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Home , Single bond

Covalent Bonding John W. Moore G.N. Lewis (1916): Conrad L. Stanitski Peter C. Jurs • Some shareshare e- to form bonds.

http://academic.cengage.com/chemistry/moore

Chapter 8 Attraction Stable bond Repulsion Covalent Bonding • Number of bonds = Number shared e- pairs.

• Shared e- molecules are covalentcovalent compounds with covalentcovalent bonds. Stephen C. Foster • Mississippi State University

Covalent Bonding Single Covalent Bonds Lewis structures: dot = 1 e-. Line = 1 pair of e- ) l

o H H H − H m / J K (

y Single bond: one shared pair of e-. g r e n E

l a i

t Octet ruleruleOctet n e

t -

o To form bonds, elements gain, lose, or share e to P achieve 8 valence e- (doesn’t apply to H)

Distance between nuclei (pm)

Single Covalent Bonds Single Covalent Bonds Bonding pairspairsBonding = e- pairs shared between atoms. 2F = 2(7) = 14 valence e- . F F Share 2 e- to form octets. Lone pairspairsLone = unshared e- pair.

O+2H=6+2(1)=8val.e-. bonding H O H Two O-H bonds. H O H H O H bonding H N H N +3H=5+3(1)=8val.e-. lone pair 3 N-H bonds. H

1 Single Covalent Bonds Guidelines for Writing Lewis Structures

# of e- shared Group # of to form an octet Example - valence e- (8 - A group#) 1. Count the valence e in the molecule. H 2. Draw a skeleton structure. Join atoms with single 4A 4 4 C in CH4 H – C – H - H lines (pairs of e )...... 3. Add e- pairs to form octets (except H). Start with . . . F – N – F . 5A 5 3 N in NF3 .. ..

. . terminal atoms. . ..F .

-- .. Extra eeExtra ? Place around the central . 6A 6 2 O in H2O H – O.. – H ..

. --

. Too fewfewToo ee ? Convert lone pairs into multiple bonds. 7A 7 1 F in HF H – ..F

Guidelines for Writing Lewis Structures Guidelines for Writing Lewis Structures

Phosphorus trifluoride, PF3 - 3. Build octets – start with terminal atoms. 1. PF3 = 5 + 3 (7) = 26 valence e

3 x F (group 7A) P (group 5A) F P F 6 e- used in 3 bonds, 20 e- remain (10 pairs)

F - 2. Skeleton (X is central in XYn ). F P F 6 + 20 = 26 e F

Guidelines for Writing Lewis Structures Guidelines for Writing Lewis Structures 3. Add e- pairs: 3- Phosphate ion, PO4 8 e- used in 4 bonds, 1. PO 3- = 5 + 4(6) + 3 = 32 4 O 24 e- remain (12 pairs) P (grp 5A) 4 x O (grp 6A) charge (-3) 3 e- - O P O 32 e used.

3- O O O

2. Skeleton. P is central. O P O O P O Add brackets and overall charge to show this is an ion. O O

2 Single Covalent Bonds in Single Covalent Bonds in Hydrocarbons

AlkanesAlkanes = hydrocarbons with C-C singlesingle bonds. Larger : H H H H

Methane: butane H C C C C H H H H H

H ee-- densitydensity: red = high density ; blue = low density H C H H H 2-methylpropane C Saturated hydrocarbonshydrocarbonsSaturated each C is bonded to its H C C H maximum number of H. H H H

Single Covalent Bonds in Hydrocarbons Multiple Covalent Bonds

Cyclic alkanes: Too few “dots” to complete all the octets? Convert lone pairs to shared pairs.

ExampleExample methanal (formaldehyde) H2CO 1. Valence e- = 2(1) + 4 + 6 = 12

2. Skeleton.

3. 6 e- in bonds. Add the other 3 pairs to O (outer atom). H • Each H shares 2 e- • C only “has” 6 e-. H C O

Multiple Covalent Bonds Multiple Covalent Bonds Convert lone pairs to bond pairs. ExampleExample carbon dioxide CO2 - H 1. 4 + 2(6) = 16 e 2. Skeleton O C O H C O 3. 4 e- in bonds. Add 3 pairs to each O.

Each H shares 2 e- H 4. Convert lone pairs to C shares 8. O C O bond pairs. H C O O shares 8.

3 Multiple Covalent Bonds in Hydrocarbons Multiple Covalent Bonds in Hydrocarbons

AlkenesAlkenes: hydrocarbons with C=C bonds. AlkynesAlkynes: hydrocarbons with C≡C bonds. •• ––eneene ending. •• ––yneyne ending.

