5.111 L1213 Lewis Diagrams PDF

Preview

Summary

Prof Keith Nelson...

Description

5.111 Lecture Summary #12-13

October 5 & 10, 2018

LEWIS DIAGRAMS 20 years before quantum mechanics - G. N. Lewis - key to bonding was sharing of 2 e–s, one from each atom. Lewis structures share the total number of valence electrons between atoms so that each atom achieves a noble gas configuration. Valence electrons for second row elements: 2s & 2p Li 2s1 N 2s2 2p3

Be 2s2 O 2s2 2p4

B 2s2 2p1 F 2s2 2p5

C 2s2 2p2 Ne 2s2 2p6

OCTET RULE: e– distributed in such a way that each main group element surrounded by 8 e– or octet. ••

F ••

••

+

•

••

•

••

F ••

••

Cl ••

••

+

•

H

•

In case of H, special stability achieved with 2e–:

Works 90% of time! Either following procedure or solve Schrödinger equation to tell where e–s are in a molecule. All the valence electrons are counted as bonding electrons (shared) or lone pairs. PROCEDURE FOR DRAWING LEWIS STRUCTURE 1. Draw a skeleton structure. H and F are always terminal atoms. Element with lowest ionization energy goes in the middle. 2. Count the total number of valence electrons. If negative ion, add the absolute value of total charge to count of valence electrons; if positive ion, subtract. 3. Count the total number of electrons needed if each atom had its own noble gas shell. 4. Subtract number in step 2 (valence electrons) from number in step 3 (total electrons for full shells). Result is number of bonding electrons. 5. Assign 2 bonding electrons to each bond. 12-13.1

5.111 Lecture Summary #12-13

October 5 & 10, 2018

6. If bonding electrons remain, make some double or triple bonds. Double bonds form only between C, N, O, S for light atoms. Triple bonds usually restricted to C, N, O. 7. If valence electrons remain, assign as lone pairs, giving octets to all atoms except hydrogen. 8. Determine formal charge.

METHANE 1) draw skeleton structure

H H

C

H

C is an example of a CENTRAL ATOM

H 2)

total # of valence e-s

3)

total # of e-s if each atom had noble gas shell

4)

subtract # in step 2 from # in step 3 — result is number of bonding e-s.

5)

assign 2 bonding e-s to each bond.

6)

any remaining bonding e-s?

7)

any remaining valence e-s?

12-13.2

5.111 Lecture Summary #12-13

October 5 & 10, 2018

Carbon Monoxide 1)

C

2)

total # of valence e-s

3)

total # of e-s if each atom had noble gas shell

4)

# of bonding e-s

5)

assign 2 bonding e-s to each bond

6)

Any remaining bonding e-s? Yes! 2 pairs of e-s form multiple bonds. Double bonds form only between C, N, O, S. Triple bonds usually restricted to C, N, O. Any remaining valence e-s? Yes! 2 pairs of e-s. Assign as lone pairs giving octets to all atoms except H.

7)

O

Hydrogen Cyanide 1)

what’s in the middle?

2)

total # of valence e-s

3)

total # of e-s if each atom had noble gas shell.

4)

# of bonding e-s

5)

assign 2 bonding e-s

6)

remaining bonding e-s?

7)

remaining valence e-s?

12-13.3

5.111 Lecture Summary #12-13 Cyanide Ion CN–

October 5 & 10, 2018

1)

[ C

N ]─

2)

Total # of valence e-s plus the absolute value of total charge if negative ion

3)

Total # of e-s if each atom had noble gas shell

4)

# of bonding e-s

5)

assign bonding e-s

6)

multiple bonds

7)

lone pairs

Thionyl Chloride SOCl2 SOCl2 is a reagent used in organic and medicinal chemistry to convert carboxylic acid (COOH) groups to acid chloride (COCl) groups in molecules. One example of using SOCl2 in the synthesis of pharmaceuticals is for novocaine, a local anesthetic drug used in dentistry and to reduce pain in intramuscular injections of other drugs, such as antibiotics. O H 2N

