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Chemical Bonding: Padhai ka Apna Style! Hey Dosto! Chemical Bonding, jo ki JEE Advanced aur Mains ke liye super important topic hai, uski ek mast cheat sheet banate hain. Ismein hum saare concepts ko detail mein samjhenge, questions solve karne ke useful tricks dekhenge, aur exceptions ko bhi highlight karenge. Chalo shuru karte hain! 3.1 Attainment of a Stable Configuration (Stable Hone ki Kahani) Atoms combine kyu karte hain? Simple hai, stability paane ke liye! Jab atoms bond banate hain, toh unki energy kam ho jaati hai, aur wo zyada stable ho jaate hain. Noble Gases (Group 18): Ye elements (He, Ne, Ar, Kr, Xe, Rn) chemical reactions mein participate nahi karte generally. Kyun? Kyunki inke outer shells already complete hote hain (stable noble gas configuration), aur inki energy already minimum hoti hai. Octet Rule: Mostly atoms chahte hain ki unke outermost shell mein 8 electrons ho jayein (ya phir He ki tarah 2, agar wo first shell hai). Isse unhe noble gas jaisi stability milti hai. Bonding ka Objective: Atoms electrons lose karte hain, gain karte hain, ya share karte hain, taaki unhe stable electronic configuration mil sake. Potential Energy Minimization: Chemical bonding ka fundamental principle hai ki system ki potential energy minimum ho. Jab do atoms ek doosre ke kareeb aate hain, attractive forces (nucleus-electron) aur repulsive forces (nucleus-nucleus, electron-electron) dono operate karte hain. Ek specific internuclear distance par, attractive forces repulsive forces ko balance karte hain, aur system ki potential energy minimum ho jaati hai. Isi point par bond formation hota hai aur energy release hoti hai (bond enthalpy). Electronegativity Difference: Atoms ke beech electronegativity ka difference bond ke nature ko determine karta hai. Agar difference bahut zyada hai (e.g., $> 1.7$), toh ionic bond banta hai (complete transfer of electrons). Agar difference kam hai, toh covalent bond banta hai (sharing of electrons). Agar difference zero hai, toh non-polar covalent bond banta hai (equal sharing). Agar difference non-zero hai lekin kam hai, toh polar covalent bond banta hai (unequal sharing, partial charges develop hote hain). Hybridization and VSEPR Theory: Ye concepts molecules ki geometry aur shape ko explain karte hain. Hybridization: Atomic orbitals combine hokar naye hybrid orbitals banate hain jo bonding ke liye suitable hote hain aur molecule ko specific geometry dete hain (e.g., $sp^3$ for tetrahedral, $sp^2$ for trigonal planar, $sp$ for linear). VSEPR (Valence Shell Electron Pair Repulsion) Theory: Electron pairs (bond pairs aur lone pairs) ek doosre ko repel karte hain aur space mein aise arrange hote hain jisse repulsion minimum ho aur stability maximum ho. Isse molecule ki exact geometry determine hoti hai. Bond Energy/Enthalpy: Ek chemical bond ko todne ke liye jitni energy chahiye hoti hai, ya bond banne par jitni energy release hoti hai, use bond energy kehte hain. Higher bond energy ka matlab hai stronger bond aur greater stability. Resonance: Jab ek single Lewis structure molecule ke sabhi properties ko explain nahi kar pata, toh multiple contributing structures (resonance structures) use kiye jaate hain. Actual molecule in sabhi structures ka resonance hybrid hota hai, jo ki individual structures se zyada stable hota hai. Resonance delocalization of electrons ki wajah se hoti hai. 3.2 Types of Bonds (Bandhan ke Prakar) Chemical bonds teen main types ke hote hain, jo depend karta hai atoms ke electropositive ya electronegative character par: Ionic Bond (Electrostatic Attraction): Complete transfer of one or more electrons from an electropositive atom (metal) to an electronegative atom (non-metal). Result: Positive and negative ions (cations and anions) banate hain, jo electrostatic forces se attract hote hain. Example: NaCl (Na electron lose karke Na$^+$ banta hai, Cl electron gain karke Cl$^-$ banta hai). Covalent Bond (Sharing is Caring): Sharing of electron pairs between two electronegative atoms. Neither atom wants to lose electrons, so they share to achieve noble gas configuration. Example: Cl$_2$ (do Cl atoms ek-ek electron share karke ek electron pair banate hain). Ismein Lewis Structures use karte hain outer electrons ko dots se dikhane ke liye. Metallic Bond (Free Electrons ka Mela): Valency electrons are delocalized and free to move throughout the whole crystal lattice of positive metal ions. Metals ki high electrical aur thermal conductivity isi wajah se hoti hai. Important Note: Most bonds pure ionic, covalent, ya metallic nahi hote. Wo in types ke beech mein kahin exist karte hain. Jaise LiCl ionic hai, but alcohol mein dissolve hota hai, matlab thoda covalent character bhi hai. 3.3 Transitions between the Main Types of Bonding (Bond Types ka Sangam) Jaise life mein sab kuch pure nahi hota, waise hi bonds bhi! Zyadatar bonds, ionic, covalent aur metallic ke beech ka koi form hote hain. Isko ek triangle se samjhaya ja sakta hai (jaise Figure 3.1 mein hai). Ionic Bonds Formation: Example NaCl Na (electropositive) aur Cl (electronegative) jab react karte hain: Na (1s² 2s² 2p⁶ 3s¹) → Na⁺ (1s² 2s² 2p⁶) + e⁻ Cl (1s² 2s² 2p⁶ 3s² 3p⁵) + e⁻ → Cl⁻ (1s² 2s² 2p⁶ 3s² 3p⁶) Na$^+$ aur Cl$^-$ ions electrostatic attraction se crystal lattice mein jud jaate hain. Ye process energetically favourable hai. Na ⋅ + ⋅ Cl : → [Na]⁺ + [ : Cl : ]⁻ Similarly, for CaCl$_2$: Ca : + 2 ⋅ Cl : → [Ca]²⁺ + 2 [ : Cl : ]⁻ Dost, ionic bond formation ko aur gehrayi se samajhne ke liye kuch important concepts hain jinse JEE mein bohot questions aate hain: Ionization Energy (IE): Kisi isolated gaseous atom se ek electron nikalne ke liye required energy. Jitni kam IE hogi, utna easily cation banega. Alkali metals (Group 1) ki IE sabse kam hoti hai. Electron Gain Enthalpy (EGE) / Electron Affinity (EA): Kisi isolated gaseous atom mein ek electron add karne par jo energy release ya absorb hoti hai. Jitni zyada negative EGE hogi (ya jitni zyada EA hogi), utna easily anion banega. Halogens (Group 17) ki EGE sabse negative hoti hai. Lattice Energy ($U$): Ek mole ionic solid ko uske constituent gaseous ions mein alag karne ke liye required energy. Ya fir, gaseous ions se ek mole ionic solid banane par jo energy release hoti hai. Iski value jitni zyada negative hogi (ya magnitude jitna zyada hoga), ionic compound utna hi stable hoga. Lattice energy depend karti hai: Charge on Ions ($q_1, q_2$): $U \propto |q_1 q_2|$. Higher charges, higher lattice energy. Jaise, NaCl se zyada lattice energy MgS ki hogi. Size of Ions ($r_+, r_-$): $U \propto \frac{1}{r_+ + r_-}$. Smaller ions, higher lattice energy. Jaise, LiF ki lattice energy CsI se zyada hogi. Born-Haber Cycle: Ye ek experimental method hai lattice energy calculate karne ka, jo Hess's Law par based hai. Isme enthalpy changes ko use karke indirectly lattice energy nikali jaati hai. Is cycle mein ye steps involved hote hain: Sublimation of metal (solid to gas) Ionization of gaseous metal (gas to gaseous ion) Dissociation of non-metal (if diatomic, e.g., Cl$_2$ to 2Cl) Electron gain by gaseous non-metal (gaseous atom to gaseous anion) Formation of ionic solid from gaseous ions (this is lattice energy) Overall enthalpy of formation of the ionic solid Formula kuch aisa dikhta hai: $$ \Delta H_f^\circ = \Delta H_{sub} + IE + \frac{1}{2} \Delta H_{diss} + EGE + U $$ Is formula ko rearrange karke $U$ nikal sakte hain. JEE mein ispe direct numericals aate hain. Factors Affecting Ionic Bond Formation: Low Ionization Energy (IE) for Cation Formation: Metal atom ki IE jitni kam hogi, utna easily wo electron lose karke cation banayega. Group 1 aur Group 2 elements ki IE kam hoti hai. High Electron Gain Enthalpy (EGE) for Anion Formation: Non-metal atom ki EGE jitni zyada negative hogi, utna easily wo electron accept karke anion banayega. Halogens ki EGE sabse zyada negative hoti hai. High Lattice Energy: Ionic solid ki stability lattice energy par depend karti hai. High lattice energy tab hoti hai jab ions par charge zyada ho aur ions ka size chhota ho. Ye overall process ko energetically favourable banata hai. Properties of Ionic Compounds: Physical State: Mostly solid hote hain, crystalline nature ke. Melting and Boiling Points: High hote hain kyunki strong electrostatic forces ko overcome karne ke liye bohot energy chahiye hoti hai. Solubility: Polar solvents (jaise water) mein soluble hote hain, kyunki water molecules ions ko hydrate karke stabilize karte hain. Non-polar solvents mein insoluble hote hain. Solubility sirf lattice energy par nahi, hydration energy par bhi depend karti hai. Agar hydration energy lattice energy se zyada ho, toh compound soluble hoga. Electrical Conductivity: Solid state mein non-conductors hote hain kyunki ions fixed positions par hote hain. Molten state ya aqueous solution mein good conductors hote hain kyunki ions free to move hote hain. Brittleness: Ionic solids brittle hote hain. Jab ek layer dusri layer par slide karti hai, toh like charges ek doosre ke saamne aa jaate hain, jisse strong repulsion hota hai aur crystal break ho jaata hai. Fajan's Rules (Covalent Character in Ionic Bonds): Dost, koi bhi bond 100% ionic ya 100% covalent nahi hota. Har ionic bond mein thoda covalent character hota hai. Fajan's rules batate hain ki kab ionic bond mein covalent character zyada hoga: Small Cation Size: Chhota cation electron cloud ko zyada effectively polarize karta hai. Large Anion Size: Bada anion easily polarize ho jaata hai kyunki uske outer electrons loosely held hote hain. High Charge on Cation/Anion: Ions par charge jitna zyada hoga, polarization utna zyada hoga. Cation with Pseudo Noble Gas Configuration: Transition metal cations (jaise Cu$^+$, Ag$^+$, Zn$^{2+}$) jinme $(n-1)d^{10}$ configuration hoti hai, wo noble gas configuration wale cations (jaise Na$^+$, K$^+$) se zyada polarizing hote hain. Zyada covalent character matlab lower melting point, lower solubility in polar solvents, aur sometimes colour formation (jaise AgCl white, AgBr pale yellow, AgI yellow). Hydration Energy: Dost, jab ek ionic compound water mein dissolve hota hai, toh uske ions water molecules se ghir jaate hain. Is process mein jo energy release hoti hai use hydration energy kehte hain. Hydration energy bhi stability aur solubility mein bohot important role play karti hai. Factors Affecting Hydration Energy: Charge on Ions: Hydration energy $\propto |q|$. Ions par charge jitna zyada hoga, water molecules utna strong attract honge aur hydration energy utni zyada hogi. Jaise, Mg$^{2+}$ ki hydration energy Na$^+$ se zyada hogi. Size of Ions: Hydration energy $\propto \frac{1}{r}$. Ions ka size jitna chhota hoga, charge density utni zyada hogi, aur water molecules utna strong attract honge, jisse hydration energy zyada hogi. Jaise, Li$^+$ ki hydration energy Cs$^+$ se zyada hogi. Solubility ka Criterion: Koi ionic compound water mein tab soluble hoga jab uski hydration energy uski lattice energy se zyada ho. Yani, ions ko alag karne mein jo energy (lattice energy) lagti hai, usse zyada energy ions ke hydrate hone par release ho. $$ \Delta H_{solution} = U + \Delta H_{hydration} $$ Agar $\Delta H_{solution}$ negative ya small positive ho, toh compound soluble hoga. Questions kahan se aate hain: Lattice energy ka order compare karna based on charge and size. Born-Haber cycle ke steps aur unki enthalpy changes ko use karke kisi unknown enthalpy change (aksar lattice energy) ko calculate karna. Ionic character ka order determine karna (Fajan's rule se related, jo covalent character explain karta hai). Melting point, solubility jaise physical properties ko lattice energy aur hydration energy se relate karna. Higher lattice energy usually means higher melting point and lower solubility (but hydration energy ka role bhi hota hai solubility mein). Fajan's rules apply karke covalent character ka order batana aur uske basis par properties (melting point, solubility, colour) explain karna. Ionic compounds ki conductivity par questions (solid vs molten/aqueous state). Stability of ionic compounds ko IE, EGE, aur Lattice Energy ke terms mein explain karna. Hydration energy ka order compare karna based on charge and size. Solubility ko lattice energy aur hydration energy ke basis par predict karna aur compare karna. Jaise, Group 2 sulfates ki solubility down the group decrease hoti hai (Lattice energy ka decrease hydration energy ke decrease ko dominate karta hai). Group 2 hydroxides ki solubility down the group increase hoti hai (Hydration energy ka decrease lattice energy ke decrease se zyada hota hai). Ye trends bohot important hain. Covalent Bonds Formation: Example Cl$_2$ Jab do electronegative atoms react karte hain, toh wo electrons share karte hain. : Cl ⋅ + ⋅ Cl : → : Cl : Cl : Har Cl atom apna ek electron share karta hai, aur dono ke outer shell mein 8 electrons (octet) complete ho jaate hain. Isko Lewis structure mein ek line (Cl—Cl) se dikhate hain. Covalent Bond Formation ke Types: Nonpolar Covalent Bond: Jab do identical atoms (jaise Cl$_2$, O$_2$) ke beech electrons equally share hote hain. Electronegativity difference zero hota hai. Polar Covalent Bond: Jab do different atoms (jaise HCl, H$_2$O) ke beech electrons unequally share hote hain. Ek atom zyada electronegative hota hai aur shared electrons ko apni taraf zyada khinchta hai, jisse partial positive ($\delta+$) aur partial negative ($\delta-$) charges develop hote hain. Factors Affecting Covalent Bond Strength: Bond Length: Jitni bond length kam hogi, utni bond strength zyada hogi. Bond Order: Single bond (C-C) se zyada strong double bond (C=C) hota hai, aur double bond se zyada strong triple bond (C≡C) hota hai. Electronegativity Difference: Zyada electronegativity difference se bond polar ho jaata hai, jo kabhi-kabhi bond strength ko affect karta hai. Example CCl$_4$: Carbon ke outer shell mein 4 electrons hote hain, aur 4 Cl atoms ke saath share karke 4 bonds banata hai, jisse carbon ka octet complete hota hai. Ismein carbon $sp^3$ hybridised hota hai aur geometry tetrahedral hoti hai. Cl | Cl — C — Cl | Cl Example NH$_3$: Nitrogen 3 H atoms ke saath 3 bonds banata hai, aur uske paas ek lone pair of electrons bhi hota hai. Nitrogen $sp^3$ hybridised hota hai, lekin lone pair ki wajah se geometry pyramidal hoti hai. H | H : N : H H Lone Pair of Electrons: Wo electron pair jo bond formation mein participate nahi karta, lekin molecule ki geometry aur polarity ko bohot affect karta hai. Important Concepts for JEE: Lewis Dot Structures: Molecules ke valence electrons aur bonding ko represent karne ka tareeka. Octet rule aur exceptions (hypervalent, hypovalent molecules) ko samajhna zaroori hai. Formal Charge: Atom par hypothetical charge, jo uske bonding pattern se determine hota hai. Stability of resonance structures compare karne mein help karta hai. Formula: $FC = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$. VSEPR Theory (Valence Shell Electron Pair Repulsion Theory): Lone pair-lone pair, lone pair-bond pair, aur bond pair-bond pair repulsions ke order ko samajhkar molecule ki geometry predict karna. Repulsion order: LP-LP > LP-BP > BP-BP. Hybridization: Atomic orbitals ka intermixing naye hybrid orbitals banane ke liye, jo bonding mein participate karte hain. $sp, sp^2, sp^3, dsp^2, sp^3d, sp^3d^2$ types aur unki corresponding geometries. Bond Energy/Enthalpy: Ek mole covalent bonds ko todne ke liye required energy. Iska use reaction enthalpy calculate karne mein hota hai. Dipole Moment ($\mu$): Polarity ka quantitative measure. Formula: $\mu = q \times r$. Net dipole moment molecule ki geometry aur individual bond dipoles par depend karta hai. Symmetrical molecules ka net dipole moment zero ho sakta hai (e.g., CO$_2$, CCl$_4$). Resonance Structures: Jab ek single Lewis structure molecule ki saari properties explain nahi kar paati, tab multiple structures (resonance structures) banate hain. Actual molecule in sabka hybrid hota hai. Resonance stability, bond order, aur formal charge se related questions aate hain. Example: Benzene, Carbonate ion ($CO_3^{2-}$). Molecular Orbital Theory (MOT): Atomic orbitals combine hokar molecular orbitals banate hain. Bonding aur anti-bonding molecular orbitals. Isse bond order, magnetic properties (paramagnetic/diamagnetic), aur stability of molecules/ions (e.g., $O_2, N_2, O_2^{2-}$) predict karte hain. Energy level diagrams banana aur fill karna seekhna. Back Bonding (p$\pi$-p$\pi$ or p$\pi$-d$\pi$): Jab ek atom ke paas lone pair ho aur adjacent atom ke paas vacant orbital ho (p ya d), toh lone pair back donate ho jaata hai. Isse bond order badhta hai aur bond length kam hoti hai. Example: $BF_3$, Silylamines ($R_3Si-NR_2$). Bridge Bonding (3-center-2-electron bond): Boron compounds mein common, jaise Diborane ($B_2H_6$). Ismein 3 atoms 2 electrons share karte hain. Ye electron deficient compounds mein hota hai. Hydrogen Bonding: Jab Hydrogen atom high electronegative atom (F, O, N) se attached ho, toh wo dusre electronegative atom ke saath weak electrostatic attraction banata hai. Types: Intermolecular (paani mein) aur Intramolecular (o-nitrophenol mein). Physical properties (boiling point, solubility) par iska bohot impact hota hai. Metallic Bonding: Metals mein atoms ke beech ka bond, jahan valence electrons delocalized hote hain "sea of electrons" mein. Iski wajah se metals electrical conductivity, malleability, aur ductility show karte hain. Van der Waals Forces: Weak intermolecular forces jo non-polar molecules aur noble gases mein bhi exist karti hain. Types: London Dispersion Forces (LDF), Dipole-Dipole interactions, Dipole-Induced Dipole interactions. Ye forces physical properties jaise boiling point, melting point aur solubility ko affect karti hain. Bond Angle: Molecules mein atoms ke beech ka angle. VSEPR theory se predict kiya jaata hai. Lone pairs ki presence bond angles ko modify karti hai (e.g., $CH_4$ vs $NH_3$ vs $H_2O$). Bond Length Order: Single, double, triple bonds ki length ka comparison. Triple bond sabse chhota aur single bond sabse lamba hota hai. Resonance structures mein bond length intermediate hoti hai. Isoelectronic Species: Molecules ya ions jinke paas same number of electrons hote hain. Inki bonding aur structure mein similarities ho sakti hain. Example: $CO, N_2$. In concepts par questions bohot aate hain, especially VSEPR, Hybridization, Dipole Moment, Resonance, aur MOT se. Advanced mein Back Bonding aur Bridge Bonding par bhi questions ban sakte hain. 3.4 The Covalent Bond (Covalent Bond ki Gehrai) Covalent bond ko samjhane ke liye alag-alag theories hain. Har theory ke apne pros and cons hain. The Lewis Theory (Octet Rule) Lewis theory kehti hai ki atoms bonds banate hain taaki unke outer shell mein 8 electrons ho jayein (octet rule). Hydrogen ke liye 2 electrons (duplet rule). Cl$_2$: Each Cl atom (7 outer electrons) shares one electron to form Cl:Cl, completing octet for both. CCl$_4$: Carbon (4 outer electrons) forms 4 bonds with 4 Cl atoms to achieve octet. NH$_3$: Nitrogen (5 outer electrons) forms 3 bonds with 3 H atoms, leaving one lone pair. Octet complete. Double and Triple Bonds: Double Bond: 4 electrons shared (2 pairs), jaise C=C (Ethene). Triple Bond: 6 electrons shared (3 pairs), jaise C≡C (Ethyne). Exceptions to Octet Rule (Ye bohot important hai JEE ke liye!): Incomplete Octet: Kuch elements ke outer shell mein 8 se kam electrons hote hain, even after bonding. Li, Be, B ke compounds mein. Jaise, LiCl (2e-), BeH$_2$ (4e-), BF$_3$ (6e-). Ye compounds electron-deficient hote hain aur Lewis acids ki tarah act karte hain. Expanded Octet (Hypervalent Compounds): Third period aur uske baad ke elements (jinme d-orbitals hote hain) 8 se zyada electrons accommodate kar sakte hain. Jaise, PF$_5$ (10e- around P), SF$_6$ (12e- around S), H$_2$SO$_4$ (S ke around 12e-). Ye d-orbitals ki involvement se hota hai. Odd-Electron Molecules: Jin molecules mein total number of valence electrons odd hota hai, unka octet complete nahi ho sakta. Jaise, NO (5+6=11e-), NO$_2$ (5+2*6=17e-). Ye molecules paramagnetic hote hain. Formal Charge (Ye bhi bohot kaam aata hai!): Kisi atom par formal charge calculate karna important hai stability aur preferred Lewis structure determine karne ke liye. Formula: $FC = V - L - \frac{1}{2}B$ $V$ = Valence electrons of the atom. $L$ = Non-bonding (lone pair) electrons. $B$ = Bonding (shared) electrons. Example: O$_3$ (Ozone) mein formal charge calculate karna. Resonance Structures (Stability ka raaz!): Jab ek single Lewis structure molecule ki saari properties explain nahi kar paati, tab hum multiple Lewis structures draw karte hain, jinhe resonance structures kehte hain. Actual structure in sabka hybrid hota hai. Resonance structures mein atoms ki position same rehti hai, sirf electrons ki placement change hoti hai. Jitne zyada stable resonance structures hote hain, molecule utna hi zyada stable hota hai. Resonance energy = (Energy of actual molecule) - (Energy of most stable canonical form). Example: CO$_3^{2-}$ , SO$_4^{2-}$ , Benzene . Lewis Structures Draw Karne Ke Steps (Ye basic hai, par bohot zaroori!): Total valence electrons count karo (anions ke liye negative charge add karo, cations ke liye subtract). Central atom choose karo (usually least electronegative atom, ya jo sabse kam hai). Hydrogen aur Fluorine kabhi central atom nahi hote. Single bonds banao central atom aur surrounding atoms ke beech. Bache hue electrons ko surrounding atoms par lone pairs ki tarah rakho, unka octet complete karte hue. Agar electrons bach gaye hain, toh unhe central atom par lone pairs ki tarah rakho. Agar central atom ka octet complete nahi hua hai, toh surrounding atoms se lone pairs ko double ya triple bonds mein convert karo. Har atom ka formal charge calculate karo aur check karo ki structure stable hai ya nahi (kam formal charge, zyada stable). JEE Advanced perspective: Octet rule exceptions par based questions, especially expanded octet, bohot common hain. Formal charge calculation aur most stable resonance structure identify karna aata hai. Lewis structures draw karna aur unse geometry (VSEPR theory se link karke) predict karna. Bond order calculation bhi important hai, especially resonance structures ke context mein. Stability of Resonance Structures: Neutral structures charged structures se zyada stable hoti hain. Less formal charge on atoms, zyada stable. Negative charge zyada electronegative atom par ho, toh zyada stable. Positive charge zyada electropositive atom par ho, toh zyada stable. Octet complete ho sab atoms ka, toh zyada stable (even if it means some formal charge). Applications: Lewis structures se hum molecule ki polarity, bond length, bond strength aur reactivity ka idea laga sakte hain. Bond Order (Resonance ke context mein): Jab resonance hota hai, toh single aur multiple bonds ka character mix ho jata hai. Bond order nikalne ka formula hai: $Bond\ Order = \frac{Total\ number\ of\ bonds\ between\ two\ atoms\ in\ all\ resonance\ structures}{Total\ number\ of\ resonance\ structures}$ Example: CO$_3^{2-}$ mein C-O bond order 1.33 hota hai, jo single aur double bond ke beech ki value hai. Isse pata chalta hai ki saare C-O bonds equivalent hain aur intermediate length ke hain. Limitations of Lewis Theory (Ye bhi pata hona chahiye!): Molecule ki exact geometry nahi batati (uske liye VSEPR theory hai). Magnetic properties explain nahi kar paati (jaise O$_2$ paramagnetic hai, jo Lewis structure se explain nahi hota). Bond angles aur bond lengths ki accurate values nahi deti. Odd-electron molecules ki stability explain nahi kar paati. Transition metal complexes ke bonding ko explain nahi kar paati. Dative bonds (coordinate bonds) ko bhi Lewis theory mein explicitly dikhana thoda tricky hota hai, although unhe bhi shared pair of electrons hi mana jata hai. Exceptions to the Octet Rule (Jahan Octet Rule Toota) Incomplete Octet: Atoms with less than 4 outer electrons (Be, B) cannot attain octet even after forming bonds. Be + 2 : F : → : F : Be : F : B + 3 : F : → : F : B : F : : F : Ye cases bohot common hain, especially Boron aur Beryllium ke compounds mein. Yaad rakhna ki ye stable hote hain despite having less than 8 electrons. Inki Lewis structures banate waqt is baat ka dhyaan rakhna. JEE Focus: Incomplete octet wale compounds (jaise BF$_3$, AlCl$_3$) Lewis Acids hote hain, kyunki inke paas electron pair accept karne ki tendency hoti hai (electron deficient hote hain). Is property pe direct questions aate hain. Example: BF$_3$ + NH$_3$ $\rightarrow$ F$_3$B $\leftarrow$ NH$_3$. Yahan BF$_3$ Lewis acid hai aur NH$_3$ Lewis base. Advanced Concept: Incomplete octet wale molecules ki back-bonding bhi ek important concept hai. Jaise BF$_3$ mein, Fluorine ke lone pair electrons Boron ke vacant p-orbital mein donate ho sakte hain, jisse B-F bond mein partial double bond character aa jaata hai. Isse BF$_3$ ki Lewis acidity kam ho jaati hai compared to BCl$_3$ (kyunki Cl ki back-bonding tendency kam hoti hai). Is back-bonding se bond length, bond strength aur stability pe bhi asar padta hai. Expanded Octet: Elements in 3rd period onwards (like P, S) can use d-orbitals, allowing them to have more than 8 electrons in their valence shell. Jaise PF$_5$ (10 electrons), SF$_6$ (12 electrons). Ye wala concept bohot important hai, especially jab aap PCl$_5$, SF$_6$, H$_2$SO$_4$ jaise molecules ki structures banate ho. D-orbitals ka involvement hi inko 8 se zyada electrons accommodate karne deta hai. Isse hybridization aur geometry bhi affect hoti hai, toh isko acche se samajhna. JEE Focus: Expanded octet wale molecules ki hybridization aur geometry (VSEPR theory) pe bohot questions aate hain. Jaise PCl$_5$ ki trigonal bipyramidal geometry hoti hai (sp$^3$d hybridization), aur SF$_6$ ki octahedral (sp$^3$d$^2$ hybridization). Inki bond angles aur axial/equatorial positions pe bhi questions ban sakte hain. Yaad rakhna ki 3rd period ke elements hi d-orbital use kar sakte hain, 2nd period ke nahi. Isliye NCl$_5$ exist nahi karta jabki PCl$_5$ karta hai. Advanced Concept: Expanded octet mein lone pairs ki position bohot critical hoti hai. VSEPR theory ke according, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. Isliye lone pairs hamesha aisi position pe aate hain jahan repulsion minimum ho. Jaise SF$_4$ mein lone pair equatorial position pe aata hai, jisse molecule ki geometry seesaw ho jaati hai, na ki trigonal bipyramidal. Similarly, XeF$_2$ mein lone pairs equatorial positions pe aate hain, jisse linear geometry milti hai. Ispe direct questions aate hain! Odd Electron Molecules: Molecules like NO, ClO$_2$ mein odd number of electrons hote hain, toh octet complete nahi ho pata. Ye free radicals hote hain aur bohot reactive bhi. Inki Lewis structures mein ek unpaired electron dikhta hai. JEE mein inke paramagnetic nature pe questions aate hain. JEE Focus: Odd electron species hamesha paramagnetic hote hain kyunki inke paas unpaired electron hota hai. Inki stability kam hoti hai aur ye dimerize karke stable ho sakte hain (jaise NO$_2$ $\rightleftharpoons$ N$_2$O$_4$). Inke Lewis structures banate waqt unpaired electron ko sahi jagah pe dikhana important hai. Kabhi-kabhi formal charge ko minimize karne ke liye unpaired electron ko less electronegative atom par rakha jata hai. Advanced Concept: Odd electron molecules ki reactivity aur dimerisation pe questions aate hain. Jaise NO$_2$ dimerize karke N$_2$O$_4$ banata hai, jo diamagnetic hota hai. Is equilibrium pe temperature aur pressure ka kya effect hoga, aise questions bhi ban sakte hain (Le Chatelier's Principle). Inki bond lengths aur bond angles pe bhi asar padta hai dimerisation se. Other Exceptions: O$_2$ is paramagnetic (2 unpaired electrons), which octet rule explain nahi karta. O$_2$ ka paramagnetic behavior Molecular Orbital Theory (MOT) se explain hota hai, octet rule se nahi. Ye ek classic example hai jahan octet rule fail ho jaata hai. Similar cases B$_2$ aur N$_2^+$ mein bhi dekhne ko milte hain. JEE Focus: O$_2$ aur B$_2$ jaise molecules ka paramagnetic nature octet rule explain nahi kar pata. Inke liye Molecular Orbital Theory (MOT) ka use hota hai. MOT se bond order, magnetic nature aur stability predict ki jaati hai. JEE mein direct questions aate hain ki 'Which of the following is paramagnetic?' aur options mein O$_2$, N$_2$, F$_2$ jaise molecules hote hain. N$_2$ ka diamagnetic hona bhi octet rule se explain nahi hota, MOT se hota hai. Advanced Concept (MOT Connection): MOT se hum O$_2$ ka electronic configuration likhte hain: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Yahan do unpaired electrons anti-bonding $\pi$ orbitals mein hote hain, jo iske paramagnetic nature ko justify karte hain. Bond order bhi MOT se hi nikalta hai: $BO = \frac{1}{2} (N_b - N_a)$. Ispe bohot questions aate hain, jaise O$_2$, O$_2^+$, O$_2^-$ ka bond order aur stability compare karna. Important Note for JEE: Formal Charge: Jab octet rule ke exceptions wale molecules ki Lewis structures banate ho, toh formal charge calculate karna mat bhoolna. Ye stability predict karne mein help karta hai. Minimum formal charge wali structure zyada stable hoti hai. Resonance: Kuch exceptions mein resonance structures bhi hoti hain (jaise O$_3$, NO$_2$). Isse molecule ki actual structure aur stability ko better explain kiya ja sakta hai. Resonance hybrid ki concept ko acche se samajhna. Bond Order: Octet rule ke exceptions wale molecules mein bond order calculate karna bhi important hai, especially jab MOT apply hoti hai. Higher bond order matlab zyada stable molecule. Hybridization and Geometry: Expanded octet wale molecules ki hybridization aur geometry (VSEPR theory se) pe bohot questions aate hain. Jaise PCl$_5$ ki trigonal bipyramidal geometry. Lone pair ki presence geometry ko kaise affect karti hai, ye bhi important hai (e.g., SF$_4$ seesaw). Stability: Incomplete octet wale compounds (BF$_3$) Lewis acids hote hain kyunki unhe electron pair ki zaroorat hoti hai. Ye bhi ek common question type hai. Stability order pe bhi questions aa sakte hain. Isostructural/Isoelectronic: Kabhi-kabhi exceptions wale molecules ke isostructural ya isoelectronic species pe questions aate hain. Jaise CO$_2$ aur N$_2$O isoelectronic hain. Bond Lengths and Angles: Back-bonding, resonance, aur lone pair repulsions ki wajah se bond lengths aur bond angles mein deviations aate hain. In deviations pe bhi conceptual questions ban sakte hain. Jaise BF$_3$ mein B-F bond length single bond se kam hoti hai due to back-bonding. Oxoacids of P, S, Cl: In sab mein expanded octet ka concept use hota hai. H$_3$PO$_4$, H$_2$SO$_4$, HClO$_4$ jaise molecules ki structures, oxidation states, aur acidity pe questions aate hain. Yahan bhi formal charge minimization aur d-orbital involvement important hai. Inert Pair Effect: Heavy p-block elements (jaise Tl, Pb, Bi) mein lower oxidation states zyada stable hoti hain due to inert pair effect. Iski wajah se bhi octet rule ke deviations dekhne ko milte hain. Jaise TlCl$_3$ se zyada stable TlCl hota hai. Ye bhi ek important concept hai jo chemical bonding aur p-block dono mein aata hai. Hypervalent Compounds: Expanded octet wale compounds ko hypervalent compounds bhi kehte hain. Inki stability aur formation conditions pe questions aa sakte hain. Jaise XeF$_2$, XeF$_4$, XeF$_6$ (noble gas compounds) bhi expanded octet ke examples hain. Inki structures aur hybridization bohot common questions hain. Bridge Bonding (3c-2e bonds): Boron compounds (jaise B$_2$H$_6$) mein incomplete octet ko compensate karne ke liye bridge bonding hoti hai. Isme 3 atoms 2 electrons share karte hain. Ye bhi octet rule ka ek major exception hai aur iski structure, bonding, aur properties pe questions aate hain. In sab points ko dimaag mein rakhna, kyunki in exceptions se direct aur indirect questions bohot aate hain JEE mein! Octet rule ki limitations ko samajhna chemical bonding ka ek crucial part hai. Is topic se har saal 2-3 questions expect kar sakte ho, especially advanced mein. Oxidation Numbers (Oxidation State ka Hisab) Covalent compounds mein oxidation number calculate karne ke liye, shared electrons ko more electronegative atom ko assign karte hain, aur phir theoretical charge count karte hain. H$_2$O: O ka oxidation state -2. SF$_4$: S ka oxidation state +4. SF$_6$: S ka oxidation state +6. Chalo, oxidation numbers ko aur detail mein samajhte hain, kyunki ye JEE Advanced aur Mains mein bohot kaam aata hai! Free State Elements: Koi bhi element jab apne free ya uncombined state mein hota hai, uska oxidation number 0 hota hai. Jaise H$_2$, O$_2$, Na, Fe, Cl$_2$. Monatomic Ions: Monatomic ion ka oxidation number uske charge ke barabar hota hai. Jaise Na$^+$ ka +1, Cl$^-$ ka -1, Mg$^{2+}$ ka +2. Group 1 Elements (Alkali Metals): Inka oxidation number hamesha +1 hota hai compounds mein (Li, Na, K, Rb, Cs). Group 2 Elements (Alkaline Earth Metals): Inka oxidation number hamesha +2 hota hai compounds mein (Be, Mg, Ca, Sr, Ba). Hydrogen (H): Non-metals ke saath +1 (jaise H$_2$O, HCl, NH$_3$). Metals ke saath -1 (metal hydrides, jaise NaH, CaH$_2$). Oxygen (O): Most compounds mein -2 (jaise H$_2$O, CO$_2$). Peroxides mein -1 (jaise H$_2$O$_2$, Na$_2$O$_2$). Superoxides mein -1/2 (jaise KO$_2$). Oxygen difluoride (OF$_2$) mein +2. Dioxygen difluoride (O$_2$F$_2$) mein +1. Halogens (F, Cl, Br, I): Fluorine (F) hamesha -1 hota hai compounds mein. Baaki halogens (Cl, Br, I) generally -1 hote hain, lekin jab oxygen ya more electronegative halogens ke saath hote hain, toh positive oxidation states bhi show karte hain (jaise HClO, KClO$_3$, ICl$_3$). Polyatomic Ions: Polyatomic ion mein sabhi atoms ke oxidation numbers ka sum us ion ke total charge ke barabar hota hai. Jaise SO$_4^{2-}$ mein S ka oxidation number nikalne ke liye, S + 4(-2) = -2 solve karte hain. Important Concepts & Questions Jahan Se Aate Hain: Redox Reactions Identify Karna: Oxidation numbers ki help se hum easily identify kar sakte hain ki kaunsa species oxidize ho raha hai (oxidation number increase) aur kaunsa reduce ho raha hai (oxidation number decrease). Balancing Redox Reactions: Oxidation number method se redox reactions balance karna ek common question type hai. Disproportionation Reactions: Ye wo reactions hain jahan ek hi element oxidize bhi hota hai aur reduce bhi. Oxidation number se inko pehchanna aasan ho jaata hai. Jaise $2H_2O_2 \rightarrow 2H_2O + O_2$. Yahan oxygen $-1$ se $-2$ (reduction) aur $0$ (oxidation) mein ja raha hai. Fractional Oxidation Numbers: Kuch compounds mein average oxidation number fractional ho sakta hai, jaise C$_3$O$_2$ mein carbon ka oxidation number +4/3. Ye tab hota hai jab atoms chemically non-equivalent hote hain. Structure bana ke individual atoms ka nikalna padta hai. Example: Fe$_3$O$_4$ (Magnetite) ko FeO.Fe$_2$O$_3$ ki tarah dekho. Yahan do Fe atoms ka +3 aur ek Fe atom ka +2 hota hai, average (2*3 + 1*2)/3 = 8/3. Example: Br$_3$O$_8$ (Tribromine Octoxide). Iska structure dekhoge toh do terminal Br ka +6 aur central Br ka +4 hota hai. Average (2*6 + 1*4)/3 = 16/3. Oxidizing and Reducing Agents: Jo species khud reduce hota hai aur doosron ko oxidize karta hai, wo oxidizing agent hota hai. Jo species khud oxidize hota hai aur doosron ko reduce karta hai, wo reducing agent hota hai. Oxidation numbers se inki strength ka idea lagaya ja sakta hai. Structure-Based Oxidation Number: Jab simple rules kaam na karein, especially organic compounds ya complex inorganic compounds mein, toh structure bana kar electronegativity differences ke basis par oxidation number calculate karte hain. Har bond ko break karte hain aur electrons more electronegative atom ko assign karte hain. H-O-Cl H (+1) - O (-2) - Cl (+1) (O is more electronegative than H and Cl) H-C=O | H (Formaldehyde) C ka oxidation number: C-H bond: C is slightly more electronegative than H, so C gets -1 from each H. (-2 total) C=O bond: O is much more electronegative than C, so O gets both electrons from each bond. C loses 2 electrons from each bond. (+4 total) Net for C = -2 + 4 = +2 Maximum and Minimum Oxidation States: Kisi bhi element ka maximum oxidation state uske group number ke barabar hota hai (except for F, O). Jaise Nitrogen (Group 15) ka max +5. Kisi bhi element ka minimum oxidation state uske group number minus 8 ke barabar hota hai. Jaise Nitrogen (Group 15) ka min 15-8 = -3. Element: N (Group 15) Max O.S. = +5 (e.g., HNO$_3$) Min O.S. = -3 (e.g., NH$_3$) Agar koi element apne maximum oxidation state mein hai, toh wo sirf oxidizing agent ki tarah act kar sakta hai (kyunki wo sirf reduce ho sakta hai). Agar koi element apne minimum