Chemical Bonding & Molecular Structure CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure

PREDICTING THE GEOMETRY OF MOLECULES 1. derive Lewis structure of the molecule 2. discriminate between bonding and non-bonding electron pairs O H 3. VALENCE SHELL ELECTRON PAIR REPULSION

2. electrons repel each other VALENCE SHELL ELECTRON PAIR REPULSION VSEPR 1. identify in a compound the central atom 2. electrons repel each other 3. valence electron pairs stay as far apart as possible 4. non-bonding electrons repel more than bonding electrons

central atom no non-bonding pairs non-bonding pairs

TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM AB2 BeCl2 Cl Be Cl TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM

LINEAR ARRANGEMENT BEST Cl Be Be 180° Be 90° 270° LINEAR ARRANGEMENT BEST IT PUTS ELECTRON PAIRS FURTHEST APART

THREE ELECTRON PAIRS AROUND THE BORON ATOM AB3 BF3 F F B F THREE ELECTRON PAIRS AROUND THE BORON ATOM

THREE ELECTRON PAIRS AROUND THE BORON ATOM F B THREE ELECTRON PAIRS AROUND THE BORON ATOM TRIGONAL PLANAR ARRANGEMENT BEST B 120°

MOLECULAR SHAPE F B F F THE SHAPE OF BF3 IS TRIGONAL PLANAR.

AB4 CH4 H C four electron pairs expect square planar 90° C

better arrangement for four electron pairs TETRAHEDRAL C 109.5° bigger than 90 ° in square planar tetrahedral 4 electron pairs put on the H-atoms

TETRAHEDRAL H C C 109.5° H H H shape of CH4 is tetrahedral

AB5 PF5 FIVE ELECTRON PAIRS AROUND PHOSPHORUS P F P 5 electron pairs trigonal bipyramidal

shape of PF5 is trigonal bipyramidal Bond angle F 900 F P F P F 1200 F shape of PF5 is trigonal bipyramidal two of the F atoms different from the others

AXIAL Bond angle F 900 F F P F 1200 EQUATORIAL F

AB6 six electron pairs around the sulfur atom SF6 F F F S S F F F octahedral

900 F F F S S F F F shape of SF6 is octahedral

central atom no non-bonding pairs non-bonding pairs

AB2E AB3 SeO2 O Se O

VSEPR treats double bonds like a single bond O Se THREE ELECTRON PAIRS AROUND SELENIUM ELECTRON PAIR GEOMETRY Se TRIGONAL PLANAR

THE MOLECULAR SHAPE IS THE POSITION OF THE ATOMS Se O Se ADD OXYGENS SeO2 IS V-SHAPED (OR BENT) THE MOLECULAR SHAPE IS THE POSITION OF THE ATOMS

electron pairs around the nitrogen atom AB3E AB4 H N H NH3 H electron pairs around the nitrogen atom

NH3 is trigonal pyramidal PUT ON THE 3 H ATOMS N H H H NH3 is trigonal pyramidal

four electron pairs around the oxygen atom AB2E2 AB4 H O H four electron pairs around the oxygen atom PUT ON THE 2 H-ATOMS O O H H shape of H2O is V-shaped or bent

AB4E AB5 SF4 S F F F S F TRIGONAL BIPYRAMID

WHERE DOES LONE PAIR GO? OR F S F OR lone pairs occupy the trigonal plane (the “equator”) to minimize the number of 90° repulsions

AB4E AB3E2 AB2E3 SF4 1 lone pair See-saw shaped ClF3 2 lone pairs T-shaped XeF2 3 lone pairs Linear F F F F Xe S Cl F F F F F F lone pairs occupy the trigonal plane (the “equator”) first to minimize the number of 90° repulsions

AB5E AB6 BrF5 Br F Br Square pyramidal

AB6 AB4E2 XeF4 Xe F : Xe Xe F : Xe F lone pairs MUST BE AT 1800

Total valence electron pairs Electron Pair Geometry Summary of Molecular Shapes Total valence electron pairs Electron Pair Geometry Lone electron pairs Shape of Molecule 2 Linear Linear Trigonal planar Trigonal planar 3 1 V-shaped Tetrahedral 4 Tetrahedral 1 Trigonal pyramid 2 V-shaped

Total valence electron pairs Electron Pair Geometry Lone electron pairs Shape of Molecule Trig. bipyramid. 1 See-saw Trigonal bipyramidal 5 2 T-shaped 3 Linear Octahedral 6 Octahedral 1 Square pyramid 2 Square planar

POLYATOMICS molecules with no single central atom we apply our VSEPR rules to each atom in the chain Example: ETHANOL

ETHANOL C2H5OH O C H The atoms around the carbons form a. tetrahedral arrangement The atoms around the oxygen form a V-shaped structure.

