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Chemical Bonding
- 1. CHEMICAL BONDING B y D r . B I N U S R E E J A Y A N A S S I S T A N T P R O F E S S O R D E P A R T M E N T O F C H E M I S T R Y K S M D B C O L L E G E , S A S T H A M C O T T A , K O L L A M
- 2. SCHRODINGER EQUATION • The fundamental equation of Quantum Mechanics. 𝐻𝜓=𝐸𝜓 • 𝐻 - 𝑡ℎ𝑒 𝐻𝑎𝑚𝑖𝑙𝑡𝑜𝑛𝑖𝑎𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 ( 𝑎𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟), • 𝜓 - 𝑤𝑎𝑣𝑒− 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑟 𝑒𝑖𝑔𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑟 𝑠𝑡𝑎𝑡𝑒 function • E - Total energy of the system 𝑯• mathematical command • The total energy operator or quantum mechanical operator of energy. • Sum of kinetic energy operator ( 𝑻) and potential energy operator ( 𝑽). 𝐻𝜓=( 𝑻 + 𝑽 )=𝐸𝜓 𝑻 ≠0; 𝑽 may or may not be zero Cap indicates An operator
- 3. Schrödinger equation for a single particle of mass m with zero potential energy • Moving in one dimension (x direction). • 𝐻𝜓 = ( 𝑻 + 𝑽 )𝜓 =𝐸𝜓 • Given, 𝑽 =0. • 𝐻𝜓 = 𝑻x 𝜓 =𝐸𝜓 −ℎ2 8𝜋2 𝑚 . 𝜕2 𝜓 𝜕𝑥2 =𝐸𝜓 𝑻x= −ℎ2 8𝜋2 𝑚 . 𝜕2 𝜕𝑥2 • Moving in two dimension (xy plane) • 𝐻𝜓 = ( 𝑻 + 𝑽 )𝜓 =𝐸𝜓 • Given, 𝑽 =0. • 𝐻𝜓 = ( 𝑻x + 𝑻y )𝜓 =𝐸𝜓 • −ℎ2 8𝜋2 𝑚 ( 𝜕2 𝜕𝑥2 + 𝜕2 𝜕𝑦2)𝜓 =𝐸𝜓 • Moving in three dimension (xyz space). • 𝐻𝜓 = ( 𝑻 + 𝑽 )𝜓 =𝐸𝜓 • Given, 𝑽 =0. • 𝐻𝜓 = ( 𝑻x + 𝑻y + 𝑻z )𝜓 =𝐸𝜓 −ℎ2 8𝜋2 𝑚 ( 𝜕2 𝜕𝑥2 + 𝜕2 𝜕𝑦2 + 𝜕2 𝜕𝑧2)𝜓 =𝐸𝜓
- 4. Writing Schrodinger equation for chemical system • For a given chemical system with “n” nuclei and “e” electrons, the total number of terms (kinetic energy and potential energy terms) 𝑁𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝐻 𝒕𝒆𝒓𝒎𝒔= (𝒏+𝒆)(𝒏+𝒆+𝟏) 2 (𝑛 + 𝑒)(𝑛 + 𝑒 + 1) 2 Kinetic energy terms Potential energy terms 𝑛 + 𝑒 Electron-nuclear attraction 𝑛𝑒 Nuclear- Nuclear repulsion 𝑛(𝑛 − 1) 2 Electron electron repulsion 𝑒(𝑒 − 1) 2