oxidation state mein hai, toh wo sirf reducing agent ki tarah act kar sakta hai (kyunki wo sirf oxidize ho sakta hai). Agar koi element intermediate oxidation state mein hai, toh wo oxidizing aur reducing agent dono ki tarah act kar sakta hai (jaise H$_2$O$_2$ mein O ka -1). Peroxy Linkage Detection: Jab kisi compound mein central atom ka calculated oxidation number uske maximum possible oxidation number se zyada aaye, toh samajh jao ki usmein peroxy linkage ($-\text{O}-\text{O}-$) present hai. Example: H$_2$SO$_5$ (Caro's acid). Sulfur (Group 16) ka max O.S. +6 hai. Agar hum calculate karein: $2(+1) + S + 5(-2) = 0 \Rightarrow 2 + S - 10 = 0 \Rightarrow S = +8$. Ye +6 se zyada hai, matlab peroxy linkage hai. Structure: H-O-S(=O)$_2$-O-O-H Yahan ek O-O bond hai, jismein dono O ka oxidation state -1 hai. Correct calculation: $2(+1) + S + 3(-2) + 2(-1) = 0 \Rightarrow 2 + S - 6 - 2 = 0 \Rightarrow S = +6$. Example: CrO$_5$ (Chromium pentoxide). Chromium (Group 6) ka max O.S. +6 hai. Agar hum calculate karein: $Cr + 5(-2) = 0 \Rightarrow Cr = +10$. Ye +6 se zyada hai, matlab peroxy linkage hai. Structure: Butterfly structure, jismein 4 oxygen atoms peroxy linkage mein hote hain (-1 each) aur ek oxygen atom double bond se जुड़ा होता hai (-2). Correct calculation: $Cr + 1(-2) + 4(-1) = 0 \Rightarrow Cr - 2 - 4 = 0 \Rightarrow Cr = +6$. Complex Ions aur Coordination Compounds mein Oxidation State: Ligands ka charge pata hona bohot zaroori hai. Neutral ligands (jaise H$_2$O, NH$_3$, CO) ka charge 0 hota hai. Anionic ligands (jaise Cl$^-$, CN$^-$, OH$^-$) ka charge unke ion ke barabar hota hai. Complex ion ka total charge uske andar ke sabhi atoms aur ligands ke oxidation states ka sum hota hai. Example: [Fe(CN)$_6$]$^{4-}$ Fe ka oxidation state nikalne ke liye: Fe + 6(CN$^-$ ka charge) = -4 Fe + 6(-1) = -4 Fe - 6 = -4 Fe = +2 Example: [Cr(H$_2$O)$_4$Cl$_2$]$^+$ Cr + 4(H$_2$O ka charge) + 2(Cl$^-$ ka charge) = +1 Cr + 4(0) + 2(-1) = +1 Cr - 2 = +1 Cr = +3 Organic Compounds mein Oxidation State: Organic compounds mein carbon ka oxidation state nikalna thoda tricky ho sakta hai kyunki carbon carbon ke saath bhi bond banata hai. Iske liye structure-based method sabse best hai. Har C-H bond ke liye C ko -1 assign karo (kyunki C, H se zyada electronegative hai). Har C-O, C-N, C-X (halogen) bond ke liye C ko +1 assign karo (kyunki O, N, X, C se zyada electronegative hain). Double ya triple bond ke liye accordingly +2 ya +3. Har C-C bond ke liye 0 assign karo (kyunki electronegativity difference 0 hai). Example: CH$_3$COOH (Acetic Acid) Carbon 1 (CH$_3$ group): 3 C-H bonds = 3 * (-1) = -3 1 C-C bond = 0 Total for C1 = -3 Carbon 2 (COOH group): 1 C-C bond = 0 1 C=O bond = +2 1 C-O bond = +1 Total for C2 = +3 Is tarah, organic compounds mein har carbon atom ka individual oxidation state alag ho sakta hai. In sab points par dhyan dena, especially fractional oxidation numbers, structure-based calculations, maximum/minimum oxidation states, peroxy linkage detection, aur complex ions/organic compounds mein oxidation state calculation, kyunki ye JEE mein twist karke pooche jaate hain. Practice se hi ye concepts strong honge! Coordinate Bonds (Dative Bonds) (Sharing with a Twist) Ye special type ka covalent bond hai jismein electron pair sirf ek atom donate karta hai, aur dusra atom accept karta hai. Once formed, ye normal covalent bond jaisa hi hota hai. H | H : N : + H⁺ → [ H : N : H ]⁺ | | H H NH$_3$ ka lone pair H$^+$ ko donate karke ammonium ion banata hai. Similarly, NH$_3$ BF$_3$ ko lone pair donate karta hai. H F H F | | | | H : N + B : F → H — N — B — F | | | | H F H F Important Concepts for JEE Advanced & Mains: Lewis Acids and Bases: Coordinate bonds Lewis acid-base reactions ka core hain. Lewis Acid: Electron pair acceptor (e.g., H$^+$, BF$_3$, AlCl$_3$, Fe$^{3+}$, CO$_2$). Inke paas vacant orbitals hote hain. Lewis Base: Electron pair donor (e.g., NH$_3$, H$_2$O, OH$^-$, Cl$^-$, R-NH$_2$). Inke paas lone pairs hote hain. Formation Conditions: Donor atom ke paas at least ek lone pair hona chahiye. Acceptor atom ke paas vacant orbital hona chahiye (usually p, d, ya hybrid orbital). Properties: Coordinate bonds ki strength normal covalent bonds se thodi weaker ho sakti hai, but once formed, unhe differentiate karna mushkil hota hai. Inmein directionality hoti hai, jaise covalent bonds mein. Ye complex compounds (coordination compounds) ke formation mein crucial role play karte hain. Examples & Applications (Jahan se Questions Aate Hain): Ammonium Ion (NH$_4^+$): NH$_3$ (Lewis base) H$^+$ (Lewis acid) ko lone pair donate karta hai. Iski geometry tetrahedral hoti hai. Hybridisation sp$^3$. Hydronium Ion (H$_3$O$^+$): H$_2$O (Lewis base) H$^+$ (Lewis acid) ko lone pair donate karta hai. Iski geometry pyramidal hoti hai. Hybridisation sp$^3$. Complex Formation: Transition metals (d-block elements) ke ions (e.g., Cu$^{2+}$, Fe$^{3+}$, Ag$^+$) ligands (Lewis bases like NH$_3$, CN$^-$, H$_2$O) ke saath coordinate bonds banakar stable complex compounds banate hain. [Cu(NH₃)₄]²⁺ Yahan, NH$_3$ ligands hain jo Cu$^{2+}$ ion ko electron pair donate kar rahe hain. Boron Compounds: BF$_3$ (electron deficient, Lewis acid) amines (Lewis bases) ke saath adducts banata hai. BF₃ + NH₃ → F₃B ← NH₃ Ismein Boron ka octet complete ho jata hai. Aluminium Chloride (AlCl$_3$): Ye dimer (Al$_2$Cl$_6$) banata hai jismein coordinate bonds hote hain. Cl atoms bridge banate hain. Cl Cl Cl \ / \ / Al — Al / \ / \ Cl Cl Cl Yahan, Cl ke lone pairs Al ke vacant orbitals ko donate hote hain. Carbon Monoxide (CO) in Metal Carbonyls: CO ek strong ligand hai jo transition metals ke saath coordinate bonds banata hai, jaise Ni(CO)$_4$, Fe(CO)$_5$. Ye metal-carbon sigma bond coordinate type ka hota hai. $\pi$-Acid Ligands: CO, CN$^-$, NO$^+$ jaise ligands metal ke vacant d-orbitals mein electron density donate karte hain (sigma bond) aur metal ke filled d-orbitals se electron density ligand ke vacant anti-bonding $\pi^*$ orbitals mein accept karte hain (pi-backbonding). Isse metal-ligand bond aur strong ho jaata hai. JEE Advanced Specific: Back Bonding: Kuch cases mein, coordinate bond formation ke baad, donor atom ke paas vacant d-orbital ho toh, acceptor atom se electron density wapas donor atom ko bhi mil sakti hai (pi-backbonding). Example: BF$_3$ mein F se B ko backbonding hoti hai, but jab NH$_3$ se coordinate bond banta hai, toh backbonding nahi hoti. Back bonding se bond length kam hoti hai aur bond strength badhti hai. Geometry aur reactivity par bhi asar padta hai. Stability of Complexes: Ligand ki basicity, metal ion ka charge, size, aur chelation effect complex ki stability ko affect karte hain. Stronger Lewis base, stronger coordinate bond. Chelate effect: Jab ek ligand ek se zyada donor atoms se metal ion ke saath coordinate karta hai (polydentate ligand), toh chelate ring banti hai. Ye complexes non-chelated complexes se zyada stable hote hain. Example: Ethylenediamine (en) ya EDTA. Hybridisation and Geometry: Coordinate bond formation ke baad central atom ki hybridisation aur molecule ki geometry determine karna. VSEPR theory aur hybridisation concepts apply hote hain. Example: [Co(NH$_3$)$_6$]$^{3+}$ mein Co$^{3+}$ ki hybridisation d$^2$sp$^3$ hoti hai aur geometry octahedral. Isomerism in Coordination Compounds: Coordinate bonds ki wajah se coordination compounds mein structural aur stereoisomerism (geometrical, optical) dikhti hai. Isomerism se related questions JEE Advanced mein common hain. Crystal Field Theory (CFT) / Ligand Field Theory (LFT): Ye theories coordination compounds ke electronic structure, magnetic properties, aur color ko explain karti hain. Ligands metal ke d-orbitals ko split karte hain, jisse electron transitions aur magnetic behavior determine hota hai. Synergic Bonding: Metal carbonyls mein, metal-carbon sigma bond (ligand to metal donation) aur metal-carbon pi bond (metal to ligand back-donation) ek doosre ko strengthen karte hain. Isse synergic effect kehte hain, aur ye M-C bond ko bahut strong banata hai. In topics par questions structure, bonding, Lewis acid-base nature, aur complex formation se related aate hain. Geometry, hybridisation, aur stability bhi important hain. 3.5 Valence Bond Theory (VBT) (Bonds ki Gehrai ka Theory) Initially, VBT kehti thi ki sirf unpaired electrons aur pure atomic orbitals hi bonding mein participate karte hain. But, ismein kuch molecules (BeH$_2$, CH$_4$) ko explain karna mushkil tha. Hybridization: Is problem ko solve karne ke liye hybridization introduce kiya gaya. Hybridization matlab comparable energy ke pure atomic orbitals ka mix hona taaki naye hybrid orbitals ban sakein. Why Hybridization? Methane (CH$_4$) mein saare C-H bonds identical hote hain (bond length aur strength, aur 109°28' ke bond angles). Pure atomic orbitals se ye explain nahi hota tha. Features of Hybrid Orbitals: Number of hybrid orbitals formed = Number of atomic orbitals mixed. All hybrid orbitals are identical in energy and shape. Ek lobe small, dusra large hota hai. % s-character increase hone se orbital bulkier aur shorter hota hai; % p-character increase hone se longer aur thinner. % s-character increase hone se hybrid orbital ki energy decrease hoti hai. Hybrid orbitals sirf sigma ($\sigma$) bonds banate hain. Pi ($\pi$) bonds pure atomic orbitals (p-p, p-d, d-d overlap) se bante hain. Hybrid orbitals hamesha stable orientation mein rehte hain taaki repulsion kam ho aur maximum stability mile. Hybridization ka Calculation: Hybridization calculate karne ka sabse common aur easy tareeka hai Steric Number (SN) method. Steric Number (SN) = Number of sigma bonds + Number of lone pairs on central atom SN = 2 $\implies$ sp hybridization (linear geometry) SN = 3 $\implies$ sp$^2$ hybridization (trigonal planar geometry) SN = 4 $\implies$ sp$^3$ hybridization (tetrahedral geometry) SN = 5 $\implies$ sp$^3$d hybridization (trigonal bipyramidal geometry) SN = 6 $\implies$ sp$^3$d$^2$ hybridization (octahedral geometry) SN = 7 $\implies$ sp$^3$d$^3$ hybridization (pentagonal bipyramidal geometry) Important Note: Coordinate bonds ko bhi sigma bond ki tarah count karte hain hybridization calculation mein. Types of Hybridization: Hybridization Type % s % p % d Geometry Bond Angle (ideal) Example sp 50 50 - Linear 180° BeCl$_2$, C$_2$H$_2$ sp$^2$ 33.33 66.67 - Trigonal Planar 120° BF$_3$, C$_2$H$_4$ sp$^3$ 25 75 - Tetrahedral 109.5° CH$_4$, NH$_3$, H$_2$O sp$^3$d 20 60 20 Trigonal Bipyramidal 90°, 120° PCl$_5$ sp$^3$d$^2$ ~16.5 50 ~33 Octahedral 90° SF$_6$ sp$^3$d$^3$ ~14 ~43 ~43 Pentagonal Bipyramidal 72°, 90° IF$_7$ VSEPR Theory ka Role: Hybridization se hum molecule ki basic geometry predict karte hain. Lekin actual shape aur bond angles ko predict karne ke liye VSEPR (Valence Shell Electron Pair Repulsion) Theory ka use karte hain. VSEPR theory kehti hai ki electron pairs (bond pairs aur lone pairs) ek doosre ko repel karte hain aur space mein aise arrange hote hain jisse repulsion minimum ho. Lone Pair-Lone Pair (LP-LP) Repulsion > Lone Pair-Bond Pair (LP-BP) Repulsion > Bond Pair-Bond Pair (BP-BP) Repulsion Lone pairs ki wajah se ideal bond angles deviate ho jaate hain aur molecule ki geometry distort ho jaati hai. Example: CH$_4$ (sp$^3$, 4 BP, 0 LP): Tetrahedral, bond angle 109.5° NH$_3$ (sp$^3$, 3 BP, 1 LP): Pyramidal, bond angle ~107° (LP-BP repulsion ki wajah se angle kam ho gaya) H$_2$O (sp$^3$, 2 BP, 2 LP): Bent/V-shape, bond angle ~104.5° (do LP-BP repulsion ki wajah se angle aur kam ho gaya) Important Points for JEE Advanced/Mains: Hybridization of Central Atom: Sabse pehle central atom identify karo. Phir uske valence electrons, surrounding atoms aur charge (agar koi ho) ko consider karke Steric Number nikalo. Hybridization of Individual Atoms in Complex Molecules: Organic compounds mein har carbon atom ka hybridization alag ho sakta hai. Count sigma bonds around that specific atom. C with 4 single bonds: sp$^3$ C with 1 double bond, 2 single bonds: sp$^2$ C with 1 triple bond, 1 single bond OR 2 double bonds: sp Back Bonding: Kuch cases mein, hybridization expected se alag ho sakta hai due to back bonding. Jaise, BF$_3$ mein B sp$^2$ hybridized hota hai, lekin B-F bond mein partial double bond character hota hai due to back bonding. Isse bond length chhoti ho jaati hai. Back bonding tab hoti hai jab ek atom ke paas filled orbital ho (jismein lone pair ho) aur doosre atom ke paas adjacent empty orbital ho (usually p-orbital). Ye stability badhata hai. Drago's Rule: Third period aur uske baad ke elements (jaise P, S, As, Se) mein, agar central atom less electronegative atom se attached ho aur uske paas lone pairs hon, toh woh unhybridized reh sakte hain (e.g., PH$_3$, H$_2$S). Isse bond angles 90° ke aas-paas hote hain. Ye rule tab apply hota hai jab central atom ka electronegativity difference surrounding atom se bahut kam ho. Drago's rule ke according, agar central atom third period ya uske neeche ka ho, aur surrounding atoms ki electronegativity 2.5 se kam ho, toh hybridization nahi hota. Iska reason hai ki large size ke atoms mein hybridize hone ke liye energy difference bahut zyada hota hai. Odd Electron Species: Free radicals mein odd electron ko lone pair ki tarah treat karte hain hybridization calculation ke liye. Jaise, NO$_2$ mein N sp$^2$ hybridized hota hai. Yahan odd electron ek hybrid orbital mein rehta hai. Odd electron ki presence se molecule paramagnetic ho sakta hai. Resonance ka Effect: Agar lone pair resonance mein participate kar raha hai, toh use hybridization mein count nahi karte. Kyunki woh lone pair delocalized ho jaata hai aur hybridization ke liye available nahi hota. Jaise, Pyrrole mein N ka lone pair aromaticity mein participate karta hai, isliye N sp$^2$ hybridized hota hai, not sp$^3$. Similarly, Aniline mein N ka lone pair benzene ring ke saath resonance mein hota hai, isliye N sp$^2$ hybridized hota hai. Isse molecule ki stability badhti hai. Bond Lengths aur Bond Strengths: % s-character increase hone se bond length decrease hoti hai aur bond strength increase hoti hai. Kyunki s-orbital nucleus ke zyada close hota hai aur electrons ko zyada tightly hold karta hai. sp > sp$^2$ > sp$^3$ (bond strength order for C-H bonds, for example) sp Electronegativity: % s-character increase hone se atom ki electronegativity increase hoti hai. Isi wajah se sp hybridized carbon zyada electronegative hota hai sp$^2$ aur sp$^3$ se. Iska reason hai ki s-orbital nucleus ke zyada close hota hai, aur ismein electron density zyada hoti hai. Bent's Rule: Ye rule kehta hai ki more electronegative substituents prefer karte hain hybrid orbitals mein zyada p-character (aur kam s-character), jabki lone pairs aur less electronegative substituents prefer karte hain hybrid orbitals mein zyada s-character. Isse molecule ki stability increase hoti hai. Example: PCl$_3$F$_2$ mein, more electronegative F atoms axial position par hote hain (jahan p-character zyada hota hai), aur less electronegative Cl atoms equatorial position par hote hain (jahan s-character zyada hota hai). Example: SF$_4$ mein lone pair equatorial position par aata hai (jahan s-character zyada hota hai) aur F atoms axial aur equatorial positions par distribute hote hain, jisse seesaw shape banti hai. Isostructural Species: Jin molecules ya ions ka hybridization aur number of lone pairs same hota hai, unki geometry bhi same hoti hai. Ye concept bahut important hai comparison-based questions ke liye. Example: NH$_3$ aur H$_3$O$^+$ dono sp$^3$ hybridized hain aur dono mein 1 lone pair hai, isliye dono ki geometry pyramidal hai. CO$_2$ aur N$_2$O dono linear hain (dono sp hybridized). XeF$_2$ aur I$_3^-$ dono linear hain (dono sp$^3$d hybridized with 3 lone pairs). Hybridization in Ions: Ions ka hybridization calculate karte waqt, charge ko bhi consider karte hain. Anions: Negative charge ko central atom ke valence electrons mein add karte hain. Jaise, CO$_3^{2-}$ mein C ka hybridization. Total valence electrons = 4 (C) + 3*6 (O) + 2 (charge) = 24. Sigma bonds = 3, Lone pairs = 0 (on C). SN = 3, so sp$^2$. Cations: Positive charge ko central atom ke valence electrons mein se subtract karte hain. Jaise, NH$_4^+$ mein N ka hybridization. Total valence electrons = 5 (N) + 4*1 (H) - 1 (charge) = 8. Sigma bonds = 4, Lone pairs = 0 (on N). SN = 4, so sp$^3$. Non-equivalent Hybrid Orbitals: sp$^3$d aur sp$^3$d$^3$ hybridization mein sabhi hybrid orbitals equivalent nahi hote. sp$^3$d (Trigonal Bipyramidal): Axial bonds aur equatorial bonds ki bond lengths alag hoti hain. Equatorial bonds mein s-character zyada hota hai (sp$^2$ jaise), aur axial bonds mein p-character zyada hota hai (pd jaise). Isliye lone pairs aur more electronegative atoms prefer karte hain equatorial position (zyada s-character) aur axial position (zyada p-character) respectively. sp$^3$d$^3$ (Pentagonal Bipyramidal): Yahan bhi axial aur equatorial bonds non-equivalent hote hain. Bonding in Xe Compounds: Xenon (noble gas) bhi compounds banata hai, aur unmein hybridization ka concept apply hota hai. XeF$_2$: sp$^3$d, linear (3 lone pairs equatorial) XeF$_4$: sp$^3$d$^2$, square planar (2 lone pairs axial) XeO$_3$: sp$^3$, pyramidal (1 lone pair) XeOF$_4$: sp$^3$d$^2$, square pyramidal (1 lone pair axial) Bridge Bonding (e.g., Diborane, Al$_2$Cl$_6$): Kuch molecules mein bridge bonds hote hain jahan ek atom do doosre atoms ko connect karta hai. Inmein hybridization aur geometry thodi alag hoti hai. Diborane (B$_2$H$_6$): Ismein 2 bridging H atoms hote hain jo "banana bonds" banate hain. Har Boron sp$^3$ hybridized hota hai. Ye 3-center 2-electron (3c-2e) bonds hote hain. Al$_2$Cl$_6$: Ismein 2 bridging Cl atoms hote hain. Har Al atom sp$^3$ hybridized hota hai. Planarity: Molecule planar hai ya non-planar, ye bhi hybridization se decide hota hai. sp$^2$ hybridized atoms usually planar geometry dete hain (e.g., Benzene, BF$_3$). sp hybridized atoms linear geometry dete hain (e.g., C$_2$H$_2$, CO$_2$). sp$^3$ hybridized atoms tetrahedral hote hain, toh agar saare atoms sp$^3$ hain toh molecule non-planar hoga (e.g., CH$_4$). Lekin agar usmein double bond ya lone pair ho toh planarity aa sakti hai (e.g., NH$_3$ pyramidal hai, H$_2$O bent hai). In sab points ko dhyan mein rakhoge toh hybridization aur VSEPR theory se related koi bhi question solve kar paoge! Especially resonance, Drago's Rule, Bent's Rule aur isostructural species wale questions JEE mein frequently puche jaate hain. Bridge bonding aur planarity par bhi questions aate hain, toh unko bhi achhe se dekh lena. Calculation of Steric Number (S.N.) (Geometry ka Formula) Central atom ki geometry predict karne ke liye Steric Number use karte hain: Calculate total valence shell electrons (n) of all atoms (H ko 7 mano for calculation ease) + negative charges - positive charges. Divide n by 8: $n = 8Q + R$ (Q = bond pairs, R = unshared electrons). Number of lone pairs = $R/2$. Steric Number ($S.N.