial lone pair faces only two 90° lp–bp repulsions, whereas an axial lone pair would suffer three severe 90° lp–bp repulsions. Also distinguish electron arrangement (tetrahedral in NH₃/H₂O) from molecular shape (pyramidal / bent)!

Topic 4.5 – 4.6: Valence Bond Theory (σ vs π) & Hybridization (sp, sp², sp³, sp³d, sp³d²)

Wait—Carbon's valence shell has one spherical 2s orbital and three 2p orbitals sitting at 90° to one another. If those pure orbitals simply overlapped with four Hydrogen 1s orbitals, methane (CH₄) would have three 90° bonds and one oddball bond! Yet real methane has four 100% identical bonds at 109.5°. How did Carbon solve it? Like pouring one cup of red paint and three cups of white paint into a blender to get four identical cups of pink paint—Carbon blends its orbitals first through Hybridization!

4.5 Valence Bond Theory (Heitler–London, 1927; Pauling)

Energetics of H₂ Formation: As two H atoms approach, attractive forces (nucleus–other electron) overcome repulsive forces until minimum potential energy is reached at equilibrium bond length 74 pm, releasing 435.8 kJ mol⁻¹ (bond enthalpy).
Orbital Overlap: Only positive (in-phase) overlap forms a bond. Stronger overlap → stronger covalent bond.
Sigma (σ) vs Pi (π) Bonds:
- σ bond: Head-on (axial) overlap along internuclear axis (s–s, s–p, p–p axial). Larger overlap → stronger; allows free rotation.
- π bond: Sidewise (lateral) overlap of parallel p orbitals perpendicular to internuclear axis. Smaller overlap → weaker; two lobes above & below molecular plane; restricts rotation; forms only alongside a σ bond in double (1σ + 1π) and triple (1σ + 2π) bonds.

4.6 Hybridization (Linus Pauling)

Intermixing of valence atomic orbitals of slightly different energies to form an equal number of identical, directional hybrid orbitals with minimum repulsion.

Hybridization	Orbitals Mixed & % s	Geometry & Angles	Key NCERT Molecules & Bond Data
sp	1 s + 1 p (50% s)	Linear (180°)	BeCl₂, C₂H₂ (ethyne: C≡C 120 pm; 3σ + 2π total)
sp²	1 s + 2 p (33.3% s)	Trigonal planar (120°)	BCl₃, C₂H₄ (ethene: C=C 134 pm, C–H 108 pm, ∠HCH 117.6°, ∠HCC 121°; 5σ + 1π)
sp³	1 s + 3 p (25% s)	Tetrahedral (109.5°)	CH₄, C₂H₆ (C–C 154 pm, C–H 109 pm), NH₃ (107°), H₂O (104.5°)
dsp²	1 d + 1 s + 2 p	Square planar (90°)	[Ni(CN)₄]²⁻, [PtCl₄]²⁻
sp³d	1 s + 3 p + 1 d	Trigonal bipyramidal	PF₅, PCl₅ (3 equatorial 120°, 2 axial 90°)
sp³d² / d²sp³	1 s + 3 p + 2 d	Octahedral (90°)	SF₆ (sp³d²), BrF₅ (sq. pyramidal), [CrF₆]³⁻, [Co(NH₃)₆]³⁺

Top Exam Questions:
(1) Why are axial P–Cl bonds in PCl₅ longer and weaker than equatorial ones? Because each axial bond pair suffers 3 repulsions at 90° from the equatorial pairs, whereas an equatorial pair suffers only 2 at 90°—making PCl₅ thermally reactive!
(2) If the x-axis is the internuclear axis, 2p_y + 2p_y forms a π bond, NOT a σ bond!
(3) Electron promotion is NOT mandatory before hybridization (filled lone-pair orbitals in NH₃ and H₂O also hybridize!).

Topic 4.7 – 4.8: Molecular Orbital Theory (MOT), Energy Ordering & Paramagnetism of O₂

Every theory we studied so far hits an embarrassing wall with ordinary Oxygen gas (O₂). Lewis dots and Valence Bond Theory pair up every single electron in O=O and swear Oxygen must be repelled by a magnet. Yet pour pale-blue liquid Oxygen between the poles of a strong magnet—and it sticks right in the magnetic field! Only Molecular Orbital Theory solved this mystery by placing O₂'s final two electrons unpaired inside two degenerate antibonding π* orbitals!

