1 SYSTEMATIC METHOD FOR DETERMINING LEWIS DOT STRUCTURES Do the following steps in order. 1. Sum the number of valence electrons from all atoms in the molecule. 2. Add or subtract appropriate number of electrons, if molecule is charged.

(Polyatomic ion) 3. Identify the central atom and decide how other atoms are bonded to it. - least electronegative atom is usually central atom - hydrogen is never central atom 4. Draw bonds between atoms. (Subtract 2 e- for each bond from total number

of valence e-) 5. Complete octets of peripheral atoms. (Subtract 1 e- for each electron used

from total number of valence e-) 6. Put remaining electrons in pairs on central atom. (if possible) 7. If central atom has too few electrons to complete octet, change lone pairs on

peripheral atoms to bonds between central atom and peripheral atom. 8. Write Lewis structure as to minimize formal charges. (More about formal

charges in a bit.) - Most correct Lewis structure will have negative formal charge on most

electronegative atom. (Also, any positive charge must be on least electronegative atom.)

Examples Draw the Lewis structure for PCl3. 1. # of e- = 5 + 3(7) = 26 e- 3. Central atom is P. Draw the Lewis Structure of SO3. 1. # of e- = 6 + 3(6) = 24 e- 3. Central atom is S. 7. Central atom has too few electrons, use multiple bond.

2 Draw the Lewis Structure of NH4

+. 1. # of e- = 5 + 4(1) – 1 = 8 e- 2. Note: one electron is subtracted. 3. Central atom is N. Draw the Lewis Structure of SF4. 1. # of e- = 6 + 4(7) = 34 e- 3. Central atom is S. 6. Put remaining electrons on central atom. - more about exceptions to octet rule later

3 Lewis Structures With More Than One Central Atom. Recall the reminders given previously concerning polyatomic bonding. 1. Carbon will form 4 covalent bonds per atom. 2. Oxygen will form 2 covalent bonds per atom. 3. Nitrogen will form 3 covalent bonds per atom. 4. Hydrogen will form 1 covalent bond per atom. 5. Halogens will form 1 covalent bond per atom. Also keep in mind that hydrogen will never be a central atom and the halogens will be a central atom generally in oxyacid anions such BrO3

-. Most difficult detail of constructing Lewis structures is deciding how central atoms are connected. Often the chemical formula gives hints as how to make structure. Example: Draw the Lewis dot structure for methanol, CH3OH.

Note the chemical formula implies that three hydrogens are bonded to carbon and one hydrogen is bonded to oxygen.

⇒ Carbon and oxygen must be bonded to each other. # of e- = 4 + 6 + 4(1) = 14 e- Example: Draw the Lewis dot structure for formaldehyde, CH2O. Example: Draw the Lewis dot structure for propane, CH3CH2CH3.

4 Example: Draw the Lewis dot structure for acetamide, CH3C(O)NH2.

We look to the chemical formula for help in deciding how to arrange atoms. The O atom is put in parenthesis to clue us that the oxygen atom is above the chain of central atoms.

Thus central atom skeleton looks like

Putting in the rest of the hydrogens and the electrons yields the structure:

Note that one carbon is deficient of electrons. Multiple bond must be used.

5 Formal Charge Each atom within a Lewis structure can be assigned a charge based on the number

of valence electrons it normally has versus the number of valence electrons actually has.

Rules for Determining Formal Charge 1. Start with normal # of valence e- for atom. 2. Subtract number of nonbonding e- surrounding atom. 3. Subtract ½ number of bonding electrons surrounding atom. Example: Which atom is assigned the 1+ charge in the ammonium ion? Therefore, the N atom has positive charge in NH4

+ ion. Example: What are the formal charges of the atoms within a water molecule? Example: What are the formal charges of the atoms in the thiocyanate ion given the

Lewis structure below? Formal Charge of S: 6 – 4 – ½(2) = 0 Formal Charge of C: 4 – 0 – ½(8) = 0 Formal Charge of N: 5 – 4 – ½(4) = -1 The negative charge of the thiocyanate ion lies on the nitrogen atom. - Note: This is sensible, since the nitrogen atom is the most electronegative of

the three atoms.

N H

H

H

H

+

O

H

H

C

S N

6 Using Formal Charge to Determine Lewis Structures Recall rule 8 of our procedure to construct Lewis structures.

8. Write Lewis structure as to minimize formal charges. - Most correct Lewis structure will have negative formal charge on most

electronegative atom. (Also, any positive charge on least electronegative atom.)

Example: Which of the following Lewis structures is correct for carbon dioxide? Note both Lewis structures satisfy octet rule. But the correct one has

minimized formal charge. Using multiple bonds in Lewis structures also minimizes formal charge. Consider NO3

- Use double bond to satisfy octet rule as well as minimize formal charges.