$) = Number of atoms bonded to central atom + Number of lone pairs. $$S.N. = Q + \frac{R}{2}$$ S.N. se hybridization aur geometry pata chalti hai: Steric Number Hybridization Electronic Geometry Bond Angles 2 sp Linear 180° 3 sp$^2$ Trigonal Planar 120° 4 sp$^3$ Tetrahedral 109°28' 5 sp$^3$d Trigonal Bipyramidal (TBP) 120° & 90° 6 sp$^3$d$^2$ Octahedral 90° 7 sp$^3$d$^3$ Pentagonal Bipyramidal (PBP) 72° & 90° Ab thoda aur detail mein jaate hain, jo JEE mein bohot kaam aayega: VSEPR Theory (Valence Shell Electron Pair Repulsion Theory): Ye theory kehti hai ki central atom ke around jo electron pairs (bond pairs aur lone pairs) hote hain, woh ek doosre ko repel karte hain. Is repulsion ko minimize karne ke liye, woh maximum possible distance par arrange ho jaate hain, jisse molecule ki geometry decide hoti hai. Lone Pair-Lone Pair (LP-LP) Repulsion > Lone Pair-Bond Pair (LP-BP) Repulsion > Bond Pair-Bond Pair (BP-BP) Repulsion: Ye repulsion ka order hai. Iski wajah se lone pairs ki presence molecular geometry ko distort kar deti hai, electronic geometry se alag. Kuch important cases jahan se questions aate hain: Steric Number = 4 (sp$^3$ Hybridization): Agar 4 bond pairs hon (e.g., CH$_4$), toh geometry Tetrahedral hoti hai, bond angle 109°28'. Agar 3 bond pairs aur 1 lone pair ho (e.g., NH$_3$), toh electronic geometry toh tetrahedral hi rehti hai, par molecular geometry Trigonal Pyramidal ho jaati hai. Lone pair ke repulsion ki wajah se bond angle 107° ho jaata hai. Agar 2 bond pairs aur 2 lone pairs hon (e.g., H$_2$O), toh electronic geometry tetrahedral, par molecular geometry Bent ya V-shaped ho jaati hai. Do lone pairs ke strong repulsion ki wajah se bond angle aur kam ho kar 104.5° ho jaata hai. Steric Number = 5 (sp$^3$d Hybridization - Trigonal Bipyramidal (TBP) Geometry): TBP mein do tarah ki positions hoti hain: Axial (90° par) aur Equatorial (120° par). Rule: Lone pairs hamesha equatorial position par aate hain, kyunki wahan repulsion kam hota hai (do 90° repulsions ke bajaye teen 120° repulsions). Agar 4 bond pairs aur 1 lone pair ho (e.g., SF$_4$), toh geometry See-Saw hoti hai. Agar 3 bond pairs aur 2 lone pairs hon (e.g., ClF$_3$), toh geometry T-shaped hoti hai. Agar 2 bond pairs aur 3 lone pairs hon (e.g., XeF$_2$), toh geometry Linear hoti hai. Steric Number = 6 (sp$^3$d$^2$ Hybridization - Octahedral Geometry): Octahedral geometry mein saari positions equivalent hoti hain. Agar 5 bond pairs aur 1 lone pair ho (e.g., BrF$_5$), toh geometry Square Pyramidal hoti hai. Agar 4 bond pairs aur 2 lone pairs hon (e.g., XeF$_4$), toh geometry Square Planar hoti hai. Yahan do lone pairs ek doosre ke opposite (180°) par hote hain, taaki repulsion minimize ho. Important points to remember: Double aur Triple bonds ko VSEPR mein single bond pair ki tarah treat karte hain, par unka repulsive effect thoda zyada hota hai. Electronegativity ka bhi bond angles par asar hota hai. Agar central atom se attached atoms ki electronegativity badhti hai, toh bond angle kam hota hai (kyunki bonding electrons central atom se door chale jaate hain, unka repulsion kam ho jaata hai). Resonance structures mein average steric number nikalna pad sakta hai. Bent's Rule: Ye rule kehta hai ki more electronegative substituents prefer karte hain hybrid orbitals jismein p-character zyada ho, aur less electronegative substituents (ya lone pairs) prefer karte hain hybrid orbitals jismein s-character zyada ho. Iska practical application TBP geometry mein dikhta hai, jahan more electronegative atoms axial position par jaate hain (kyunki axial orbitals mein p-character zyada hota hai), aur lone pairs ya bulky groups equatorial position par aate hain (kyunki equatorial orbitals mein s-character zyada hota hai). Back Bonding: Kuch cases mein back bonding ki wajah se expected geometry change ho sakti hai. Jaise, SiH$_3$NCS mein Nitrogen linear hota hai, jabki expected bent hona chahiye. Ye Silicon ke empty d-orbitals aur Nitrogen ke lone pair ke beech back bonding ki wajah se hota hai. Bridge Bonding: Boron hydrides (diborane, etc.) mein bridge bonding hoti hai, jisse unki geometry aur bonding normal VSEPR se alag hoti hai. Ye 'banana bonds' ya 3-center-2-electron bonds hote hain. Isoelectronic Species: Jin species ka total number of electrons same hota hai, unki geometry aksar similar hoti hai. Ye ek quick check ho sakta hai. Jaise, CO$_2$ aur N$_2$O (linear), NO$_2^+$ aur N$_3^-$ (linear). Bond Angle Comparison: Same central atom, different surrounding atoms: Jaise H$_2$O (104.5°) vs H$_2$S (92°). Electronegativity difference ki wajah se bond angle kam hota hai (kyunki more electronegative surrounding atom bonding electrons ko apni taraf kheench leta hai, jisse bond pair-bond pair repulsion kam ho jaata hai). Different central atom, same surrounding atoms: Jaise NH$_3$ (107°) vs PH$_3$ (93.5°). Central atom ka size badhne par bond angle kam hota hai, kyunki lone pair ka repulsion effect kam ho jaata hai (as electron density spreads over a larger volume). Drago's Rule: Ye rule un hydrides par apply hota hai jahan central atom third period ya uske baad ka ho, aur uski electronegativity 2.5 se kam ho. Aise cases mein hybridization nahi hoti aur bond angles approximately 90° hote hain. Example: PH$_3$, AsH$_3$, H$_2$S, H$_2$Se. Yahan lone pairs pure s-orbitals mein hote hain aur bonding p-orbitals se hoti hai. Steric Number 7 (sp$^3$d$^3$ Hybridization - Pentagonal Bipyramidal (PBP) Geometry): Ismein bhi axial aur equatorial positions hoti hain. Lone pairs equatorial position par hi aate hain. Example: IF$_7$ (no lone pairs, perfect PBP), XeF$_5^-$ (2 lone pairs, pentagonal planar). Odd Electron Species: Jaise NO$_2$, ClO$_2$. Inmein ek unpaired electron hota hai. Is unpaired electron ko bhi ek lone pair ki tarah treat karte hain, par iska repulsive effect thoda kam hota hai as compared to a full lone pair. Iski wajah se bond angles thode alag aate hain. Influence of $\pi$-bonds: VSEPR theory primarily $\sigma$-bonds aur lone pairs par based hai. $\pi$-bonds ka geometry par direct effect nahi hota, par woh bond length aur bond energy ko affect karte hain. Ye saare concepts aur unke examples acche se practice kar lena, JEE mein inhi se direct questions aate hain! 3.6 Valence Shell Electron Pair Repulsion (VSEPR) Theory (Repulsion Se Shape) VSEPR theory molecules ke shapes aur bond angles ko predict karti hai, Sidgwick-Powell theory ko improve karte hue. Electron Pair Repulsion: Electron pairs (bond pairs and lone pairs) repel each other, aur space mein maximum distance par rehna chahte hain. Repulsion ka order: Lone Pair - Lone Pair (LP-LP) > Lone Pair - Bond Pair (LP-BP) > Bond Pair - Bond Pair (BP-BP) . Double Bond Repulsion: Double bond single bond se zyada space occupy karta hai. Double Bond - Double Bond > Double Bond - Single Bond > Single Bond - Single Bond . Electronegativity Effect: Surrounding atom ki electronegativity increase hone se bond angle decrease hota hai. Central atom ki electronegativity increase hone se bond angle increase hota hai (agar lone pair ho). Back Bonding: Sometimes lone pair filled shell se unfilled shell mein transfer ho jaata hai. Effect of Lone Pair (LP) on Shape: Agar LP nahi hai, toh shape electronic geometry jaisa hi hoga (e.g., CH$_4$ - tetrahedral). Agar LP present hai, toh bond angles distort ho jaate hain aur shape irregular ho jaati hai (e.g., NH$_3$ - pyramidal, H$_2$O - bent). CH$_4$: 109°28' NH$_3$: 107° (LP-BP repulsion > BP-BP) H$_2$O: 104°5' (LP-LP repulsion > LP-BP) Shapes of Species with Different Hybridizations: sp (S.N. 2): 2 BP, 0 LP: Linear (BeH$_2$, BeCl$_2$) 1 BP, 1 LP: Linear (CO) sp$^2$ (S.N. 3): 3 BP, 0 LP: Trigonal Planar (BF$_3$, AlCl$_3$) 2 BP, 1 LP: Bent/V-shaped (SnCl$_2$, SO$_2$) sp$^3$ (S.N. 4): 4 BP, 0 LP: Tetrahedral (CH$_4$, NH$_4^+$) 3 BP, 1 LP: Pyramidal (NH$_3$, PCl$_3$) 2 BP, 2 LP: Angular/V-shaped/Bent (H$_2$O, Cl$_2$O) sp$^3$d (S.N. 5): 5 BP, 0 LP: Trigonal Bipyramidal (TBP) (PCl$_5$, PF$_5$) 4 BP, 1 LP: See-saw (SF$_4$, XeO$_2$F$_2$) 3 BP, 2 LP: T-shaped (ClF$_3$, BrF$_3$) 2 BP, 3 LP: Linear (XeF$_2$, ICl$_2^-$) Bent's Rule: More electronegative atom prefers orbitals with less s character (axial position in TBP), while lone pairs prefer orbitals with more s character (equatorial position). sp$^3$d$^2$ (S.N. 6): 6 BP, 0 LP: Octahedral (SF$_6$, XeOF$_4$) 5 BP, 1 LP: Square Pyramidal (IF$_5$, XeOF$_4$) 4 BP, 2 LP: Square Planar (XeF$_4$, ICl$_4^-$) sp$^3$d$^3$ (S.N. 7): 7 BP, 0 LP: Pentagonal Bipyramidal (IF$_7$) 6 BP, 1 LP: Distorted Octahedral (XeF$_6$) 5 BP, 2 LP: Pentagonal Planar (XeF$_5^-$) Effect of Double Bond: Double bonds single bonds se zyada space lete hain, isliye repulsion pattern change hota hai. For example, SO$_4^{2-}$ mein sab OSO angles identical hote hain, lekin H$_2$SO$_4$ mein nahi, kyunki resonating structures identical nahi hain. Effect of Electronegativity: NF$_3$ ka bond angle NH$_3$ se kam hota hai (102° vs 107°). F ki high electronegativity N se bonding electrons ko apni taraf pull karti hai, jisse BP-BP repulsion kam ho jaati hai. H$_2$O ka bond angle F$_2$O se zyada hota hai (104.5° vs 103.1°). Exceptions: CH$_4$ aur CF$_4$ ke bond angles same hote hain (109°28'), kyunki unmein lone pair nahi hai. PH$_3$ ka bond angle (93.8°) NH$_3$ (107°) se kam hai. P third period ka element hai, isliye hybridization kam hoti hai, aur bond formation mein p-orbitals zyada use hote hain, jisse bond angle 90° ke close hota hai. Steric Number (S.N.) Calculation: $S.N. = \frac{1}{2} (V + M - C + A)$ Jahan: $V$ = Central atom ke valence electrons $M$ = Monovalent atoms ki sankhya (H, F, Cl, Br, I) $C$ = Cationic charge (agar positive charge ho) $A$ = Anionic charge (agar negative charge ho) Isse S.N. nikal ke hybridization aur geometry pata chalti hai. Important Points for JEE Advanced/Mains: Isostructural Species: Jin molecules ya ions ka hybridization aur number of lone pairs same hota hai, unka shape bhi same hota hai. Jaise, NH$_3$ aur H$_3$O$^+$ dono pyramidal hain. Bond Angle Trends: Central atom ki electronegativity badhne par (agar lone pair ho), bond angle badhta hai (e.g., H$_2$O > H$_2$S). Side atom ki electronegativity badhne par, bond angle kam hota hai (e.g., NH$_3$ > NF$_3$). Size of central atom badhne par, bond angle kam hota hai (e.g., NH$_3$ > PH$_3$ > AsH$_3$). Ye effect hybridization ke kam hone ki wajah se hota hai, jahan p-character badhta hai. Multiple Bonds ka Effect: Double ya triple bonds single bonds se zyada space lete hain, isliye woh zyada repulsion create karte hain aur bond angles ko affect karte hain. Resonance ka Impact: Agar molecule mein resonance ho rahi hai, toh bond angles average ho jaate hain. Jaise CO$_3^{2-}$ mein sab C-O bond lengths aur O-C-O bond angles (120°) identical hote hain. Back Bonding aur Bond Angles: Agar back bonding hoti hai, toh central atom par electron density badh jaati hai, jisse repulsion badh sakta hai aur bond angles change ho sakte hain. Jaise BF$_3$ mein back bonding hoti hai, jisse B-F bond mein partial double bond character aata hai. Steric Hindrance: Bade groups ke beech mein repulsion ki wajah se bhi bond angles change ho sakte hain. Hybridization aur Bond Angle ka Relation: S-character jitna zyada hoga, bond angle utna hi zyada hoga. sp (50% s-character) $\rightarrow$ 180° sp$^2$ (33.3% s-character) $\rightarrow$ 120° sp$^3$ (25% s-character) $\rightarrow$ 109.5° Isliye, bond angles ko predict karte waqt hybridization ek important factor hai. Lone Pair ke Effect ka Detail: Lone pair electrons sirf ek nucleus se attract hote hain, isliye woh zyada space occupy karte hain aur bond pairs ko zyada repel karte hain. Is repulsion ki wajah se bond angles compress ho jaate hain. Example: Water (H$_2$O) mein do lone pairs hote hain, jo O-H bonds ko push karte hain, jisse bond angle 104.5° ho jaata hai (tetrahedral ke 109.5° se kam). Distorted Geometries: Jab lone pairs present hote hain, toh molecular geometry electronic geometry se alag ho jaati hai. Electronic geometry mein lone pairs aur bond pairs dono count hote hain, jabki molecular geometry mein sirf atoms ki arrangement dekhi jaati hai. Multiple Central Atoms: Agar molecule mein ek se zyada central atoms hain, toh har central atom ke around geometry alag se determine karni padti hai. Jaise CH$_3$COOH mein C=O carbon sp$^2$ hybridized hai aur CH$_3$ carbon sp$^3$ hybridized hai. Drago's Rule: Ye rule un molecules par apply hota hai jahan central atom third period ya uske baad ka ho (P, As, Sb, S, Se, Te) aur surrounding atoms ki electronegativity 2.5 se kam ho (e.g., H). Aise cases mein, central atom hybridization show nahi karta aur bond angles 90° ke close hote hain. Iska reason hai ki large size ke atoms mein s-p mixing energy difference zyada hone ki wajah se kam hoti hai, aur p-orbitals directly bond formation mein participate karte hain. Example: PH$_3$, AsH$_3$, H$_2$S, H$_2$Se. Hybridization aur Bond Length ka Relation: S-character badhne se bond length kam hoti hai. Kyunki s-orbital nucleus ke zyada close hota hai, toh s-character badhne se electron density nucleus ke paas aati hai, jisse bond chhota aur strong banta hai. Example: C-C bond length: sp-sp (shortest) Polarity aur Dipole Moment: VSEPR theory se hum molecule ki geometry predict kar sakte hain, jo dipole moment determine karne mein help karti hai. Symmetric molecules (jaise CH$_4$, CO$_2$) ka dipole moment zero hota hai, bhale hi individual bonds polar hon. Asymmetric molecules (jaise H$_2$O, NH$_3$) ka net dipole moment hota hai. Lone pairs bhi dipole moment mein contribute karte hain. Isoelectronic Species: Jin species mein same number of electrons hote hain, unki chemical properties similar ho sakti hain, lekin unki geometry VSEPR theory se hi determine hoti hai, jo ki different ho sakti hai agar central atom ya lone pairs alag hon. Example: CO$_2$ aur N$_2$O. Dono isoelectronic hain, lekin CO$_2$ linear hai aur N$_2$O bhi linear hai but resonance structures different hain. Coordinate Bonds ka Effect: Coordinate bonds ko bhi normal covalent bonds ki tarah treat kiya jaata hai VSEPR theory mein. Example: NH$_4^+$ ion mein N-H bonds sab identical hote hain aur tetrahedral geometry hoti hai. Valence Shell Electron Pair Repulsion (VSEPR) Theory ke Limitations: Ye theory transition metal complexes ki geometry explain nahi kar paati, kyunki unmein d-orbitals involved hote hain aur ligand field effects bhi hote hain. Kuch molecules jahan lone pairs hote hain, unki geometry predict karne mein bhi ye theory kabhi-kabhi fail ho jaati hai. Jaise, Li$_2$O linear hota hai, jabki VSEPR ke hisaab se bent hona chahiye. Ye theory bond angles ki exact value predict nahi karti, bas unki relative order aur general shape batati hai. Odd electron molecules (jaise NO$_2$) ki geometry ko accurately predict karna mushkil hota hai. Molecular Orbital Theory (MOT) ka Role: Jab VSEPR theory limitations dikhati hai, tab MOT zyada accurate predictions deti hai, khaaskar magnetic properties aur bond order ke liye. Lekin VSEPR simple molecules ke shapes ke liye abhi bhi bohot useful hai. 3.7 The Extent of d Orbital Participation in Molecular Bonding (d-Orbitals ka Khel) PCl$_5$ (sp$^3$d) ya SF$_6$ (sp$^3$d$^2$) jaise molecules mein d-orbitals participate karte hain. Lekin d-orbitals generally bade aur high energy ke hote hain, toh s aur p se mix hona mushkil hota hai. Factors Affecting d-orbital participation: Charge on Atom: Agar central atom par positive charge ho, toh electrons nucleus ki taraf pull hote hain, aur orbitals contract ho jaate hain. D-orbitals zyada contract hote hain, jisse unki energy s aur p ke close aa jaati hai. Electronegative Ligands: Highly electronegative elements (F, O, Cl) jab central atom se judte hain, toh wo electrons ko apni taraf kheench lete hain, jisse central atom par partial positive charge aata hai, aur d-orbitals contract hote hain. Size of Central Atom: Jaise jaise period mein neeche jaate hain, atoms ka size badhta hai. Bade atoms mein d-orbitals ki energy s aur p ke zyada close hoti hai, aur unka participation easier ho jaata hai. Isliye 3rd period elements (P, S, Cl) ke liye d-orbital participation common hai, jabki 2nd period elements (N, O, F) mein nahi. Example: PCl$_5$ exist karta hai, lekin PH$_5$ nahi. Kyunki Cl electronegative hai aur P ke d-orbitals ko contract karta hai, lekin H nahi. Isi tarah, SF$_6$ exist karta hai, lekin OF$_6$ nahi. Oxygen 2nd period ka element hai aur uske paas accessible d-orbitals nahi hote. d-orbital size: Table 3.5 shows that for S atom, average radial distance of 3d orbital (2.4 Å) is much larger than 3s (0.88 Å) and 3p (0.94 Å). But with +0.6 charge, 3d contracts to 1.4 Å. Important Concept: Inert Pair Effect: Heavier p-block elements (especially Group 13, 14, 15) mein, outermost s-electrons (ns$^2$) bond formation mein participate nahi karte. Ye electrons "inert pair" banate hain. Iska reason hai ki higher atomic number wale elements mein ns-electrons nucleus ke zyada close hote hain aur relativistic effects ki wajah se unki energy bahut kam ho jaati hai, jisse unhe excite karna mushkil hota hai. Iski wajah se, higher oxidation states ki stability kam ho jaati hai, aur lower oxidation states zyada stable hote hain. Example: Group 13: Thallium (Tl) mein +1 oxidation state +3 se zyada stable hai. Group 14: Lead (Pb) mein +2 oxidation state +4 se zyada stable hai. Group 15: Bismuth (Bi) mein +3 oxidation state +5 se zyada stable hai. JEE Advanced Connection: Inert pair effect se related stability order, reducing/oxidizing nature, aur compound formation par questions aate hain. Jaise, "Why is PbCl$_2$ more stable than PbCl$_4$?" ya "Why does Bi show +3 oxidation state predominantly?" Back Bonding (p$\pi$-d$\pi$ bonding): Kuch