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Bond types

Bond types. Comparing ionic and covalent compounds. Metallic bonds. Metals’ valence electrons can move about between metal ions. These nonlocalized electrons give metals their conductivity, malleability and strength. weaker. stronger. Intermolecular forces (van der Waals forces).

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Bond types

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  1. Bond types

  2. Comparing ionic and covalent compounds

  3. Metallic bonds • Metals’ valence electrons can move about between metal ions. • These nonlocalized electrons give metals their conductivity, malleability and strength

  4. weaker stronger Intermolecular forces(van der Waals forces) • London dispersion-instantaneous dipole moment -increases with mass -found between all molecules • Dipole-dipole- increases with polarity • Hydrogen bonding The stronger the intermolecular forces, the greater the melting and boiling points and viscosity will be. Viscosity- resistance to flow.

  5. Lewis Dot Diagrams H2 HhHhbond to form H:H Each bond uses an electron from each atom. To create a Lewis dot diagram. . . • Count the total valence electrons available. • Ensure each atom has an octet (or a pair for H)

  6. Exceptions to the Octet Rule • Less than 8: B & Be • Expanded valence: elements in period 3 or higher bonding with highly electronegative elements like F, O, and Cl. • Odd no. of valence electrons

  7. Valence Shell Electron Pair Repulsion theory- electron pairs repel each other, so are oriented as far apart as possible. > > Unshared-unshared repulsion Unshared-shared repulsion Shared-shared repulsion

  8. Total no. of electron pairs No. of shared pairs No. of unshared pairs Molecular shape Molecule

  9. Total no. of electron pairs No. of shared pairs No. of unshared pairs Molecule Molecular shape

  10. Bent 104.5° Trigonal pyramidal 107.3°

  11. Polarity Molecule is polar if. . . there is a polar bond AND it is asymmetrical H+ O - +H C H+ H H + H+ polar nonpolar

  12. If a central atom is symmetrically surrounded by identical atoms, it will be nonpolar. • Ex: linear AB2 trigonal planar AB3 tetrahedral AB4 square planar AB4 trigonal bipyramidal AB5 octahedral AB6

  13. Hybridization • The mixing of 2 or more orbitals of similar energies on the same atom to produce new orbitals of equal energies. • C EEhh_ 1s2s 2p Ehhhh 1s sp3 tetrahedral 3 p orbitals were used

  14. How many bonds and unshared pairs are around the central atom? This is the number of equal energy orbitals needed. • sp has 2 electron domains • sp2 has 3 electron domains • sp3 has 4 electron domains • sp3d has 5 electron domains • sp3d2 has 6 electron domains

  15. Sigma bonds are along the bond axis.Pi bonds are sideways with parallel overlap.

  16. Single bond: • 1 sigma bond • Double bond: • 1 sigma, 1 pi bond • Triple bond: • 1 sigma, 2 pi bonds

  17. Coordinate covalent bonds • The shared electrons are supplied by a single atom H H + H N H + H+ H N H H

  18. Heat of vaporization Heat of fusion Heating Curve for Water

  19. Calculate the enthalpy change for each stage, then total them. To get the ice to 0◦C: ∆H= (10.0g)(2.09J/g·K)(25◦K)=515J To melt the ice: ∆H= (10.0g)(334 J/g)=3340J To heat the water to 100◦C: ∆H= (10.0g)(4.18J/g·K)(100◦K)=4180J To vaporize the water: ∆H= (10.0g)(2260 J/g)=22600J To heat the water to 125◦C: ∆H= (10.0g)(2.02 J/g·K)(25◦K)=505J

  20. Phase Diagram Critical point- beyond this, the gas and liquid phases become indistinguishable. Triple point- all three phases are at equilibrium

  21. The effect of increasing temperature and Activation Energy As the temperature increases, the peak for the most probable KE is reduced, and more significantly, moves to the right to higher values so more particles have the highest KE values. At the higher temperature T2, a greater fraction of particles has the minimum KE to react.

  22. Enthalpy of solution- ∆Hsol The energy change during dissolving. For solids in liquids, usually endothermic (>0) For gases in liquids, usually exothermic (<0)

  23. Henry’s Law • The amount of a gas that will dissolve in a liquid at a given temperature varies directly with the partial pressure of that gas.

  24. Concentrations • Molarity (M) moles solute L solution • Molality (m) moles solute kg solvent • % by mass g solute g solution • Mole fraction moles of solute moles of solution

  25. Colligative properties • Properties determined by the number of particles in solution (rather than the type) • Vapor pressure decreases • Boiling point increases • Freezing point decreases • Osmotic pressure

  26. vapor pressure • Decreases (with a nonvolatile solute) • Psolution = xsolvent ∙P solvent Vapor pressure = (mole fraction) (vapor pressure) solution solvent solvent Raoult’s Law: vapor pressure of a solution varies directly as the mole fraction of the solvent.

  27. Determining freezing/boiling point changes ∆Tfp = m∙Kfp molality x freezing point constant ∆Tbp = m∙Kbp molality x boiling point constant Kfp for water is 1.853 C°/1m water Kbp for water is .515 C°/1m water For other substances, see table A-8 on p.860.

  28. Determining molecular mass If 8.02g of solute in 861g water lowers the freezing point to -0.430°C, calculate the molecular mass of the solute. ∆Tfp = m∙Kfp so m = ∆Tfp = -0.430°C = .232m Kfp 1.853°C/m 8.02g solutex1 kg water= 40.1 g/mol .861kg water .232 mol solute

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