VKC Notes · Chemical bonding · Civil Services P-I · IFoS P-I · Geo-Scientist Prelims and P-I

2.2 Covalent Bonding: VSEPR, Hybridisation and Bent's Rule

★★★ Mains 2021–2025: 14 parts, 134 marks (Civil Services 9 parts / 95 m · IFoS 4 / 29 m · Geo-Scientist 1 / 10 m). Geo-Scientist Prelims 2021–2026: 21 questions, three or four every year.

Syllabus lines. Civil Services and IFoS: “covalent bond and its general characteristics, polarities of bonds in molecules and their dipole moments; valence bond theory, concept of resonance and resonance energy”; shapes of interhalogen and noble-gas compounds come under main-group chemistry. Geo-Scientist Prelims: chemical bonding (VSEPR, hybridisation, resonance, dipole moment). Ionic bonding and MO theory get notes of their own.

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CSE 2023 Q3(a) = asked there; tap to open the solved answerSN = line kept in the short notes

2.2.1 Lewis structures and formal charge ★★

Count valence electrons; add one per negative charge, remove one per positive charge.
Put the least electronegative atom in the centre (never H or F); join atoms by single bonds.
Complete octets on the outer atoms, then put what is left on the centre. Atoms of period 3 and below may hold more than eight.
If the centre is short of an octet, turn outer lone pairs into multiple bonds.

(2.2.1)

V = valence electrons of the free atom, N = non-bonding electrons, B = bonding electrons on that atom. The sum of formal charges equals the charge on the species.SN

The best structure has the fewest and smallest formal charges, with any negative charge on the more electronegative atom and no like charges on adjacent atoms.SN
[Br₃]⁻: 22 electrons; the central Br has 2 bond pairs and 3 lone pairs (expanded octet), formal charges 0, −1, 0 CSE 2023 Q1(b) 10m.

Solved PYQ CSE 2022 Q3(a) 10m

Which is least stable: OCN⁻, ONC⁻, SCN⁻?

All three have 16 valence electrons (like CO₂) and are linear; they differ in the middle atom.
Cyanate and thiocyanate: carbon in the middle, formal charge 0 on C, a single −1 on an end atom shared by resonance.
Fulminate: nitrogen in the middle with four bonds is +1, and the best structure still puts −1 on carbon, the less electronegative end: three formal charges in all.
So ONC⁻ (fulminate) is least stable; SCN⁻ ≈ OCN⁻ ≫ ONC⁻. Mercury fulminate is a primary explosive.SN

Fig. 2.2.1 Best Lewis structures of the three 16-electron anions.

Draw it in 60 s: write O–C≡N, S–C≡N, O–N≡C in a column; put the charges (−, 0, 0), (−, 0, 0), (−, +, −); circle the −1 on carbon.

2.2.2 Resonance and resonance energy ★★★ Prelims every year

Contributing structures differ only in where the electrons are: same positions of nuclei, same total electron count, same net charge, same number of unpaired electrons.SN
A structure contributes more when its energy is close to the others, its octets are complete, its formal charges are few, and its negative charge sits on the electronegative atom Geo Pre 2023 Q26.
The hybrid is one real structure, lower in energy than any contributor. Resonance energy = energy of the most stable contributor − energy of the real molecule; benzene ≈ 150 kJ mol⁻¹ (3 × 120 expected for hydrogenation − 208 found).SN
Bond order from resonance = total bonds between the two atoms ÷ number of contributors over which they spread: NO₃⁻ and CO₃²⁻ 4/3, SO₄²⁻ (octet picture) 3/2, benzene 3/2.SN
In valence-bond language H₂ itself is a resonance of covalent and ionic forms: Ψ = Ψ_cov + λ(Ψ_H⁺H⁻ + Ψ_H⁻H⁺), ionic–covalent resonance Geo Pre 2024 Q25.

Solved PYQ IFoS 2025 Q1(a) 8m

All N–O bonds in nitrate are 1.21 Å, between N–O (1.36 Å) and N=O (1.15 Å). Account for this.

Three equivalent contributors: one N=O and two N–O⁻, N formal charge +1; the double bond is placed on each O in turn.
Bond order of each N–O = (2 + 1 + 1)/3 = 4/3, so all three bonds are equal and lie between single and double.
The ion is trigonal planar (120°); the −1 charge is spread over the three oxygens.

Fig. 2.2.2 Nitrate: three equivalent contributors.

