Class 11 Chemistry · Chapter 4
Short answer:
Chemical Bonding and Molecular Structure Class 11 Chemistry ka Chapter 4 hai — rationalised 2026-27 syllabus me ye Part I ka aakhri chapter hai (States of Matter jaisa chapter is session me delete ho chuka hai, isliye Chapter 3 Classification of Elements ke turant baad seedha ye chapter aata hai). Is chapter me: Kössel-Lewis approach, ionic bond, Lewis structures + formal charge, bond parameters, VSEPR theory (molecule ka shape), Valence Bond Theory, hybridisation (sp/sp²/sp³/sp³d/sp³d²), Molecular Orbital Theory (bond order, paramagnetism), aur hydrogen bonding — sab cover hota hai. Neeche is chapter ke saare intext + exercise questions ka step-by-step solution hai — chaho to ise apna class 11 chemistry ncert solutions pdf substitute samajh lo.
Chapter 4 poore Class 11 Chemistry ki neev hai — jaise Chapter 1 poore Class 10 Chemistry ki neev tha. Chapter 2 (Structure of Atom) me tumne electrons, orbitals aur quantum numbers padhe the — agar rusty lag raha hai to pehle structure of atom class 11 formulas aur class 11 chemistry chapter 2 structure of atom important questions ek baar revise kar lo, kyunki yahan wahi electrons milkar bond banate hain. Chapter 3 (Classification of Elements and Periodicity) ka electronegativity trend bhi seedha yahan kaam aayega.
Bina Chemical Bonding samjhe, aage kuch tikega nahi: Chapter 6 Equilibrium ka acid-base theory isi par tika hai, Chapter 7 me redox reactions class 11 important questions solve karte waqt bond-polarity ka concept use hoga, aur sabse zyada — Part II ki Organic Chemistry (Chapter 8: class 11 chemistry organic chemistry basic principles and techniques notes, aur Chapter 9: hydrocarbons class 11 ncert solutions) poori ki poori hybridisation ke concept par khadi hai. Ye chapter lamba zaroor hai, par rata-marne wala nahi hai — concept-heavy hai. Ek baar VSEPR, hybridisation aur MOT ka logic clear ho gaya, teeno cheezein easy lagengi. Agar poore class 11 chemistry syllabus 2026-27 rationalised ka overview chahiye ya class 11 chemistry all chapters list ncert dekhni hai, README/syllabus doc pehle check kar lo.
Chapter 4 Summary — 5 Minute Revision
Chapter 4 ke 9 sections ka ek-line recap:
- 4.1 Kössel-Lewis approach: atoms octet (8 valence e⁻) pane ke liye bond banate hain — electron transfer (ionic) ya sharing (covalent) se.
- 4.2 Ionic bond: low ionisation enthalpy (cation) + high electron gain enthalpy (anion) + high lattice enthalpy — teeno favourable ho to ionic bond banta hai.
- 4.3 Bond parameters: bond length, bond angle, bond enthalpy, bond order, resonance, polarity (dipole moment), formal charge.
- 4.4 VSEPR theory: molecule ka shape electron-pair (bond pair + lone pair) repulsion minimise karke decide hota hai; repulsion order lp-lp > lp-bp > bp-bp.
- 4.5 Valence Bond Theory: bond atomic orbitals ke overlap se banta hai (H₂ ka example).
- 4.6 Hybridisation: same energy ke hybrid orbitals banane ke liye atomic orbitals ka intermixing — sp (linear), sp² (trigonal planar), sp³ (tetrahedral), sp³d (trigonal bipyramidal), sp³d² (octahedral).
- 4.7 Molecular Orbital Theory: bond order = (Nb−Na)/2; O₂ paramagnetic nikalta hai (VBT isse explain nahi kar pata).
- 4.8 Homonuclear diatomic molecules: H₂, He₂, Li₂, N₂, O₂, F₂ ki MO configuration aur stability.
- 4.9 Hydrogen bonding: H (bonded to F/O/N) aur doosre electronegative atom ke beech weak attraction — intermolecular (HF, H₂O) aur intramolecular (o-nitrophenol) types.
In-Text Questions — Solutions
4.1 Explain the formation of a chemical bond.
Chemical bond ek attractive force hai jo do ya zyada atoms/ions ko aapas me jodkar rakhti hai aur system ko stabilise karti hai. Kössel-Lewis approach ke according, atoms noble gas jaisi stable electronic configuration (octet, 8 valence electrons) pane ki koshish karte hain — ya to electron transfer karke (ionic bond) ya electron pair share karke (covalent bond). Jab bond banta hai to system ki potential energy kam ho jaati hai isolated atoms ke comparison me; bond tabhi banega jab is process se energy release ho (system stable ho).
4.2 Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻.
Lewis symbol me sirf valence electrons dots ke form me symbol ke around dikhaye jaate hain:
- S: 6 valence e⁻ → symbol ke around 6 dots.
- S²⁻: S ne 2 electron gain kiye → 8 dots, square bracket me likha jaata hai with 2− charge outside — [S]²⁻.
- Al: 3 valence e⁻ → symbol ke around 3 dots.
- Al³⁺: Al ne 3 electron lose kiye → koi dot nahi, [Al]³⁺.
- H: 1 valence e⁻ → symbol ke paas 1 dot.
- H⁻: H ne 1 electron gain kiya (duplet complete) → 2 dots, [H]⁻.
