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Molecular Bonds: Part II

Molecular Bonds: Part II. This part contains information on the following:. Polarity Van der Waals and London Dispersion Forces Intra/Intermolecular Forces (IMF) Setting Up Lewis Dot Diagrams for Molecules Determining the Structure of a Molecule Resonance.

adrienne-conway
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Molecular Bonds: Part II

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  1. Molecular Bonds: Part II

  2. This part contains information on the following: • Polarity • Van der Waals and London Dispersion Forces • Intra/Intermolecular Forces (IMF) • Setting Up Lewis Dot Diagrams for Molecules • Determining the Structure of a Molecule • Resonance

  3. Let’s tackle polarity first. . . • This is measuring the electrostatic charge (the negative or positive charge) of a molecule and ends side(s) are which

  4. Polarity: • Determining the + and – ends of a molecule • A molecule’s polarity is measured by its dipole moment (μ) • The dipole moment is measured in Debyes (D) which is a measure of charge

  5. Polarity’s SI unit is the Coulomb Meter (C m) - However, the Debye is used for molecules - D is in units of 10-10 statcoulombs - Distance is measured in Angstroms (A) - And 1 C m = 2.9979 x 1029 D

  6. Polarity results from uneven sharing of electrons and means a partial charge distribution in a molecule • The higher the EN – tendency to δ- • The lower the EN – tendency to δ+ • The separation of these two partial charges creates a dipole molecule (two poles)

  7. A non-polar covalent bonds means the electrons are shared equally between the two atoms • This is usually diatomic molecules of the same element (nonmetal) • EN Difference is 0 to 0.5 on Pauling Scale • A polar covalent bond means an unequal sharing of the electrons among the atoms • This means a partial ionic charge • Two or more different nonmetals make up these molecules • EN Difference is 0.6 to 1.6

  8. Polarity leads to Magnetism • If a molecule is diamagnetic – all of the e- are paired up and the substance is unaffected by magnetic fields • If the substance has unpaired e-’s, it can become magnetized while in a magnetic field – but stops when removed • This is paramagnetism and includes Al, Pt • These substances have permanent dipole moments

  9. Ferromagnetic materials remain magnetic when the original field is removed – like Fe and Ni • The e-’s line up parallel with each other in a region called a domain • Ferromagnetic materials remaining magnetized is called hysteresis • The magnetic domains will remain aligned until they become randomized again by thermal, electrical or some other force or agitation

  10. Predicting Molecular Polarity • When there are no polar bonds in a molecule, there is no permanent charge difference between one part of the molecule and another, and the molecule is nonpolar. • For example, the Cl2 molecule has no polar bonds because the electron charge is identical on both atoms. It is therefore a nonpolar molecule. None of the bonds in hydrocarbon molecules, such as hexane,C6H12, are significantly polar, so hydrocarbons are nonpolar molecular substances.

  11. A molecule can possess polar bonds and still be nonpolar. If the polar bonds are evenly (or symmetrically) distributed, the bond dipoles cancel and do not create a molecular dipole. For example, the three bonds in a molecule of BF3 are significantly polar, but they are symmetrically arranged around the central fluorine atom. No side of the molecule has more negative or positive charge than another side, and so the molecule is nonpolar.

  12. A water molecule is polar because: • (1) its O-H bonds are significantly polar, and • (2) its bent geometry makes the distribution of those polar bonds asymmetrical. The side of the water molecule containing the more electronegative oxygen atom is partially negative, and the side of the molecule containing the less electronegative hydrogen atoms is partially positive.

  13. Predicting Molecular Polarity • Tip-off – You are asked to predict whether a molecule is polar or nonpolar; or you are asked a question that cannot be answered unless you know whether a molecule is polar or nonpolar. (For example, you are asked to predict the type of attraction holding the particles together in a given liquid or solid.) • General Steps - • Step 1: Draw a reasonable Lewis structure for the substance. • Step 2: Identify each bond as either polar or nonpolar. (If the difference in electronegativity for the atoms in a bond is greater than 0.5, we consider the bond polar. If the difference in electronegativity is less than 0.5, the bond is essentially nonpolar.) • If there are no polar bonds, the molecule is nonpolar. • If the molecule has polar bonds, move on to Step 3.

  14. Step 3: If there is only one central atom, examine the electron groups around it. • If there are no lone pairs on the central atom, and if all the bonds to the central atom are the same, the molecule is nonpolar. (This shortcut is described more fully in the Example that follows.) • If the central atom has at least one polar bond and if the groups bonded to the central atom are not all identical, the molecule is probably polar. Move on to Step 4. • Step 4: Draw a geometric sketch of the molecule.

  15. Step 5: Determine the symmetry of the molecule using the following steps. • Describe the polar bonds with arrows pointing toward the more electronegative element. Use the length of the arrow to show the relative polarities of the different bonds. (A greater difference in electronegativity suggests a more polar bond, which is described with a longer arrow.) • Decide whether the arrangement of arrows is symmetrical or asymmetrical • If the arrangement is symmetrical and the arrows are of equal length, the molecule is nonpolar. • If the arrows are of different lengths, and if they do not balance each other, the molecule is polar. • If the arrangement is asymmetrical, the molecule is polar.