O

SOCl2

C

H2N OH

C

O H2N

C

Cl

O

novocaine

Cl S

1)

Cl

O

2)

total # valence e-s

3)

total # of e-s if each atom had noble gas shell

4)

# of bonding e-s

5)

assign bonding e-s 12-13.4

C2H5 N C2H5

5.111 Lecture Summary #12-13 6)

multiple bonds

7)

lone pairs

October 5 & 10, 2018

FORMAL CHARGE Formal charge is a measure of the extent to which an atom has gained or lost an electron in the process of forming an covalent bond. To assign a formal charge to an atom, we have to decide how many e-s an atom has in a molecule and compare it to how many e-s it has when it is a free atom.

To assign a formal charge: 1) number of valence electrons on a free atom, V 2) 3)

subtract from it the # of e-s that are lone pairs, L V–L subtract from that 1/2 of # of e-s that it shares in molecule V – L – S/2 = FC

For an electrically neutral molecule, the sum of the formal charges on the individual atoms must be zero. For a molecule with a net charge, the sum of the formal charges equals the net charge. Example: CN–

C

N

-1

C

N

-1

FC on C = V – L - (½)S = _____ - _____ - (½) _____ = _____ FC on N = V – L - (½)S = _____ - _____ - (½) _____ = _____

Significance: Measures extent to which atoms have gained or lost an e– in the process of forming a covalent bond.

FORMAL CHARGE ≠ OXIDATION NUMBER 12-13.5

5.111 Lecture Summary #12-13

October 5 & 10, 2018

Structures with lower absolute values of formal charge are the more stable structures. Formal charge is useful in deciding between different possible Lewis structures. The Lewis structure with lowest set of absolute values of FC is likely to be lowest energy.

Example: NCS– thiocyanate ion

•• •• • N •• C • •S • • •• •• •

-1

IEC = 1090

IES = 1000 IEN = 1400 kJ/mole

3 possible structures 1

–1

+

0

+

0

••N• •• C • ••S•• • •• •• •

= –1 OK

-1

FCN = 5 – 4 – 2 = –1 FCC = 4 – 0 – 4 = 0 FCS = 6 – 4 – 2 = 0

2

••C• •• S • ••N•• • •• •• •

-1

FCC = FCS = FCN =

3

••S• •• N • ••C•• • •• •• •

-1

FCC = FCN = FCS = 12-13.6

5.111 Lecture Summary #12-13

October 5 & 10, 2018

SKELETAL STRUCTURE Formal charge useful for deciding between structures CH3 usually represents a methyl group. A methyl group is always terminal.

H

-1

• • •• • •• • • H ••C NO • • • • • • • ••

CH3NHO-

For “chain” molecules, atoms usually written in order. Atoms that are attached usually follow the atom to which they are bonded.

H• -1 • • — • • •• • H •• •C• •• N OH • ••• • •

—

H

H H

Zero FC on all other atoms

Zero FC on all other atoms Negative charge on most electronegative atom F: F > O > N > C

RESONANCE STRUCTURES FC useful for deciding when resonance structures are appropriate

O O N O –

NO3 1) 2) 3) 4) 5) 6) 7)

skeletal structure valence e–s: 3(6) + 5 + 1 = 24 inert shell e–s: 4(8) = 32 bonding e–s: 32 – 24 = 8 assign bonding e–s remaining bonding e–s assigned to multiple bonds remaining valence e–s assigned as lone pairs

12-13.7

• • ••• •• ••••• • • • • • • • • • • • • valence e–s

5.111 Lecture Summary #12-13 –1

+ –1

+

0

+

1

October 5 & 10, 2018

= –1 TOTAL CHARGE

FCO FCODB FCN All structures appear equivalent because FC same. Might expect one short N=0 bond and 2 longer NO bonds but all 3 bonds equivalent experimentally. Better model is a blend of all 3 structures denoted by arrows. – ••O•• • • • •O• • • • • •• • • • • • • N • •O•

– • • •O • • •• • • •O•• •N• •O • • • • • •• ••

• • •• O • • • ••• •••• • • N •O O • •• • • ••

–

RESONANCE HYBRID Electrons in resonance structure are delocalized — sharing of e– pair over several atoms, not just 2.