Developed by Hund and Mulliken (1932), MOT treats electrons as occupying polycentric molecular orbitals (MOs) formed by Linear Combination of Atomic Orbitals (LCAO):

Constructive interference (ψ_A + ψ_B) → Bonding MO (σ, π): electron density builds between nuclei → lower energy, stabilizes molecule.
Destructive interference (ψ_A − ψ_B) → Antibonding MO (σ*, π*): nodal plane of zero density between nuclei → higher energy, destabilizes molecule.
3 LCAO Conditions: (1) Combining AOs must have similar energy, (2) same symmetry about the z molecular axis (2p_z combines with 2p_z, never 2p_x/2p_y), and (3) maximum overlap.

1. Energy Order WITH 2s–2p mixing (≤ 14 e⁻: Li₂, Be₂, B₂, C₂, N₂):
σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z

2. Energy Order WITHOUT 2s–2p mixing (> 14 e⁻: O₂, F₂, Ne₂):
σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z

Bond Order (BO) = ½ (N_b − N_a) | Positive BO → Stable; Zero/Negative BO → Does not exist!

Species (Total e⁻)	Valence MO Configuration	Bond Order	Magnetic Nature (Unpaired e⁻)
H₂ (2) / He₂ (4)	(σ1s)² / (σ1s)²(σ*1s)²	H₂: 1 \| He₂: 0	H₂: Dia (0) \| He₂: Does not exist!
Li₂ (6) / Be₂ (8)	KK(σ2s)² / KK(σ2s)²(σ*2s)²	Li₂: 1 \| Be₂: 0	Li₂: Dia (0) \| Be₂: Does not exist!
B₂ (10)	KK(σ2s)²(σ*2s)²(π2p_x¹ = π2p_y¹)	1	Paramagnetic (2 unpaired e⁻)
C₂ (12)	KK(σ2s)²(σ*2s)²(π2p_x² = π2p_y²)	2	Diamagnetic (Both bonds are π bonds!)
N₂ (14)	KK(σ2s)²(σ*2s)²(π2p⁴)(σ2p_z)²	3	Diamagnetic (945 kJ mol⁻¹, 110 pm)
O₂ (16)	KK(σ2s)²(σ2s)²(σ2p_z)²(π2p⁴)(π2p_x¹=π*2p_y¹)	2	Paramagnetic (2 unpaired e⁻ in π*2p)
O₂⁺ / O₂⁻ / O₂²⁻	π*2p has 1 / 3 / 4 electrons	2.5 / 1.5 / 1.0	O₂⁺ & O₂⁻: Para (1) \| O₂²⁻: Dia (0)
F₂ (18) / Ne₂ (20)	π2p⁴ / π2p⁴ σ*2p_z²	F₂: 1 \| Ne₂: 0	F₂: Dia (0) \| Ne₂: Does not exist!

Three High-Frequency MOT Traps:
1. Up to 14 electrons (B₂, C₂, N₂), π2p is LOWER than σ2p_z—that single switch explains why B₂ is paramagnetic and C₂ has two π bonds and no σ bond!
2. Stability order of Oxygen species follows Bond Order: O₂⁺ (2.5) > O₂ (2.0) > O₂⁻ (1.5) > O₂²⁻ (1.0), while bond length reverses.
3. He₂, Be₂, Ne₂ all have Bond Order = 0, so they do not exist!

Topic 4.9: Hydrogen Bonding: Intermolecular vs Intramolecular & Why Ice Floats

Look at two close chemical cousins from Group 16: H₂S is a foul-smelling gas at room temperature, yet H₂O—which is lighter!—is a life-giving liquid that boils only at 100 °C. Even stranger, when water freezes into ice, it expands and floats on top of lakes, insulating fish beneath during freezing winters! Both miracles come from one special electrostatic bridge: the Hydrogen Bond.

When hydrogen is covalently bonded to a strongly electronegative atom F, O, or N, the shared electron pair shifts far away from H. The partially positive hydrogen (H^δ+) then forms an attractive bridge with another electronegative atom (F, O, or N), shown by a dotted line (---).

⋯ H^δ+—F^δ− ⋯ H^δ+—F^δ− ⋯ H^δ+—F^δ−
Strength: Stronger than van der Waals forces, much weaker than covalent bonds. Magnitude is maximum in solids and minimum in gases!