4.3 Draw the Lewis structures of the following: H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH.
H₂S (total valence e⁻ = 2×1 + 6 = 8): H–S–H, S par 2 lone pairs.
SiCl₄ (4 + 4×7 = 32 e⁻): Si central atom, 4 Si–Cl single bonds (tetrahedral), har Cl par 3 lone pairs — total 4×2 + 12×2 = 32 e⁻ match.
BeF₂ (2 + 2×7 = 16 e⁻): F–Be–F, linear, Be par koi lone pair nahi (incomplete octet — Be ke around sirf 4 electrons), har F par 3 lone pairs.
CO₃²⁻ (4 + 3×6 + 2 = 24 e⁻): central C, ek C=O double bond aur do C–O single bond (dono par − charge), teeno O equivalent hain resonance ki wajah se.
HCOOH (formic acid, 4 (C) + 2×1 (H) + 2×6 (O) = 4+2+12 = 18 e⁻): H–C(=O)–O–H, yaani C, H se single bond, ek O se double bond, doosre O se single bond jo aage H se juda hai.
4.4 Define octet rule. Write its significance and limitations.
Octet rule: atoms generally electron lose/gain/share karke apni valence shell me 8 electrons (nearest noble gas jaisi configuration) achieve karne ki koshish karte hain, jisse extra stability milti hai.
Significance: main group elements ke liye ionic aur covalent bond formation, Lewis structures, aur basic reactivity samajhne me kaam aata hai.
Limitations:
- Incomplete octet: BeCl₂, BF₃, AlCl₃ me central atom ke around 8 se kam electrons.
- Odd electron molecules: NO, NO₂ me total valence e⁻ hi odd hai, sabka pairing nahi ho sakta.
- Expanded octet: PF₅, SF₆, H₂SO₄ me central atom ke around 8 se zyada electrons (d-orbitals involve hote hain).
- Molecule ka shape explain nahi karta.
- Relative stability (energy) ke baare me kuch nahi batata.
- Noble gas compounds (XeF₂, XeF₄) ki existence explain nahi karta — jabki unka octet pehle se hi complete hai.
4.5 Write the favourable factors for the formation of an ionic bond.
(i) Metal atom ki low ionisation enthalpy — asaani se electron lose karke cation bane.
(ii) Non-metal atom ki high (more negative) electron gain enthalpy — asaani se electron accept karke anion bane.
(iii) Resulting compound ki high lattice enthalpy — jab gaseous cation aur anion crystal lattice me pack hote hain to zyada energy release honi chahiye, tabhi compound stable banega. Jitni zyada (more negative) lattice enthalpy, utna stronger ionic bond.
4.6 Discuss the shape of BeCl₂, BCl₃, CH₄, NH₃ aur H₂O using VSEPR model.
| Molecule | Bond pairs | Lone pairs | Shape | Bond angle |
|---|---|---|---|---|
| BeCl₂ | 2 | 0 | Linear | 180° |
| BCl₃ | 3 | 0 | Trigonal planar | 120° |
| CH₄ | 4 | 0 | Tetrahedral | 109.5° |
| NH₃ | 3 | 1 | Pyramidal (distorted tetrahedral) | ~107° |
| H₂O | 2 | 2 | Bent/angular (distorted tetrahedral) | ~104.5° |
↔ Table ko side me swipe karein
4.7 NH₃ aur H₂O dono distorted tetrahedral geometry rakhte hain, phir bhi H₂O ka bond angle NH₃ se kam kyu hai?
Repulsion ka order hota hai: lone pair–lone pair (lp-lp) > lone pair–bond pair (lp-bp) > bond pair–bond pair (bp-bp).
NH₃ me sirf 1 lone pair hai → 3 lp-bp repulsions bond pairs ko thoda paas dhakelti hain, angle 109.5° se ghatkar ~107° ho jaata hai.
H₂O me 2 lone pairs hain → zyada lp-bp (aur lp-lp bhi) repulsion, bond pairs aur zyada compress hote hain, angle ~104.5° tak ghat jaata hai. Yaani jitne zyada lone pairs, utna zyada repulsion aur utna chhota bond angle.
4.8 Explain the important aspects of resonance with reference to the CO₃²⁻ ion.
Experimentally CO₃²⁻ ke teeno C–O bonds ki length equal (136 pm) paayi gayi hai, jo C–O single bond (143 pm) aur C=O double bond (121 pm) ke beech me hai. Ek single Lewis structure (1 C=O + 2 C–O⁻) ye galat predict karega ki ek bond chhota aur do bade honge — jo experiment se match nahi karta.
Isliye CO₃²⁻ ko teen equivalent canonical (contributing) structures ka resonance hybrid maana jaata hai, jisme double bond alag-alag position pe hota hai. Actual structure in sabka weighted average hai — kisi ek structure se nahi, balki hybrid se represent hota hai, aur ye hybrid kisi bhi single canonical form se zyada stable (extra resonance-stabilised) hota hai.
4.9 O₃ (ozone) ke resonance structures draw karo formal charge ke saath.
O₃ me total valence e⁻ = 18. Central O, do terminal O se juda hota hai — ek O=O double bond aur ek O–O single bond, jismein double bond alag-alag terminal O par ho sakta hai (2 canonical structures).
Formal charges: double-bonded terminal O = 0, single-bonded terminal O = −1, central O = +1.
Actual O–O bond length equal (128 pm) paayi jaati hai, jo O–O single (148 pm) aur O=O double (121 pm) ke beech me hai — ye resonance hybrid ko confirm karta hai.