H C O

EXAMPLES BF4- ICl4- Cl2O SO2Cl2 Cl2CO Cl2SO N2F2 NH4+ NH2OH

1. Lewis structures 2. VSEPR model WHY DO MOLECULES FORM?

two H-atoms approach each other and the electron waves interact simplest molecule H2 two H-atoms 1s1 two H-atoms approach each other and the electron waves interact OVERLAP to form a region of increased electron density between the atoms

chemical bond with electron density in between the nuclei is called

VALENCE BOND THEORY a covalent bond is formed by an overlap of two valence atomic orbitals that share an electron pair the better the overlap the stronger the bond the orbitals need to point along the bonds

hydrogen atoms bond using their 1s orbitals C H CH4 What orbitals are used? hydrogen atoms bond using their 1s orbitals carbon needs four orbitals to bond with. [He] 2s22p2 2s, 2px , 2py, 2pz

1. The electronic configuration of carbon is [He] 2s22p2 The orbital diagram is:

1. The electronic configuration of carbon is [He] 2s22p2 the orbital diagram is: [He] C . the Lewis dot structure is necessary to promote one 2s electron

PROMOTE AN ELECTRON [He] [He] [He] 2s22p2 [He] 2s12p3 excited state (valence state) C Lewis dot structure four unpaired electrons we can use these to form chemical bonds

experiment shows that methane has 109.5° bond angles 1. a covalent bond is formed by an overlap of two valence atomic orbitals that share an electron pair 2. bonds formed with s orbitals will be different to bonds formed with p orbitals Experiment shows that all four bonds are identical 3. three p orbitals are mutually perpendicular, suggesting 90° bond angles experiment shows that methane has 109.5° bond angles combining the orbitals

we need four orbitals pointing to the vertices of a tetrahedron orbitals are just mathematical functions C H we can combine them HYBRIDIZATION

COMBINING ORBITALS TO FORM HYBRIDS HYBRIDIZATION number of atomic orbitals that are combined IS EQUAL TO the number of resulting hybrid orbitals

HYBRIDIZATION + + 2s+ 2p Combine one s and one p a sp- hybrid ADD the orbitals + 2s+ 2p

HYBRIDIZATION + + 2s+ 2p Combine one s and one p a sp- hybrid s + p What do we get? + The positive part cancels negative part DESTRUCTIVE INTERFERENCE The positive part adds to positive part CONSTRUCTIVE INTERFERENCE

HYBRIDIZATION + 2s+ 2p Combine one s and one p to give a sp- hybrid Where is the nucleus? + REMEMBER IF WE MIX TWO WE MUST GET TWO BACK The other combination is s - p

HYBRIDIZATION + + 2s- 2p Combine one s and one p a sp- hybrid SUBTRACT the orbitals 2s- 2p

SUBTRACTING THE p ORBITAL CHANGES ITS PHASE HYBRIDIZATION Combine one s and one p a sp- hybrid SUBTRACTING THE p ORBITAL CHANGES ITS PHASE + + SUBTRACT the orbitals 2s- 2p

SUBTRACTING THE p ORBITAL CHANGES ITS PHASE HYBRIDIZATION Combine one s and one p a sp- hybrid SUBTRACTING THE p ORBITAL CHANGES ITS PHASE + + SUBTRACT the orbitals 2s- 2p

HYBRIDIZATION + + 2s- 2p Combine one s and one p a sp- hybrid s - p What do we get? + The positive part cancels negative part DESTRUCTIVE INTERFERENCE The positive part adds to positive part CONSTRUCTIVE INTERFERENCE

We get two equivalent sp orbitals HYBRIDIZATION Combine one s and one p a sp- hybrid s - p 2s- 2p Where is the nucleus? + The positive part cancels negative part We get two equivalent sp orbitals ORIENTED AT 1800

sp-HYBRIDIZATION s and p orbitals two sp-hybrids

COMBINE one s-orbital and two p-orbitals Get three sp2 - orbitals oriented at 1200 s and p orbitals three sp2-hybrids directed at 1200

COMBINE one s-orbital and three p-orbitals three sp3- orbitals oriented at 109.50

What happens to the energies of the orbitals? C H METHANE: CH4 four hybrid orbitals needed to form four bonds s + px + py + pz 4 sp3 hybrids an atom with sp3 hybrid orbitals is said to be sp3 hybridized The four sp3 hybrid orbitals form a tetrahedral arrangement. EPG of 4 pairs sp3 hybridization What happens to the energies of the orbitals?

