molecular bonds: part ii

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Molecular Bonds: Part II

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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. - PowerPoint PPT Presentation

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Page 1: Molecular Bonds:  Part II

Molecular Bonds: Part II

Page 2: 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

Page 3: Molecular Bonds:  Part II

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

Page 4: Molecular Bonds:  Part II

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

Page 5: Molecular Bonds:  Part II

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

Page 6: Molecular Bonds:  Part II

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)

Page 7: Molecular Bonds:  Part II

• 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

Page 8: Molecular Bonds:  Part II

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

Page 9: Molecular Bonds:  Part II

• 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

Page 10: Molecular Bonds:  Part II

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.

Page 11: Molecular Bonds:  Part II

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.

Page 12: Molecular Bonds:  Part II

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.

Page 13: Molecular Bonds:  Part II

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.

Page 14: Molecular Bonds:  Part II

• 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.

Page 15: Molecular Bonds:  Part II

• 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.

Page 16: Molecular Bonds:  Part II

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.             

Page 17: Molecular Bonds:  Part II

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

Page 18: Molecular Bonds:  Part II

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

Page 19: Molecular Bonds:  Part II

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).

Page 20: Molecular Bonds:  Part II
Page 21: Molecular Bonds:  Part II

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

Page 22: Molecular Bonds:  Part II
Page 23: Molecular Bonds:  Part II

Lewis Diagrams of Molecules

• Steps1. Determine the type of molecule – ionic or

covalent

2. 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

Page 24: Molecular Bonds:  Part II

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 indicated10. Resonance structures means there can

be more than 1 correct diagram

Page 25: Molecular Bonds:  Part II

Example with CO2:

1. Σ ve = 4 for C and (2 x 6) for O = 16 ve

2. 16 ve divided by 2 = 8 BP

3. Carbon is CA due to lower EN and fewest number of atoms in molecule

4. Place BP around C until octet reached

5. Fill in remaining LP around O’s

Page 26: Molecular Bonds:  Part II

Let's try a negative polyatomic ion:  PO4-3 (phosphate)

1) Add up the total number of electrons:

2) Draw the skeleton structure:

Page 27: Molecular Bonds:  Part II

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.  

Page 28: Molecular Bonds:  Part II

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.

Page 29: Molecular Bonds:  Part II

5) Remember, P can have an expanded octet or hypervalence – so there are two possible structures…

Resonant Structures for PO4-3

Page 30: Molecular Bonds:  Part II

Comparison of Ionic and Covalent CompoundsIonic 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

Page 31: Molecular Bonds:  Part II

The End