4.10 Formal charge kya hai? NH₄⁺ me N atom ka formal charge calculate karo.
Formal charge = (free atom ke valence e⁻) − (non-bonding e⁻) − ½(bonding e⁻).
FC(N) = 5 − 0 − 8/2 = 5 − 0 − 4 = +1
N ke around non-bonding electron 0 hai aur 4 N–H bonds ke total 8 bonding electrons hain. FC = +1 aata hai, jo ion ke overall +1 charge se match karta hai (har H ka FC = 1 − 0 − 2/2 = 0, sum = +1) — isse Lewis structure verify hota hai.
4.11 Hybridisation kya hai? CH₄ me sp³ hybridisation explain karo.
Hybridisation ek atom ke slightly different-energy atomic orbitals ka intermixing hai, jisse equivalent energy aur shape ke naye hybrid orbitals bante hain — jo stronger, directional bonds banate hain.
CH₄ me carbon ki ground state config 2s² 2p² hai. Ek 2s electron 2p me promote hota hai → 2s¹ 2p³ (4 unpaired electrons). Ye ek 2s aur teen 2p orbitals mix hokar 4 equivalent sp³ hybrid orbitals banate hain, jo tetrahedrally (109.5° apart) arranged hote hain — har ek 1s(H) ke saath overlap karke 4 equivalent C–H σ bonds banata hai.
4.12 AlCl₃ + Cl⁻ → AlCl₄⁻ reaction me Al ke hybridisation me kya change hota hai?
AlCl₃ me Al sp² hybridised hai (3 bond pairs, planar shape, incomplete octet — Al ke around sirf 6 electrons).
Jab Al, Cl⁻ ka lone pair accept karta hai (coordinate bond), tab Al ka hybridisation sp² se sp³ me badal jaata hai (ab 4 bond pairs, tetrahedral shape, octet complete).
4.13 NH₃ aur NF₃ me se kiska dipole moment zyada hai aur kyu?
NH₃ (1.47 D) ka dipole moment NF₃ (0.24 D) se zyada hai.
NH₃ me N, H se zyada electronegative hai → N–H bond dipole N ki taraf point karta hai, aur N ka lone pair bhi usi general direction me hota hai — dono add hokar net dipole moment badhate hain.
NF₃ me F, N se zyada electronegative hai → N–F bond dipole F ki taraf point karta hai (yaani lone pair ki opposite direction me) — lone pair moment aur bond moments ek doosre ko partially cancel karte hain, isliye net dipole moment kam ho jaata hai.
4.14 MOT se O₂, O₂⁺, O₂⁻ aur N₂ ka bond order calculate karo aur O₂ ka magnetic behaviour predict karo.
Bond order = (Nb − Na)/2
O₂ (16 e⁻): σ1s²σ*1s²σ2s²σ*2s²σ2pz²π2px²=π2py²π*2px¹=π*2py¹ → Nb=10, Na=6, BO = (10−6)/2 = 2. π* orbitals me 2 unpaired electrons hain → O₂ paramagnetic hai.
O₂⁺ (15 e⁻, π* se 1 e⁻ hataya): Nb=10, Na=5, BO = (10−5)/2 = 2.5.
O₂⁻ (17 e⁻, π* me 1 e⁻ add): Nb=10, Na=7, BO = (10−7)/2 = 1.5.
N₂ (14 e⁻): σ1s²σ*1s²σ2s²σ*2s²π2px²=π2py²σ2pz² → Nb=10, Na=4, BO = (10−4)/2 = 3 (triple bond, diamagnetic, sab electrons paired).
4.15 Hydrogen bond kya hai? Iske do types example ke saath explain karo.
Hydrogen bond ek weak electrostatic attraction hai H atom (jo highly electronegative atom F/O/N se covalently juda ho) aur doosre highly electronegative atom (lone pair wale) ke beech — kyunki polar bond ki wajah se us H par partial positive charge (δ+) hota hai.
(i) Intermolecular H-bond: do alag molecules ke beech — jaise HF me F–H···F–H chains, ya H₂O me extensive network (isi se paani ka boiling point abnormally high hota hai).
(ii) Intramolecular H-bond: ek hi molecule ke andar, jab geometry allow kare — jaise o-nitrophenol me –OH aur ortho-position ke –NO₂ group ke beech. Isse H internally 'use' ho jaata hai, isliye o-nitrophenol ka boiling point/water solubility p-nitrophenol se kam hoti hai.
Poore Class 11 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q25)
1. Electronegativity define karo. Ye electron gain enthalpy se kaise different hai?
Electronegativity ek atom ki tendency hai ki wo covalent bond me shared electron pair ko apni taraf kitna attract karta hai (ye ek relative, unit-less number hai, bonded state me define hoti hai).
Electron gain enthalpy ek measurable quantity hai — jab isolated gaseous atom electron gain karke anion banta hai to jo energy change hota hai. Electronegativity ek periodic property/trend hai jo bond ke context me hoti hai, jabki electron gain enthalpy isolated gaseous atom ki thermodynamic property hai jo experimentally measure ki jaa sakti hai.