What happens to orbital energies when the are hybridized?? Orbitals in free C atom

When orbitals are hybridized they have the same energy: 2p sp3 E 2s E Orbitals in free C atom Hybridized orbitals of C atom in methane The FOUR sp3 hybrids are DEGENERATE.

x y z x y z x y z Combine one s and three p orbitals….. x y z

Now form the bonds to the H-atoms……... sp3 HYBRIDS C sp3 orbitals Now form the bonds to the H-atoms……...

Each bond in methane results from the overlap of a hydrogen 1s orbital and a carbon sp3 orbital. Carbon sp3 orbitals H C Form a chemical bond by sharing a pair of electrons. H H H Each hybrid ready to overlap with H 1s orbitals

VALENCE BOND MODEL Hybrid orbital model Step 1: Draw the Lewis structure(s) Step 2: Determine the geometry of the electron pairs around each atom using VSEPR OR preferably use the EXPERIMENTAL GEOMETRY Step 3: Specify the hybrid orbitals needed to accommodate the electron pairs on each atom

OTHER MOLECULES USING sp3 HYBRIDS sp3 hybrids are also employed in …... all molecules that have a 4 pair EPG…. NH3, H2O, NH4+ , CCl4 AMMONIA…..

Nitrogen electronic configuration AMMONIA: NH3 VSEPR H N Valence shell has four pairs EPG is TETRAHEDRAL Need sp3 hybrids Nitrogen electronic configuration 2s 2p N [He] HYBRIDIZE

When orbitals are hybridized they have the same energy: 2p sp3 E 2s E Orbitals in free N atom Hybridized orbitals of N atom in ammonia The FOUR sp3 hybrids are DEGENERATE. sp3 hybridization…….

sp3 hybrids on N in AMMONIA . . . 2s 2p N Now form a bond H H Overlap H 1s….. H

AMMONIA N . . . 2s 2p N H Three  bonds One lone pair in an sp3 hybrid

AMMONIUM ION NH4+ four  bonds. H+ 2p 2s . N . . N H ISOELECTRONIC WITH ? CH4

WATER O : . . . FOUR PAIRS 2s 2p EPG? O TETRAHEDRAL!! sp3 HYBRIDIZATION?

WATER. Overlap of two of oxygen sp3 hybrids with ….. Lone pairs in two of the sp3 hybrids. O H atom 1s orbitals. To form two  bonds. H H Think about H3O+ !!!

HYDRONIUM ION. Overlap of Oxygen sp3 hybrids containing a lone pair ISOELECTRONIC WITH? O NH3 H+ ion empty 1s orbitals. H+ H H

QUESTION Which of the following molecules is uses sp3 hybrids in the valence bond description of its bonding? 1 C and D A CO2 2 B and E B NF3 3 A and D C O3 4 B and C D NO2+ 5 B and A E F2O ANSWER…….

QUESTION Which of the following molecules is uses sp3 hybrids in the valence bond description of its bonding? C O 1 C and D A CO2 N F 2 B and E B NF3 3 A and D C O3 O O N [ ]+ 4 B and C D NO2+ 5 B and A O F E F2O WHAT ABOUT OTHER EPG’S …….

VALENCE BOND THEORY FOR OTHER ELECTRON PAIR GEOMETRIES A four electron pair EPG uses sp3 hybrids The three electron pair EPG uses sp2 hybrids The two electron pair EPG uses sp hybrids

EPG’s sp3 sp sp2 4 2 3 X X X HYBRIDS 109.5° 3 X 120° X 180° HYBRIDS sp3 sp sp2 lets look at a molecule that needs sp2

C H Ethylene: C2H4 trigonal planar EPG around each C-atom. VSEPR a HCH angle of 1200. three hybrid orbitals on each carbon for the trigonal planar EPG. s + px + py 3 sp2 hybrids The CARBON is sp2 hybridized The 3 sp2 hybrid orbitals form a trigonal planar arrangement. 3 effective electron pairs sp2 hybridization

FORMATION OF sp2 hybrids VALENCE STATE C atom E 2p 2s GROUND STATE C atom

FORMATION OF sp2 hybrids VALENCE STATE C atom E 2p 2s E 2p 2s GROUND STATE C atom HYBRIDIZE

FORMATION OF sp2 hybrids VALENCE STATE C atom E 2p 2s E 2p 2s GROUND STATE C atom HYBRIDIZE E sp2 2p sp2 hybridized orbitals of C This leaves one p orbital unhybrized…….

