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![Lewis dot structure:
Represents the total no. of (e–)
present in the outermost shell
Molecule with complete
octet and donor atom should
house at least one lone pair
Acceptor atom molecule it
should be electron-deficient
Coordinate bond
Example:
NH3. BF3; NH4Cl; K4 [Fe (CN)6]
Ek tarfa pyaar](https://image.slidesharecdn.com/chemicalbondingclassnotes-251124173932-41e253ee/75/Chemical-Bonding-_-Class-Notes-pdf-for-jee-7-2048.jpg)
![Formal Charge:
F.C [V – L –
𝟏
𝟐
S]
V = Total no. of valence (e–) for
atom in free state
L = Total non-bonded or lone pair
electrons
S Total no. of shared (e–)
Resonance:
Delocalization of () electrons
. . . .
:O:
||
C
:O–: :O–:](https://image.slidesharecdn.com/chemicalbondingclassnotes-251124173932-41e253ee/75/Chemical-Bonding-_-Class-Notes-pdf-for-jee-8-2048.jpg)
![Bonding Parameter
1.Bond length: Internuclear distance
Factor affecting Bond length
(i) Atomic size: bond length size [No. of shell]
(ii) EN, Bond length
1
EN
(iii) Bond order: Bond length
1
B.O
(iv) Hybridisation : Bond length
1
%age of s−character
dA−B = rA + rB − 0.09 × ΔENÅ
Bond Angle : Angle between two adjacent bond is
known as bond angle.
bonded atom
bond angle
bond angle
central atom
Factors Affecting Bond Angle
(i) Hybridisation:
Bond angle %age of s-character
(ii)No. of lp/bp:
[when hybridisation is same]
Bond angle
1
lp
Eg. : CH4 > NH3 > H2O:
(iii) Type of Central atom: Applicable when :
➢ hybridisation same
➢ No. of Ip/bp same.
Bond angle EN of central atom
Eg. NH3 > PH3 > AsH3 > SbH3
:
:](https://image.slidesharecdn.com/chemicalbondingclassnotes-251124173932-41e253ee/75/Chemical-Bonding-_-Class-Notes-pdf-for-jee-42-2048.jpg)
![ANPARA THERMAL POWER STATION[1] sangam.pdf](https://cdn.slidesharecdn.com/ss_thumbnails/anparathermalpowerstation1sangam-251121115219-9261cde4-thumbnail.jpg?width=640&height=640&fit=bounds)
Chemical Bonding _ Class Notes.pdf for jee
- 1. INORGANIC CHEMISTRY CHEMICAL BONDING BY –SAGAR VARJATIA SIR
- 2. Covalent bond &Ionic bond V.S.E.P.R Theory Hybridisation Questions
- 3. Force of attractionwhich hold two or more than two species together is know as bond. Chemical Bonding Reason To attain the state of maximum stability species have tendency of bonding Classification of bonds Oo this basis of type of species getting bonded bond can be classify into following categories. H — H H2O H2O Na+ Cl– Bond Interatomic bond between two atom strong bond 200-400 kJ/mol Intermolecular force between molecules weak bond 2-40 kj/mol ➢ Non-metal + Non-metal-covalent bond ➢ Non-metal + Metal – Ionic bond ➢ Metal + Metal-Metallic bond H-Bond 1. Ion-dipole 2. Dipole-Dipole 3. Ion-Induced dipole 4. Dipole-Induced dipole 5. London force ❑ Coordination bond is type of covalent bond Atoms undergoes sharing of electron. Sharing of electron leads to the formation of covalent bond. Valence bond theory Formation of covalent bond explained by three theories Covalent bond Lewis octet theory Valence bond theory Molecular orbital theory As per lewis octet theory Bonding for stability Stability by achieving Nobel gas configuration Lewis octet theory Covalent bond Equal sharing-Covalent bond Unequal sharing-Coordinate bond Sharing
- 4. Incomplete octet molecules Lewis-kosselapproach To acquire inert gas configuration atoms loose or gain (e–) or share Octet Rule: Tendency of atoms to achieve eigtht (e–) in their outermost shell Exceptions to octet rule Expanded octet molecules Odd (e–) molecules Shape of molecules XeF2; XeF4 etc.