C

O O

C

O O

C

O O

+1 0 -1 C

O O

0 0 0

correct Lewis structure

N O

O

O

0 -1 +1

-1

7 RESONANCE - Sometimes more than one correct Lewis structure can be drawn. - In that case, actual structure is a blend of correct structures. Consider the nitrate ion again - Double-headed arrows indicate resonance structures. Resonance structure implies the following: - Double bond is not confined to a single O – N pair. - Double bond is distributed over all three O – N pairs. Consider oxalate ion, C2O4

2- - resonance adds to the stability of a compound or polyatomic ion.

N O

O

O

NO

O

O

NO

O

O

C C

OO

O O

C C

OO

O O

C C

OO

O O

C C

OO

O O

8 BENZENE An important application of the idea of resonance is the molecule benzene, C6H6. To illustrate that the double bonds form an uninterrupted circle, often benzene is written as The resonance in benzene makes it an exceptionally stable molecule. It is extremely important in the chemistry of carbon (organic chemistry).

--C C

O O

O O

Double bond is distributed over both oxygen atoms.

C

CC

C

CC

H

H

H

H

H

HC

CC

C

CC

H

HH

HH

H

C

CC

C

CC

H

H H

H

H

H

9 EXCEPTIONS TO OCTET RULE Three usual exceptions to Octet Rule 1. Molecules with odd number of e- (rare) Example: ClO2 - note chlorine atom has 7 e-, not 8 e- 2. Molecules where atom has less than octet - common in beryllium and boron compounds Example: BF3 - Hmm! Why don’t we create double bond to satisfy octet rule. Look at

formal charge. 3. Molecules where atom has more than octet - relatively common - only atom in third row or below can “expand its octet”.

- valence shell is larger, more electrons can fit around the central atom

Recall previous example of SF4 - note sulfur has ten electrons - sulfur has “expanded its octet”

Cl

O

O

B

F

F

F

B F

F

F

+1 0

0

-1

S F

F

F

F

10 Example: I3

- BOND ENTHALPIES When two atoms bond together, the chemical energy of the system decreases. Consider an energy level diagram of before bonding and after bonding. The energy of the bonded system is lower than the unbonded system. The energy released when two unbonded atoms become bonded is called the

bond enthalpy. Aside: Enthalpy is another word for heat. We learn more about heat

and enthalpy in Chapter 6. The bond enthalpy increases as atoms are more strongly bonded together. As the strength of the bond increases, the distance between the atoms decrease. Bond enthalpies are an experimentally found quantity; i. e., we can’t predict bond enthalpies from periodic table. Bond Enthalpies and Chemical Changes **All chemical changes involve the breaking and creation of bonds.** If we can understand what bonds are breaking and what bonds are forming, then we can use bond enthalpies to estimate the energy (technically, enthalpy) change of the reaction.

To break a bond, we input (add) the bond enthalpy. When a bond is formed, the bond enthalpy is released (subtracted).

H H

H – H

before after

11 Example: Given the table of bond enthalpies below, calculate the energy change,

when two molecules of hydrogen and one molecule of oxygen change into two molecules of water.

TABLE OF BOND ENTHALPIES

Bond E (kJ/mol) Bond E (kJ/mol) C – H 413 H – H 436 C – C 348 N – N 163 C – O 358 N = N 418 C = C 614 N ≡ N 941 C ≡ C 839 O – H 463 C = O 1072 O = O 495

Breaking two H – H bonds means inputting 2 x 436 kJ/mol. Breaking one O = O bond means inputting 495 kJ/mol. Forming four O – H bonds means releasing 4 x 463 kJ/mol. Overall the energy change is

2 (436 kJ/mol) + 495 kJ/mol – 4 (463 kJ/mol) = – 485 kJ/mol SHAPES OF MOLECULES (VSEPR MODEL) Valence Shell Electron-Pair Repulsion model - Electron domains surrounding atom spread out as to minimize repulsion. - Electron domains can be bonding pairs (including multiple bonds) or

nonbonding pairs. - Arrangement of all the atoms surrounding central atom depends on electron

domains surrounding central atom. Two similar, but different geometries 1. Electron domain geometry - arrangement of e- domain around central atom - remember: multiple bonds count as a single e- domain

2. Molecular geometry - arrangement of atoms around central atom **A molecular geometry is decided only after an electron domain geometry has been

determined.** *- need to write Lewis structure to determine number of electron domains.*

H – H

H – H + O O

––

O

H

H

O

H

H

→

12 Geometries with two e- domains about central atom. 1. electron domain geometry – linear - angle between e- domains is 180 ° 2. possible molecular geometries a) Linear - only linear geometry is possible with two electron domains Example: BeCl2 Example: CO2 - Note: Only two electron domains around central atom since multiple bonds

count as a single domain. Example: CO Geometries with three e- domains about central atom. 1. electron domain geometry – trigonal planar