- 5. EXAMPLES…. System n e Total number of terms in Hamiltonian H 1 1 3 He+ 1 1 3 He 1 2 6 He2 2 4 1 C 1 6 28 CH4 5 10 120
- 6. Set Up SWE • 𝐻, 𝑡h 𝑒𝐻𝑎𝑚𝑖𝑙𝑡𝑜𝑛𝑖𝑎𝑛𝑜𝑝𝑒𝑟𝑎𝑡or: The total energy operator or quantum mechanical operator of energy. • Sum of kinetic energy operator ( 𝑻) and potential energy operator ( 𝑽). 𝐻𝜓=( 𝑻 + 𝑽 ) 𝜓 =𝐸𝜓 𝑻 = −ℏ 𝟐 𝟐𝒎 𝛁 𝟐 𝑽= attraction between e-n + repulsion between e-e + repulsion between n-n 𝑽 = 𝒒 𝟏 𝒒 𝟐 𝒓 𝐻= −ℏ 𝟐 𝟐𝒎 𝛁 𝟐+ 𝒒 𝟏 𝒒 𝟐 𝒓 r 𝒒 𝟏 𝒒 𝟐
- 7. 𝐻= −ℏ 𝟐 𝟐𝒎 𝛁 𝟐 − 𝒁𝒆 𝟐 𝒓 For one e system, 𝐻= −ℏ 𝟐 𝟐𝒎 𝛁 𝟐 − 𝒆 𝟐 𝒓 In au, 𝐻= −𝟏 𝟐 𝛁 𝟐 − 𝟏 𝒓 For two e system, 𝐻= −ℏ 𝟐 𝟐𝒎 𝛁 𝑨 𝟐 + −ℏ 𝟐 𝟐𝒎 𝛁 𝑩 𝟐 − 𝟐𝒆 𝟏 𝟐 𝒓 𝟏 − 𝟐𝒆 𝟐 𝟐 𝒓 𝟐 + 𝒆 𝟐 𝒓 𝟏𝟐 In au, 𝐻= −𝟏 𝟐 𝛁 𝑨 𝟐 + −𝟏 𝟐 𝛁 𝑩 𝟐 − 𝟐 𝒓 𝟏 − 𝟐 𝒓 𝟐 + 𝟏 𝒓 𝟏𝟐 −𝒆 𝟏 −𝒆 𝟐 𝒓 𝟏𝟐 +𝟐𝒆+Z e e 𝒓 𝟏 𝒓 𝟐 For n electron system, 𝐻𝜓= [ −ℏ 𝟐 𝟐𝒎 𝒊=𝟏 𝒏 𝛁𝒊 𝟐 − 𝒊=𝟏 𝒏 𝒁𝒆 𝟐 𝒓 𝒊 + 𝟏 𝟐 𝒊=𝟏 𝒏 𝒆 𝟐 𝒓 𝒊𝒋 ]𝜓 = E𝜓
- 8. APPROXIMATION METHODS Why approximation methods are needed??? Complete solution of Schrodinger equation is possible only for a few systems…. 1. Electron moving under a constant potential field. 2. Rigid Rotor 3. Harmonic Oscillator 4. Hydrogen atom ----- Real sytem SWE exactly soluble True 𝛹 & E All properties
- 9. E 𝜓 = −ℏ 𝟐 𝟐𝒎 𝛁𝟏 𝟐 + −ℏ 𝟐 𝟐𝒎 𝛁𝟐 𝟐 − 𝟐𝒆 𝟏 𝟐 𝒓 𝟏 − 𝟐𝒆 𝟐 𝟐 𝒓 𝟐 + 𝒆 𝟐 𝒓 𝟏𝟐 = ( −ℏ 𝟐 𝟐𝒎 𝛁𝟏 𝟐 − 𝟐𝒆 𝟏 𝟐 𝒓 𝟏 ) +( −ℏ 𝟐 𝟐𝒎 𝛁𝟐 𝟐 − 𝟐𝒆 𝟐 𝟐 𝒓 𝟐 ) + 𝒆 𝟐 𝒓 𝟏𝟐 OR 𝐻12 = 𝐻1 + 𝐻2 + 𝒆 𝟐 𝒓 𝟏𝟐 VARIATION Method PERTURBATION Method For He…..
- 10. Approximation methods…. • Variational Approximations From The (Rayleigh-ritz) Variational Principle • Time-independent Perturbation Theory For Schr¨odinger Eigenvalue Problem • Self Consistent Field [SCF] Method… • Time-dependent Perturbation Theory And Fermi’s Golden Rule • Semiclassical Approximation And The WKB Method. • Bohr-sommerfeld Approximation For Excited States • Partial Wave Approximation In Potential Scattering……..