- 5. Ionic or electrovalent bond NaCl;KCl; MgCl2 Electrovalency: Total no. of (e–) gain or lose by non-metal and metal to form ion bond Electrostatic force of attraction between metal and non-metal Non-directional bond No hybridization No Isomerism Only empirical formula
- 6. Ionic Bond Q. Highlattice Enthalpy? (i) NaCl MgCl2 AlCl3 (ii) NaCl MgO Energy released when two oppositively charged ions combine to form a crystal lattice Conductivity (Molten/Aquacous) Hard crystalline; brittle High M.P and B.P Solubility {in polar solvents} 1. Low I.E. of metal 2. High electron hain enthalpy of non-metal 3. High lattice Enthalpy Conditions
- 7. Lewis dot structure: Representsthe total no. of (e–) present in the outermost shell Molecule with complete octet and donor atom should house at least one lone pair Acceptor atom molecule it should be electron-deficient Coordinate bond Example: NH3. BF3; NH4Cl; K4 [Fe (CN)6] Ek tarfa pyaar
- 8. Formal Charge: F.C [V – L – 𝟏 𝟐 S] V = Total no. of valence (e–) for atom in free state L = Total non-bonded or lone pair electrons S Total no. of shared (e–) Resonance: Delocalization of () electrons . . . . :O: || C :O–: :O–:
- 10. Mutual sharing ofelectrons Formation of bond is done with a release of energy Covalent Bond Polar covalent bond Non-Polar covalent bond
- 11. Covalent Bond: ➢ Physicalstate: In all F2; Cl2; Br2; I2 gas liquid solid. ➢ Conductivity: Poor ➢ Solubility: Non Polar ➢ Isomerism ➢ Directional Bonds ➢ Slow Reactions Covalency: ➢ Total of bonds made by central atom through sharing ➢ Total no. of half filled atomic orbitals. ➢ Maximum covalency is generally shown in exited state. ➢ PCl5 exists but NCl5 doesnot.
- 13. Polarization in ionicbond Small size of cation Large size of anion High charge on anion or cation FAZAN'S RULE No Bond is 100% Ionic. Cation: Polarising power Anion: Polarizability {Anion gets distorted} Pseuto Inert gas electronic configuration
- 14. Applications of Fazan'sRule: Polyatomic ions {CO3 2–; SO4 2–; HCO3 –, NO3 –} I. Thermal decomposition of Metal carbonates Li2CO3 𝛥 li2O + CO2; But Na2CO3; K2CO3; and rest can't Increasing thermal li2CO3 < Na2CO3 < K2CO3 < RbCO3 < Cs2CO3. stability order For metal Halides: {Small anions: F–; Cl–; O2–; OH–; S2–; N3– etc.} We see lattice energy rather than polarization concept LiF > LiCl > LiBr > LiI 1 Polarization Covalent character Thermal
- 15. Intensity of Colour: AgClAgBr AgI Ag2S White Pale yellow Dark yellow Black Cl– < Br– < I– Large size 𝜶 Polarization 𝜶 Intensity of anion Solubility of Ionic Compounds: {H.E. > L.E.} Soluble Case-I: {When rc ra} {For small anions like: O2–; S2–; N3–; F–; Cl–; OH– etc.} ➢ Lattice energy is the deciding factor: H.E. > L.E. Solubility () Top (As) () (L.E.) ( ) Bottom H.E. ()
- 16. Ex. liF <NaF < KF < RbF < CsF MgO < CaO < SrO < BaO increasing solubility NaCl > KCl > RbCl > CsCl > liCl lattice energy () Melting point () Case-II: {When rc << ra} {SO4 2–; CO3 2–; PO4 3–; NO3 –; I–; Br– } Top L.E.() L.E. > H.E. { Solubility decrease} H.E. ( ) Bottom
- 17. Ex.: BeCO3 >MgCO3 > CaCO3 > SrCO3 > BaCO3 BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4 Decreasing solubility