molecules mein, central atom ke paas vacant d-orbitals hote hain aur surrounding atoms ke paas lone pair electrons hote hain. Ye lone pair electrons central atom ke vacant d-orbitals mein donate ho sakte hain, jisse p$\pi$-d$\pi$ bond banta hai. Isse molecule ki stability badhti hai aur bond length kam hoti hai. Conditions for Back Bonding: Central atom ke paas vacant d-orbital ho. Surrounding atom ke paas lone pair electrons ho. Atoms ka size comparable ho (usually 2nd period element se 3rd period element). Examples: SiF$_4$ mein Si ke vacant 3d-orbitals aur F ke lone pairs ke beech p$\pi$-d$\pi$ back bonding hoti hai. Iski wajah se Si-F bond length kam hoti hai aur SiF$_4$ stable hota hai. Trimethylamine ($(\text{CH}_3)_3\text{N}$) mein back bonding nahi hoti, lekin Trisilylamine ($(\text{SiH}_3)_3\text{N}$) mein N ke lone pair aur Si ke vacant 3d-orbitals ke beech back bonding hoti hai. Iski wajah se Trisilylamine planar hota hai (sp$^2$ hybridisation for N) aur Trimethylamine pyramidal (sp$^3$ hybridisation for N). Phosphine (PH$_3$) aur Ammonia (NH$_3$) ki basicity compare karte waqt bhi d-orbital participation ka role aata hai. PH$_3$ mein P ke vacant d-orbitals lone pair ko accommodate kar sakte hain, jisse basicity kam hoti hai. JEE Advanced Connection: Back bonding se related geometry, bond length, bond angle, basicity, aur acidity ke questions frequently पूछे जाते hain. Hypervalency and d-orbital participation: Hypervalent compounds wo hote hain jahan central atom ka valence shell 8 electrons se zyada accommodate karta hai. Jaise PCl$_5$, SF$_6$, ICl$_4^-$. Pehle mana jata tha ki isme d-orbitals ka participation hota hai (octet rule expand karne ke liye). Lekin modern theories (jaise 3-center-4-electron bond model) suggest karti hain ki d-orbitals ka role utna significant nahi hota jitna pehle socha jata tha, especially for 3rd period elements. Phir bhi, JEE context mein, d-orbital participation ko hypervalency ka reason mana jaata hai, khaaskar jab hybridisation sp$^3$d ya sp$^3$d$^2$ dikhaya jata hai. JEE Advanced Connection: Questions aate hain ki kaunsa molecule hypervalent hai aur uski geometry kya hai. Hybridisation nikaalte waqt d-orbital involvement ko consider karna padta hai. Relativistic Effects: Heavy elements mein, inner shell electrons ki speed light ki speed ke comparable ho jaati hai, jisse unka mass badh jata hai (relativistic mass increase). Iski wajah se s-orbitals (jo nucleus ke sabse close hote hain) contract ho jaate hain aur unki energy kam ho jaati hai. Yehi reason hai inert pair effect ka. D-orbitals aur f-orbitals par iska ulta asar hota hai, wo expand ho sakte hain aur unki energy badh sakti hai. JEE Advanced Connection: Relativistic effects directly questions mein kam aate hain, lekin ye inert pair effect, gold ka colour (yellowish), aur mercury ka liquid state mein hona jaise phenomena ko explain karte hain. Understanding ke liye important hai. Bonding in Xenon Compounds: Xenon (Xe) ek noble gas hai, lekin ye Flouorine aur Oxygen jaise highly electronegative elements ke saath compounds banata hai (jaise XeF$_2$, XeF$_4$, XeF$_6$, XeO$_3$, XeOF$_4$). Isme bhi Xe ke vacant d-orbitals ka participation hota hai. Xe ke electrons excite hokar d-orbitals mein jaate hain aur bonding karte hain. JEE Advanced Connection: Xenon compounds ki geometry, hybridisation, aur stability par questions bahut common hain. Jaise XeF$_4$ ki square planar geometry (sp$^3$d$^2$ hybridisation with two lone pairs) ya XeO$_3$ ki pyramidal geometry (sp$^3$ hybridisation with one lone pair). Hybridisation Involving d-orbitals: Jab central atom ke paas vacant d-orbitals hote hain aur wo bond formation mein participate karte hain, toh hybridisation mein d-orbitals bhi shaamil hote hain. Isse molecule ki geometry aur bond angles determine hote hain. Common Hybridisations: sp$^3$d: Trigonal Bipyramidal (e.g., PCl$_5$) sp$^3$d$^2$: Octahedral (e.g., SF$_6$) sp$^3$d$^3$: Pentagonal Bipyramidal (e.g., IF$_7$) d-orbitals ke involvement se geometry mein changes aate hain, khaaskar lone pairs ki presence mein. Jaise XeF$_4$ mein sp$^3$d$^2$ hybridisation hai, lekin do lone pairs ki wajah se geometry square planar hoti hai. JEE Advanced Connection: Hybridisation nikalna, geometry predict karna, aur bond angles compare karna bahut common questions hain. VSEPR theory ke saath d-orbital involvement ko samajhna zaroori hai. Molecular Orbital Theory (MOT) and d-orbitals: MOT, jo bonding ko zyada accurately describe karti hai, d-orbitals ke role ko bhi explain karti hai. Transition metals mein d-orbitals ligand field theory aur crystal field theory mein bahut important hote hain, jahan wo metal-ligand bonding aur complex compounds ke properties ko determine karte hain. JEE Advanced Connection: Coordination compounds mein d-orbitals ki splitting (t$_{2g}$ aur e$_g$) aur uske effects (colour, magnetic properties) par questions aate hain. Although ye direct d-orbital participation in p-block se thoda alag hai, d-orbitals ki general understanding ke liye important hai. Acidity of Oxoacids: Central atom ke d-orbitals ki involvement oxoacids ki acidity ko bhi affect karti hai. Jaise H$_3$PO$_4$ mein P ke vacant d-orbitals oxygen ke lone pairs ke saath back bonding kar sakte hain, jisse P-O bond mein double bond character aata hai aur H-O bond weak ho jata hai, acidity badhti hai. Example: H$_3$PO$_4$ is a stronger acid than H$_3$PO$_3$ because in H$_3$PO$_4$, the phosphorus atom is in a higher oxidation state and can form more p$\pi$-d$\pi$ bonds with oxygen, stabilizing the conjugate base. JEE Advanced Connection: Oxoacids ki relative acidity, bond order, aur bond length par questions aate hain, jahan d-orbital participation ek key factor hota hai. 3.8 Types of Covalent Bonds (Sigma ($\sigma$) and Pi ($\pi$) Bonds) (Covalent Bonds ke Do Roop) Covalent bonds do main types ke hote hain: Sigma ($\sigma$) Bond (Head-on Overlap): Orbitals ka head-on (axial) overlap hota hai. Electron density dono atoms ke beech, internuclear axis par concentrated hoti hai. Types of $\sigma$ overlap: s-s overlap (H$_2$) s-p overlap (HX) p-p overlap (X$_2$) Bond Strength Order: s-s Har single bond ek $\sigma$ bond hota hai. $\sigma$ bonds free rotation allow karte hain around the internuclear axis. Pi ($\pi$) Bond (Sideways Overlap): Orbitals ka sideways (lateral) overlap hota hai. Electron density internuclear axis ke upar aur niche hoti hai (nodal plane axis par). $\pi$ bonds $\sigma$ bonds se weaker hote hain, aur bond length ko shorten karte hain. Types of $\pi$ overlap: p-p overlap (C=C, C≡C) p-d overlap (SO$_2$) d-d overlap (transition metal complexes) Relative Strength of $\pi$ bonds: $2p\pi-2p\pi > 2p\pi-3d\pi > 2p\pi-3p\pi > 3p\pi-3p\pi$. Double bond mein ek $\sigma$ aur ek $\pi$ bond hota hai. Triple bond mein ek $\sigma$ aur do $\pi$ bonds hote hain. $\pi$ bonds free rotation restrict karte hain, leading to geometric isomerism (e.g., cis-trans isomers). Key Point: Molecule ka shape $\sigma$ bonds aur lone pairs se determine hota hai, $\pi$ bonds se nahi. Important Concepts for JEE: Hybridization aur Bond Types: Hybridization mein sirf $\sigma$ bonds aur lone pairs count hote hain. $\pi$ bonds hybridization mein participate nahi karte. Example: C$_2$H$_4$ (ethene) mein har carbon sp$^2$ hybridized hai. Ismein ek C-C $\sigma$, do C-H $\sigma$ aur ek C-C $\pi$ bond hota hai. Example: C$_2$H$_2$ (ethyne) mein har carbon sp hybridized hai. Ismein ek C-C $\sigma$, ek C-H $\sigma$ aur do C-C $\pi$ bonds hote hain. Bond Strength aur Reactivity: $\sigma$ bonds generally $\pi$ bonds se strong hote hain, kyunki head-on overlap zyada effective hota hai. Lekin, multiple bonds (double/triple) single bonds se strong hote hain overall, due to the presence of both $\sigma$ and $\pi$ bonds. $\pi$ bonds zyada exposed hote hain aur easily attack ho sakte hain electrophiles dwara, isliye compounds with $\pi$ bonds (alkenes, alkynes) zyada reactive hote hain addition reactions ke liye. Bond Length: Single bond > Double bond > Triple bond. $\pi$ bonds ki presence bond length ko shorten karti hai. Resonance aur $\pi$ bonds: Resonance mein $\pi$ electrons delocalize hote hain. Yeh delocalization molecule ko stability provide karta hai. Example: Benzene mein 6 carbon atoms ke beech delocalized $\pi$ electrons hote hain. Iski wajah se benzene bahut stable hota hai. Nodal Planes: $\sigma$ bonds mein internuclear axis par koi nodal plane nahi hota. $\pi$ bonds mein internuclear axis par ek nodal plane hota hai. Bond Order Calculation: Bond order = (No. of bonds between two atoms) / (No. of resonating structures) (for resonating structures). Example: Benzene mein C-C bond order 1.5 hota hai. Higher bond order matlab shorter bond length aur greater bond strength. Molecular Orbital Theory (MOT) aur $\pi$ bonds: MOT mein $\pi$ bonds ko bonding aur anti-bonding molecular orbitals ke terms mein describe kiya jata hai. HOMO-LUMO concept organic reactions ki reactivity explain karta hai, jismein $\pi$ bonds ka role crucial hota hai. Back Bonding ($p\pi-p\pi$ or $p\pi-d\pi$): Jab ek atom ke paas filled p-orbital ho aur doosre atom ke paas adjacent vacant p-orbital ya d-orbital ho, toh unke beech sideways overlap ho sakta hai. Yeh extra stability provide karta hai aur bond length ko shorten karta hai. Example: BF$_3$ mein boron ke paas vacant p-orbital hota hai aur fluorine ke paas filled p-orbital. Isse $p\pi-p\pi$ back bonding hoti hai, jisse B-F bond mein partial double bond character aata hai aur BF$_3$ ek weak Lewis acid banta hai. Example: SiCl$_4$ mein silicon ke paas vacant d-orbitals hote hain aur chlorine ke paas filled p-orbitals. Isse $p\pi-d\pi$ back bonding hoti hai. Hyperconjugation (No Bond Resonance): Yeh $\sigma$ electrons ka delocalization hota hai adjacent $\pi$ system ya vacant p-orbital mein. Alkyl groups ki stability aur carbocations ki stability explain karta hai. Example: Propene mein methyl group ke C-H $\sigma$ bonds adjacent C=C $\pi$ system ke saath hyperconjugation karte hain. Bond Angle aur Repulsion: VSEPR theory ke according, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. $\pi$ bonds bond angles ko affect karte hain, kyunki multiple bonds single bonds se zyada space occupy karte hain. Example: H$_2$O mein bond angle 104.5° hai, jabki CH$_4$ mein 109.5°. Yeh lone pair repulsion ki wajah se hai. Dipole Moment: $\pi$ bonds ki presence molecule ke dipole moment ko affect kar sakti hai. Cis-trans isomers mein dipole moment alag-alag hota hai due to different spatial arrangements of $\pi$ bonds. Example: cis -1,2-dichloroethene ka dipole moment hota hai, jabki trans -1,2-dichloroethene ka zero hota hai (symmetrical cancellation). Aromaticity: Aromatic compounds cyclic, planar hote hain aur unmein complete delocalization of $\pi$ electrons hota hai. Hückel's Rule follow karte hain: $(4n+2)$ $\pi$ electrons (jahan n = 0, 1, 2...). Aromaticity ki wajah se compounds exceptionally stable hote hain aur electrophilic substitution reactions dete hain. Example: Benzene (6 $\pi$ electrons, n=1). Anti-aromaticity: Cyclic, planar compounds jinmein complete delocalization of $\pi$ electrons ho, lekin $(4n)$ $\pi$ electrons ho. Yeh compounds highly unstable hote hain. Example: Cyclobutadiene (4 $\pi$ electrons, n=1). Non-aromaticity: Compounds jo aromatic ya anti-aromatic nahi hote. Ya toh cyclic nahi hote, ya planar nahi hote, ya complete delocalization nahi hota. Example: Cyclohexene. Conjugation: Conjugation tab hota hai jab multiple bonds (double/triple) single bonds se alternate karte hain, ya jab multiple bond ke adjacent vacant p-orbital ya lone pair ho. Yeh $\pi$ electrons ke delocalization ko allow karta hai, jisse molecule ki stability badhti hai aur uski reactivity change hoti hai. Example: Buta-1,3-diene ($CH_2=CH-CH=CH_2$) mein do double bonds single bond se conjugated hain. Conjugation ki wajah se UV-Vis absorption spectra mein shifts aate hain. Inductive Effect: Yeh $\sigma$ bonds ke through electron density ka transmission hota hai, jo electronegativity differences ki wajah se hota hai. Electron-donating groups (EDG) jaise alkyl groups (+I effect) electron density push karte hain, jabki electron-withdrawing groups (EWG) jaise halogens, nitro groups (-I effect) electron density pull karte hain. Inductive effect bond polarity aur molecule ki reactivity ko affect karta hai. Electrophilic/Nucleophilic Attack: $\pi$ bonds electron-rich hote hain, isliye woh electrophiles (electron-deficient species) ke liye attractive sites hote hain. Organic reactions mein, jaise addition reactions of alkenes/alkynes, $\pi$ bond par electrophilic attack hota hai. Nucleophiles (electron-rich species) electrophilic centers par attack karte hain, jahan $\pi$ bonds ki polarity important role play karti hai. Example CO$_2$: O=C=O. Carbon sp hybridized hai, do $\sigma$ bonds aur do $\pi$ bonds banata hai. Linear shape (180°). Yahan, har C=O bond mein ek $\sigma$ aur ek $\pi$ bond hai. Example SO$_2$: O=S=O. Sulfur sp$^2$ hybridized hai, do $\sigma$ bonds aur ek lone pair. Iski wajah se bent/V-shaped (119°30') hota hai. Ismein do S-O $\sigma$ bonds aur ek delocalized $\pi$ bond hota hai, jisse har S-O bond ka bond order 1.5 hota hai (resonance ki wajah se). Ismein $p\pi-d\pi$ bonding bhi hoti hai. JEE Advanced Tip: Kabhi-kabhi d-orbitals bhi $\pi$ bonding mein participate karte hain, jaise SO$_2$, SO$_3$, H$_3$PO$_4$ mein. Isko $p\pi-d\pi$ bonding kehte hain. Yeh concept back-bonding aur hyperconjugation se bhi related hai. JEE Advanced Tip 2: Aromaticity ka concept bhi $\pi$ electron delocalization par based hai. Huckel's Rule ($4n+2$ $\pi$ electrons) aromatic compounds ki stability explain karta hai. Tricks for Fast Solving: $\sigma$ aur $\pi$ bonds count karna: Single bond = 1 $\sigma$ Double bond = 1 $\sigma$ + 1 $\pi$ Triple bond = 1 $\sigma$ + 2 $\pi$ Pehle saare single bonds ko $\sigma$ count kar lo, phir har extra bond ko $\pi$ count kar lo. Example: CH$_3$CH=CH$_2$ (Propene) C-C single bonds: 2 $\sigma$ C=C double bond: 1 $\sigma$ + 1 $\pi$ C-H bonds: 3 (CH$_3$) + 1 (CH) + 2 (CH$_2$) = 6 $\sigma$ Total $\sigma$ = 2 + 1 + 6 = 9 $\sigma$ bonds. Total $\pi$ = 1 $\pi$ bond. Hybridization quickly identify karna: Hybridization = (No. of $\sigma$ bonds around atom) + (No. of lone pairs around atom). 2 = sp, 3 = sp$^2$, 4 = sp$^3$. Example: H$_2$O mein Oxygen ke paas 2 $\sigma$ bonds (O-H) aur 2 lone pairs hain. So, 2+2=4, matlab sp$^3$ hybridized. Example: CO$_2$ mein Carbon ke paas 2 $\sigma$ bonds (C=O mein ek-ek $\sigma$) aur 0 lone pairs hain. So, 2+0=2, matlab sp hybridized. Bond Order (Resonance wale compounds ke liye): Bond Order = $1 + \frac{\text{Number of } \pi \text{ bonds}}{\text{Number of resonating positions}}$ Example: Carbonate ion (CO$_3^{2-}$). Ek C=O double bond aur do C-O single bonds, jo resonance mein hain. Total $\pi$ bonds = 1 (initially in one C=O). Resonating positions = 3 (teen C-O bonds). Bond Order = $1 + \frac{1}{3} = 1.33$. Aromaticity Check (Hückel's Rule): Cyclic? Planar? Conjugated (continuous $\pi$ system)? $(4n+2)$ $\pi$ electrons? Agar saare 'yes' hain, toh aromatic. Agar $(4n)$ $\pi$ electrons hain, toh anti-aromatic (highly unstable). Baki sab non-aromatic. Trick: $\pi$ electrons count karte waqt, negative charge ko 2 $\pi$ electrons count karo agar wo conjugation mein hai. Positive charge ko 0 $\pi$ electrons. Lone pair ko 2 $\pi$ electrons agar wo conjugation mein hai. Geometric Isomerism (Cis-Trans) Check: Double bond ke around free rotation restricted hoti hai. Agar double bond ke har carbon atom par do different groups lage hon, toh geometric isomerism possible hai. Example: But-2-ene ($CH_3-CH=CH-CH_3$). Har double bonded carbon par ek -CH$_3$ aur ek -H hai, jo different hain. Isliye cis aur trans isomers banenge. Example: Propene ($CH_3-CH=CH_2$). Ek double bonded carbon par do -H hain (same groups). Isliye geometric isomerism nahi hogi. Bond Length Comparison: Bond order jitna zyada, bond length utni kam. Resonance ki wajah se partial double bond character aata hai, jisse bond length single bond se kam aur double bond se zyada hoti hai. Example: C-C (1.54 Å) > C=C (1.34 Å) > C≡C (1.20 Å). Benzene mein C-C bond length 1.39 Å hoti hai, jo single aur double bond ke beech ki hai, resonance ki wajah se. Bridge Bonding (Bridge Banake Bonding) Kuch compounds mein atoms bridge banate hain, jaise B$_2$H$_6$ (diborane), BeCl$_2$ (polymeric). In B$_2$H$_6$: 2 BH$_3$ molecules dimerize karke B$_2$H$_6$ banate hain. Ismein 3-center-2-electron (3c-2e) Banana Bonds hote hain, jahan hydrogen atoms do boron atoms ke beech bridge banate hain. Hybridization: Both B atoms are sp$^3$ hybridized. Non-planar structure. Bond Lengths: $d_{B-H (terminal)} Har Boron atom 2 terminal H atoms se aur 2 bridging H atoms se attached hota hai. Total 4 terminal B-H bonds aur 2 bridging B-H-B bonds hote hain. Terminal B-H bonds regular 2c-2e bonds hote hain. Bridging B-H-B bonds 3c-2e bonds hote hain (banana bonds). Iska structure electron deficient hota hai (total 12 valence electrons, but 14 electrons ki requirement hoti hai if all were 2c-2e bonds). Trick: Electron deficient compounds (like diborane) generally bridge bonding show karte hain to achieve stability. Important Point (JEE): Diborane mein koi B-B bond nahi hota. Sirf B-H terminal aur B-H-B bridge bonds hote hain. Reactivity: Diborane Lewis acids ki tarah behave karta hai kyunki iske paas vacant orbitals hote hain aur ye electron deficient hota hai. In BeH$_2$ and BeCl$_2$: BeH$_2$ polymeric solid mein exist karta hai, jahan Be atoms 3c-2e bonds banate hain. BeCl$_2$ dimer (Be$_2$Cl$_4$) aur polymeric solid mein exist karta hai. Ismein 3-center-4-electron (3c-4e) bonds hote hain, jahan Cl atoms bridge banate hain. BeCl$_2$ mein, Be atom sp$^3$ hybridized hota hai. Har Be atom do terminal Cl atoms aur do bridging Cl atoms se attached hota hai. Bridging Cl atoms apne lone pair electrons ko Be atoms ke vacant orbitals mein donate karte hain, isliye ye 3c-4e bonds hote hain. Important: BeCl$_2$ gas phase mein linear monomer (sp hybridized) hota hai, but solid phase mein polymeric (sp$^3$ hybridized) hota hai. Trick: Be compounds (Group 2) mein bridge bonding tab hoti hai jab wo electron deficient ho aur bridging atom ke paas lone pairs ho (e.g., Cl) ya wo smaller atom ho (e.g., H). In