H2OH2O... O... 2s 2p y 2p x 2p z 2p y 2p x 2F-1 (of 14)

H2OH2O... O... 2s 2p y 2p x 2p z 2p y 2p x HH 1s This predicts the shape of a water molecule to be bent with a 90º angle However, the molecule is bent with a 105º angle Each bond is formed by combining a H 1s atomic orbital (with 1 e - ) with an O 2p atomic orbital (with 1 e - ) 2F-2 (of 14)

1939 LINUS PAULING Showed that linear combinations of an atom’s valence atomic orbitals produce another set of equivalent valence atomic orbitals HYBRIDIZATION – The combining of 2 or more orbitals of different sublevels to make an equal number of HYBRID ORBITALS of equivalent energy (said to be DEGENERATE) 2F-3 (of 14)

E 2s 2p hybridization These 4 hybrid orbitals are called sp 3 orbitals sp 3 2F-4 (of 14)

2p z 2p x 2p y 2s sp 3 hybridization The sp 3 hybrid orbitals are arranged in a tetrahedron, with an angle of 109.5º between them 2F-5 (of 14)

2p z 2p x 2p y 2s sp 3 hybridization The sp 3 hybrid orbitals are arranged in a tetrahedron, with an angle of 109.5º between them 2F-6 (of 14)

sp 3 This forms a bent molecule with a theoretical bond angle of 109.5º Each bond is formed by combining an O sp 3 atomic orbital (with 1 e - ) with a H 1s atomic orbital (with 1 e - ) 2F-7 (of 14)

Each bond in H 2 O is completely symmetrical around its internuclear axis SIGMA BOND (σ) – A bond that is completely symmetrical around its internuclear axis Each bond is named:σ(sp 3 +1s) 2F-8 (of 14)

NH 3... N.. 2s 2p y 2p x 2p z 2p x 2p y 2F-9 (of 14)

NH 3... N.. 2s 2p y 2p x 2p z 2p x HH 1s This predicts the shape of an ammonia molecule to be trigonal pyramidal with a 90º angle However, the molecule is trigonal pyramidal with a 107º angle 2p y 1s.H.H 2F-10 (of 14)

E 2s 2p hybridization sp 3 2F-11 (of 14)

sp 3 This forms a trigonal pyramidal molecule with a bond angle of 109.5º Each sigma bond in NH 3 is formed by combining a N sp 3 atomic orbital (with 1 e - ) with a H 1s atomic orbital (with 1 e - ) Each bond is named:σ(sp 3 +1s) 2F-12 (of 14)

CH 4... C. 2s 2p y 2p x 2p z.. C.. sp 3 E 2s 2p hybridization sp 3 2F-13 (of 14)

sp 3 This forms a tetrahedral molecule with a theoretical bond angle of 109.5º Each bond is named:σ(sp 3 +1s) 2F-14 (of 14) Central atoms with SN = 4 always undergo sp 3 hybridization when bonding

BH 3 2G-1 (of 19) H H B H SN = 3 Trigonal Planar H HH B If the B undergoes sp 3 hybridization: Central atoms with SN = 3 only need to hybridize 3 valance atomic orbitals Trigonal pyramidal sp 3

BH 3 E 2s 2p hybridization sp 2 2p... B 2s 2p y 2p x 2p z.. B. sp 2 2p z 2G-2 (of 19)

2p z 2p x 2p y 2s hybridization The angle between the sp 2 hybrid orbitals is 120º sp 2 2p z sp 2 The 2p z orbital is 90º from the plane of the sp 2 hybrid orbitals 2G-3 (of 19)

2p z 2p x 2p y 2s hybridization The angle between the sp 2 hybrid orbitals is 120º sp 2 2p z sp 2 The 2p z orbital is 90º from the plane of the sp 2 hybrid orbitals 2G-4 (of 19)