An sp2 hydridized C atom x y z sp2 - hybrid orbital UNHYBRIDIZED p- orbital The unhybridized p orbital is perpendicular to sp2 plane. Lets put it all together…….

Now put the orbitals on…... C H DRAW TWO C-ATOMS x z y y x z C C Now put the orbitals on…...

C H BONDING IN ETHYLENE z z y y C C x x

s bond BONDING IN ETHYLENE z z y y C C x x PUT THE ELECTRONS IN AND….. OVERLAP the sp2 hybrids from the two carbons to form a sigma bond between them. PUT THE ELECTRONS IN AND…..

The two unhybridized p orbitals are left over to form….. overlap two sp2 hybrids on each carbon with hydrogen 1s orbitals to form sigma bonds and... C H z z H H y y H H x x The two unhybridized p orbitals are left over to form…..

pi bond (p bond) The second part of the carbon-carbon double bond ! z y y H H x x The two unhybridized p orbitals are left over to form a ….. pi bond (p bond)

pi bond (p bond) The second part of the carbon-carbon double bond ! z y y H H x x Electrons are shared between the unhybridized p orbitals in an area above and below the line between nuclei.

pi bond (p bond) sigma bonds (s bond) THE COMPLETE PICTURE!!!!!!! z z y y H H x x SUMMARY... sp2 sigma bonds (s bond)

p:C(2p)-C(2p) s:C(sp2)-C(sp2) s: H(1s)-C(sp2) s: H(1s)-C(sp2) s bonding Now look at p bond

p:C(2p)-C(2p) s:C(sp2)-C(sp2) s: H(1s)-C(sp2) s: H(1s)-C(sp2) p bonding Now look at ethyne (acetylene)

BONDING SCHEME IN ETHYNE p:C(2p)-C(2p) TWO OF THESE!! s: H(1s)-C(sp) s: H(1s)-C(sp) C H s:C(sp)-C(sp) What does this look like????

Now put the sp-orbitals on…... DRAW TWO C-ATOMS C H x z y y x z C C Now put the sp-orbitals on…...

Put in the unhybridized p orbitals OVERLAP the sp hybrids from the two carbons to form a sigma bond between them. z z y y C C x x Put in the unhybridized p orbitals

OVERLAP the hydrogen 1s orbitals OVERLAP the sp hybrids from the two carbons to form a sigma bond between them. z z y y C C x x OVERLAP the hydrogen 1s orbitals

OVERLAP the C sp hybrids with H 1s to form sigma bonds z z y y H H C C x x

sigma framework of s bonds OVERLAP the sp hybrids from the two carbons to form a sigma bond between them. z z y y H H C C x x sigma framework of s bonds pi bonding?

two pi bonds (p bonds) LATERAL OVERLAP of p orbitals to form pi bonds. z z y y H H C C x x two pi bonds (p bonds)

two pi bonds (p bonds) LATERAL OVERLAP of p orbitals to form pi bonds. z z y y H H C C x x SO….. two pi bonds (p bonds)

SUMMARY Single bond: One s bond Double bond: One s bond, one p bond Triple bond: One s bond, two p bonds

VALENCE BOND THEORY Lewis Dot Structure Step 1 Step 2 Get Molecular Geometry VSEPR EXPERIMENTAL Step 3 Choose hybrids Describe bonds…... Step 4

What about molecules with more than an octet around the central atom? Examples: PCl5, or SF4 or SiF62- Four pairs needs Four orbitals Five orbitals Five pairs needs six pairs needs six orbitals

PCl5 We ignore the chlorine atoms and just describe central atom. Need five hybrid orbitals on the phosphorus to fit the trigonal bipyramidal EPG. d + s + px + py + pz 5 dsp3 hybrids 5 effective electron pairs dsp3 hybridization Five equivalent orbitals……..

dsp3 - hybrid orbitals x y z TRIGONAL BIPYRAMID EPG 5 PAIRS 900 1200 SIX PAIRS….. overlap with orbitals on chlorine to form 5 s bonds.