- 18. V.S. E. P.R. Theory: ➢ Valence shell electron pair repulsion. ➢ Covalent molecule → Geometry shape find → Valence shell ke electron Repulsion ko Minimum Rakhna hoga. ➢ Minimum Repulsion → Side atom maximum distance apart → Maximum bond angle → Molecule stable → Geometry L.P. – L.P > L.P. – B.P > B.P. – B.P
- 19. Case-I: Molecules withRegular Geometry: Molecular type Geometry and Bond angle Examples 1. AX2 2. AX3 3. AX4
- 20. Molecular type Geometryand Bond angle Examples 4. AX5 5. AX6 6. AX7
- 21. Case-II: Irregular Geometry: Moleculartype Expected geometry Actual shape Examples 1. AB2L 2. AB2L2 3. AB3L
- 22. Molecular type Expectedgeometry Actual shape Examples 4. AB4L 5. AB3L2 6. AB2L3
- 23. Molecular type Expectedgeometry Actual shape Examples 7. ABs2 8. AB4L2 9. AB6L
- 24. (i) It iskind of (Intermolecular) lewis acid-base interaction. (ii) Partial pi bond or coordinate pi bond. (iii) Types: (A) p – p Back bonding (B) p – d Back bonding (iv) Conditions: Donor atom Acceptor atoms 2nd period 2nd or 3rd period lone pair vacant orbital Back Bonding
- 26. 1. Valence bondtheory Valence bond theory Proposed by Heitler and London as per VBT bonding takes place for attaining stability. 1 Stability Potential energy ➢ Bond formation is an exothermic process. ➢ During this process some extent of electron cloud merge into each other, this part is known as overlapped region and this process is known as overlapping. ➢ Only those orbitals of valence shell can exhibit overlapping which has ➢ Unpaired electron For example: H-Cl bond form by overlapping of 1s-3p orbitals H → 1s1 Cl → 1s2 2s2 2p6 3s2 3p5 ➢ Opposite spin Strength of covalent bond Strength of covalent bond extent of overlapping. 1. Nature of orbitals (a) No. of shell: Lower the number of shell higher overlapping. Bond strength /size of orbitals 1-1 > 1-2 > 2-2 > 2-3 ➢ Exception: Cl2 > Br2 > F2 > I2 due to lip-lp O–O < S–S repulsion N–N < P–P 1 No. of shell
- 27. S-non-directional P-directional (B) Type ofsub-shell Valence shell contain subshell s and p Directional orbital has higher extent of overlapping. Possible combination and strength of overlapping s-s < s-p < p-p This factor is applicable when number of shell is same otherwise shell factor prominent 2s - 2s < 2s - 2p < 2p - 2p sub-shell factor 1s - 1s > 1s - 2s > 1s - 3s shell factor (a) Axial overlapping: Along the internuclear axis; form sigma () bond, strong bond. (b) Co-lateral overlapping: Side wise overlapping has less extent of overlapping form -bond ➢ In case of multiple bond between two atom one bond is sigma and rest are pi-bonds. ➢ VBT was not able to define geometry of molecule therefore a new concept came into existence known as hybridization. Hybridisation ➢ Intermixing of atomic orbitals and formation of new orbital these orbitals are known as hybrid orbital and this concept is known as hybridisation. ➢ It is hypothetical concept. ➢ Only those orbitals can participate in hybridisation which has slight difference in energy. ➢ No. of hybrid orbitals: No. of atomic orbitals participate in intermixing. ➢ Hybrid orbitals oriented at maximum possible distance three dimensionally ➢ On the basis of type of orbitals participating in hybridization, we can divide hybridization into following categorise.