2p z 2p x 2p y 2s hybridization sp 2 2p z sp 2 Rotating the top 90º towards you sp 2 2G-5 (of 19)

sp 2 2p z sp 2 Rotating the top 90º towards you 2G-6 (of 19)

Rotating the top 90º towards you 2G-7 (of 19)

This forms a trigonal planar molecule with a bond angle of 120º Each bond is named:σ(sp 2 +1s) 2G-8 (of 19)

BeH 2 H Be H SN = 2 Linear If the B undergoes sp 3 hybridization: Central atoms with SN = 2 only need to hybridize 2 valance atomic orbitals Bent, 109.5º H Be H If the B undergoes sp 2 hybridization:Bent, 120º 2G-9 (of 19)

BeH 2 E 2s 2p hybridization sp 2p.. Be 2s 2p y 2p x 2p z.. Be sp 2p y sp 2p z 2G-10 (of 19)

2p z 2p x 2p y 2s hybridization The sp hybrid orbitals are linear, with an angle of 180º between them sp 2p y 2p z sp The 2p y and 2p z orbitals are 90º from the sp hybrid orbitals 2G-11 (of 19)

sp 2p y 2p z sp This forms a linear molecule with a bond angle of 180º Each bond is named:σ(sp+1s) 2G-12 (of 19)

PH (1) = 10 valence e - s HPHP HH HH 2G-13 (of 19) Central atoms with SN = 5 need to hybridize 5 valance atomic orbitals

PH 5 E 3s 3phybridization sp 3 d 3d These 5 hybrid orbitals are called 3d sp 3 d orbitals 2G-14 (of 19)

sp 3 d This forms a trigonal bipyramidal molecule Each bond is named:σ(sp 3 d+1s) PH 5 2G-15 (of 19)

SH 6 HSHHSH H H H H 6 + 6(1) = 12 valence e - s 2G-16 (of 19) Central atoms with SN = 6 need to hybridize 6 valance atomic orbitals

SH 6 E 3s 3phybridization sp 3 d 2 3d These 6 hybrid orbitals are called 3d sp 3 d 2 orbitals 2G-17 (of 19)

sp 3 d 2 SH 6 This forms an octahedral molecule Each bond is named:σ(sp 3 d 2 +1s) 2G-18 (of 19)

SNHybrid sp sp 2 sp 3 sp 3 d sp 3 d 2 Hybrid Orbital Geometry Linear Trigonal Planar Tetrahedral Trigonal Bipyramidal Octahedral sp 2 sp 3 sp 3 d sp 3 d 2 sp 2G-19 (of 19)

HYDROCARBON – Molecules composed of only carbon and hydrogen SATURATED HYDROCARBON – Hydrocarbons with only single bonds between the carbon atoms The ending -ane indicates only single bonds between the carbon atoms 1meth- 2eth- 3prop- 4but- 5pent- 6hex- 7hept- 8oct- 9non- 10dec- The prefix tells the number of carbon atoms in the molecule 2H-1 (of 10)

methane: CH 4 H H C H H SN of C = 4 Tetrahedral Nonpolar Each bond: σ(sp 3 +1s) HC HC H H H  Hybridization = sp 3 sp 3 2H-2 (of 10)

ethane: C 2 H 6 H H C C H H SN of each C = 4  Hybridization = sp 3 sp 3 C-C bond: σ(sp 3 +sp 3 ) Each C-H bond: σ(sp 3 +1s) HC HC H H H H H C 2H-3 (of 10)

UNSATURATED HYDROCARBON – Hydrocarbons with at least one double or triple bond between carbon atoms The ending -ene indicates a double bond between carbon atoms A double bond consists of a SIGMA BOND and a PI BOND PI BOND (  ) – A bond that is only symmetrical upon a 180º rotation around its internuclear axis 2H-4 (of 10)

ethene: C 2 H 4 H H C C H SN of each C = 3  Hybridization = sp 2 C-C bonds: σ(sp 2 +sp 2 )  (2p+2p) Each C-H bond: σ(sp 2 +1s) sp 2 2p z sp 2 C H H C H H 2H-5 (of 10)