I$_2$Cl$_6$: Planar molecule. I atoms sp$^3$d hybridized hote hain. Cl atoms bridge banate hain. General Concepts and Tricks for Bridge Bonding: Electron Deficiency: Agar compound electron deficient hai (jaise Group 13 elements ke hydrides ya alkyls), toh bridge bonding ki high possibility hoti hai, especially 3c-2e bonds. Isse stability badhti hai. Vacant Orbitals: Central atom ke paas vacant orbitals hone chahiye bridge bond banane ke liye (electron acceptor). Ye vacant orbitals hi bridge bond formation ko facilitate karte hain. Lone Pairs: Bridging atom ke paas lone pairs hone chahiye agar wo 3c-4e bond bana raha hai (electron donor). Ye lone pairs hi bridge bond mein electrons provide karte hain. Size of Bridging Atom: Smaller atoms (like H) generally 3c-2e bonds banate hain. Larger atoms (like Cl, Br, I) 3c-4e bonds banate hain jab unke paas lone pairs hote hain. Iska reason steric hindrance aur orbital overlap ki efficiency hai. Hybridization: Bridge bonding se hybridization change ho sakti hai. Jaise BeCl$_2$ monomer mein sp, polymer mein sp$^3$. Ye change molecule ki geometry aur stability ko affect karta hai. Planarity: Bridge bonding se molecule planar ya non-planar ho sakta hai. Generally, 3c-2e bonds non-planar structures dete hain (e.g., B$_2$H$_6$), jabki 3c-4e bonds planar ya non-planar ho sakte hain depending on the atoms involved (e.g., I$_2$Cl$_6$ is planar, Al$_2$Cl$_6$ is non-planar). JEE Advanced Tip: Bridge bonding wale compounds ki stability, reactivity, aur physical properties (boiling point, melting point) par questions aate hain. Dimerization ya polymerization se boiling point increase hota hai kyunki intermolecular forces badh jaati hain. Common Bridge Bonding Examples: 3c-2e (Banana Bonds): B$_2$H$_6$, Al$_2$(CH$_3$)$_6$, (AlH$_3$)$_n$, BeH$_2$. Ye sab electron deficient hote hain. 3c-4e (Halogen Bridges): Al$_2$Cl$_6$, Be$_2$Cl$_4$, I$_2$Cl$_6$. Yahan bridging atoms (halogens) lone pairs donate karte hain. Trick for counting 3c-2e and 2c-2e bonds in Diborane (B$_2$H$_6$): Total valence electrons = $2 \times 3 (B) + 6 \times 1 (H) = 12$ electrons. Terminal B-H bonds (2c-2e) = 4 bonds. These use $4 \times 2 = 8$ electrons. Remaining electrons = $12 - 8 = 4$ electrons. These 4 electrons form 2 bridge B-H-B bonds. Since each bridge bond is 3c-2e, $2 \times 2 = 4$ electrons are used. So, 4 terminal 2c-2e bonds and 2 bridging 3c-2e bonds. Important Distinction (JEE): Diborane (B$_2$H$_6$) mein H atoms bridge banate hain. Aluminium chloride (Al$_2$Cl$_6$) mein Cl atoms bridge banate hain. Diborane electron deficient hai, Al$_2$Cl$_6$ nahi. Al$_2$Cl$_6$ mein Al ka octet complete hota hai bridge bonding se. Stability Order: Bridge bonding ki stability bridging atom ke size aur electronegativity par depend karti hai. Generally, smaller aur less electronegative atoms (like H) 3c-2e bonds mein zyada stable hote hain. Halogen bridges mein, stability Cl > Br > I hoti hai due to effective orbital overlap. Question Type (JEE): Identify the type of bridge bonding (3c-2e or 3c-4e). Predict hybridization of central atoms. Compare bond lengths (e.g., terminal vs. bridge). Determine planarity/non-planarity. Relate bridge bonding to physical properties (e.g., why AlCl$_3$ exists as dimer in vapor phase). Draw structures of bridge bonded compounds. Molecules Nature of bonds (3c-2e / 3c-4e) Planar (P) and Non-planar (NP) Al$_2$Cl$_6$ 3c-4e NP Al$_2$Br$_6$ 3c-4e NP (AlH$_3$)$_n$ 3c-2e NP Al$_2$I$_6$ 3c-4e NP Al$_2$(CH$_3$)$_6$ 3c-2e NP I$_2$Cl$_6$ 3c-4e P Be$_2$Cl$_4$ 3c-4e NP B$_2$H$_6$ 3c-2e NP 3.9 Molecular Orbital Method (MO Method) (Electrons ka Naya Ghar) VBT ke opposite, MO theory mein electrons poore molecule ke orbitals mein rehte hain, na ki individual atoms ke orbitals mein. Atomic orbitals combine karke Molecular Orbitals (MOs) banate hain. 3.10 LCAO Method (MOs Banane ka Tarika) Linear Combination of Atomic Orbitals (LCAO) method se MOs banate hain. Jab do atomic orbitals (AO) ($\Psi_A$ aur $\Psi_B$) combine karte hain, toh do MOs bante hain: Bonding Molecular Orbital (BMO): $\Psi_{AB} = N(c_1\Psi_A + c_2\Psi_B)$. Electron density atoms ke beech increase hoti hai, energy kam hoti hai, stability badhti hai. Antibonding Molecular Orbital (ABMO): $\Psi_{AB}^* = N(c_1\Psi_A - c_2\Psi_B)$. Electron density atoms ke beech decrease hoti hai (nodal plane banti hai), energy zyada hoti hai, stability kam hoti hai. Important Points for LCAO: Combine hone wale AOs ki energy comparable honi chahiye. Combine hone wale AOs ki symmetry same honi chahiye (internuclear axis ke respect mein). Combine hone wale AOs ka maximum overlap hona chahiye. Types of Overlap (JEE perspective se bohot important): s-s Combinations: s-orbitals combine karke $\sigma$ (bonding) aur $\sigma^*$ (antibonding) MOs banate hain. $\sigma_g$ (gerade) and $\sigma_u^*$ (ungerade). Example: H$_2$ molecule. s-p Combinations: s-orbital aur p-orbital (jo internuclear axis par ho, usually pz) combine karke $\sigma$ aur $\sigma^*$ MOs banate hain. Example: HF molecule (H ka 1s aur F ka 2pz). p-p Combinations: p-orbitals (jo internuclear axis par ho, usually pz) head-on overlap karke $\sigma$ aur $\sigma^*$ MOs banate hain. p-orbitals (jo internuclear axis ke perpendicular ho, px aur py) sideways overlap karke $\pi$ aur $\pi^*$ MOs banate hain. $\pi$ bonds mein nodal plane hoti hai jo internuclear axis se pass karti hai. Example: N$_2$, O$_2$ molecules. p-d Combinations: p-orbital aur d-orbital combine karke $\pi$ aur $\pi^*$ MOs banate hain. Ye short bonds aur transition metal complexes mein important hai. d-d Combinations: d-orbitals combine karke $\delta$ aur $\delta^*$ MOs banate hain. Ye complex compounds mein dekhe jaate hain, jahan four lobes overlap karte hain. Energy Level Diagram (MOED) - Tricks aur Concepts: MOED se hum Bond Order, Magnetic Properties, aur Stability predict karte hain. Iske liye do main cases hain: Case 1: Total electrons $\le$ 14 (e.g., B$_2$, C$_2$, N$_2$) $\sigma_{1s}$ $\sigma_{1s}^*$ $\sigma_{2s}$ $\sigma_{2s}^*$ $\pi_{2px} = \pi_{2py}$ $\sigma_{2pz}$ $\pi_{2px}^* = \pi_{2py}^*$ $\sigma_{2pz}^*$ Yahan $\pi_{2p}$ orbitals ki energy $\sigma_{2p}$ se kam hoti hai due to s-p mixing. Case 2: Total electrons > 14 (e.g., O$_2$, F$_2$, Ne$_2$) $\sigma_{1s}$ $\sigma_{1s}^*$ $\sigma_{2s}$ $\sigma_{2s}^*$ $\sigma_{2pz}$ $\pi_{2px} = \pi_{2py}$ $\pi_{2px}^* = \pi_{2py}^*$ $\sigma_{2pz}^*$ Yahan $\sigma_{2p}$ orbital ki energy $\pi_{2p}$ se kam hoti hai, s-p mixing kam ho jaati hai. Bond Order (BO) Calculation: $BO = \frac{1}{2} (N_b - N_a)$ $N_b$: Number of electrons in bonding MOs. $N_a$: Number of electrons in antibonding MOs. Trick: Higher BO = Higher Stability, Shorter Bond Length, Higher Bond Energy. BO = 0 matlab molecule exist nahi karta (e.g., He$_2$). Magnetic Properties: Paramagnetic: Agar MOs mein unpaired electrons hain. (Attracted by magnetic field) Diamagnetic: Agar sabhi electrons paired hain. (Repelled by magnetic field) Trick: O$_2$ paramagnetic hai (do unpaired electrons $\pi_{2p}^*$ mein), jo VBT explain nahi kar paata tha. Ye MO theory ka bada success hai. Stability Comparison (JEE Advanced favourite): Same molecule ke ions ki stability compare karne ke liye, sabse pehle Bond Order nikalo. Agar BO same hai, toh antibonding electrons dekho. Jis mein $N_a$ kam, woh zyada stable. Example: O$_2$, O$_2^+$, O$_2^-$, O$_2^{2-}$ O$_2$ (16e): BO = 2, 2 unpaired e- (paramagnetic) O$_2^+$ (15e): BO = 2.5, 1 unpaired e- (paramagnetic) O$_2^-$ (17e): BO = 1.5, 1 unpaired e- (paramagnetic) O$_2^{2-}$ (18e): BO = 1, 0 unpaired e- (diamagnetic) Stability Order: O$_2^+$ > O$_2$ > O$_2^-$ > O$_2^{2-}$ Heteronuclear Diatomic Molecules (e.g., CO, NO): Energy level diagrams thode different hote hain kyunki more electronegative atom ke AOs ki energy kam hoti hai. Bond order aur magnetic properties ka concept same rehta hai. Example: CO (14e) - BO = 3, diamagnetic. Example: NO (15e) - BO = 2.5, paramagnetic. Shortcuts for Bond Order (Total electrons ke basis par): Ye trick sirf diatomic species (homo/hetero) ke liye hai: Total Electrons Bond Order 14 3.0 13 2.5 12 2.0 11 1.5 10 1.0 9 0.5 8 0.0 14 electrons se upar ya neeche jaane par, har ek electron change ke liye BO 0.5 se change hota hai. Example: N$_2$ (14e) -> BO = 3. N$_2^+$ (13e) -> BO = 2.5. O$_2$ (16e) -> 14e se 2 electron zyada, toh BO = 3 - (2 * 0.5) = 2. C$_2$ (12e) -> 14e se 2 electron kam, toh BO = 3 - (2 * 0.5) = 2. Ye trick bohot time bachati hai exam mein! 3.11 Rules for Linear Combination of Atomic Orbitals (MOs Banane ke Niyam) MOs banane ke teen rules hote hain: Comparable Energy: Combining AOs ki energy comparable honi chahiye. Maximum Overlap: AOs ka overlap maximum hona chahiye. Same Symmetry: AOs ki symmetry same honi chahiye (internuclear axis ke respect mein). Quantum Numbers for MOs: n (principal) and l (subsidiary) atomic orbitals jaise hi hote hain. Magnetic quantum number (m) ki jagah $\lambda$ use karte hain, jo angular momentum dikhata hai. $\lambda=0$ for $\sigma$ orbitals, $\lambda=\pm 1$ for $\pi$ orbitals, $\lambda=\pm 2$ for $\delta$ orbitals. Spin quantum number ($s = \pm 1/2$) same rehta hai. Pauli Exclusion Principle: Koi bhi do electrons ke charon quantum numbers same nahi ho sakte MO mein bhi. Energy Order of MOs: For O$_2$, F$_2$, Ne$_2$ (heavier elements): $\sigma1s For B$_2$, C$_2$, N$_2$ (lighter elements, due to s-p mixing): $\sigma1s Bond Order (BO): Bond order se humein pata chalta hai ki do atoms ke beech kitne bonds hain. Iska formula hai: $$BO = \frac{1}{2} (N_b - N_a)$$ $N_b$ = Bonding electrons ki sankhya (electrons in bonding MOs) $N_a$ = Antibonding electrons ki sankhya (electrons in antibonding MOs) Important Points: Agar BO = 0, toh molecule exist nahi karega (jaise He$_2$). Higher BO ka matlab hai stronger bond aur shorter bond length. BO fractional bhi ho sakta hai (jaise O$_2^+$ ka 2.5). Magnetic Properties: Paramagnetic: Agar MOs mein unpaired electrons hain, toh molecule paramagnetic hoga (magnetic field se attract hoga). Example: O$_2$. Diamagnetic: Agar sabhi electrons paired hain, toh molecule diamagnetic hoga (magnetic field se repel hoga). Example: N$_2$. Trick: Agar total electrons odd hain, toh molecule hamesha paramagnetic hoga. Agar total electrons even hain, toh diamagnetic ho sakta hai, but O$_2$ (16 electrons) paramagnetic hai. Isliye, even electrons ke liye MO diagram banana padega ya yaad rakhna padega. Stability of Molecules/Ions: Higher bond order ka matlab hai greater stability. Agar do species ka BO same hai, toh jismein antibonding electrons kam honge, woh zyada stable hoga. Bond Length: Bond order badhne se bond length kam hoti hai. Stronger bond = shorter bond length. Tricks for JEE: Total Electrons (8 se 20 tak) ke liye BO ka Quick Trick: Total Electrons Bond Order 14 3.0 13, 15 2.5 12, 16 2.0 11, 17 1.5 10, 18 1.0 9, 19 0.5 8, 20 0.0 Note: Ye trick homonuclear diatomic molecules aur unke ions ke liye zyada useful hai. 14 electrons (N$_2$) ka BO 3 hota hai, isko reference point maan kar, har ek electron add/subtract karne par BO 0.5 se kam hota hai. Magnetic Nature Quick Check: Total electrons odd hain? -> Paramagnetic. Total electrons even hain? -> Diamagnetic (except 10 and 16 electrons, jo paramagnetic hote hain - jaise B$_2$ aur O$_2$). 3.12 Examples of Molecular Orbital Treatment for Homonuclear Diatomic Molecules (Apne Jaise Atoms ke MOs) Electrons ko MOs mein Aufbau principle (lowest energy first) aur Hund's rule (degenerate orbitals mein single occupy first) ko follow karte hue fill karte hain. H$_2^+$ (1 electron): $\sigma1s^1$. Bond order = 0.5. Exists. H$_2$ (2 electrons): $\sigma1s^2$. Bond order = 1. Exists. He$_2^+$ (3 electrons): $\sigma1s^2 \sigma^*1s^1$. Bond order = 0.5. Exists, but less stable. He$_2$ (4 electrons): $\sigma1s^2 \sigma^*1s^2$. Bond order = 0. Does not exist. Li$_2$ (6 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2$. Bond order = 1. Exists in vapor state. Be$_2$ (8 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2$. Bond order = 0. Does not exist. B$_2$ (10 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \pi2p_y^1 \pi2p_z^1$. Bond order = 1. Paramagnetic (2 unpaired electrons). Exists. C$_2$ (12 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \pi2p_y^2 \pi2p_z^2$. Bond order = 2. Diamagnetic. N$_2$ (14 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \pi2p_y^2 \pi2p_z^2 \sigma2p_x^2$. Bond order = 3. Diamagnetic. O$_2$ (16 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_x^2 \pi2p_y^2 \pi2p_z^2 \pi^*2p_y^1 \pi^*2p_z^1$. Bond order = 2. Paramagnetic (2 unpaired electrons). MO theory ki sabse badi safalta! F$_2$ (18 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_x^2 \pi2p_y^2 \pi2p_z^2 \pi^*2p_y^2 \pi^*2p_z^2$. Bond order = 1. Diamagnetic. Ne$_2$ (20 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_x^2 \pi2p_y^2 \pi2p_z^2 \pi^*2p_y^2 \pi^*2p_z^2 \sigma^*2p_x^2$. Bond order = 0. Does not exist. Bond Order: $BO = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons})$. MO Theory ke Important Points (JEE Advanced/Mains ke liye) Dost, MO theory se har saal questions aate hain. Isko achhe se samajh le, khaas kar ke stability, magnetic properties aur bond order pe. 1. Energy Level Diagram (ELD) Ye sabse important hai! Do alag-alag ELD hote hain: For N$_2$ and molecules with total electrons $\le$ 14: $\sigma1s Yahan $\pi2p$ orbitals $\sigma2p_x$ se pehle aate hain due to s-p mixing. For O$_2$, F$_2$ and molecules with total electrons $>$ 14: $\sigma1s Yahan $\sigma2p_x$ orbital $\pi2p$ orbitals se pehle aata hai . 2. Stability of Molecules/Ions Bond Order (BO) jitna zyada, molecule utna hi stable. Agar BO = 0, toh molecule exist nahi karta. Example: He$_2$, Be$_2$, Ne$_2$. Bonding electrons ($N_b$) aur Antibonding electrons ($N_a$) ka role: Agar $N_b > N_a$, toh stable. Agar $N_b Agar $N_b = N_a$, toh unstable (BO = 0). Ek aur cheez: Agar do species ka BO same hai, toh jisme antibonding electrons ($N_a$) kam honge, wo zyada stable hoga . Example: N$_2^+$ (BO=2.5, $N_a$=2) aur O$_2^+$ (BO=2.5, $N_a$=3). N$_2^+$ zyada stable hai O$_2^+$ se. 3. Magnetic Properties Paramagnetic: Agar MOs mein unpaired electrons hon. Ye magnetic field se attract hote hain. Example: O$_2$, B$_2$, NO, O$_2^+$, O$_2^-$. Diamagnetic: Agar MOs mein saare electrons paired hon. Ye magnetic field se repel hote hain. Example: N$_2$, C$_2$, F$_2$, CO, N$_2^{2-}$. Trick for Magnetic Nature (Total Electrons): Agar total electrons odd hain, toh Paramagnetic hoga (except for some rare cases like B$_2$). Agar total electrons even hain, toh generally Diamagnetic hota hai, lekin 10 aur 16 electrons wale species (B$_2$ aur O$_2$) Paramagnetic hote hain. Ye exceptions yaad rakhna! 4. Bond Length Bond Order jitna zyada, bond length utni hi kam. Example: N$_2$ (BO=3) ki bond length O$_2$ (BO=2) se kam hoti hai. Bond length order for isoelectronic species: Agar BO same hai, toh bond length bhi same hogi. Bond length order for different species: BO calculate karo, phir inverse relation apply karo. 5. Tricks for Bond Order (Fast Calculation) JEE mein time bachane ke liye ye trick bahut kaam aayegi: Total electrons count karo, phir is table se BO nikal lo (valid for diatomic species): Total Electrons Bond Order 8 0 9 0.5 10 1 11 1.5 12 2 13 2.5 14 3 15 2.5 16 2 17 1.5 18 1 19 0.5 20 0 Example: N$_2$ = 14 electrons $\implies$ BO = 3 O$_2$ = 16 electrons $\implies$ BO = 2 O$_2^+$ = 15 electrons $\implies$ BO = 2.5 O$_2^-$ = 17 electrons $\implies$ BO = 1.5 Note: Ye trick sirf BO nikalne ke liye hai, magnetic nature ke liye electronic configuration likhna padega ya upar wali magnetic nature trick use karni padegi. 6. Heteronuclear Diatomic Molecules (e.g., CO, NO, CN$^-$, OCl) Inmein bhi MO theory apply hoti hai, bas thoda sa difference hota hai energy levels mein kyunki atoms alag hote hain. Lekin basic principles (Aufbau, Hund's, Pauli) same rehte hain. ELD symmetric nahi hota, more electronegative atom ke atomic orbitals lower energy par hote hain. CO (14 electrons): BO = 3. Diamagnetic. (Isostructural and isoelectronic with N$_2$) NO (15 electrons): BO = 2.5. Paramagnetic (1 unpaired electron). CN$^-$ (14 electrons): BO = 3. Diamagnetic. (Isostructural and isoelectronic with N$_2$) Isostructural/Isoelectronic Species: Jin species mein total electrons same hote hain, unka BO, bond length, aur magnetic nature (mostly) same hota hai. Ye trick heteronuclear molecules ke liye bhi kaam aati hai. N$_2$ (14e), CO (14e), CN$^-$ (14e) $\implies$ Sabka BO = 3, sab diamagnetic. O$_2$ (16e), S$_2$ (16e) $\implies$ Sabka BO = 2, sab paramagnetic. 7. Questions kahan se aate hain? Stability order: Different species (e.g., O$_2$, O$_2^+$, O$_2^-$) ki stability compare karna. Jiska BO zyada, wo zyada stable. Agar BO same, toh $N_a$ kam wala zyada stable. Magnetic nature: Paramagnetic ya Diamagnetic identify karna. Unpaired electrons hain ya nahi, ye check karna. Total electrons wali trick bahut useful hai. Bond length order: BO se inversely proportional. Direct BO calculation: Kisi species ka BO pucha ja sakta hai. Electronic configuration: Kisi species ka MO electronic configuration likhna. Ionization Energy (IE): MO theory se IE ka trend bhi explain hota hai. Highest Occupied Molecular Orbital (HOMO) se electron nikalta hai. Bond dissociation energy: BO ke directly proportional. Toh bhai, in sab points pe focus karna, MO theory se questions pakke solve honge! 