• CnH2n (unless cyclic) • CnH2n-2 • Unsaturated (can add more H) • Unsaturated ethyne (acetylene) propyne ethene C2H4 propene C3H6 butene C4H8 H H H H H H H H-C≡C-H H-C≡C-C-H H C C H H C C C H H C C C C H H H H H H H H

(ethylene) (propylene) (1 isomer shown)

C2H2 C3H4

Double Bonds and Isomerism Double Bonds and Isomerism Bond rotation: 1,21,2--dichloroethenedichloroethene (Cl on carbon 1 and 2)

same opposite side sides

CHCl=CHCl is “locked” - Two isomersisomers occur. m.p.= −81°C, b.p.= 60°C m.p.= −50°C, b.p.= 48°C.

Double Bonds and Isomerism Double Bonds and Isomerism

1,11,1--dichloroethenedichloroethene does not have isomeric forms. Unsubstituted alkenes can have cis-trans isomers. (both Cl on carbon 1) 22--butenebutene (C=C starts on carbon 2) Cl H H Cl H C CH H C H C=C C=C 3 3 3 C=C C=C Cl H H Cl H H H CH3

2,22,2--dichloroethene? NO!dichloroethene?NO! Same molecule (lowest numbers used) ciscis--22--butenebutene transtrans--22--butenebutene Methyls on the same side. Methyls on opposite sides. m.p.= −139°C, b.p.= 4°C. m.p.= −106°C, b.p.= 1°C.

4 Bond Properties: Bond Length Atom size and bond type are important:

Bond Enthalpies Bond Enthalpies Bond Enthalpy (Bond (DD)) Shorter bond = stronger bond. Estimate ΔH for the following reaction from bond E.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

H ...... O = O H – O – H

.. .. .

...... H – C – H + .. .. → . O = C = O + .. H ..O = O.. H – O.. – H

ΔH = {sum of bonds broken} – {sum of bonds formed}

= {4 DC-H + 2 DO=O} – {2 DC=O + 4 DH-O} = {4(416) + 2(498)} – {2(803) + 4(467)} = −814 kJ (experimental value is −802 kJ)

Bond Polarity and Electronegativity Bond Polarity and Electronegativity

Bonding pairs are not always equally shared. Polar bond: one atom is more electronegativeelectronegative. NonpolarNonpolar = equally shared (identical atoms). ElectronegativityElectronegativity = ability of an atom in a F F to attract shared e- to itself. • Linus Pauling developed the first scale PolarPolar = unequal sharing (different attraction for e-) • Based on bond energies.

F has stronger e- attraction. • Pauling’s scale: F = 4.0 (arbitrary). d- Polar bond. d+ H F • Unitless.

5 Bond Polarity and Electronegativity Bond Polarity and Electronegativity Differences, ΔEN, determine bond polarity:

increasing electronegativity

Bond Polarity and Electronegativity Bond Polarity and Electronegativity

Consider the bond pairs (i) Cl–F & Br–F and (ii) Si–Br & (ii) Si–Br & C–Br. More polar? δ+ and δ- ? C–Br. Which bond is more polar in each pair? Show δ+ and δ-. increasing electronegativity increasing ClF vs BrFClF BrF electronegativity SiBr vs CBrSiBr CBr EN > EN > EN . F Cl Br ENSi < ENC < ENBr. Si is δ+ in SiBr, C is δ+ in CBr. In each case F is δ-, the other end is δ+ . Largest ΔEN between Si and Br ΔEN (F and Br) > ΔEN (F and Cl) SiBr is more polar than SiC. BrF is more polar than ClF. (Electronegativities: Si = 1.8, C = 2.5, Br = 2.8 (Electronegativities: Br = 2.8, Cl = 3.0, F = 4.0

Formal Charge Formal Charge

Used to study charge distribution in a molecule. []O C N – MethodMethod 1. Find the number of e- assigned to each atom: O C N - - - e “on” an atom = (lone pair e ) + ½½ (bonding e ) Valence e- 6 4 5 Lone pair e- 6 0 2 2. Formal charge of each atom ½ shared e- 1 4 3 - - = (# of valence e ) – (e “on” the atom). Formal Charge -1 0 0

Note: sum of formal charges = charge on species Check:  (formal charges) = ion charge = -1

6 Formal Charge Formal Charge

Given a choice between Lewis structures: - Which ClO2 structure is preferred? • Smaller formal charges are favored. • Negative formal charges should be on the most EN atoms • Like charges should not be on adjacent atoms O Cl O O Cl O

ExampleExample Formal charges: -1 0 0 -1 +1 -1

Which N2O structure is preferred? PreferredPreferred (Smaller charges) O N N O N N

Formal charges: -1 +1 0 0 +1 -1

PreferredPreferred. ENO > ENN

Lewis Structures and Lewis Structures and Resonance

Ozone has 2 equivalent structures: Resonance structuresstructuresResonance are used to show O3 is a mixture of both:

O O O O O O O O O O O O

• They differ only in e- pair positions. Both: • Atom positions must not change. • obey the octet rule • have the same number and types of bonds Resonance hybridshybridsResonance are also used: O O O • have the same formal charges

Experiments show that the OO bonds are identical. Each is 1½ bonds. O3 does NOTNOT “flip” back & forth.