12-13.8

5.111 Lecture Summary #12-13

October 5 & 10, 2018

BREAKDOWN OF OCTET RULE Case 1 : Odd # of valence electrons

CH3

H H C H 1) Draw skeletal structure 2) 3(1) + 4 = 7 valence e-s 3) 3(2) + 8 = 14 e-s needed for octet 4) 14 – 7 = 7 bonding e-s Radical species: molecule with unpaired e-, usually very reactive

A few radicals are very stable: NO

N O 1) 2) 3) 4)

Draw skeletal structure 5 + 6 = 11 valence e-s 8 + 8 = 16 e-s needed for octet 16 – 11 = 5 bonding e-s

O2

O O 2) 3) 4) 5) 6) 7)

6 + 6 = 12 valence e-s 8 + 8 = 16 e-s needed for octet 16 – 12 = 4 bonding e-s 2 bonding e-s remaining make double bond 8 valence e-s – make lone pairs

Lewis method seems to work but reality is biradical!

O O

Need MO theory

Case 2 : Octet deficient molecules

12-13.9

One atom cannot have an octet bc octet rule works by pairing electrons. If odd, one electron left over ! !

5.111 Lecture Summary #12-13 Some molecules are stable even though there are too few e-s Group 13 elements, B and Al do this

October 5 & 10, 2018

F F B F 2) 3) 4) 5) 6) 7)

3 + 3(7) = 24 valence e-s 8 + 3(8) = 32 e-s needed for octet 32 – 24 = 8 bonding e-s 2 bonding e-s remaining make double bond 24 – 8 = 16 valence e-s remaining – make lone pairs

FCB 3 – 0 – 4 = -1 FCFDB 7 – 4 – 2 = 1 FCF 7 – 6 – 1 = 0 But observation is that all three B-F bonds have same length typical of a single bond!

F F B F FCB 3 – 0 – 3 = 0 FCF 7 – 6 – 1 = 0 Formal charges are more favorable for this structure. Case 3 : Valence shell expansion Elements with n ≥ 3 have empty d orbitals, so more than an octet can go around central atom. Expanded valence shells are more common when central atom is large and is bonded to small, highly electronegative atoms such as O, F, Cl. PCl5 2) 5(7) + 5 = 40 valence e-s Cl 3) 5(8) + 8 = 48 e-s needed for octet 4) 48 – 40 = 8 bonding e-s! Cl P Cl Need 10 shared e-s to make 5 single bonds. Then… 7) 40 – 10 = 30 valence e-s remaining – make lone pairs Cl Cl So each Cl has an octet and P has an expanded shell. FCP 5 – 0 – 5 = 0 FCCl 7 – 6 – 1 = 0

12-13.10

5.111 Lecture Summary #12-13 CrO42-

October 5 & 10, 2018

2-

O O Cr O O 6 + 4(6) + 2 = 32 valence e-s 8 + 4(8) = 40 e-s needed for octet 40 – 32 = 8 bonding e-s 32 – 8 = 24 valence e-s remaining – make lone pairs

2) 3) 4) 7)

FCCr 6 – 0 – 4 = 2 FCO 6 – 6 – 1 = -1 2 + 4(-1) = -2 total charge But experimentally, Cr-O bond is between a single and a double bond!

O

O Cr O

2-

O

O

O

2-

Cr O

O

O

O

FCCr 6 – 0 – 6 = 0 FCO 6 – 6 – 1 = -1 FCODB 6 – 4 – 2 = 0 0 + 2(-1) + 0 = -2 total charge Valence shell expansion around Cr IF4-

-

F F I

F

O Cr O

2) 3) 4)

F FC I = FC F =

12-13.11

23 other structures...