Two Types of Hydrogen Bonding

Type	Definition	NCERT Examples & Consequences
1. Intermolecular H-Bond	Forms between different molecules of the same or different compounds.	HF, H₂O, NH₃, alcohols, p-nitrophenol. Raises boiling point dramatically (H₂O liquid vs H₂S gas; HF > HCl). In ice, each H₂O forms 4 tetrahedral H-bonds creating an open cage structure with lower density than liquid water (max density of water at 4 °C)!
2. Intramolecular H-Bond	Forms when H sits between two F, O, or N atoms within the SAME molecule (forming a ring).	o-Nitrophenol (H between phenolic O and nitro O). Prevents intermolecular association → lower boiling point & steam-volatile compared to p-nitrophenol!

Board Exam Classic: Why is o-nitrophenol steam volatile while p-nitrophenol is not? Because o-nitrophenol shows intramolecular H-bonding (stays as independent molecules with low b.p.), whereas p-nitrophenol forms intermolecular H-bonded chains (high b.p.)! Also remember: H-bonds only form significantly with F, O, and N—not Cl or S (except rare cases like chloral hydrate), so HCl and H₂S lack strong H-bonding.

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Classic) • 2 Marks): Both NH₃ and NF₃ have trigonal pyramidal shape, yet the dipole moment of NH₃ (1.47 D) is greater than that of NF₃ (0.23 D). Explain.

In both NH₃ and NF₃, nitrogen carries a lone pair whose orbital dipole points upward away from the three bond pairs.

In NH₃, nitrogen is more electronegative than H, so the resultant of the three N–H bond dipoles points in the same direction as the lone-pair orbital dipole → vectors reinforce (μ = 1.47 D).
In NF₃, fluorine is more electronegative than N, so the resultant of the three N–F bond dipoles points opposite to the lone-pair orbital dipole → partial cancellation lowers μ to 0.23 D.

Q2 (Kerala HSE March / SAY (Very Important) • 3 Marks): Draw the molecular orbital configuration of O₂ molecule. Calculate its bond order and explain why O₂ is paramagnetic whereas N₂ is diamagnetic.

O₂ (16 e⁻): (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p_z)²(π2p_x² = π2p_y²)(π*2p_x¹ = π*2p_y¹)
Bond Order = ½(N_b − N_a) = ½(10 − 6) = 2. Presence of 2 unpaired electrons in degenerate π*2p orbitals makes O₂ paramagnetic.

N₂ (14 e⁻): KK(σ2s)²(σ*2s)²(π2p_x² = π2p_y²)(σ2p_z)² → Bond Order = ½(10 − 4) = 3. All electrons are paired → diamagnetic.

Q3 (Kerala HSE Model / NCERT Ex 4.7 & 4.8 • 3 Marks): Explain why H₂O has a smaller bond angle (104.5°) than NH₃ (107°), even though both have distorted tetrahedral electron pair arrangements.

According to VSEPR theory, the order of electron pair repulsion is: lp–lp > lp–bp > bp–bp.

NH₃ (AB₃E): Central N has 3 bond pairs and 1 lone pair; lp–bp repulsion compresses the H–N–H angle from 109.5° to 107°.
H₂O (AB₂E₂): Central O has 2 bond pairs and 2 lone pairs; strong lp–lp repulsion compresses the H–O–H angle further down to 104.5°.

Q4 (Kerala HSE March • 2 Marks): Explain the hybridization of Phosphorus in PCl₅. Why are the axial P–Cl bonds longer and weaker than the equatorial bonds?

Phosphorus (Z=15) in excited state uses one 3s, three 3p, and one 3d orbital to undergo sp³d hybridization, giving trigonal bipyramidal geometry with three equatorial bonds (120°) and two axial bonds (90° to equatorial plane).

Reason: Each axial bond pair suffers three 90° repulsive interactions from equatorial bond pairs, whereas each equatorial pair suffers only two 90° interactions. Hence axial bonds are slightly longer and weaker, making PCl₅ reactive.

Q5 (Kerala HSE SAY (High Yield) • 2 Marks): Use Molecular Orbital Theory to prove that He₂ molecule does not exist, whereas Li₂ exists in the vapour phase.

He₂ (4 e⁻): (σ1s)²(σ*1s)² → N_b = 2, N_a = 2 → Bond Order = ½(2 − 2) = 0. Since BO = 0, He₂ is unstable and does not exist.