Lewis Structures and Resonance Lewis Structures and Resonance

Benzene, C H , is a ring compound best described Resonance in CO 2- 6 6 3 using resonance ideas:

Experiment: All three CO bonds = 129 pm.

Typical bond lengths: C-O 143 pm Experimental: All C-C bonds are identical. C=O 122 pm

7 Lewis Structures and Resonance Exceptions to the Octet Rule

Benzene is often represented by line-angle structures: Fewer than 8 valence : Be and B compounds:

H Be H 2 + 2(1) = 4 valence e-

F or F B 3 + 3(7) = 24 valence e- Where a circle shows: • The C=C π bond e- are delocalized.delocalized. F • The π e- spread out around the ring.

Fewer than Eight Valence Electrons Odd Number of Valence Electrons

Some stable molecules have an odd number of e-. Often very reactive: ExamplesExamples NO 5+6=11valencee- N O F H F H F B + N H F B N H NO 5 + 2(6) = 17 valence e- F H F H 2 O N O

Free radicalradicalFree = atom or molecule with unpaired e-. Very reactive. Most stable molecules have paired e-.

More Than Eight Valence Electrons More Than Eight Valence Electrons

“Expanded octets” are relatively common. - ClF3 7 + 3(7) = 28 val. e F Low-lying d orbitals can accept extra e- Make octets on F. F Cl (onlyonly 3rd period and beyond).

24 e- used, 4 remain. F ExamplesExamples [bonds (3 x 2); Lone pairs (3 x 6)] - • 5 bonds (5 e pairs) around P in PF5

. NF5 does not exist - Add 2 lone pairs to Cl – the 3rd period element. • 4 bonds and 1 lone-pair (5 e pairs) around S in SF4

. OF4 does not exist.

8 Theory

Bond formation by overlap of atomic orbitals and • Valence atomic orbitals (AOs) are combined to e--pair sharing is called valence bond theorytheoryvalence . form MOs. • MOs can extend over an entire molecule. It works well, but it fails to reliably predict magnetic • They are not confined to pairs of atoms. properties.

∙ ∙∙ ∙∙∙ ∙ ∙ - • O2 prediction: O=O (all e paired = diamagnetic). • Number of AOs mixed = number of MOs formed • Experiment: O2 is paramagnetic. • Valence e- fill the MOs following the Pauli exclusion principle. Molecular orbital (MO) theorytheoryMolecular corrects this problem. • Hund’s rule is obeyed.

Molecular Orbitals for Diatomics Molecular Orbitals for Diatomics MOs produced by combining 1s AOs: Add valence e- (obey Pauli):

- - H2 2 e He2 4 e

σ*1s σ*1s ↓↑

1s 1s 1s 1s

↓↑ ↓↑

σ1s σ1s

Bond order = ½ {(bonding e-) – (antibonding e-)} • Bonding MOs are more stable (lower E) than the original AOs. H2: = ½ { 2 – 0 } = 1 bond

• Antibonding (*) MOs are less stable. He2 does not exist! (It would have ½ {2 – 2} = 0 bond)

Molecular Orbitals for Diatomics Molecular Orbitals for Diatomics

2s AOs form σ2s and σ*2s. 2p AOs form: MO predictions:

Occupancy of Orbitals

σ2s σ*2s π2p π2p σ2p π*2p π*2p σ*2p

Li2 (↑↓) Be (↑↓) (↑↓) One set of σ and σ* 2 2p 2p B (↑↓) (↑↓) ( ↑ ) ( ↑ ) MOs 2 C2 (↑↓) (↑↓) (↑↓) (↑↓)

N2 (↑↓) (↑↓) (↑↓) (↑↓) (↑↓)

O2 (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) ( ↑ ) ( ↑ )

Two degenerate sets of π2p F2 (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) and π*2p MOs Ne2 (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (↑↓) (2py AOs not shown) Degenerate Degenerate (Equal E) (Equal E)

9 Molecular Orbitals for Diatomics Polyatomics; Delocalized π-Electrons

2- Consider CO3 Predicted Properties Observed Properties Lewis dot + resonance view: Unpaired e- Unpaired e- Bond E (kJ/mol)

Li2 0 1 0 105

Be2 0 0 0 Unstable

B2 2 1 2 289

C2 0 2 0 598

N2 0 3 0 946 2- MO view of CO3 O2 2 2 2 498 • The skeleton involves three σ bonds. F2 0 1 0 158 Ne2 0 0 0 Nonexistent • A delocalized π MO holds the 4th e- pair.

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