2. Fajans' rules explain karo — LiCl aur LiI me se kaunsa zyada covalent character rakhta hai?
Fajans' rules ke according, ionic bond me covalent character tab zyada hota hai jab:
- Cation chhota ho aur anion bada ho (high charge density → zyada polarisation)
- Anion easily polarisable ho (bada, high charge)
- Cation ka charge zyada ho
Li⁺ same hai dono me, lekin I⁻ (iodide) Cl⁻ se bada aur zyada polarisable hai. Isliye Li⁺ ka high charge density I⁻ ki electron cloud ko zyada distort karega → LiI, LiCl se zyada covalent character rakhta hai.
3. Lewis structures me formal charge ke plus aur minus sign ka kya significance hai?
Formal charge molecule ka actual charge nahi hota — ye ek bookkeeping tool hai jo batata hai ki electrons kitne 'equally' distribute hue maane gaye hain agar bonding electrons dono atoms ke beech barabar baante jaayen. Plus sign matlab us atom ne apne free-atom valence electrons se kam electrons 'apne paas' rakhe hain (relatively electron-deficient), minus sign matlab zyada rakhe hain (relatively electron-rich). Formal charges Lewis structure ki plausibility check karne aur multiple possible structures me se most stable structure choose karne me help karte hain (generally jis structure me formal charges minimum ho aur negative charge zyada electronegative atom par ho, wo zyada stable maana jaata hai).
4. Polar covalent bond ko example ke saath explain karo.
Jab do different electronegativity wale atoms covalent bond banate hain, to shared electron pair equally share nahi hota — zyada electronegative atom electron density apni taraf zyada kheenchta hai, jisse us par partial negative charge (δ−) aur doosre atom par partial positive charge (δ+) aa jaata hai. Ye polar covalent bond kehlata hai.
Example: HCl me Cl (electronegativity 3.0) H (2.1) se zyada electronegative hai, isliye Hδ+–Clδ− — bond me dipole moment aa jaata hai (1.03 D).
5. N–H, F–H, C–H, O–H bonds ko increasing ionic character ke order me arrange karo.
Ionic character electronegativity difference (ΔEN) badhne ke saath badhta hai. Electronegativity: H = 2.1, C = 2.5, N = 3.0, O = 3.5, F = 4.0.
ΔEN: C–H = 0.4, N–H = 0.9, O–H = 1.4, F–H = 1.9.
Isliye increasing order: C–H < N–H < O–H < F–H.
6. CH₃COOH (acetic acid) ka correct Lewis structure draw karo (bonds galat dikhaye gaye the skeleton me).
Acetic acid me carbon skeleton CH₃–C(=O)–O–H hota hai. Correct structure:
Pehla C, 3 H se single bond + doosre C se single bond (total 4 bonds, sp³) — is C par koi lone pair nahi.
Doosra C (carboxyl C), pehle C se single bond, ek O se double bond (C=O), aur doosre O se single bond (sp² hybridised) — 3 sigma + 1 pi bond, koi lone pair nahi is C par.
Terminal –OH ka O, C se single bond aur H se single bond, 2 lone pairs. Carbonyl O (=O), C se double bond, 2 lone pairs.
Total valence electron count: 2×4 (C) + 4×1 (H) + 2×6 (O) = 8+4+12 = 24 e⁻. Structure me isko is tarah distribute karke verify karte hain: bonding pairs — 3 (C1–H) + 1 (C1–C2) + 2 (C2=O, double bond) + 1 (C2–O) + 1 (O–H) = 8 bonding pairs = 16 electrons; non-bonding — carbonyl O ke 2 lone pairs + hydroxyl O ke 2 lone pairs = 4 lone pairs = 8 electrons. Total = 16 + 8 = 24 e⁻, jo match karta hai — isse confirm hota hai ki structure sahi hai aur sab octet complete hai.
7. NO₃⁻ ion ke liye resonance ke important aspects explain karo (CO₃²⁻ jaisa hi analysis).
NO₃⁻ me total valence e⁻ = 5 + 3×6 + 1 = 24. Experimentally teeno N–O bonds equal length (~124 pm) rakhte hain, jo N–O single bond aur N=O double bond ke beech me hai. Single Lewis structure (1 N=O + 2 N–O⁻) ye galat predict karega ki bonds unequal honge.
Isliye NO₃⁻ teen equivalent canonical structures ka resonance hybrid hai, jismein double bond alag-alag N–O position pe hota hai. Actual structure in teeno ka weighted average hai, jo kisi bhi single structure se zyada stable hai (extra resonance stabilisation energy).
8. BF₃, CO, CN⁻, N₂, C₂H₄, C₂H₂ ke Lewis structures draw karo.
BF₃ (24 e⁻): B central, 3 B–F single bonds (trigonal planar), B par koi lone pair nahi — incomplete octet (6 e⁻).
CO (10 e⁻): C≡O triple bond, C par 1 lone pair, O par 1 lone pair (both satisfy octet via coordinate + normal bonding).
CN⁻ (10 e⁻): C≡N triple bond, C par 1 lone pair (with − charge), N par 1 lone pair.
N₂ (10 e⁻): N≡N triple bond, har N par 1 lone pair.
C₂H₄ (12 e⁻): H₂C=CH₂, C=C double bond, har C, 2 H se single bond (sp² carbons, planar).
C₂H₂ (10 e⁻): H–C≡C–H, C≡C triple bond, har C ek H se single bond (sp carbons, linear).
9. Bond order kya hai? N₂, O₂, C₂, He₂ aur N₂⁺ ka bond order calculate karo.
Bond order bataata hai molecule me kitne bonds hain, aur us bond ki strength kitni hai — jitna zyada bond order, utna chhota bond length aur zyada bond strength.
