Lecture 11 Chemical Bond: Force between nuclei Which is spontaneous? a) Na (g) + Cl (g) → Na⁺ (g) + Cl⁻ (g) b) O (g) + Cl (g) → O⁺ (g) + Cl⁻ (g) c) H (g) + O (g) → H⁺ (g) + O⁻ (g) d) Li (g) + F (g) → Li⁺ (g) + F⁻ (g) IE (Na) = 496 kJ/mol (required to break bond) EA(Cl) = -349 kJ/mol (formed from creating one new e⁻) ΔE = EA - IE = nonspontaneous no two neutral atoms spontaneously transfer e⁻s no separation of opposite charges must consider sharing e⁻s (even in ionic bonds) chemical bonds btw different types of atoms will form provided that E gained in bond formation > E required to break the elemental bonds intramolecular bonding (btw atoms in a molecule) → chemical properties intermolecular bonding (btw molecules) → physical properties H₂ (2 p and 1 e-) H atom and distant 2nd nucleus (MO) most stable: 2 H nuclei (p') with an e- between them e⁻ = 1 - cos x / 10⁻¹⁰ pc PE (attraction) = (2/r) k PE = 1/2r * (2e⁻) = (2e⁻)/r * k (MO) least stable: 2 H nuclei (p") greater distance with an e⁻ outside them (worse than no bond at all) e⁻ pushes apart PE = 1/2r * e⁻ - e⁻/r - e⁻/3r = 3/2r * e⁻ - k molecular orbital theory goal: find allowed states (orbitals) for the e-s that involve the entire molecule (both nuclei, not just one) bonding orbital: results from 2 atomic orbitals (AOs) added in-phase; increased e- (probability) density between the nuclei ex adding two 1s orbitals is an allowed e- state ao from atom A ao from atom B Sum Squared (probability) large probability e- will be btw nuclei e- spends a lot of time in btw nuclei molecular orbitals (MO) more stable than original AOs 2 e-s go in 1s, then σ* (other combination of 1s orbitals), which energetically more stable than 2 1s two atomic orbitals overlapping to form a bonding molecular orbital antibonding orbital: results from two AOs added out-of-phase; decreased e- (prob) density btw the nuclei contains a nodal plane perpendicular to the bonding axis antibonding orbital (*) e- prob; repulsion of nuclei destabilizing ya-yb N low prob of e-s btw nuclei e- prob on outside of nuclei two atomic orbitals overlapping to form an antibonding molecular orbital ↑ length, BDE weakens H₂ σ1s² bond order: ½ (σ bond) bond energy: 436 kJ/mol bond length: 74 pm He₂ σ1s² σ*1s² bond order: 0 (σ bond) - (σ* bond) / 2 = 0 bond energy: 0 kJ/mol bond length: 0 pm (He₂ doesn't exist) H₂⁺ σ1s¹ bond order: ½ bond energy: 269 kJ/mol bond length: 106 pm He₂⁺ σ1s² σ*1s¹ bond order: ½ (σ bond) - ½ (σ* bond) = ½ bond energy: 247 kJ/mol bond length: 108 pm antibonding orbitals are bad for the bond (destabilizing) but still a good place to put e-s b/c e-s still attracted to nuclei chemical bond: 2 e-s shared by 2 nuclei A bond is good: e-s tgt; too many is bad (pushes apart) from H₂ to H₂⁺, more "glue" bc 2x as many e-s but BDE not doubled bc of e- repulsion energies of MOs depend on amt of overlap of AOs. distance ↓ overlap ↑, E of σ ↓ (less antibonding)* as R (distance) approaches E of σ is pulled apart* orbitals as H-H is pulled apart less bonding ... distance of minimum E btw H nuclei σ orbitals are symmetrical about the bonding axis (z-axis) π orbitals have one nodal plane containing the bonding axis (z-axis) σ orbitals have 2 nodal planes combined.* end view