- 29. Dipole Moment Measurement ofPolarity in a molecule μ = q × d debye = esu-cm 1D = 10-18 esu.cm (A) Identification of polar or Non-polar molecule. Molecule: Symmetrical distribution of electron cloud- Non-polar. Molecule: Unsymmetrical distribution of electron cloud- Polar. Diatomic Molecule: (a) Homoatomic ΔΕΝ = 0 → μ = 0 → Non-polar H2, F2, Cl2, N2 etc. (b) Heteroatomic EN ≠ 0 = 0 → μnet = 0 → polar HF > HCI > HBr > HI Polyatomic molecule: R → Vector sum of bond moment R → Important Order: NH3 > NF3 H2O > H2S CH3Cl > CH3F > CH3Br > CH3I CH3CI > CH2CI2 > CHCI3 > CCI4 μ1 2 + μ2 2 + 2μ1μ2cosθ Applications: Predict shape and polarity of molecule Symmetrical geometry → = 0 → non- polar Unsymmetrical geometry → µ ≠ 0 → polar Distinguish between cis & trans form Dipole moment in Aromatic Compounds Maleic acid 0 fumaric acid = 0 μ ∝ 1 bond angle
- 30. Strength Intermolecular H-bond >Intramolecular H-bond ➢ Intramolecular H-bonding takes place in ortho derivatives only Hydrogen Bonding Electrostatic force of attraction between hydrogen & highly electronegative atoms. It is dipole-dipole type of attraction Hydrogen should be covalently bonded with highly electronegative elements. Like : F, O & N. Strength of H-bond Electronegativity of electronegative elements should Type of Hydrogen Bond Intermolecular Intramolecular Between molecule Within molecule ➢ It is not an intermolecular force ➢H₂O is liquid while H2S is gas. ➢HF is liquid while HCl is gas. EXAMPLES Applications of H-bonding Physical State (densile nature) H-bond Melting Point (mp) H-bond Boiling Point (bp) H-bond Viscosity H-bond Surface Tension H-bond Volatility . 1/H-bond Vapour Pressure . 1/H-bond
- 31. Molecular Orbital Theory Givenby Hund & Mulliken given Given To explain: ➢O2: Paramagnetic nature. ➢Existence of species like H2 +, H2 – As per MOT bond form by combination of atomic orbitals & interference of electron wave interference of electron wave leads to formation of molecular orbitals. atomic orbitals electron waves - interference constructive interference destructive interference Constructive Interference same phase wave - bonding molecular orbital (BMO) Destructive Interference opposite phase wave - anti-bonding molecular orbital (ABMO) Therefore by the combination of two atomic orbitals there is formation of two molecular orbitals; BMO & ABMO. ➢ Energy level BMO < ABMO, ➢ All atomic orbitals of an atom participate in combination and form molecular orbitals with atomic orbitals of another atom. ➢ Energy level of molecular orbital 1s *1s 2s *2s 2px = 2py 2pz *2px *2py 2px Total electron <14 Internuclear axis = z ➢ Bonding electron (N2) → No. of electrons present in BMO. ➢ Anti-bonding electron (N2) → No. of electrons present in ABMO. ➢ Bond order: No. of Bonds present between two atoms. ➢ Paramagnetic nature: If any molecule orbital contain unpaired electron either bonding molecular orbital or anti-bonding molecular orbital it would be paramagnetic otherwise diamagnetic. ➢ Isoelectronic species: Same bond order - same magnetic properties. Significant of MOT B. O. = N2 − N2 2 Stability of molecule B.O. Bond strength B.O. Bond length 1 B. O.
- 32. Formation of molecularorbitals by linear combination of atomic orbitals Formation of bonding () and antibonding (*) molecular orbitals by the linear combination of atomic orbitals yA and yB centred on two atoms A and B respectively.
- 34. Electronic configuration ofMolecules with more than 14 e– : 2p 2p 2s 2s 1s 1s
- 35. Electronic configuration ofMolecules with less than or equal to 14 e– : 2p 2p 2s 2s 1s 1s
- 36. M.O.T for O2: 2p2p 2s 2s 1s 1s Increasing Energy
- 37. Types of Questions: 1.Bond Order 2. Stability of O2 molecule 3. Electronic Configuration: 4. Magnetic Character Paramagnetic Diamagnetic
- 38. 5. Magnetic Moment: 6.H.O.M.O : L.U.M.O :
- 39. For N2 : 2p2p 2s 2s 1s 1s
- 40. 1. Bond Order 2.Stability of O2 molecule 3. Electronic Configuration: 4. Magnetic Character 5. Magnetic Moment
- 41. N2 → N2 +NO → NO+ O2 → O2 2– O2 → O2 + In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic? QUESTION_2019 A C B D
- 42. Bonding Parameter 1.Bond length:Internuclear distance Factor affecting Bond length (i) Atomic size: bond length size [No. of shell] (ii) EN, Bond length 1 EN (iii) Bond order: Bond length 1 B.O (iv) Hybridisation : Bond length 1 %age of s−character dA−B = rA + rB − 0.09 × ΔENÅ Bond Angle : Angle between two adjacent bond is known as bond angle. bonded atom bond angle bond angle central atom Factors Affecting Bond Angle (i) Hybridisation: Bond angle %age of s-character (ii)No. of lp/bp: [when hybridisation is same] Bond angle 1 lp Eg. : CH4 > NH3 > H2O: (iii) Type of Central atom: Applicable when : ➢ hybridisation same ➢ No. of Ip/bp same. Bond angle EN of central atom Eg. NH3 > PH3 > AsH3 > SbH3 : :