Draw it in 60 s: draw N⁺ with three O at 120°, one double bond; copy twice moving the double bond round; curly arrow from an O⁻ lone pair to its N–O bond and from the N=O π bond to that O; ⟷ between.

Prelims: which is not a resonance form of OCN⁻ Geo Pre 2021 Q18 or SCN⁻ Geo Pre 2022 Q24 (check the electron count and the net charge first) · CO₃²⁻ has 3 contributors Geo Pre 2025 Q24 · equal S–O bonds in SO₃ come from delocalised π bonding, not from hybrids Geo Pre 2024 Q16 · HN₃ bond lengths: H–N 101 < N≡N end 113 < middle N–N 124 pm Geo Pre 2026 Q24.

2.2.3 VSEPR: the rules ★★★

Electron pairs in the valence shell of the central atom, bonded and lone pairs both, repel one another and get as far apart as possible Geo Pre 2023 Q24.SN The steric number fixes the arrangement of pairs; the lone pairs then decide what shape the atoms make.

(2.2.2)

Fig. 2.2.3 The five arrangements of electron pairs.

Draw it in 60 s: draw a centre five times; give it 2, 3, 4, 5, 6 lines; for 4, 5, 6 make one line thick (towards you) and one dashed (away) → write the angle under each.

Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. A lone pair is held by one nucleus only, so it is fatter and sits closer to the centre.SN
Only 90° contacts matter when comparing arrangements; contacts at 120° and above are small.
Multiple bonds count as one pair for the shape but take more room: H–C–H in H₂C=O is 116.5°, not 120°.
Electronegative ligands pull bond pairs away from the centre and shrink the angles between them (NF₃ 102° < NH₃ 107°). A more electronegative centre pulls bond pairs in and opens the angles (H₂O 104.5° > H₂S 92°).SN

Steric no.	Arrangement	Lone pairs: shape (example)
2	linear	0: linear (BeCl₂, CO₂)
3	trigonal planar	0: trigonal planar (BF₃, SO₃) · 1: bent (SO₂, O₃)
4	tetrahedral	0: tetrahedral (CH₄, NH₄⁺) · 1: pyramidal (NH₃) · 2: bent (H₂O)
5	trigonal bipyramidal	0: TBP (PF₅) · 1: see-saw (SF₄) · 2: T-shaped (ClF₃) · 3: linear (XeF₂, I₃⁻)
6	octahedral	0: octahedral (SF₆) · 1: square pyramidal (BrF₅, XeOF₄) · 2: square planar (XeF₄, ICl₄⁻)
7	pentagonal bipyramidal	0: IF₇ · 1: distorted octahedron (XeF₆)

2.2.4 Shapes with lone pairs: where the lone pairs go ★★★ Civil Services favourite

Fig. 2.2.4 Trigonal bipyramid family.

Draw it in 60 s: draw PF₅ (vertical axis + three equatorial) four times; replace one, two, three equatorial bonds by red lobes → name: see-saw, T, linear.

In a trigonal bipyramid an axial site has three neighbours at 90°, an equatorial site only two. So lone pairs take equatorial sites.SN
SF₄: see-saw, F_ax–S–F_ax 173°, F_eq–S–F_eq 102° (both closed up by the lone pair). ClF₃: T-shaped, 87.5°. XeF₂, I₃⁻, ICl₂⁻: three equatorial lone pairs, exactly linear.SN

Solved PYQ CSE 2024 Q8(b)(i) 10m

Draw all possible arrangements of ClF₃ and decide which is favoured.

Cl: 7 + 3 = 10 electrons, 5 pairs (3 bond, 2 lone): trigonal bipyramid, sp³d.
(I) both lone pairs equatorial; (II) both axial; (III) one axial, one equatorial.
Count 90° contacts. III has a lone pair–lone pair contact at 90°: rejected first. I has 4 lp–bp contacts at 90°, II has 6. So I wins: T-shaped.
The lone pairs bend the axial F towards the equatorial one: F–Cl–F 87.5°; axial Cl–F 170 pm > equatorial 160 pm. The molecule is polar.

Fig. 2.2.5 Octahedral family.

Draw it in 60 s: draw the octahedron (vertical, horizontal, one thick, one dashed) three times; remove the bottom bond for BrF₅, top and bottom for XeF₄; put red lobes there.

All six octahedral sites are alike, so one lone pair can go anywhere: square pyramidal (BrF₅, XeOF₄). Two lone pairs go trans, 180° apart: square planar XeF₄, ICl₄⁻.SN
BrF₅: the lone pair pushes the four basal F up, F_ap–Br–F_bas = 84.8° CSE 2023 Q3(a) 10m.
XeF₆ (seven pairs) is a distorted octahedron: the lone pair is stereochemically active and pokes through a face. [SbCl₆]³⁻ and [TeCl₆]²⁻ are regular octahedra: there the lone pair sits in a spherical s orbital and is inactive.