SF6 We ignore the chlorine atoms and just describe central atom. We need six hybrid orbitals on the sulfur to allow for the octahedral EPG and six bonds. 6 d2sp3 hybrids d + d + s + px + py + pz d2sp3 hybridization 6 effective electron pairs SIX equivalent orbitals……..

d2sp3 - hybrid orbitals x y z 900 900 EXAMPLE overlap with orbitals on flourine to form 6 s bonds.

Three lone pairs in equatorial hybrids EXAMPLES Describe the molecular structure and bonding in XeF2 and XeF4 Xe F EPG 5 pairs Linear dsp3 hybrids Two axial s bonds at 1800 Three lone pairs in equatorial hybrids

Three lone pairs in equatorial hybrids Two lone pairs in axial hybrids EXAMPLES Describe the molecular structure and bonding in XeF2 and XeF4 Xe F Xe F EPG 5 pairs EPG 6 pairs Linear Square planar dsp3 hybrids d2sp3 hybrids Two axial s bonds at 1800 four s bonds at 900 in a plane Three lone pairs in equatorial hybrids Two lone pairs in axial hybrids

electrons occupy orbitals each of which spans the entire molecule MOLECULAR ORBITAL THEORY electrons occupy orbitals each of which spans the entire molecule molecular orbitals each hold up to two electrons and obey Hund’s rule, just like atomic orbitals

H2 molecule: 1s orbital on Atom A 1s orbital on Atom B the H2 molecule’s molecular orbitals can be constructed from the two 1s atomic orbitals 1sA + 1sB = MO1 constructive interference 1sA – 1sB = MO2 destructive interference

ADDITION OF ORBITALS A B combine them by addition builds up electron density in overlap region 1sA + 1sB = MO1 A B combine them by addition

ADDITION OF ORBITALS A B what do we notice? builds up electron density in overlap region. 1sA + 1sB = MO1 A B what do we notice? electron density between atoms

SUBTRACTION OF ORBITALS results in low electron density in overlap region.. 1sA – 1sB = MO2 A B subtract

SUBTRACTION OF ORBITALS results in low electron density in overlap region.. 1sA – 1sB = MO2 A B what do we notice? no electron density between atoms

COMBINATION OF ORBITALS 1sA + 1sB = MO1 builds up electron density between nuclei

COMBINATION OF ORBITALS 1sA – 1sB = MO2 ANTI-BONDING results in low electron density between nuclei 1sA + 1sB = MO1 BONDING builds up electron density between nuclei

THE MO’s FORMED BY TWO 1s ORBITALS

s1s* s1s sigma anti-bonding = s1s* sigma bonding = s1s 1sA – 1sB = MO2

COMBINING TWO 1s ORBITALS Energy of a 1s orbital in a free atom Energy of a 1s orbital in a free atom A B E

s1s E 1sA+1sB MO Energy of a 1s orbital in a free atom

s1s* s1s 1sA-1sB MO E 1sA+1sB MO Energy of a 1s orbital in a free atom

COMBINING TWO 1s ORBITALS s1s* 1sA A B 1sB E s1s

bonding in H2 H H2 H s1s* E 1s 1s s1s

H H2 H s1s* E 1s 1s s1s the electrons are placed in the s1s molecular orbitals

H H2 H s1s* E 1s 1s s1s H2: (s1s)2

He2 atomic configuration of He 1s2 He He2 He s1s* E 1s 1s s1s

He2: (s1s)2(s1s*)2 s1s* s1s He He2 He E 1s 1s bonding effect of the (s1s)2 is cancelled by the antibonding effect of (s1s*)2

BOND ORDER net number of bonds existing after the cancellation of bonds by antibonds He2 the electronic configuration is…. (s1s)2(s1s*)2 the two bonding electrons were cancelled out by the two antibonding electrons BOND ORDER = 0

= BOND ORDER measure of bond strength and molecular stability If # of bonding electrons > # of antibonding electrons the molecule is predicted to be stable Bond order =

high bond order indicates high bond energy and short bond length measure of bond strength and molecular stability If # of bonding electrons > # of antibonding electrons the molecule is predicted to be stable Bond order { } = # of bonding electrons(nb) # of antibonding electrons (na) 1/2 – = 1/2 (nb - na) high bond order indicates high bond energy and short bond length H2+,H2,He2+

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 H2+ He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- ½ 225 106 He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- ½ 225 106 He2+ Para- ½ 251 108 He2 E

First row diatomic molecules and ions s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- ½ 225 106 He2+ Para- ½ 251 108 He2 — E

second period HOMONUCLEAR DIATOMICS Li2 Li : 1s22s1 both the 1s and 2s overlap to produce s bonding and anti-bonding orbitals