  16. For an example, the Lewis structure for CH2Cl2 is • The electronegativities of hydrogen, carbon, and chlorine are 2.20, 2.55, and 3.16. The 0.35 difference in electronegativity for the H-C bonds tells us that they are essentially nonpolar. The 0.61 difference in electronegativity for the C-Cl bonds shows that they are polar.

  17. Therefore, the molecular structure for this molecule would look like this:

  18. Molecular Polarity is the Σ of all bond polarities in the molecule • μ (dipole moment) is a vector with magnitude and direction Example: H C = O O = C = O H μ = 0.0 D μ = 2.2 D Arrow points to negative charge

  19. Van der Waals Forces • The attractive or repulsive force between molecules (or between parts of the same molecule) other than those due to covalent bonds or to the electrostatic interaction of ions with one another or with neutral molecules. This includes: • permanent dipole forces • induced dipole forces • instantaneous induced dipole-induced dipole (aka London dispersion force).

  20. Van der Waals Forces are also called Intermolecular Forces. . . . • These are attractive forces between NEUTRAL molecules • The main types: • London Dispersion Forces – exist between nonpolar molecules and are the result of the + charged nucleus of one atom attraction the e- of the other atom • Dipole-Dipole Forces exist between polar molecules where the + end of one molecule attracts the – end of another molecule • Hydrogen Bonds where the H is bonded to a high EN atom such as O, N, F • Induced Dipole Forces – due to the fluctuation of e- clouds around 2 atoms’ nuclei – a dipole force is created

  21. Lewis Diagrams of Molecules • Steps • Determine the type of molecule – ionic or covalent • Count the # of ve for entire formula unit *Remember to add 1 for every – charge; *Subtract 1 for every + charge 3. Divide the Σ ve by 2 to determine # BP 4. Select CA – it has the lowest EN; is usually the fewest atom; and can never be H

  22. 5. Do a skeleton equation by placing BP around the CA connecting it to TA (ligand or peripheral atoms) 6. Place remaining ve around CA until octet reached 7. Place any LP e- around TA(s) 8. Calculate FC FC = ve – (n + b/2) 9. If ionic – place entire structure in [ ] with the charge indicated 10. Resonance structures means there can be more than 1 correct diagram

  23. Example with CO2: • Σ ve = 4 for C and (2 x 6) for O = 16 ve • 16 ve divided by 2 = 8 BP • Carbon is CA due to lower EN and fewest number of atoms in molecule • Place BP around C until octet reached • Fill in remaining LP around O’s

  24. Let's try a negative polyatomic ion:  PO4-3 (phosphate) • Add up the total number of electrons: 2) Draw the skeleton structure:

  25. 3) Distribute the valence electrons around the atoms until each atom has a complete valence shell (eight valence electrons). (4) Since all 32 electrons have been used, we double check the central atom (P) to make sure that it has a complete octet.  It does, so now we calculate the formal charge on the Lewis Structure.  

  26. a) Divide the covalent bonds and distribute the electrons. b) Add up the electrons surrounding each atom. • Each O atom has 7 electrons.  The central atom (P) has 4 electrons. c) Compare the total number of electrons around each individual atom in the Lewis Structure to the number of valence electrons in each respective neutral atom. • A neutral O atom has 6 valence electrons, which is one less than the O atoms in the Lewis Structure.  Therefore, the formal charge on each of the O atoms is -1. • A neutral P atom has 5 valence electrons, which is one more than the N atom in the Lewis Structure.  Therefore, the formal charge on the N atom is +1. d) Find the net formal charge by adding up all the formal charges of each atom in the Lewis Structure. • The four O atoms each have a -1 FC totaling -4.  The P atom has a +1 FC.  The net FC is -3 which is correct because the charge on PO4-3 is -3.

  27. 5) Remember, P can have an expanded octet or hypervalence – so there are two possible structures… Resonant Structures for PO4-3

  28. Comparison of Ionic and Covalent Compounds Ionic Compounds Covalent Compounds -Form crystal lattice solids (salts) Form g, l and/or s -Exothermic Rxns Mostly Endothermic Rxns -High melt/boil points due to strong Low melt/boil points since atoms bonds (e- transfer) remain somewhat independent -High D0 Energy needed to break bonds Low Do needed to break bonds -Conductors of thermal/electrical energy Insulators – no free e- to leave due to Electron Sea -Properties result from electrostatic Properties result from IMF attraction among atoms -Hard, Brittle Soft, Malleable, Flammable -Soluble in H2O (like dissolves like) Nonsoluable in nonpolar solutions -Metal cation + Nonmetal anion Nonmetal + Nonmetal -Monatomic + Mon/Polyatomic Ions Monatomic + Mon/Polyatomic -Large EN Difference (1.7+) Lower to Zero EN Difference

  29. The End

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