The ending -yne indicates a triple bond between carbon atoms A triple bond consists of a 1 SIGMA BOND and 2 PI BONDS 2H-6 (of 10)

sp ethyne: C 2 H 2 H C C H SN of each C = 2  Hybridization = sp C-C bonds: σ(sp+sp)  (2p z +2p z )  (2p y +2p y ) Each C-H bond: σ(sp+1s) sp 2p z sp 2p y 2p z sp 2p z 2p y C HCH 1s 2H-7 (of 10)

BOND ORBITAL MODELS (BOM’s) – Geometric diagrams of the bonding in molecules To draw a BOM for a molecule: 1 – Draw the Lewis structure and predict the hybridizations of each atom 2 – Draw the correct geometries for the hybrid atomic orbitals of each atom, starting with atoms involved in double or triple bonds 3 – Have atoms bond by overlapping atomic orbitals, and label each bond 2H-8 (of 10)

H H C H H C H H C Draw a BOM for CH 3 CHCH 2 C 1 sp 3 C 2 sp 2 C 3 sp 2 C H C H H H H H C C 2 -C 3 bonds: C 2 -H & C 3 -H bonds: C 1 -C 2 bond: C 1 -H bonds: 2H-9 (of 10) σ(sp 2 +sp 2 )  (2p+2p) σ(sp 2 +1s) σ(sp 3 +sp 2 ) σ(sp 3 +1s)

H H C C H C H C 1 sp 2 C 2 sp C 3 sp 2 Draw a BOM for CH 2 CCH 2 C C H H H C C-C bonds: C-H bonds: H 2H-10 (of 10) σ(sp 2 +sp)  (2p+2p) σ(sp 2 +1s)

Valence Bond Theory cannot predict (1)magnetism of molecules (2)the e - arrangement of molecules with an odd number of e - s (3)the e - arrangement in molecules with resonance MOLECULAR ORBITAL THEORY – Electrons in molecules exist in molecular orbitals that extend over the entire molecule 2I-1 (of 7)

H2H2 The number of valence MO’s in the H 2 molecule equals the number of valence AO’s in the 2 individual H atoms 2 valence AO’s in the atoms,  2 valence MO’s in the molecule 1s AO This MO is the result of the addition of the 2 AO’s: 1s + 1s This MO is the result of the subtraction of the 2 AO’s: 1s - 1s 2I-2 (of 7)

SIGMA MOLECULAR ORBITAL – An orbital that is completely symmetrical around its internuclear axis MO 1 MO 2 BONDING MOLECULAR ORBITAL – An orbital lower in energy than the original atomic orbitals because most of the e - density is between the 2 nuclei σ σ b * ANTIBONDING MOLECULAR ORBITAL – An orbital higher in energy than the original atomic orbitals because most of the e - density is outside the 2 nuclei 2I-3 (of 7)

MO Energy Level Diagram for H 2 Eatom 1atom 2molecule 1s σ 1s b σ 1s * 2I-4 (of 7)

Bond Order = Bonding e - s – Antibonding e - s _______________________________________ 2 Bond Order for H 2 = 2 – 0 ______ 2 = 1 PARAMAGNETIC – A substance that is magnetic in a magnetic field Due to unpaired e - s DIAMAGNETIC – A substance that is never magnetic Due to no unpaired e - s H 2 is diamagnetic Electron configuration for H 2 :  σ 1s b ) 2 2I-5 (of 7)

He 2 The valence AO’s of each He atom combine to form MO’s in the He 2 molecule MO Energy Level Diagram for He 2 E 1s σ 1s b σ 1s * Bond Order for He 2 = (2 – 2) / 2= 0  He 2 does not exist 2I-6 (of 7)