Bond order = (Nb − Na)/2
N₂ (14 e⁻): Nb=10, Na=4 → BO = 3.
O₂ (16 e⁻): Nb=10, Na=6 → BO = 2.
C₂ (12 e⁻): configuration σ1s²σ*1s²σ2s²σ*2s²π2px²π2py² → Nb=8, Na=4 → BO = (8−4)/2 = 2.
He₂ (4 e⁻): σ1s²σ*1s² → Nb=2, Na=2 → BO = (2−2)/2 = 0 — isliye He₂ exist hi nahi karta.
N₂⁺ (13 e⁻, N₂ se 1 e⁻ σ2pz se hataya): Nb=9, Na=4 → BO = (9−4)/2 = 2.5.
10. Bond length, bond angle aur bond enthalpy define karo. Bond length ko affect karne wale factors likho.
Bond length: do bonded atoms ke nuclei ke beech ki equilibrium (average) distance.
Bond angle: ek hi central atom pe do bonds ke beech ka angle.
Bond enthalpy: ek mole gaseous covalent bonds ko homolytically break karne ke liye required energy (kJ/mol).
Bond length ko affect karne wale factors: (i) bond order — zyada bond order, chhoti length; (ii) atomic size/radii jinse bond banta hai — bade atoms lambi bond banate hain; (iii) hybridisation state — s-character zyada (sp > sp² > sp³) to bond chhoti aur strong hoti hai.
11. NH₃ aur NF₃ me se kiska bond angle zyada hai aur kyu?
NH₃ ka bond angle (~107°) NF₃ (~102°) se zyada hai.
N zyada electronegative hai H se, isliye N–H bond pair N ke close/central atom ki taraf zyada shifted rehta hai — bond pairs ek doosre ke close hote hain, unke beech repulsion thoda zyada hota hai, angle thoda expand hota hai.
Ulta, N–F me F zyada electronegative hai, isliye bond pairs F ki taraf khinch jaate hain (N se door), N ke around bond pairs ek doosre se relatively door ho jaate hain, unke beech repulsion kam ho jaata hai, aur lone pair ka effect zyada dominant ho jaata hai — isliye angle NF₃ me compress hokar NH₃ se kam ho jaata hai.
12. Bond strength ko bond order ke terms me kaise express karte hain?
Bond order bond strength ke directly proportional hota hai — jitna zyada bond order, utni zyada bond strength (bond enthalpy) aur utni chhoti bond length. Example: N₂ (BO=3, bond enthalpy ~945 kJ/mol) O₂ (BO=2, ~498 kJ/mol) se kaafi zyada strong hai, aur N₂ ki bond length (110 pm) O₂ (121 pm) se chhoti hai — dono trends bond order se consistent hain.
13. Hybridisation define karo. Iske main characteristics/rules kya hain?
Hybridisation: same atom ke slightly-different-energy atomic orbitals ka intermixing, jisse equivalent shape/energy ke naye hybrid orbitals banate hain.
Main rules/characteristics:
- Utne hi hybrid orbitals banenge jitne atomic orbitals mix hue hain (orbital count conserve hota hai).
- Hybrid orbitals ka size/shape/energy identical hota hai (equivalent).
- Hybrid orbitals directional hote hain — sirf ek lobe overlapping ke liye use hota hai (strong bond).
- Hybridisation me involved orbitals same principal energy level ke honi chahiye, jab tak d-orbitals bhi involve na ho.
- Half-filled orbitals bond formation me involve hote hain, filled orbitals lone pairs banate hain.
14. PCl₅ me hybridisation describe karo. Axial bonds equatorial se lambi kyu hote hain?
PCl₅ me P, sp³d hybridised hai — 3s, 3p (3), aur 3d ek orbital mix hote hain, 5 equivalent-ish hybrid orbitals bante hain jo trigonal bipyramidal geometry me arranged hote hain (3 equatorial, 90° apart from axial, 120° apart from each other; 2 axial, 180° apart).
Axial bonds equatorial se longer hote hain kyunki axial bond pairs, teeno equatorial bond pairs se 90° angle par repulsion face karte hain (3 close-range 90° repulsions), jabki equatorial bond pairs sirf 2 axial se 90° pe aur 2 equatorial se 120° pe repulsion face karte hain (kam close-range repulsion). Zyada repulsion face karne wale axial bonds thode 'stretched'/weaker aur isliye longer ho jaate hain.
15. CH₃–CH=CH₂ (propene) me carbon atoms ka hybridisation batao, aur total sigma + pi bonds count karo.
Propene me: C1 (CH₃) — sp³ hybridised (4 sigma bonds — 3 C–H + 1 C–C). C2 aur C3 (CH=CH₂) — dono sp² hybridised (double bond ke carbons).
Sigma bonds: 3 (C1–H) + 1 (C1–C2) + 1 (C2–H) + 1 (C2–C3) + 2 (C3–H) = 8 sigma bonds.
Pi bonds: C2=C3 ke beech 1 pi bond.
Total = 9 bonds (8σ + 1π).
16. C₂H₂ (ethyne/acetylene) ki formation hybridisation concept se explain karo, aur sigma + pi bonds count karo.
Ethyne (H–C≡C–H) me har carbon sp hybridised hota hai — ek 2s aur ek 2p orbital mix hokar 2 sp hybrid orbitals (180° apart, linear) banate hain; bache hue 2 unhybridised p-orbitals (perpendicular to sp axis aur ek doosre se bhi perpendicular) side-by-side overlap karke 2 pi bonds banate hain.