- 11. VARIATION METHOD… • Applicable to system whose 𝜓 can be guessed. • Guessed 𝜓 ---- Trial 𝜓 • Adjustable parameters ----- Variation parameters • Trial 𝜓 gives Ea Based on Eckart’s Theorem / Variation Theorem 𝑬 𝒂 ≥ 𝑬 𝟎 𝐻 𝜓= E 𝜓 𝐸 = 𝜓∗ 𝐻 𝜓𝑑𝜏 𝜓𝜓∗ 𝑑𝜏 𝐸 = < 𝜓∗/H/𝜓 > < 𝜓/𝜓∗ > 𝐸0 = 𝜓0 ∗ 𝐻 𝜓0 𝑑𝜏 𝜓0 𝜓0 ∗ 𝑑𝜏 𝐸 𝑎 = 𝜓∗ 𝐻 𝜓𝑑𝜏 𝜓𝜓∗ 𝑑𝜏
- 12. Variation Theorem – Definition The energy evaluated by trial wavefunction should be always greater than or equal to true energy of the system in its ground state. OR The variation theorem says that no approximate wave function can have lower energy than the exact ground state energy of the system OR If the ground state function 𝜓0 𝑓𝑜r a system is approximated by a trial wave function 𝜓, has the same boundary conditions of the problem, then the calculated energy (Ea) must be greater than or equal to the actual ground state energy. 𝐸0 = 𝜓0 ∗ 𝐻 𝜓0 𝑑𝜏 𝜓0 𝜓0 ∗ 𝑑𝜏 𝐸 𝑎 = 𝜓∗ 𝐻 𝜓𝑑𝜏 𝜓𝜓∗ 𝑑𝜏 𝑬 𝒂 ≥ 𝑬 𝟎 𝑖𝑓, 𝜓0 ≠ 𝜓
- 13. PROOF The trial wave function satisfies the boundary conditions as the exact wave function Linear combination of orthonormal set of independent function. 𝜓 = 𝑖=0 ∞ 𝑐𝑖 𝜓𝑖 Consider Trial wave function (not normalised) 𝜓 = 𝑐1 𝜓1+ 𝑐2 𝜓2 𝐸 𝑎 = 𝜓∗ 𝐻 𝜓𝑑𝜏 𝜓𝜓∗ 𝑑𝜏 𝐸 𝑎 = (𝑐1 𝜓1+ 𝑐2 𝜓2) 𝐻 (𝑐1 𝜓1+ 𝑐2 𝜓2)𝑑𝜏 (𝑐1 𝜓1+ 𝑐2 𝜓2)(𝑐1 𝜓1+ 𝑐2 𝜓2)𝑑𝜏 𝐸 𝑎 = 𝑐1 2(𝜓1 𝐻 𝜓1) 𝑑 𝜏 + 𝑐1 𝑐2( 𝜓1 𝐻 𝜓2)𝑑𝜏+ 𝑐1 𝑐2( 𝜓2 𝐻 𝜓1)dτ + 𝑐2 2(𝜓2 𝐻 𝜓2) 𝑑 𝜏 (𝑐1 𝜓1+ 𝑐2 𝜓2)(𝑐1 𝜓1+ 𝑐2 𝜓2)𝑑𝜏
- 14. 𝐸 𝑎 = 𝑐1 2(𝜓1 𝐸1 𝜓1) 𝑑 𝜏 + 𝑐2 2(𝜓2 𝐸2 𝜓2) 𝑑 𝜏 (𝑐1 𝜓1+ 𝑐2 𝜓2)2 𝐸 𝑎 = 𝑐1 2 𝐸1+𝑐2 2 𝐸2 𝑐1 2+𝑐2 2 𝐸 𝑎- 𝐸0 = 𝑐1 2(𝐸1−𝐸0)+𝑐2 2(𝐸2−𝐸0) 𝑐1 2+𝑐2 2 RHS is +ve or 0, since 𝑐1 2 and 𝑐2 2 is always positive. So 𝐸 𝑎- 𝐸0 ≥ 0 𝑬 𝒂 ≥ 𝑬 𝟎. Hence variation theorem is verified. Since 𝜓1 and 𝜓2 are normalised and orthogonal wave function of 𝐻 … 𝐻 𝜓1 = 𝐸1 𝜓1 𝐻 𝜓2 = 𝐸2 𝜓2 𝜓1 𝐻𝜓1 𝑑𝜏 = 𝜓1 𝐸1 𝜓1 𝑑𝜏 = 𝐸1 𝜓1 𝜓1 𝑑𝜏 = 𝐸1 𝜓2 𝐻𝜓2 𝑑𝜏 = 𝜓2 𝐸2 𝜓2 𝑑𝜏 = 𝐸2 𝜓2 𝜓2 𝑑𝜏 = 𝐸2 𝜓1 𝐻𝜓2 𝑑𝜏 = 𝜓1 𝐸2 𝜓2 𝑑𝜏 =𝐸2 𝜓1 𝜓2 𝑑𝜏=0 𝜓2 𝐻𝜓1 𝑑𝜏 = 𝜓2 𝐸1 𝜓1 𝑑𝜏 = 𝐸1 𝜓2 𝜓1 𝑑𝜏 = 0
- 15. THANK YOU….