Solved PYQ CSE 2023 Q3(a) 10m

Electronic and molecular geometry of BrF₅ and [ICl₂]⁻.

BrF₅: 7 + 5 = 12 electrons, 6 pairs (5 bond + 1 lone): electron pairs octahedral, molecule square pyramidal, sp³d².
[ICl₂]⁻: 7 + 2 + 1 = 10 electrons, 5 pairs (2 bond + 3 lone): electron pairs trigonal bipyramidal, ion linear, sp³d.

Always write both: the “electronic geometry” (arrangement of pairs) and the “molecular geometry” (shape of the atoms). The question asks for both by name.

Also asked: Geo 2025 Q5(a) 10m [CuCl₅]³⁻ trigonal bipyramidal, [Ni(CN)₅]³⁻ square pyramidal, SF₆ octahedral, I₃⁻ linear.

2.2.5 Bond-angle trends ★★★ Prelims

Series	Angles	Reason
NH₄⁺ > NH₃ > NH₂⁻	109.5 · 107 · 104.5°	0, 1, 2 lone pairs on an sp³ N Geo Pre 2021 Q16
H₂O > H₂S > H₂Se > H₂Te	104.5 · 92 · 91 · 90°	Larger, less electronegative centre; bonds become almost pure p (Drago) Geo Pre 2024 Q11
NH₃ > PH₃ > AsH₃	107 · 93.5 · 92°	Same reason down the group
NH₃ > NF₃	107 · 102°	F pulls bond pairs away from N Geo Pre 2024 Q17
Cl₂O > H₂O > OF₂	111 · 104.5 · 103°	Electronegativity of ligand, plus Cl…Cl crowding in Cl₂O
PF₃ > PH₃ (exception)	97.8 · 93.5°	Against the electronegativity rule: F lone pairs give P–F partial π character (F → P back-donation), and the shorter, fatter bonds repel more

Drago's rule: if the central atom is from period 3 or below, carries a lone pair, and the ligand electronegativity is ≤ 2.5, hybridisation is negligible and the angle is near 90° (PH₃, H₂S, AsH₃).SN
NF₃ is a very poor Lewis base: the three F atoms pull charge away from the lone pair Geo Pre 2024 Q17.

2.2.6 Hybridisation ★★★

Hybridisation is a description of the bonding once the shape is known, not the cause of the shape. Number of hybrids = steric number: 2 sp, 3 sp², 4 sp³, 5 sp³d, 6 sp³d².SN
Count σ bonds + lone pairs; π bonds do not count. An atom whose lone pair is in conjugation is sp², not sp³: amide N, enolate C, carboxylate O, pyrrole N, benzylic carbanion Geo Pre 2021 Q83.SN
Diborane: both B are sp³ (two terminal B–H and two 3c–2e bridges) Geo Pre 2022 Q17. Boric acid: B sp² (planar), O sp³ by count Geo Pre 2025 Q23. Imine N: sp² Geo Pre 2022 Q83.
Cytosine and thymine: every ring carbon is sp² (three σ bonds + one π), planar; only the CH₃ of thymine is sp³ CSE 2023 Q5(c) 10m.

Solved PYQ CSE 2021 Q2(b) 20m

Draw sp³d, sp³d², dsp² and sd³ arrangements and name the d orbitals used.

Hybrid	Shape	d orbital(s)	Example
sp³d	trigonal bipyramidal	d_z² (axial bonds lie along z)	PCl₅, [Fe(CO)₅]
sp³d²	octahedral	d_x²−y², d_z² (the e_g pair, along the axes)	SF₆, [CoF₆]³⁻
dsp²	square planar	d_x²−y²	[Ni(CN)₄]²⁻, [PtCl₄]²⁻
sd³	tetrahedral	d_xy, d_yz, d_zx (between the axes)	MnO₄⁻, CrO₄²⁻

The rule that earns the marks: a hybrid can point along a bond only if the d orbital mixed in has lobes in that direction.SN

2.2.7 Bent's rule and Coulson's relation ★★★ Prelims 2024, 2025

Bent's rule: an atom directs hybrids richer in s character towards electropositive substituents, and hybrids richer in p character towards electronegative ones.SN Lone pairs count as the most electropositive “substituent”: they take the most s.