- angle between e- domains is 120 ° 2. possible molecular geometries a) Trigonal Planar - all three electron domains are bonding pairs Examples: BF3 and NO3

-

A A – generic atom

Be Cl

Cl

C O O

C O

A

N

O

O

O

B

F

F

F

A – generic atom

13 b) Bent (V-shaped) - two bonding pairs and one nonbonding pair Example: dichlorocarbene *Nonbonding e- pairs take up more room than bonding pairs. Therefore bond angle

between chlorine atoms is slightly less than 120°.* Geometries with four e- domains about central atom. 1. electron domain geometry – tetrahedral - tetrahedron is three dimensional object - angle between electron domains is 109.4° 2. possible molecular geometries a) Tetrahedral - all four electron domains are bonding pairs Example: CH4 Example: PO4

3-

C Cl

Cl

A

H

C H

H

H

CHH

H H

Caution! Representation is 2-D, not 3-D, bond angle between e- pairs is not 90°

2-D picture 3-D picture

P O

O

O

O

PO O

O O

14 b) Trigonal Pyramidal - 3 bonding pairs and 1 nonbonding pair Example: NH3 - Bond angle is 107°. (specifically for NH3) - Bond angle is less than 109.4° because nonbonding pair takes more room

than bonding pair.

Example: ClO3-

c) Bent (V-shaped) - two bonding pairs and two nonbonding pairs Example: H2O - Bond angle is 104.5°. Redraw H2O to show tetrahedral angle. Example: SF2

N H

H

H N

H HH

O

H

H

Cl O

O

O

O

H H

ClO OO

F

F

S

15 Geometries with five e- domains about central atom. 1. electron domain geometry – trigonal bipyramidal - two different positions in a trigonal bipyramid - axial – two position along axis - equatorial – three positions along equator - angle between equatorial positions is 120° - axial positions are 90° from equator * - nonbonding pairs prefer equatorial position*

2. possible molecular geometries a. Trigonal Bipyramidal - all five electron domains are bonding pairs Example: PF5

Note: axial bond lengths usually longer than equatorial bond lengths b. Seesaw - four electron domains are bonding pairs and one nonbonding pair Example: SF4 - remember lone pairs prefer equatorial position

P F F

F F

F

S F

F

F

F

P

F

F

FFF

S

F

F

FF

16 c. T-shaped - three electron domains are bonding pairs and two are nonbonding pairs Example: ClF3 - both lone pairs occupy an equatorial position d. Linear - two electron domains are bonding pairs and three are nonbonding pairs Example: I3

- (triiodide ion) - all three lone pairs occupy equatorial positions Geometries with six e- domains about central atom. 1. electron domain geometry – octahedral - angle between electron domains is 90° 2. possible molecular geometries a.) Octahedral - all six electrons domains are bonding pairs Example: SF6 b.) Square pyramidal - five electrons domains are bonding pairs with one lone pair Example: BrF5 - note angles will be slightly less than 90°

Cl F

F

F

II

I

Cl

F

F

F

S

F

FF F

F F

F

BrF F

FF

17 c.) Square planar - four electrons domains are bonding pairs with two lone pairs Example: XeF4 - What are angles between fluorine atoms? POLAR MOLECULES A polar molecule has one side slightly positive and the other slightly negative. Two conditions must be met in a polar molecule. 1.) Polar covalent bonds 2.) Correct geometry To emphasize necessity of correct geometry, compare two examples. BeCl2 H2O χBe = 1.5 χH = 2.1 χCl = 3.0 χO = 3.5 |χBe – χCl| = 1.5 |χH – χO| = 1.4 δ+ δ- δ+ δ- Be – Cl H – O Be – Cl bond is more polar than H – O bond; however, BeCl2 is nonpolar molecule and H2O is a polar molecule. Question: How can this be? Answer: BeCl2 has a linear geometry and H2O has a bent geometry.

XeF F

FF

indicates positive end of bond

Be Cl

Cl

total polarity adds to zero

total polarity is nonzero

O

H H

δ+

δ-

18 Example: Is either ammonia or methane a polar molecule? Answer: Ammonia is a polar molecule, but methane is a nonpolar molecule. Nonequivalent polar bonds can affect overall polarity. CF4 is a nonpolar molecule. CF3Cl is a polar molecule. Dipole Moment - µ When equal and opposite charges, ±Q, are separated by a distance, d; the dipole

moment is defined as µ = Q × d

Since a polar molecule has a separation of charge, it has a dipole moment. Polarity of molecule is usually considered via its dipole moment.

NH HH

δ+

δ-

CHH

H H

C

Cl

F FFC

F

F FF