3.13 Examples of Molecular Orbital Treatment for Heteronuclear Diatomic Molecules (Alag-Alag Atoms ke MOs) Heteronuclear molecules mein AO energies different hoti hain. Zyada electronegative atom ke AOs ki energy kam hoti hai. NO (15 electrons): $\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \sigma2p_x^2 \pi2p_y^2 \pi2p_z^2 \pi^*2p_y^1$. Bond order = 2.5. Paramagnetic. HOMO ($\pi^*$) ki energy N aur O ke AOs se zyada hoti hai, isliye NO ka ionization energy N aur O se kam hota hai. Important Trick: NO mein N aur O ki electronegativity mein difference hai. Isliye, MO diagram symmetric nahi hota. Zyada electronegative atom (Oxygen) ke atomic orbitals (AOs) ki energy kam hoti hai, aur wo bonding MOs mein zyada contribute karte hain. Kam electronegative atom (Nitrogen) ke AOs ki energy zyada hoti hai, aur wo anti-bonding MOs mein zyada contribute karte hain. Isse HOMO ($\pi^*$) mainly Nitrogen ke AO character ka hota hai. JEE Advanced Concept: NO ka HOMO ($\pi^*$) ek anti-bonding orbital hai. Jab NO se electron nikalta hai (NO $\rightarrow$ NO$^+$), toh ye electron anti-bonding orbital se nikalta hai. Isse bond order badhta hai (2.5 se 3.0 ho jata hai), aur bond length decrease hoti hai. NO$^+$ ka bond order 3 hota hai aur ye diamagnetic hota hai. Ye NO ki stability explain karta hai, kyunki electron nikalne se molecule aur stable ho jata hai. CO (14 electrons): C aur O ki electronegativity difference zyada hoti hai. S-p mixing bhi hoti hai. CO mein HOMO C atom ke non-bonding orbital se hota hai. Electron removal C atom se hota hai, bond length decrease hoti hai. Detailed Concept: CO mein, Oxygen zyada electronegative hai, isliye uske AOs ki energy kam hoti hai. Bonding MOs mein Oxygen ka contribution zyada hota hai. HOMO (Highest Occupied Molecular Orbital) mainly Carbon ke non-bonding character ka hota hai. Jab CO se electron nikalta hai (CO $\rightarrow$ CO$^+$), toh ye electron HOMO se nikalta hai jo ki C-atom par localized hai. Isse C-O bond order badhta hai (kyunki anti-bonding electron nahi nikal raha), aur bond length decrease hoti hai. Ye ek common misconception hai ki electron nikalne se bond order hamesha kam hota hai, but CO ke case mein ulta hota hai. JEE Advanced Tip: CO ka MO diagram N$_2$ jaisa hi hota hai s-p mixing ki wajah se, lekin energy levels asymmetric hote hain due to electronegativity difference. Iska HOMO ek sigma orbital hota hai jo mainly Carbon ke lone pair character ka hota hai. Important Trick for CO: CO ek bahut hi stable molecule hai. Iski bond dissociation energy bahut high hoti hai. Iska reason hai ki iska HOMO Carbon ke lone pair character ka hai. Ye property CO ko metals ke saath complex banane mein help karti hai, jahan CO ligand ki tarah act karta hai aur metal ko electron donate karta hai (synergic bonding). Isse related questions aksar coordination chemistry mein bhi aate hain. HF (10 electrons): Hydrogen aur Fluorine mein bahut zyada electronegativity difference hai. F zyada electronegative hai. Detailed Concept: HF mein, H ka 1s orbital aur F ka 2p$_z$ orbital overlap karte hain (assuming internuclear axis is z-axis). F ka 2s orbital non-bonding rehta hai kyunki uski energy bahut kam hoti hai. Do MOs bante hain: ek bonding $\sigma$ aur ek anti-bonding $\sigma^*$. F ke 2p$_x$ aur 2p$_y$ orbitals bhi non-bonding rehte hain kyunki unke paas overlap karne ke liye H ke paas koi suitable orbital nahi hota. Isliye, HF mein ek single bond hota hai aur 3 lone pairs Fluorine par hote hain. JEE Advanced Tip: HF ka MO diagram banate waqt, energy differences ka dhyan rakhna. Fluorine ke AOs ki energy Hydrogen ke AOs se bahut kam hoti hai. Isliye, bonding MO mein Fluorine ka contribution zyada hota hai, aur anti-bonding MO mein Hydrogen ka contribution zyada hota hai. Isse HF ka dipole moment bhi explain hota hai. Trick: Jab bhi ek highly electronegative atom aur ek less electronegative atom ke beech bond ho, toh bonding MOs zyada electronegative atom ke character ke honge aur anti-bonding MOs kam electronegative atom ke character ke honge. Non-bonding orbitals bhi ho sakte hain agar atomic orbitals ka energy mismatch ya symmetry mismatch ho. 3.14 Dipole Moment (Charges ka Balance) Dipole moment ($\mu$) bond ki polarity measure karta hai. Ye ek vector quantity hai. $\mu = q \times d$ (q = charge, d = bond length). Direction: Low electronegative atom se high electronegative atom ki taraf (dikhaate hain $\rightarrow$ se). Bond Moment: Individual bond ka dipole moment. Molecule ka Dipole Moment: Vector sum of all bond moments aur lone pair moments. Diatomic Molecules: Homoatomic (Cl$_2$): $\mu = 0$. Heteroatomic (HCl): $\mu \neq 0$. Polyatomic Molecules: Symmetrical molecules (CH$_4$, CO$_2$): $\mu = 0$. Asymmetrical molecules (H$_2$O, NH$_3$): $\mu \neq 0$. Lone Pair Moment: Hybrid orbitals mein lone pair ka bhi dipole moment hota hai (small lobe se large lobe ki taraf). % s-character badhne se lone pair moment badhta hai, phir spherical hone par kam hota hai. Applications of Dipole Moment: Percentage Ionic Character: $$ \% \text{ionic character} = \frac{\mu_{obs}}{\mu_{calc}} \times 100 $$ Jahan $\mu_{obs}$ observed dipole moment hai aur $\mu_{calc}$ calculate kiya gaya dipole moment hai (assuming 100% ionic). Example LiF: 84.32% ionic character. Prediction of Molecular Shape: AX$_2$: $\mu=0$ (Linear, BeCl$_2$, XeF$_2$), $\mu \neq 0$ (Angular, H$_2$O, SO$_2$). AX$_3$: $\mu=0$ (Trigonal Planar, BF$_3$, AlCl$_3$), $\mu \neq 0$ (Pyramidal, NH$_3$; T-shaped, ClF$_3$). AX$_4$: $\mu=0$ (Tetrahedral, CH$_4$; Square Planar, XeF$_4$), $\mu \neq 0$ (See-saw, SF$_4$). Prediction of Cis-Trans Isomerism: Cis-isomer ka $\mu \neq 0$, Trans-isomer ka $\mu = 0$ (agar symmetrical ho). Example C$_2$H$_2$Cl$_2$. Prediction of Substituent Position in Benzene: Disubstituted benzene mein ortho, meta, para positions ke liye dipole moment order alag-alag hota hai, jo substituents ke nature par depend karta hai. Halogen atoms (X) ke liye: ortho > meta > para. Electron withdrawing (X) aur electron donating (Y) ke liye: ortho Distinguishing between Isomers: Dipole moment se hum isomers ko distinguish kar sakte hain, khaaskar organic chemistry mein. Jaise o-, m-, p-dichlorobenzene. Exceptions and Special Cases: NH$_3$ ka $\mu$ (1.47 D) NF$_3$ (0.24 D) se zyada hai, kyunki NF$_3$ mein F ki electronegativity bond moment ko lone pair moment ke oppose karti hai. CH$_3$Cl > CH$_2$Cl$_2$ > CHCl$_3$ > CCl$_4$ (dipole moment order). CO ka dipole moment bahut kam (0.112 D) hota hai, kyunki C-O bond moment aur lone pair moment ek dusre ko cancel karte hain. Trick: Agar molecule mein central atom par lone pair hain aur surrounding atoms same hain, toh generally $\mu \neq 0$ (e.g., H$_2$O, NH$_3$). Agar lone pair nahi hain aur surrounding atoms same hain, toh $\mu = 0$ (e.g., CH$_4$, CCl$_4$). Important: Symmetrical structure wale molecules ka dipole moment zero hota hai, bhale hi unmein polar bonds hon (e.g., CO$_2$, CCl$_4$, BF$_3$). Kyunki bond moments ek dusre ko cancel kar dete hain. Kya yaad rakhein: Linear molecules (CO$_2$, BeF$_2$) ka $\mu = 0$. Trigonal planar molecules (BF$_3$, AlCl$_3$) ka $\mu = 0$. Tetrahedral molecules (CH$_4$, CCl$_4$) ka $\mu = 0$. Square planar molecules (XeF$_4$) ka $\mu = 0$. Octahedral molecules (SF$_6$) ka $\mu = 0$. Jab $\mu \neq 0$ hota hai: Bent (H$_2$O, SO$_2$). Pyramidal (NH$_3$, PCl$_3$). See-saw (SF$_4$). T-shaped (ClF$_3$). Square pyramidal (BrF$_5$). Resonance aur Dipole Moment: Agar kisi molecule mein resonance hota hai, toh uska actual dipole moment resonance structures ke dipole moments ka average hota hai. Resonance ke kaaran charge separation ho sakta hai, jisse dipole moment badh ya ghat sakta hai. Jaise Aniline (C$_6$H$_5$NH$_2$) mein lone pair of electrons nitrogen par benzene ring mein delocalize hote hain, jisse dipole moment badh jaata hai. Solubility: Polar molecules polar solvents (jaise water) mein soluble hote hain, aur non-polar molecules non-polar solvents (jaise benzene) mein soluble hote hain. "Like dissolves like" ka concept yahan apply hota hai. Dipole moment se hum molecule ki polarity ka idea laga sakte hain. Boiling Point aur Melting Point: Higher dipole moment wale molecules mein strong intermolecular forces (dipole-dipole interactions) hote hain, jiske kaaran unka boiling point aur melting point zyada hota hai. Hybridization aur Dipole Moment: Central atom ki hybridization bhi dipole moment ko affect karti hai. Agar hybridization ke kaaran molecule asymmetrical ho jaata hai, toh dipole moment non-zero ho sakta hai. Example: sp hybridization (linear) -> CO$_2$ ($\mu=0$). sp$^2$ hybridization (trigonal planar) -> BF$_3$ ($\mu=0$). sp$^3$ hybridization (tetrahedral) -> CH$_4$ ($\mu=0$). Lekin agar lone pair aa jaye (jaise H$_2$O mein sp$^3$ hybridization, bent shape), toh $\mu \neq 0$. Bond Angle aur Dipole Moment: Bond angle mein change se bhi dipole moment change hota hai. Jaise H$_2$O aur H$_2$S mein, H$_2$O ka bond angle zyada hai aur iska dipole moment bhi zyada hai. Kyunki bond moments ka vector sum bond angle par depend karta hai. Trick for JEE Advanced: Agar aapko do molecules ka dipole moment compare karna hai, toh pehle unki geometry aur hybridization check karo. Phir dekho ki central atom par lone pair hain ya nahi. Agar dono mein lone pair hain, toh electronegativity differences aur bond angles ko consider karo. Agar ek mein lone pair hai aur dusre mein nahi, aur surrounding atoms same hain, toh jismein lone pair hai uska dipole moment non-zero hoga (unless symmetry cancel kar de). Organic compounds mein inductive effect aur resonance effect ko bhi consider karna padta hai. Electron withdrawing groups (EWG) dipole moment badhate hain aur electron donating groups (EDG) kam karte hain (position ke hisaab se). 3.15 The Ionic Bond (Ionic Bond ki Gehrai) Ionic solids (salts, oxides) electrostatic attraction se bane hote hain. Ions ke beech attraction tab maximum hoti hai jab oppositely charged ions surrounding mein hon. Coordination Number: Ek ion ke around oppositely charged ions ka number. Radius Ratio Rules (Size ka Hisab) Ion ke sizes aur numbers se coordination number predict kar sakte hain. Radius Ratio ($r_+/r_-$) = Cation radius / Anion radius. Limiting Radius Ratio Coordination Number Shape 2 Linear 0.155 - 0.225 3 Planar Triangle 0.225 - 0.414 4 Tetrahedral 0.414 - 0.732 4 or 6 Square Planar or Octahedral 0.732 - 0.999 8 Body-centred Cubic Calculations: C.N. 3 (Planar Triangle): $r_+/r_- = 0.155$. C.N. 4 (Tetrahedral): $r_+/r_- = 0.225$. C.N. 6 (Octahedral): $r_+/r_- = 0.414$. C.N. 8 (Body-centred Cubic): $r_+/r_- = 0.732$. 3.16 Close Packing (Atoms ka Sahi Tarike Se Jamna) Ionic solids mein large ions (usually anions) close-packed arrangements banate hain, aur small ions (cations) holes (interstitial sites) occupy karte hain. Close-Packed Arrangements: Cubic Close-Packed (ccp) / Face-Centred Cubic (fcc): ABCABC... stacking. Hexagonal Close-Packed (hcp): ABAB... stacking. Holes: Tetrahedral Holes: Four spheres se bounded, smaller. Octahedral Holes: Six spheres se bounded, larger. Har close-packed arrangement mein ek octahedral hole aur do tetrahedral holes hote hain per atom. 3.17 Ionic Compounds of the type AX (ZnS, NaCl, CsCl) (Simple Ionic Compounds) AX type compounds ke teen main structures: Zinc Sulphide (ZnS): Radius ratio ~0.40, suggests tetrahedral arrangement. Coordination Number: 4:4 (each Zn$^{2+}$ is surrounded by 4 S$^{2-}$, and vice versa). Two forms: Zinc Blende (ccp related) and Wurtzite (hcp related). Zn$^{2+}$ ions tetrahedral holes occupy karte hain. Sodium Chloride (NaCl): Radius ratio ~0.52, suggests octahedral arrangement. Coordination Number: 6:6. Cl$^-$ ions ccp array banate hain, Na$^+$ ions octahedral holes occupy karte hain. Caesium Chloride (CsCl): Radius ratio ~0.93, suggests body-centred cubic arrangement. Coordination Number: 8:8. Cl$^-$ ions cube ke corners par, Cs$^+$ ion center mein. 3.18 Ionic Compounds of the Type AX$_2$ (CaF$_2$, TiO$_2$, SiO$_2$) (Thode Complex Ionic Compounds) AX$_2$ type compounds ke main structures: Fluorite (CaF$_2$): Coordination Number: 8:4 (each Ca$^{2+}$ surrounded by 8 F$^-$, each F$^-$ by 4 Ca$^{2+}$). Radius ratio >0.73. Ca$^{2+}$ ions ccp array banate hain, F$^-$ ions saare tetrahedral holes occupy karte hain. Rutile (TiO$_2$): Coordination Number: 6:3. Radius ratio 0.41-0.73. Ti$^{4+}$ ions octahedrally surrounded by 6 O$^{2-}$, each O$^{2-}$ by 3 Ti$^{4+}$ (planar triangular). $\beta$-Cristobalite (SiO$_2$): Related to zinc blende structure. Coordination Number: 4:2. Oxygen atoms Si atoms ke beech mein hote hain, bond angle Si-O-Si 180° nahi hota. 3.19 Layer Structures (CdI$_2$, CdCl$_2$, [NiAs]) (Layers Wale Structures) Ye structures ideal ionic nahi hote, inmein covalent character bhi hota hai. Ye layers mein arranged hote hain. Cadmium Iodide (CdI$_2$): Radius ratio 0.45, C.N. 6. Cd$^{2+}$ ions I$^-$ ions ke beech sandwich banate hain. Within sandwich strong electrostatic bonding, between sandwiches weak van der Waals forces. Isliye flaky hote hain. Cadmium Chloride (CdCl$_2$): Similar to CdI$_2$, but Cl$^-$ ions ccp arrangement mein hote hain. Nickel Arsenide ([NiAs]): As atoms hcp array banate hain, Ni atoms saare octahedral holes occupy karte hain. Metallic properties bhi show karte hain. 3.20 Lattice Energy (Crystal ki Energy) Lattice energy (U) wo energy hai jo ek gram mole crystal ke formation mein release hoti hai gaseous ions se. Example: Na$_{(g)}^+$ + Cl$_{(g)}^-$ $\rightarrow$ NaCl$_{(crystal)}$. U = -782 kJ mol$^{-1}$. Lattice energy ko Born-Haber cycle se calculate karte hain (experimental) ya Born-Landé equation se (theoretical). Born-Landé Equation: $$ U = \frac{N_A A z^+ z^- e^2}{r_0} \left(1 - \frac{1}{n}\right) $$ $N_A$ = Avogadro constant. A = Madelung constant (crystal geometry par depend karta hai). $z^+, z^-$ = charges on ions. $e$ = electron charge. $r_0$ = equilibrium inter-ionic distance. $n$ = Born exponent (repulsive forces ke liye, ions ke electronic configuration par depend karta hai). Key Points: U is proportional to $z^+z^- / r_0$. $r_0$ kam hone par U zyada negative hota hai, crystal zyada stable. Ionic charges badhne par U zyada negative hota hai. Born-Landé equation se calculated U (theoretical) aur Born-Haber cycle se obtained U (experimental) match karte hain toh compound ionic hota hai. Agar match nahi karte toh covalent character hota hai. 3.21 Stoichiometric Defects (Perfect Crystal Mein Kami) Crystalline solids perfect nahi hote, unmein defects hote hain. Stoichiometric Compounds: Chemical formula ke hisab se atoms ka ratio fixed hota hai. Schottky Defects: Crystal lattice mein se equal number of cations aur anions missing hote hain, jisse vacancies create hoti hain. Density decrease hoti hai. Aise compounds mein hote hain jahan ions ka size similar hota hai aur C.N. high hoti hai (e.g., NaCl, CsCl). Frenkel Defects: Ek ion apni lattice site chhodkar interstitial position mein chala jaata hai, jisse vacancy (hole) aur interstitial ion bante hain. Density change nahi hoti. Aise compounds mein hote hain jahan ions ka size different hota hai (cations small, anions large) aur C.N. low hoti hai (e.g., ZnS, AgCl). Electrical Conductivity: Defects ki wajah se crystals mein thodi electrical conductivity hoti hai (ionic mechanism). 3.22 Nonstoichiometric Defects (Ratio Fixed Nahi) Nonstoichiometric compounds mein atoms ka ratio chemical formula ke hisab se fixed nahi hota (e.g., Fe$_{0.84}$O). Metal Excess Defects: Anion Vacancies: Negative ion lattice site se missing hota hai, aur uski jagah ek electron occupy karta hai (F-centre). F-centres ki wajah se color aata hai (e.g., NaCl yellow, KCl lilac). Paramagnetic hote hain (unpaired electron ki wajah se). n-type semiconduction show karte hain. Interstitial Cations: Extra positive ion interstitial position mein hota hai, aur electrical neutrality maintain karne ke liye ek electron bhi interstitial hota hai. n-type semiconduction show karte hain (e.g., ZnO). Metal Deficiency Defects: Cation Vacancies: Positive ion lattice site se missing hota hai. Electrical neutrality maintain karne ke liye adjacent metal ions ki oxidation state badh jaati hai (e.g., FeO mein Fe$^{2+}$ ki jagah 2 Fe$^{3+}$). p-type semiconduction show karte hain. Interstitial Anions: Bahut rare hote hain kyunki anions bade hote hain. 3.23 Born-Haber Cycle (Energy ka Chakkar) Born-Haber cycle lattice energy ko other thermochemical data se relate karta hai (Hess's Law). NaCl ke formation ke liye: Na(s) → Na(g) ΔH_sub (sublimation energy) ½ Cl₂(g) → Cl(g) ½ ΔH_diss (dissociation energy) Na(g) → Na⁺(g) + e⁻ IE (ionization energy) Cl(g) + e⁻ → Cl⁻(g) EA (electron affinity) Na⁺(g) + Cl⁻(g) → NaCl(s) U (lattice energy) Overall reaction: Na(s) + ½ Cl$_2$(g) → NaCl(s) ($\Delta H_f$, enthalpy of formation) According to Hess's Law: $$ \Delta H_f = \Delta H_{sub} + \frac{1}{2}\Delta H_{diss} + IE + EA + U $$ Uses: Unmeasured quantities (jaise EA ya U) ko calculate karne ke liye. Ionic character judge karne ke liye (theoretical U aur experimental U ko compare karke). 