Fig. 2.2.6 Bent’s rule in two molecules.

Draw it in 60 s: C with two H up and two F down; mark the H–C–H angle larger. Then a TBP with F on the axis and CH₃ on the equator; write “F axial, CH₃ equatorial”.

More s in a bond → shorter bond and wider angle. CH₂F₂: H–C–H 111.9° > F–C–F 108.3°.SN
In a trigonal bipyramid the axial sites are p-rich, the equatorial sp²-like: electronegative groups go axial (apicophilicity F > Cl > CH₃); alkyls and lone pairs go equatorial (PCl₂F₃: F axial; CH₃PF₄: CH₃ equatorial).SN
Bulky R groups open the C–N–C angle of R₃N; the C–N bonds take more s, the lone pair more p, and Lewis basicity towards a large acid falls Geo Pre 2025 Q31.

(2.2.3)

For two equivalent hybrids spᵡ making angle θ, s is the fractional s character of each (λ² = p/s). 180° → s = ½ (sp); 120° → ⅓ (sp²); 109.5° → ¼ (sp³); 90° → s = 0 (pure p).SN

Solved PYQ Geo Pre 2024 Q24

H₂O has θ = 104.5° (cos θ = −0.25). Find the p character of the non-bonding orbitals.

−0.25 = −s/(1 − s) → s = 0.20 in each O–H hybrid (p 80 %).
Two O–H hybrids use 0.40 of the 2s orbital; 0.60 s is left for the two lone-pair hybrids: 0.30 each.
p character of each lone-pair orbital = 1 − 0.30 = 70 %. The trap answer is 80 %, the bonding value.

2.2.8 Berry pseudorotation ★★

PF₅ has two kinds of F (2 axial, 3 equatorial), yet its ¹⁹F NMR shows one signal (a doublet from ³¹P), even at −100 °C.SN
One equatorial F stays as the pivot; the two axial F close to 120° while two equatorial F open to 180°, through a square-pyramidal transition state. No bond breaks, the barrier is small (about 15 kJ mol⁻¹), and the exchange is faster than the NMR time scale CSE 2024 Q8(a) 10m Geo Pre 2023 Q25.SN
It works in a trigonal bipyramid only. CH₃PF₄ pseudorotates with CH₃ kept as the pivot (one F signal). (CH₃)₂PF₃ cannot: every pseudorotation would send a CH₃ axial, so it is rigid and shows axial and equatorial F separately Geo Pre 2024 Q23.

2.2.9 Bond polarity and dipole moment ★★★ most asked · 4 Mains parts

(2.2.4)

(2.2.5)

The molecular dipole is the vector sum of bond moments and lone-pair moments, so shape matters as much as electronegativity. Two or more polar bonds can cancel: CO₂, BF₃, CCl₄ are non-polar Geo Pre 2021 Q85.SN
Pauling: ; Hannay–Smith: .
Reversed dipoles: CO is C⁻–O⁺ (0.11 D, the C lone pair wins); in H–C≡C–Cl the sp carbon is the negative end Geo Pre 2021 Q8.

Fig. 2.2.7 Why NH₃ beats NF₃.

Draw it in 60 s: two pyramids with a lobe on N; beside each, a red arrow up (lone pair) and a blue arrow: up for N–H, down for N–F → write “add” and “nearly cancel”.

Solved PYQ CSE 2025 Q1(f) 5m IFoS 2025 Q2(c)(i) 5m IFoS 2022 Q1(b) 8m

Arrange NH₃, NF₃, H₂O in increasing dipole moment.

NF₃: N–F moments point away from N, opposite to the lone-pair moment; they nearly cancel: 0.23 D.
NH₃: N–H moments point towards N, the same way as the lone pair; they add: 1.47 D.
H₂O: O–H more polar than N–H, two lone pairs along the bisector: 1.85 D.
NF₃ < NH₃ < H₂O. Draw Fig. 2.2.7: the diagram carries the marks.SN

Solved PYQ CSE 2022 Q4(e) 10m

BrCl: μ = 1.42 × 10⁻³⁰ C m, d = 214 pm. Percentage ionic character?

μ for a full charge: e × d = (1.6 × 10⁻¹⁹)(2.14 × 10⁻¹⁰) = 3.42 × 10⁻²⁹ C m.
% ionic = 1.42 × 10⁻³⁰ / 3.42 × 10⁻²⁹ × 100 = 4.15 %: a polar covalent bond, Br^δ+–Cl^δ− (χ Cl 3.16 > Br 2.96).