Li₂ (6 e⁻): (σ1s)²(σ*1s)²(σ2s)² → N_b = 4, N_a = 2 → Bond Order = ½(4 − 2) = 1 (diamagnetic), so Li₂ is stable in the vapour phase.

Q6 (Kerala HSE March / NCERT Ex 4.36 & 4.40 • 3 Marks): Compare the relative stability, bond order, and magnetic properties of O₂, O₂⁺, O₂⁻ (superoxide), and O₂²⁻ (peroxide).

All four follow the >14 e⁻ MO filling order; extra electrons enter antibonding π*2p orbitals:

O₂⁺ (15 e⁻): π*2p¹ → BO = 2.5, Paramagnetic (1 unpaired e⁻)
O₂ (16 e⁻): π*2p² → BO = 2.0, Paramagnetic (2 unpaired e⁻)
O₂⁻ (17 e⁻): π*2p³ → BO = 1.5, Paramagnetic (1 unpaired e⁻)
O₂²⁻ (18 e⁻): π*2p⁴ → BO = 1.0, Diamagnetic (0 unpaired e⁻)

Relative Stability Order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻

Q7 (Kerala HSE March • 2 Marks): What is hydrogen bonding? Distinguish between intermolecular and intramolecular hydrogen bonding with one example each.

Hydrogen bond: The attractive force binding a hydrogen atom (covalently attached to a strongly electronegative atom F, O, or N) with another electronegative atom (F, O, or N).

Intermolecular H-bond: Formed between two different molecules of the same or different substances (e.g., HF, H₂O, alcohols, p-nitrophenol).
Intramolecular H-bond: Formed when H lies between two electronegative atoms within the same molecule (e.g., o-nitrophenol).

Q8 (Kerala HSE Model / NCERT • 2 Marks): Predict the shapes of (a) ClF₃ and (b) XeF₄ using VSEPR theory.

(a) ClF₃ (AB₃E₂): Central Cl has 3 bond pairs + 2 lone pairs (5 total → trigonal bipyramidal arrangement). Both lone pairs occupy equatorial sites → T-shaped.

(b) XeF₄ (AB₄E₂): Central Xe has 4 bond pairs + 2 lone pairs (6 total → octahedral arrangement). Two lone pairs stay trans (180° opposite) → Square planar.

Q9 (Kerala HSE SAY • 2 Marks): Write any two limitations of the octet rule with suitable chemical examples.

Incomplete octet of central atom: Molecules where central atom has fewer than 8 electrons, e.g., BeH₂ (4 e⁻), BF₃ / BCl₃ (6 e⁻).
Expanded octet: Elements of Period 3 and beyond use d orbitals to hold >8 electrons, e.g., PF₅ (10 e⁻), SF₆ (12 e⁻), H₂SO₄ (12 e⁻). (Also valid: odd-electron molecules like NO, NO₂; noble gas compounds like XeF₂.)

Q10 (Kerala HSE Improvement • 1 Marks): Why is the dipole moment of BeF₂ and BF₃ equal to zero even though Be–F and B–F bonds are polar?

In linear BeF₂ (180°), two equal Be–F bond dipoles act in opposite directions and cancel out. In trigonal planar BF₃ (120°), the resultant of any two B–F bond dipoles is equal in magnitude and opposite in direction to the third, giving net μ = 0.

Q11 (Kerala HSE March • 2 Marks): Define formal charge. Calculate the formal charge on each oxygen atom in the O₃ molecule.

Formal Charge: The difference between the number of valence electrons of an atom in an isolated state and the number assigned to it in a Lewis structure: FC = V − L − ½S.

In O(2)=O(1)−O(3): Central O(1) = 6 − 2 − ½(6) = +1; Double-bonded O(2) = 6 − 4 − ½(4) = 0; Single-bonded O(3) = 6 − 6 − ½(2) = −1.

Q12 (Kerala HSE SAY / NCERT • 2 Marks): Distinguish between a sigma (σ) bond and a pi (π) bond. Why is a σ bond stronger than a π bond?

A σ bond is formed by head-on (axial) overlap along the internuclear axis and is cylindrically symmetrical, whereas a π bond is formed by sidewise (lateral) overlap of parallel orbitals perpendicular to the axis (forming electron clouds above and below the plane).

Because axial overlap occurs to a much greater extent than sidewise overlap, a σ bond is stronger than a π bond.