Sigma bonds: C–H (2) + C–C (1, sp-sp head-on overlap) = 3 sigma bonds.
Pi bonds: C≡C ke beech 2 pi bonds (do perpendicular p-p overlaps).
Total = 5 bonds (3σ + 2π) — yahi C≡C triple bond banata hai.
17. Sigma aur pi bonds ki total sankhya batao: (a) C₂H₂ (b) C₂H₄ (c) HCN
(a) C₂H₂: 3σ + 2π (2 C–H σ, 1 C–C σ, 2 C≡C π) — total 5 bonds.
(b) C₂H₄: 5σ + 1π (4 C–H σ, 1 C–C σ, 1 C=C π) — total 6 bonds.
(c) HCN (H–C≡N): 2σ (H–C, C–N) + 2π (C≡N triple bond ka doosra aur teesra bond) — total 4 bonds.
18. In molecules me carbon atoms kaunse hybrid orbitals use karte hain: CH₃–CH₃, CH₃–CH=CH₂, CH₃–CH₂–OH, CH₃–CHO, CH₃COOH?
| Molecule | Carbon(s) ka hybridisation |
|---|---|
| CH₃–CH₃ (ethane) | Dono C — sp³ (single bond only) |
| CH₃–CH=CH₂ (propene) | CH₃ ka C — sp³; CH=CH₂ ke dono C — sp² |
| CH₃–CH₂–OH (ethanol) | Dono C — sp³ |
| CH₃–CHO (acetaldehyde) | CH₃ ka C — sp³; CHO ka C (C=O) — sp² |
| CH₃COOH (acetic acid) | CH₃ ka C — sp³; COOH ka C (C=O) — sp² |
↔ Table ko side me swipe karein
19. Bond pairs aur lone pairs of electrons kya hote hain? Ek-ek example do.
Bond pair: electron pair jo do atoms ke beech share hokar covalent bond banata hai. Example: CH₄ me har C–H bond ka electron pair ek bond pair hai (total 4 bond pairs C ke around).
Lone pair: electron pair jo central atom pe hi rehta hai, kisi bond me share nahi hota. Example: NH₃ me N ke around 1 lone pair hai (jo bond nahi banata, par molecule ki shape aur basicity determine karta hai).
20. C₂H₄ aur C₂H₂ me double aur triple bond formation kaise hota hai — explain karo.
C₂H₄ (double bond): har C, sp² hybridised — 3 sp² orbitals (2 C–H sigma + 1 C–C sigma banate hain, sab ek plane me 120° apart), aur ek unhybridised p-orbital (plane ke perpendicular) bacha reh jaata hai jo doosre C ke p-orbital se sideways overlap karke 1 pi bond banata hai. Sigma (head-on) + pi (sideways) milkar C=C double bond banate hain.
C₂H₂ (triple bond): har C, sp hybridised — 2 sp orbitals (1 C–H sigma + 1 C–C sigma, linear 180°), aur 2 unhybridised p-orbitals (perpendicular to axis, aur ek doosre se bhi perpendicular) bache rehte hain, jo doosre C ke corresponding p-orbitals se overlap karke 2 pi bonds banate hain. 1 sigma + 2 pi milkar C≡C triple bond banate hain.
21. Valence Bond Theory (VBT) kya hai? Iske basis par H₂ molecule ki formation explain karo.
Valence Bond Theory ke according, covalent bond do atoms ke half-filled atomic orbitals ke overlap se banta hai — jitna zyada overlap, utna strong bond.
H₂ formation: jab do H atoms (har ek me 1s¹, ek unpaired electron) paas aate hain, unke 1s orbitals overlap karte hain. Jaise-jaise distance kam hota hai, attraction (nucleus-electron) badhta hai, potential energy kam hoti hai, jab tak equilibrium bond distance (74 pm) pe minimum energy nahi mil jaati — yahi H–H covalent bond hai. Isse aage paas laane par nuclear-nuclear repulsion dominate karega aur energy badhne lagegi.
22. Hydrogen bond kya hota hai? Kya ye van der Waals forces se strong hai ya weak?
Hydrogen bond ek weak electrostatic attraction hai H atom (jo F/O/N jaise highly electronegative atom se bonded ho) aur ek doosre electronegative atom ke lone pair ke beech.
Hydrogen bond, normal covalent/ionic bonds se kaafi weaker hai (~10–40 kJ/mol vs covalent ~200–800 kJ/mol), lekin van der Waals forces (London/dipole-dipole) se strong hai (van der Waals ~2–20 kJ/mol tak hoti hai). Isliye strength order: covalent/ionic bond > hydrogen bond > van der Waals forces.
23. O₂, O₂⁺, O₂⁻ (superoxide) aur O₂²⁻ (peroxide) ki relative stability compare karo aur unka magnetic property batao.
Bond order jitna zyada, stability utni zyada:
- O₂⁺: BO = 2.5 → sabse stable, 1 unpaired e⁻ → paramagnetic.
- O₂: BO = 2 → 2 unpaired e⁻ (π*) → paramagnetic.
- O₂⁻ (superoxide): BO = 1.5 → 1 unpaired e⁻ → paramagnetic.
- O₂²⁻ (peroxide): BO = (10−8)/2 = 1 → sab electrons paired → diamagnetic, sabse kam stable in inme se.
Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻ (bond order ke order me).
24. Molecular Orbital Theory (MOT), Valence Bond Theory (VBT) se kaise better/different hai — limitations bhi likho.
MOT ke advantages VBT se: MOT O₂ ki paramagnetic nature correctly predict karta hai (2 unpaired electrons π* orbitals me), jabki VBT ye explain nahi kar paata (VBT ke hisaab se O₂ ke sab electrons paired hone chahiye). MOT bond order jaisa quantitative concept bhi deta hai jisse bond strength/length compare kar sakte hain, aur ions (O₂⁺, O₂⁻) ki stability bhi predict karta hai.
MOT ki limitations: complex/heteronuclear molecules ke liye calculation mushkil ho jaati hai, aur ye polyatomic molecules ki geometry (shape) directly predict nahi karta — uske liye VSEPR/hybridisation zyada intuitive hote hain.
25. Covalent bonds directional kyu hote hain jabki ionic bonds non-directional?
Covalent bond specific atomic/hybrid orbitals ke overlap se banta hai, jo fixed direction me oriented hote hain (jaise sp³ orbitals 109.5° pe fixed directions me) — isliye covalent bond directional hota hai aur molecule ki definite shape/angle hoti hai.
Ionic bond electrostatic attraction hai poore-charge cation aur anion ke beech, jo sabhi directions me equally act karta hai (jaise ek point-charge sab directions me field banata hai) — isliye ionic bond non-directional hota hai, aur ionic compounds ek 3D crystal lattice banate hain jahan har ion apne around maximum opposite-charge ions se ghira hota hai, kisi fixed 'molecule shape' ki jagah.
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Formal Charge | FC = (valence e⁻ of free atom) − (non-bonding e⁻) − ½(bonding e⁻) |
| Bond Order (MOT) | Bond Order = (Nb − Na)/2, jahan Nb = bonding electrons, Na = antibonding electrons |
| Dipole moment | μ = q × d (Debye units; 1 D = 3.336 × 10⁻³⁰ C·m); vector sum of all bond dipoles |
| Repulsion order (VSEPR) | lone pair–lone pair > lone pair–bond pair > bond pair–bond pair |
| Hybridisation vs Geometry | 2 electron domains → sp → linear (180°) · 3 → sp² → trigonal planar (120°) · 4 → sp³ → tetrahedral (109.5°) · 5 → sp³d → trigonal bipyramidal (90°,120°,180°) · 6 → sp³d² → octahedral (90°) |
| Fajans' Rules (covalent character direction) | Zyada covalent character jab: cation chhota + high charge, anion bada + easily polarisable |
| Bond strength trend | Bond order ∝ Bond enthalpy ∝ 1/Bond length |
| Octet rule (limitations) | Incomplete octet (BeCl₂, BF₃) · Odd-electron (NO, NO₂) · Expanded octet (PF₅, SF₆) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Electron-pair geometry (VSEPR) aur molecular shape ko same maan lena — jaise H₂O ki electron-pair geometry tetrahedral hoti hai, par molecular shape 'bent/angular' kehlaati hai (lone pairs count hoti hain repulsion me, par naam me nahi aatin).
- NH₃ aur H₂O dono 'distorted tetrahedral' hain isliye bond angle same maan lena — jabki lone pairs ki sankhya (1 vs 2) directly bond angle kam karti hai (107° vs 104.5°).
- MOT me bond order calculate karte waqt antibonding electrons subtract karna bhool jaana, ya He₂ jaisi species ke liye bhi bond banta hai maan lena (jabki BO=0 hone se He₂ exist hi nahi karta).
- Formal charge formula me bonding electrons ko 2 se divide karna bhool jaana, ya lone pair electrons double count kar dena.
- Sirf central atom se juda atoms ginkar hybridisation decide kar dena, lone pairs ko ignore karna — jaise NH₃ ko '3 bonds hain isliye sp²' maan lena, jabki lone pair include karke ye actually sp³ hai.
- Ionic aur covalent bond ko do bilkul separate categories maan lena, jabki Fajans' rules ke according har ionic bond me kuch na kuch covalent character hota hai (aur vice versa) — ye ek spectrum hai, absolute divide nahi.
Board-Style Important Questions
- 1 mark: Assertion: BeCl₂ molecule linear hota hai. Reason: BeCl₂ me Be, sp hybridised hai aur uske around koi lone pair nahi hota. — Dono assertion aur reason correct hain, aur reason, assertion ki sahi explanation hai: Be ke 2 valence electrons 2 Cl atoms ke saath sp hybrid orbitals bana kar 2 sigma bonds banate hain, koi lone pair na hone se VSEPR ke hisaab se shape linear (180°) hoti hai.
- 2 marks: SO₂ molecule ka Lewis structure draw karo aur uske resonance ko briefly explain karo. — SO₂ me total valence e⁻ = 6+2×6=18. S central atom, ek S=O double bond aur ek S–O single bond (with lone pair based charge distribution), S par bhi 1 lone pair. Do resonance structures possible hain jismein double bond position swap hoti hai, actual structure inka hybrid hai jisme dono S–O bonds equal length (~143 pm, single aur double ke beech intermediate) rakhte hain.