3.24 Polarizing Power and Polarizability – Fajans' Rules (Ions Ki Surat Bigadna) Jab ek cation aur anion bond banate hain, toh cation anion ke electron cloud ko apni taraf kheenchta hai, jisse anion distort ho jaata hai. Is process ko polarization kehte hain. Polarization se covalent character badhta hai. Polarizing Power: Cation ki ability anion ko distort karne ki. Polarizability: Anion ki ability distort hone ki. Fajans' Rules: (Polarization Kahan Zyada Hoti Hai) Small Cation: Small cations ki charge density zyada hoti hai, isliye unki polarizing power zyada hoti hai (e.g., Li$^+$ > Na$^+$). Large Anion: Large anions ke outer electrons nucleus se kam tightly held hote hain, isliye unki polarizability zyada hoti hai (e.g., I$^-$ > Br$^-$ > Cl$^-$ > F$^-$). Large Charges: Cation ya anion par charge zyada hone par polarization badhti hai (e.g., Al$^{3+}$ > Mg$^{2+}$ > Na$^+$). Non-Noble Gas Configuration: Cations with (n-1)d$^{10}$ configuration (e.g., Cu$^+$, Ag$^+$, Zn$^{2+}$) noble gas configuration (ns$^2$np$^6$) wale cations se zyada polarizing hote hain, kyunki d-electrons shielding kam karte hain (e.g., CuCl is more covalent than NaCl). Thoda aur detail mein samjho polarization ko: Polarizing Power (Cation ki): Yeh cation ki ability hai ki woh anion ke electron cloud ko kitna distort kar sakta hai. Jitna zyada distort karega, utni zyada polarizing power. Charge density ka funda: Charge/Radius ratio. Cation ka charge zyada ho aur size chota ho, toh charge density high hogi, aur polarizing power bhi high hogi. Example: Li$^+$ (chota size, +1 charge) ki polarizing power Na$^+$ se zyada hai. Al$^{3+}$ (chota size, +3 charge) ki polarizing power Mg$^{2+}$ se zyada hai. Polarizability (Anion ki): Yeh anion ke electron cloud ki tendency hai ki woh kitna distort ho sakta hai cation ke electric field mein. Jitna zyada distort hoga, utni zyada polarizability. Size ka funda: Anion ka size jitna bada hoga, outer electrons nucleus se utne hi door honge aur kam tightly held honge. Isliye unko distort karna aasan hoga. Example: I$^-$ (bada size) ki polarizability F$^-$ (chota size) se zyada hai. Covalent Character: Jab polarization hoti hai, toh electron cloud ka overlap badhta hai, jisse ionic bond mein covalent character aa jata hai. Jitni zyada polarization, utna zyada covalent character. JEE Advanced/Mains ke liye Important Points aur Tricks: Covalent Character ka Order: Jab bhi covalent character compare karna ho, Fajans' Rules yaad karo. Example: LiCl, NaCl, KCl, RbCl. Cation size badh raha hai (Li$^+$ to Rb$^+$), toh polarizing power kam ho rahi hai. Isliye LiCl sabse zyada covalent hoga. Example: NaF, NaCl, NaBr, NaI. Anion size badh raha hai (F$^-$ to I$^-$), toh polarizability badh rahi hai. Isliye NaI sabse zyada covalent hoga. Melting Point aur Solubility: Covalent character badhne se melting point aur solubility in polar solvents (jaise water) kam hoti hai. Non-polar solvents mein solubility badhti hai. Trick: Zyada covalent = kam ionic = kam melting point = kam water solubility. Example: AgF (ionic) water soluble hai, jabki AgCl, AgBr, AgI (zyada covalent) water insoluble hain. Thermal Stability: Zyada covalent character wale compounds ki thermal stability generally kam hoti hai, kyunki covalent bonds weaker hote hain ionic bonds se (in terms of lattice energy). Par yeh hamesha true nahi hota, context pe depend karta hai. Example: Carbonates ki thermal stability: Li$_2$CO$_3$ (zyada covalent) Na$_2$CO$_3$ se kam stable hai. Colour: Transition metal compounds mein polarization se colour ki intensity badh sakti hai. Jab anion ka electron cloud distort hota hai, toh HOMO-LUMO gap change hota hai, jisse light absorption change hoti hai. Example: AgCl (white), AgBr (pale yellow), AgI (yellow). AgI mein sabse zyada polarization hai, isliye uska colour sabse intense hai. Acidic Nature of Oxides: Covalent character badhne se oxides ka acidic nature badhta hai. Non-metal oxides generally acidic hote hain, aur metal oxides basic. Lekin jab metal-oxygen bond mein covalent character badhta hai, toh uska acidic nature bhi badhta hai. Example: Al$_2$O$_3$ (amphoteric) mein Mg$^+$ ke oxide (basic) se zyada covalent character hota hai, isliye Al$_2$O$_3$ acidic properties bhi show karta hai. Hydration Energy vs. Lattice Energy: Solubility decide karne mein in dono ka role hota hai. Agar hydration energy lattice energy se zyada ho, toh compound soluble hoga. Fajans' Rules indirectly inko affect karte hain. Covalent Character aur Bond Strength: Covalent character badhne se bond strength bhi badh sakti hai, khaaskar jab directional bonding involve ho. Lekin ionic compounds ke context mein, zyada covalent character ka matlab hai ki pure ionic character kam ho raha hai, jisse lattice energy kam hoti hai. Volatility: Zyada covalent character wale compounds generally zyada volatile hote hain, kyunki unke intermolecular forces weak hote hain as compared to strong electrostatic forces in highly ionic compounds. Trick: Covalent character $\uparrow$ $\implies$ Volatility $\uparrow$ (e.g., SnCl$_4$ is liquid, SnCl$_2$ is solid). Hardness: Ionic compounds generally hard hote hain due to strong electrostatic forces. Covalent character badhne se hardness kam hoti hai. Example: AgF (hard) vs AgI (soft). Ionic Potential ($\phi$): Cation ka ionic potential = Charge of Cation / Radius of Cation. Higher the ionic potential, higher the polarizing power. Yeh charge density ka hi ek tarah se measure hai. Trick: $\phi \uparrow \implies$ Polarizing Power $\uparrow \implies$ Covalent Character $\uparrow$. Diagonal Relationship: Li aur Mg, Be aur Al, B aur Si jaise elements diagonal relationship show karte hain. Iska ek reason unki similar polarizing power hoti hai, jisse unke compounds ke properties mein similarity aati hai. Example: Li aur Mg dono covalent character wale compounds banate hain, jaise LiCl aur MgCl$_2$ dono organic solvents mein soluble hote hain. Properties Affected by Polarization: Covalent character increases. Lattice energy decreases (generally, as covalent character increases, the ionic contribution to lattice energy decreases). Melting point decreases. Hardness decreases. Solubility in polar solvent decreases, in non-polar solvent increases. Electrical conductivity decreases (in molten state or solution, due to less free ions). Intensity of colour increases (e.g., ZnS (white) $\rightarrow$ CdS (yellow) $\rightarrow$ HgS (black)). Acidic nature of oxides increases. Thermal stability generally decreases (for compounds like carbonates, hydroxides, nitrates). Volatility increases. 3.25 Melting Point of Ionic Compounds (Ionic Compounds ka Melting Point) Melting point lattice energy aur polarization par depend karta hai. When Polarization is Negligible: (Small anions like F$^-$, O$^{2-}$) Melting point order lattice energy order ko follow karta hai. Lattice energy ($U$) $\propto z^+z^- / r_0$. Example: NaF Example: BeO > MgO > CaO > SrO > BaO (melting point and lattice energy decrease with increasing $r_0$). When Polarization is Dominating: Polarization badhne se melting point decrease hota hai. Example: NaX Example: MF > MCl > MBr > MI (F$^-$ ki polarizability kam, I$^-$ ki zyada). 3.26 Solubility of Ionic Compounds (Ionic Compounds ki Ghulansheelta) Solubility lattice energy, hydration energy, polarization, etc. par depend karti hai. Hydration Energy ($\Delta H_{hyd}$): Ion jab water mein dissolve hota hai, toh energy release hoti hai. Dissolution Process: MX$_{(s)}$ $\rightarrow$ M$_{(g)}^{n+}$ + X$_{(g)}^{m-}$ ($\Delta H_{lattice}$ = -U) then M$_{(g)}^{n+}$ + X$_{(g)}^{m-}$ $\rightarrow$ M$_{(aq)}^{n+}$ + X$_{(aq)}^{m-}$ ($\Delta H_{hyd}$). Overall: MX$_{(s)}$ $\rightarrow$ M$_{(aq)}^{n+}$ + X$_{(aq)}^{m-}$ ($\Delta H_{sol}$ = $\Delta H_{lattice}$ + $\Delta H_{hyd}$) For solubility, $\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol}$ should be negative. Generally, $\Delta H_{sol}$ should be negative for good solubility. When Polarization is Dominating: Polar solvent mein, polarization badhne se solubility kam hoti hai (e.g., AgF > AgCl > AgBr > AgI). Non-polar solvent mein, covalent character badhne se solubility badhti hai (e.g., NaCl When Lattice Energy and Hydration Energy are Dominating: Agar cation ka size small ho, toh hydration energy dominant hoti hai, solubility badhti hai (e.g., LiClO$_4$ > NaClO$_4$). Agar anion ka size small ho, toh lattice energy dominant hoti hai, solubility kam hoti hai (e.g., LiF Exceptions: LiF (lattice energy high), NaI (polarization high). 3.27 Electrical Conductivity and Colour (Bijli aur Rang) Electrical Conductivity: Polarization badhne se covalent character badhta hai, ionisation tendency kam hoti hai, isliye electrical conductivity decrease hoti hai (e.g., LiCl > BeCl$_2$ in molten state). Colour: Polarization badhne se ionic compounds ka colour intensity badhti hai (e.g., AgCl (colourless) $\rightarrow$ AgI (yellow)). Ye charge transfer transitions ki wajah se hota hai. 3.28 Acidic Nature of Oxides (Oxides ka Acidic Nature) Polarization badhne se oxides ka acidic nature badhta hai. Ionic Potential ($\phi$): $\phi = \text{Charge} / \text{Radius}$. $\phi > 3.2$: Acidic oxide (e.g., N$_2$O$_5$, CrO$_3$). $2.2 $\phi Example: Li$_2$O 3.29 Thermal Stability of Ionic Compounds (Ionic Compounds ki Garmi Sehishnuta) Thermal stability lattice energy aur polarization par depend karti hai. Binary Compounds (Smallest Stoichiometric Ratio): Lattice energy order ko follow karta hai. Lattice energy badhne se thermal stability badhti hai. Example: Li$_3$N > Na$_3$N > K$_3$N (lattice energy decreases down the group). Example: BeO > MgO > CaO > SrO > BaO. Multiatomic Anion Compounds: Efficient packing of larger cation and larger anion leads to more stable compounds. Thermal stability increases down the group. Example: LiClO$_3$ Example: BeCO$_3$ 3.30 Weak Forces (Kamzor Taakatein) Molecules ke beech mein weak attractive forces hote hain, jo gas ko liquefy karte hain. Van der Waals Forces: Total attractive forces. Dipole-Dipole Interaction (Keesom forces): Polar molecules ke beech electrostatic attraction (head-tail ya anti-parallel arrangement). Jitna zyada polarity, utna strong force. Temperature increase karne par force decrease hota hai. Hydrogen bonding is a special case. Dipole-Induced Dipole Interaction (Debye forces): Polar molecule ek non-polar molecule mein dipole induce karta hai. Induced dipole ki strength polar molecule ki polarity aur non-polar molecule ki polarizability par depend karti hai. Example: Noble gases ki water mein solubility. Instantaneous Dipole-Induced Dipole Interaction (London forces): Non-polar molecules mein electrons ke random fluctuation ki wajah se instantaneous dipole banta hai, jo adjacent molecule mein dipole induce karta hai. Weakest force. Depend karta hai: Number of polarizable electrons, molecular weight, molecular volume, surface area. Larger molecules aur zyada electrons wale molecules mein London forces zyada strong hote hain. Branching se surface area kam hota hai, isliye London forces kam hote hain (e.g., n-pentane > isopentane > neopentane boiling point). Example: H$_2$, He ka liquefaction. Repulsive Intermolecular Forces: Jab molecules bahut close aa jaate hain, toh electron-electron aur nucleus-nucleus repulsion start hota hai. Lennard-Jones Potential: Net interaction energy = Attractive energy + Repulsive energy. $$V(r) = 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right]$$ $V(r)$: Potential energy between two particles at distance $r$. $\epsilon$: Depth of the potential well (energy required to separate the particles). $\sigma$: Distance at which the potential energy is zero (collision diameter). $r$: Distance between the centers of the two particles. $(\sigma/r)^{12}$: Repulsive term (short-range). $(\sigma/r)^{6}$: Attractive term (long-range, Van der Waals forces). JEE Tip: Minimum potential energy par system most stable hota hai. Is point par attractive aur repulsive forces balance ho jaate hain. Hydrogen Bonding: Special type of dipole-dipole interaction. Jab Hydrogen atom directly attached ho highly electronegative atoms (F, O, N) se. Conditions: High electronegativity of atom bonded to H. Small size of electronegative atom. Types: Intermolecular H-bonding: Different molecules ke beech (e.g., H$_2$O, HF, NH$_3$, alcohols). Boiling point increase karta hai. Intramolecular H-bonding: Same molecule ke andar (e.g., o-nitrophenol, salicylaldehyde). Boiling point decrease karta hai (chelation ki wajah se volatility badhti hai). JEE Trick: Boiling point order decide karte waqt, pehle H-bonding dekho. Agar H-bonding hai, toh boiling point significantly high hoga. Phir molecular weight aur London forces dekho. Example: H$_2$O ka boiling point H$_2$S se zyada hai, kyunki H$_2$O mein H-bonding hoti hai. Factors Affecting Intermolecular Forces (IMF): Molecular Size/Molecular Weight: Generally, larger molecules have stronger London forces (more electrons, more polarizable). Molecular Shape: More compact/spherical shapes have less surface area for interaction, so weaker London forces (e.g., n-pentane > isopentane > neopentane). Polarity: More polar molecules have stronger dipole-dipole interactions. Presence of H-bonding: Strongest IMF after covalent/ionic bonds. Applications and Questions: Boiling Point/Melting Point: Stronger IMF $\implies$ Higher boiling/melting point. (Common JEE question) Viscosity: Stronger IMF $\implies$ Higher viscosity. Surface Tension: Stronger IMF $\implies$ Higher surface tension. Vapor Pressure: Stronger IMF $\implies$ Lower vapor pressure. Solubility: "Like dissolves like" principle. Polar solutes dissolve in polar solvents (due to dipole-dipole, H-bonding). Non-polar solutes dissolve in non-polar solvents (due to London forces). Anomalous properties of water: High boiling point, high specific heat, density maximum at 4°C, ice floats – all due to extensive H-bonding. JEE Advanced Tip: Boiling point comparisons mein, pehle H-bonding dekho, phir molecular weight (London forces), phir branching. Example Question: Arrange H$_2$O, H$_2$S, H$_2$Se, H$_2$Te in increasing order of boiling points. H$_2$O mein H-bonding hai, toh highest. Baaki sab mein London forces predominant. Molecular weight badhne par London forces badhte hain. Order: H$_2$S < H$_2$Se < H$_2$Te < H$_2$O. JEE Advanced Question: Why is HF a liquid at room temperature while HCl, HBr, HI are gases? Answer: HF mein extensive intermolecular H-bonding hoti hai, jo uske molecules ko strongly hold karti hai, isliye woh liquid state mein rehta hai. HCl, HBr, HI mein H-bonding nahi hoti, sirf weaker dipole-dipole aur London forces hote hain. JEE Mains Question: Which of the following has the highest boiling point? (a) n-pentane (b) isopentane (c) neopentane Answer: n-pentane. Iska surface area sabse zyada hai, isliye London forces sabse strong hain. Branching badhne se surface area kam hota hai aur London forces weak hote hain. JEE Trick for Boiling Point Order: H-bonding: Agar H-bonding hai, toh boiling point significantly high hoga. (e.g., H$_2$O > H$_2$S) Molecular Weight: Same type ke IMF wale molecules mein, zyada molecular weight matlab zyada London forces, toh zyada boiling point. (e.g., H$_2$Te > H$_2$Se > H$_2$S) Branching: Isomers mein, unbranched chain ka boiling point branched chain se zyada hota hai, kyunki unbranched chain ka surface area zyada hota hai. (e.g., n-butane > isobutane) Polarity: Agar molecular weight similar ho, toh zyada polar molecule ka boiling point zyada hoga due to stronger dipole-dipole interactions. (e.g., CH$_3$Cl > C$_2$H$_6$) JEE Advanced Tip: Sometimes, a molecule with lower molecular weight but strong H-bonding can have a higher boiling point than a molecule with higher molecular weight but only weak London forces. Always prioritize H-bonding. Example: Ethanol (CH$_3$CH$_2$OH, MW=46) ka boiling point Diethyl ether (CH$_3$OCH$_3$, MW=46) se zyada hota hai, kyunki ethanol mein intermolecular H-bonding hoti hai. JEE Advanced Concept: Effect of IMF on Physical States: Gases: Weakest IMF (e.g., H$_2$, O$_2$, N$_2$, noble gases). Liquids: Moderate IMF (e.g., H$_2$O, alcohols, organic solvents). Solids: Strongest IMF (e.g., ionic compounds, metals, network solids). Covalent solids jaise diamond mein IMF nahi hote, covalent bonds hi pure structure ko hold karte hain. JEE Mains Trick: Volatility: Volatility inversely proportional hoti hai IMF ke. Matlab, jitne weak IMF, utni zyada volatility. Intramolecular H-bonding wale compounds (e.g., o-nitrophenol) intermolecular H-bonding wale compounds (e.g., p-nitrophenol) se zyada volatile hote hain, kyunki intramolecular H-bonding external interactions ko kam karti hai. JEE Advanced Question: Density of Ice vs Water: Answer: Water mein H-bonding ki wajah se open cage-like structure banta hai jab wo freeze hota hai (ice). Is structure mein bahut saari empty spaces hoti hain. Isliye, ice ka volume zyada hota hai same mass of water se, aur iski density kam hoti hai. JEE Advanced Tip: Dipole Moment and IMF: Higher dipole moment $\implies$ Stronger dipole-dipole interactions. Symmetry bhi important hai. Agar molecule symmetrical hai, toh uska net dipole moment zero ho sakta hai, bhale hi usme polar bonds hon (e.g., CO$_2$, CCl$_4$). Aise molecules mein sirf London forces predominate karte hain. JEE Mains Question: Solubility of Alcohols in Water: Answer: Lower alcohols (methanol, ethanol) water mein highly soluble hote hain kyunki wo water ke saath extensive H-bonding bana sakte hain. Jaise-jaise alkyl chain ki length badhti hai, non-polar hydrophobic part dominant ho jaata hai aur H-bonding ki capability relative kam ho jaati hai, isliye higher alcohols ki solubility water mein decrease hoti hai. JEE Advanced Concept: Effect of Pressure on IMF: Gases ko liquefy karne ke liye high pressure aur low temperature chahiye hota hai. High pressure se molecules paas aate hain, jisse IMF strong ho jaate hain aur gas liquid mein convert ho jaati hai. 3.31 Interactions between Ions and Covalent Molecules (Ions aur Molecules ka Mel) Ion-Dipole Interactions: Ionic compounds ki polar solvents mein solubility explain karta hai. Positive ion negative end of polar solvent ko attract karta hai, aur negative ion positive end ko. Ion-Induced Dipole Interaction: Ion ek non-polar molecule mein dipole induce karta hai. Example: Polyhalide ions (X$_3^-$) ka formation. Stability order: F$_3^-$ 3.32 The Metallic Bond (Metals ki Bonding) Metals ki unique properties (conductivity, lustre, malleability) metallic bond ki wajah se hoti hain. Free Electron Theory: Metals ko positive ions ke lattice mein free electrons ka "sea" mana jaata hai. Lustre: Free electrons light absorb aur re-emit karte hain. Malleability and Ductility: Atoms ke beech strong cohesive forces hote hain, but planes easily glide kar sakte hain. Cohesive Force: Atomization ki heat se measure karte hain. Rules: Bonding energy unpaired electrons par depend karti hai. Crystal structure s aur p orbitals ke number par depend karti hai. 3.33 Theories of Bonding in Metals (Metals ki Bonding Theories) Free Electron Theory: Metals = lattice of positive ions + sea of mobile electrons. Conductivity, malleability explain karta hai, but quantitative calculations weak hain. Valence Bond Theory: Pauling ne suggest kiya ki metallic bonding resonance ki wajah se hoti hai, jahan atoms ke beech covalent bonds rapidly shift hote hain. Ek Li atom 8 neighbors se bond bana sakta hai. Molecular Orbital or Band Theory: Atomic orbitals combine karke continuous band of molecular orbitals banate hain. Valence band (filled MOs) aur Conduction band (empty MOs). 3.34 Conductors, Insulators and Semiconductors (Bijli ke Khiladi) Band theory se electrical properties explain hoti hain. Conductors (Metals): Valence band partially filled hota hai, ya valence band aur conduction band overlap karte hain. No energy gap, electrons easily move karte hain. Example: Li (valence band half-filled), Be (bands overlap). Insulators (Non-metals): Valence band full hota hai, aur conduction band se large energy gap hota hai. Electrons move nahi kar pate. Semiconductors: Energy gap insulators se kam hota hai. At room temperature, kuch electrons valence band se conduction band mein jump kar sakte hain. Intrinsic Semiconductors: Pure substance (e.g., Si, Ge). Extrinsic Semiconductors (Doped): Impurities add karne se conductivity badhti hai. n-type: Donor impurity (extra electrons, e.g., P in Si). p-type: Acceptor impurity (holes, e.g., B in Si).