Does He 2 + exist? If yes, give bond order, magnetism, and electron configuration notation MO Energy Level Diagram for He 2 + E 1s σ 1s b σ 1s * Bond Order for He 2 + = (2 – 1) / 2= ½  He 2 + does exist He 2 + is paramagneticElectron configuration for He 2 + :  σ 1s b ) 2 (σ 1s *) 1 2I-7 (of 7)

Diatomic Homonuclear Molecules of the 2 nd Period E atom 1atom 2molecule Valence Orbitals 2s 2p 2s 2p ? 2J-1 (of 15)

The 2 2s AO’s combine the same way as 2 1s AO’s, forming a  2s b MO and a  2s * MO 2p z 2p x 2p y 2p z 2p x 2p y Each pair of p orbitals will combine to form a bonding MO and an antibonding MO 2J-2 (of 15)

2p x  2px b 2p x  2px * 2J-3 (of 15)

2p z PI MOLECULAR ORBITAL – An orbital that is symmetrical only upon a 180º rotation around the internuclear axis  2pz b  2pz * 2p z 2J-4 (of 15)

The 2 2p y AO’s combine the same way as 2 2p z AO’s, forming a  2py b MO and a  2py * MO The  2pz b MO and the  2py b MO are equal in energy The  2pz * MO and the  2py * MO are equal in energy 2J-5 (of 15)

E atom 1atom 2molecule 2s 2p 2s 2p  2s b  2s *  2px b  2px *  2pz *  2py *  2pz b  2py b Experimental data has determined the energies of the 8 MOs 2J-6 (of 15)

2s 2p 2s 2p  2s b  2s *  2px b  2px *  2pz *  2py *  2pz b  2py b C2C2 Bond Order for C 2 = (6 – 2) / 2= 2 C 2 is diamagnetic Electron configuration for C 2 :  2s b ) 2 (  2s *) 2 (  2p b ) 4 E 2J-7 (of 15)

2s 2p 2s 2p  2s b  2s *  2px b  2px *  2pz *  2py *  2pz b  2py b F2F2 Bond Order for F 2 = (8 – 6) / 2= 1 F 2 is diamagnetic Electron configuration for F 2 :  2s b ) 2 (  2s *) 2 (  2p b ) 4 (  2p b ) 2 (  2p *) 4 E 2J-8 (of 15)

C2C2 Bond Order= 2F2F2 Bond Order= 1 Highest Bond Energy: Longest Bond Length: C2C2 F2F2 2J-9 (of 15)

Diatomic Heteronuclear Molecules of the 2 nd Period 2s 2p 2s 2p  2s b  2s *  2px b  2px *  2pz *  2py *  2pz b  2py b CN Bond Order for CN= (7 – 2) / 2= 2½ CN is paramagnetic Electron configuration for C 2 :  2s b ) 2 (  2s *) 2 (  2p b ) 4 (  2p b ) 1 E N’s greater nuclear charge makes its AO energies lower that C’s 2J-10 (of 15)

Benzene, C 6 H 6 C C C C C C H H H H H H C C C C C C H H H H H H Each C is sp 2, and all atoms are planar Each C has a p orbital perpendicular to the plane 2J-11 (of 15) Molecular Orbital Theory can also explain the e - arrangement in molecules with resonance

Resonance structures assume that these p orbitals make distinct pi bonds Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6 MO’s 2J-12 (of 15)

Resonance structures assume that these p orbitals make distinct pi bonds Molecular Orbital Theory predicts the 6 2p AO’s will combine to make 6 MO’s 2J-13 (of 15)

2J-14 (of 15) DELOCALIZED PI SYSTEM – A group of pi molecular orbitals spread out over more than 2 atoms The e - s in the delocalized pi system strengthen the bonds in the ring Molecules possessing resonance always bond with delocalized pi systems, which are formed from PARALLEL P ORBITALS