- 3 marks: N₂ ke liye Molecular Orbital Theory se electronic configuration likho, bond order calculate karo, aur magnetic nature predict karo. — N₂ (14 e⁻): σ1s² σ*1s² σ2s² σ*2s² π2px²=π2py² σ2pz². Nb = 10, Na = 4. Bond order = (10−4)/2 = 3 (triple bond). Sabhi electrons paired hain (koi unpaired electron nahi) → N₂ diamagnetic hai. Ye high bond order N₂ ki extreme inertness/stability explain karta hai.
- 3 marks: NH₃, NF₃ se zyada basic kyu hai? Hybridisation aur electronegativity ke terms me explain karo. — Dono me N sp³ hybridised hai aur ek lone pair rakhta hai jo basicity ke liye available hai. NH₃ me H, N se kam electronegative hai isliye electron density N ke lone pair ki taraf zyada concentrated rehti hai, lone pair 'available'/reactive hota hai proton accept karne ke liye. NF₃ me F, N se zyada electronegative hai, isliye N ka lone pair partially F atoms ki taraf khinch jaata hai (electron density N se door), jisse lone pair kam available hota hai donation ke liye. Isliye NH₃, NF₃ se kaafi zyada basic hai.
- 5 marks: VSEPR theory kya hai? Isse BeCl₂, BCl₃, CH₄, PCl₅ aur SF₆ ki shapes predict/explain karo. — VSEPR theory kehti hai ki central atom ke around electron pairs (bonding + lone) ek doosre ko repel karte hain aur is repulsion ko minimise karne wali geometry me arrange hote hain. BeCl₂: 2 bond pairs, 0 lone pair → linear (180°). BCl₃: 3 bond pairs, 0 lone pair → trigonal planar (120°). CH₄: 4 bond pairs, 0 lone pair → tetrahedral (109.5°). PCl₅: 5 bond pairs, 0 lone pair → trigonal bipyramidal (axial 180°, equatorial 120°, axial-equatorial 90°). SF₆: 6 bond pairs, 0 lone pair → octahedral (90°). Har case me electron pairs jitna door-door (maximum separation) rahenge, molecule utni hi stable geometry adopt karega.
Aksar Poochhe Jaane Wale Sawaal
Kya Chemical Bonding chapter rationalised syllabus me delete ho gaya hai?
Nahi, bilkul nahi — Chemical Bonding and Molecular Structure (Chapter 4) poori tarah retained hai, syllabus me nahi kata. Jo actually class 11 chemistry deleted syllabus topics hain wo hain: States of Matter (Gases and Liquids), Hydrogen, s-Block Elements, Some p-Block Elements, aur Environmental Chemistry — ye 5 poore chapters/units 2026-27 rationalised session me nahi padhaye jaate. Chemical Bonding un 9 chapters me se ek hai jo bache hain.
Ab Class 11 Chemistry me total kitne chapters hain?
Rationalised 2026-27 syllabus me sirf 9 chapters hain (pehle 14 the). Poori class 11 chemistry all chapters list ncert: Part I — 1. Some Basic Concepts of Chemistry, 2. Structure of Atom, 3. Classification of Elements and Periodicity in Properties, 4. Chemical Bonding and Molecular Structure. Part II — 5. Thermodynamics, 6. Equilibrium, 7. Redox Reactions, 8. Organic Chemistry — Some Basic Principles and Techniques, 9. Hydrocarbons.
Chemical Bonding ke baad kaunsa chapter aata hai aur usme kya focus rehta hai?
Chapter 4 ke baad Part II shuru hota hai — Chapter 5 Thermodynamics (jahan class 11 chemistry chapter 5 thermodynamics numericals practice karni padti hai — ΔH, ΔS, ΔG calculations), phir Chapter 6 Equilibrium, Chapter 7 Redox Reactions, aur phir Organic Chemistry shuru hoti hai.
Chemical Bonding ke saath saath revision ke liye kya padhna chahiye?
Peeche se: some basic concepts of chemistry class 11 notes (Chapter 1 — mole concept, isi se bond ke stoichiometry questions banenge) aur classification of elements and periodicity in properties important questions (Chapter 3 — electronegativity trend, jo bond polarity ka base hai). Aage ke liye: chemical bonding and molecular structure class 11 notes pdf revise karke seedha Chapter 6 ke class 11 chemistry chapter 6 equilibrium ncert solutions aur Chapter 7 ke redox reactions class 11 important questions pe jump kar sakte ho, kyunki dono me bonding ka concept reuse hota hai.
Board exam me is chapter se numericals ya sirf theory poochi jaati hai?
Mostly conceptual/theory-based hota hai (Lewis structures, VSEPR shapes, hybridisation, MOT bond order), kabhi-kabhi choti calculation (bond order, formal charge, dipole moment) bhi aati hai. Exact marks-weightage har saal/board ke paper pattern par depend karta hai, isliye specific weightage claim yahan nahi diya ja raha — apne school/board ke latest sample paper se confirm karna best rahega.
Is chapter ka concept Organic Chemistry me kaise use hota hai?
Bahut directly — hybridisation (sp/sp²/sp³) hi decide karta hai ki carbon single/double/triple bond banayega, sigma-pi bonds kaise arrange honge, aur molecule ka shape kya hoga. Ye foundation hydrocarbons class 11 ncert solutions (Chapter 9) aur class 11 chemistry organic chemistry basic principles and techniques notes (Chapter 8) — dono me constantly reuse hota hai. Bina Chemical Bonding clear kiye, Organic Chemistry ka structure-drawing part samajhna mushkil ho jaata hai.
Class 11 Chemistry — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Chemistry ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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