Also asked: CSE 2023 Q1(b) 10m most polar bond in thiourea is N–H (Δχ 0.84), not C=S (0.03).

2.2.10 Bond energy and potential-energy curves ★

Fig. 2.2.8 C–F and C–I potential-energy curves.

Draw it in 60 s: axes E and r; a steep wall on the left, a well, a flat tail to zero → deep narrow well at short r for C–F (−484), shallow well further right for C–I (−238).

Zero of energy = separated atoms. The minimum is at the equilibrium bond length r₀, where attraction and repulsion balance (dV/dr = 0); its depth is the dissociation energy.SN
Short, strong bonds have deep wells at small r: C–F 484 kJ mol⁻¹ at 135 pm; C–I 238 kJ mol⁻¹ at 214 pm IFoS 2021 Q1(c) 8m.
Bond order up → bond shorter and stronger (C–C 154, C=C 134, C≡C 120 pm).

2.2.11 ★ Add-on beyond the syllabus

Hypervalent molecules without d orbitals. Calculations show that the d orbitals of S, P and Xe hardly take part in bonding. XeF₂ is better described by a three-centre, four-electron bond: one Xe 5p and two F 2p orbitals give a bonding, a non-bonding (on F only) and an antibonding orbital; four electrons fill the first two, so each Xe–F has bond order ½ and the F atoms carry negative charge. This explains why hypervalent molecules need electronegative ligands (SF₆ exists, SH₆ does not). One line on it in a 10-mark answer on sp³d hybrids shows depth.

2.2.12 How UPSC asks this

Question type	Asked	What earns the marks	Time
Dipole moments, NH₃ vs NF₃, % ionic	CSE 2025 IFoS 2025 IFoS 2022 CSE 2022 CSE 2023	Vector diagram with lone-pair moment; values in D; formula with units	4–8 min
Shape by VSEPR, lone-pair placement	CSE 2023 CSE 2024 Geo 2025	Electron count, steric number, both geometries named, 90° contacts compared, angle	7–8 min
Hybridisation and d orbitals	CSE 2021 20m CSE 2023	Shape drawn with axes; d orbitals named and justified by direction	7–14 min
Lewis structures, formal charge, resonance	CSE 2022 IFoS 2025 CSE 2023	Electron count, formal charges written on atoms, bond order 4/3	6–7 min
Fluxionality	CSE 2024	Berry mechanism drawn through the square pyramid; NMR time scale	7 min
Prelims MCQs	Geo Pre 2021 Q16 Geo Pre 2024 Q11 Geo Pre 2024 Q24 Geo Pre 2025 Q31 Geo Pre 2023 Q26 Geo Pre 2021 Q8	Angle order, resonance validity, Coulson, Bent's rule, dipole direction	1 min

Time at 0.72 min per mark for Civil Services and 0.8 min per mark for IFoS and Geo-Scientist Mains.

2.2.13 Traps

✗ Putting lone pairs axial in a trigonal bipyramid. They always go equatorial: SF₄ is a see-saw, not a distorted tetrahedron; ClF₃ is T-shaped, not trigonal planar.

✗ Giving only one geometry. VSEPR questions want the arrangement of electron pairs and the shape of the molecule, by name.

✗ Counting π bonds or forgetting conjugation when assigning hybridisation. An amide N or an enolate carbon is sp².

2.2.14 Where to read

Any one book is enough.

Book 1: J.D. Lee, Concise Inorganic Chemistry, 5th ed., Ch. 4 “The covalent bond”: Lewis theory and the octet rule (p. 72), Sidgwick–Powell and VSEPR with BF₃, NH₃, H₂O, PCl₅, ClF₃, SF₄, I₃⁻, SF₆, IF₇ worked one by one (pp. 74–80), valence bond theory and hybridisation (pp. 80–85), d-orbital participation (p. 85), σ and π bonds (p. 87). Best for first reading.
Book 2: J.E. Huheey, E.A. Keiter and R.L. Keiter, Inorganic Chemistry, 4th ed., Ch. 5 valence bond theory and resonance (p. 139) and electronegativity, bond polarity and dipole moments (p. 182); Ch. 6 “The structure and reactivity of molecules”: VSEPR (p. 203) and “Structure and hybridization”, where Bent's rule, Coulson's relation and apicophilicity are treated (p. 220); bond lengths (p. 232). Best for Bent's rule and the reasons behind the angle trends.

Online: LibreTexts Inorganic Chemistry (search “VSEPR” or “Bent's rule”).

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