ENERGY LEVEL DIAGRAM FOR DILITHIUM s2s* Li2 2s 2s E s2s s1s* 1s 1s s1s

ELECTRONS FOR DILITHIUM s2s* Li2 2s 2s E s2s s1s* 1s 1s s1s

Li2 Electron configuration for DILITHIUM s2s* (s1s)2(s1s*)2(s2s)2 2s Bond Order ? 1s 1s s1s

Li2 Electron configuration for DILITHIUM s2s* (s1s)2(s1s*)2(s2s)2 2s nb = 4 na = 2 E s2s Bond Order = 1 single bond. 1s 1s s1s

the s1s and s1s* orbitals can be ignored when both are FILLED! Electron configuration for DILITHIUM s2s* Li2 (s1s)2(s1s*)2(s2s)2 2s 2s the s1s and s1s* orbitals can be ignored when both are FILLED! E s2s 1s 1s omit the inner shell s1s

only valence orbitals contribute to molecular bonding Li2 (s2s)2 Li Li2 Li s2s* E 2s 2s s2s The complete configuration is: (s1s)2(s1s*)2 (s2s)2

Be2 Be Be2 Be s2s* E 2s 2s s2s

Be2 Electron configuration for DIBERYLLIUM Be Be2 Be s2s* E 2s 2s s2s Bond order = 0

Be2 Electron configuration for DIBERYLLIUM Be Be2 Be (s2s)2(s2s*)2 nb = 2 E 2s 2s na = 2 s2s Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0 Now B2... No bond!!! The molecule is not stable!

form molecular orbitals the Boron atomic configuration is 1s22s22p1 we expect B to use 2p orbitals to form molecular orbitals addition and subtraction

s-molecular orbitals

p molecular orbitals

ENERGY LEVEL DIAGRAM E 2s s2s* s2s

E 2p s2p* p2p* p2p s2p

expected orbital splitting p2p p2p* 2p E 2s s2s* s2s This pushes the s2p up

MODIFIED ENERGY LEVEL DIAGRAM 2s s2s* s2s 2p s2p* s2p p2p p2p* E Notice that the s2p and p2p have changed places!!!!

Electron configuration for B2 B is [He] 2s22p1 s2p* p2p* s2p 2p 2p p2p E Place electrons from 2s into s2s and s2s* s2s* 2s 2s s2s

s2p* p2p* s2p 2p 2p p2p E Place electrons from 2p into p2p and p2p s2s* 2s 2s Remember HUND’s RULE s2s

ELECTRONS ARE UNPAIRED s2p* s2p p2p p2p* Abbreviated configuration (s2s)2(s2s*)2(p2p)2 E Complete configuration 2s s2s* s2s (s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)2

Electron configuration for B2: s2p* (s2s)2(s2s*)2(p2p)2 p2p* na = 2 nb = 4 s2p 2p 2p p2p E Bond order 1/2(nb - na) s2s* = 1/2(4 - 2) =1 2s 2s Molecule is predicted to be stable and paramagnetic. s2s

A SUMMARY OF THE MO’s Emphasizing nodal planes

ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS Li2 B2 C2 N2 O2 F2

O2 F2 Li2 B2 C2 N2 2s s2s* s2s 2p s2p* s2p p2p p2p* E 2s s2s* s2s s2p* 2p s2p p2p p2p*

Second row diatomic molecules s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 C2 N2 O2 F2 E

Second row diatomic molecules s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 N2 O2 F2 E

Second row diatomic molecules s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 O2 F2 E

Second row diatomic molecules s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 F2 E

Second row diatomic molecules NOTE SWITCH OF LABELS s2p* p2p* p2p s2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 Para- 2 495 121 F2 E

Second row diatomic molecules NOTE SWITCH OF LABELS s2p* p2p* p2p s2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 Para- 2 495 121 F2 Dia- 1 154 143 E

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : O2+ : O2– : O22-:

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : B.O. = (8 - 4)/2 = 2 O2+ : B.O. = (8 - 3)/2 = 2.5 O2– : B.O. = (8 - 5)/2 = 1.5 O22- : B.O. = (8 - 6)/2 = 1

O2+ >O2 >O2– > O22- BOND ENERGY ORDER s2p* p2p* p2p s2p s2s*

OXYGEN How does the Lewis dot picture correspond to MOT? p2p* s2p p2p s2s* s2s E 12 valence electrons BO = 2 but PARAMAGNETIC