NO 3 - O N O O - O N O O -- O N O O ↔↔ All atoms that can form a double bond in at least 1 resonance structure must be sp 2 to have an unhybridized p orbital to form the delocalized pi system NO O O NO O O 2J-15 (of 15)

EMPIRICAL FORMULA CALCULATIONS EMPIRICAL FORMULA – The simplest whole-number ratio of the atoms of different elements in a compound C6H6C6H6 C1H1C1H1 C 6 H 12 O 6 C1H2O1C1H2O1 H2OH2O H2OH2O Molecular Formula: Empirical Formula: 2K-1 (of 8)

1)Assume you have 100 g of the compound Find the empirical formula of a compound that is 75.0% carbon and 25.0% hydrogen by mass.  75.0 g C and 25.0 g H 2)Calculate the moles of atoms of each element x 1 mol C _____________________ g C = mol C 75.0 g C x 1 mol H ____________________ 1.01 g H = mol H 25.0 g H 3)Divide each number of moles by the smallest number of moles mol C _______________________ mol H _______________________ = 1.00 mol C= 3.96 mol H 4)The integer mole ratio must be the atom ratio:CH 4 2K-2 (of 8)

Find the empirical formula of a compound that is 90.0% carbon and 10.0% hydrogen by mass. x 1 mol C _____________________ g C = mol C 90.0 g C x 1 mol H ___________________ 1.01 g H = mol H 10.0 g H mol C _______________________ mol H _______________________ = mol C= mol H If the moles of all elements are not within 0.1 moles of an integer, they must all be multiplied by an integer until they are integers  2 = 2.00 mol C= 2.64 mol H  3 = 3.00 mol C= 3.96 mol H Empirical formula: C 3 H 4 2K-3 (of 8)

MOLAR MASSES OF COMPOUNDS MOLAR MASS – The mass of one mole of molecules of a molecular substance, or one mole of formula units of an ionic substance The mass necessary to have 1 mole of Al 2 (SO 4 ) 3 formula units Al 2 (SO 4 ) 3 2 mol Al (26.98 g/mol)= g 3 mol S (32.07 g/mol)=96.21 g 12 mol O (16.00 g/mol)= g g 2K-4 (of 8)

CCl 4 1 mol C (12.01 g/mol)= g 4 mol Cl (35.45 g/mol)= g g Calculate the molar mass of carbon tetrachloride  g CCl 4 = 1 mol CCl 4 Calculate the number of chlorine atoms in 1.00 g carbon tetrachloride. x 1 mol CCl 4 __________________ g CCl 4 = 1.57 x atoms Cl 1.00 g CCl 4 x 4 mol Cl ______________ 1 mol CCl 4 x x atoms Cl _____________________________ 1 mol Cl 2K-5 (of 8)

MOLECULAR FORMULA CALCULATIONS MOLECULAR FORMULA – The actual number of the atoms of different elements in a molecule C1H1C1H1 or C 2 H 2 Empirical Formula: Molecular Formula: C1H1C1H1 or C 3 H 3 or C 4 H 4 etc. 2K-6 (of 8)

Find the molecular formula of a compound that is 43.7% phosphorus and 56.3% oxygen by mass, and has a molar mass of about 280 g/mol. x 1 mol P _____________________ g P = mol P 43.7 g P x 1 mol O _____________________ g O = mol O 56.3 g O mol P _______________________ mol O _______________________ = 1.00 mol P= 2.49 mol O Empirical formula: P 2 O 5 Find the molar mass of the empirical formula  2 = 2.00 mol P= 4.98 mol O 2K-7 (of 8)

P2O5P2O5 2 mol P (30.97 g/mol)= g 5 mol O (16.00 g/mol)=80.00 g g Divide the compound’s actual molar mass by the empirical formula’s molar mass – it should be very close to an integer 280 g/mol ____________________________ g/mol ≈ 2  The molecular formula is 2 times the empirical formula Molecular formula: P 4 O 10 2K-8 (of 8)