Lewis Structures H is always a terminal atom The atom with the lowed EN is the central atom Find the total # of valence (ions add and lose electrons) Bond each atom to the central atom with a single bond Place lone pairs around terminal atoms to complete their octet Add remaining electrons to central atom Use multiple bonds to complete the octet of the central atom

Exceptions to the Octet Rule Hydrogen only gets 2 Boron is a moron, he likes 6 electrons Expanded Valence – 3 rd energy level or higher

HYBRID ORBITALS covalent bonds are formed by overlap of atomic orbitals atomic orbitals on the central atom can mix and exchange their character – hybridization ONLY  BONDS ARE FORMED FROM OVERLAP OF HYBRIDS ORBITALS  ARE NOT FORMED FROM HYBRID ORBITALS

Number of electron structural pairs Atomic orbitals Hybrid orbitals formed Geometry 2 one s + one p two sp AX 2 : linear AX 2 : linear 3 one s + two p three sp 2 AX 3 : trigonal planar AX 3 : trigonal planar 4 one s + three p four sp 3 AX 4 : tetrahedral AX 4 : tetrahedral 5 s + three p + one d five sp 3 d AX 5 : trigonal bipyramidal AX 5 : trigonal bipyramidal 6 one s + three p + two d six sp 3 d 2 AX 6 : octahedral AX 6 : octahedral

practice

2 sp Hybrids Formed From an s and p

3 sp 2 Hybrids Formed From an s and 2 p’s

Forming sigma bonds Three ways atoms can form a sigma bond: – 2 hydrogen atoms’ 1s orbitals overlap – a hybrid orbital overlaps a hydrogen’s 1s orbital – 2 hybrid orbitals overlap end-to-end

Forming pi bonds Pi bonds form when – atoms are already bonded with a sigma bond – both atoms have a leftover p orbital Pi bonds form when two atoms’ p orbitals overlap side-by-side

NO 2 1- Resonance Structures    next formal charge

Delocalized pi bonding in CO 3 –2 In the CO 3 –2 ion, one pi bond is delocalized over three C-O connections The C atom and all three O atoms are sp 2 hybridized and have a leftover p orbital The four p orbitals overlap side-to-side to create the pi bonding

C 6 H 6, Benzene an Important Molecule With Resonance Delocalized e - due to resonance explains why the bonds in benzene are identical.

LINEAR AX 2 EXAMPLE BH 2

AX 2 WITH DOUBLE BOND CO 2

AX 3 EXAMPLE BH 3 TRIGONAL PLANAR

AX 4 EXAMPLE CH 4

AX 3 E EXAMPLE NH 3 PYRAMIDAL

AX 2 E 2 EXAMPLE H 2 S BENT

AX 5 EXAMPLE PCl 5 TRIGONAL BIPYRAMIDAL

AX 4 E EXAMPLE SF 4 SEE SAW

AX 3 E 2 EXAMPLE BF 3 T SHAPE

EXAMPLE XeF 2 LINEAR

Next resonance

EXAMPLE SeF 6 OCTAHEDRAL

EXAMPLE IF 5 SQUARE PYRAMIDAL

EXAMPLE XeF 4 SQUARE PLANAR

Polar molecules A compound’s polarity can be measured experimentally as its dipole moment Polar molecules align themselves with an electrical field  +  – – – – – – –

Polar molecules The C=O bond is polar But the dipole moment of CO 2 is zero CO 2 is not a polar molecule

Examples SO 2 S-O bonds are polar Geometry is bent (120°) sp Bond dipoles don’t cancel SO 2 is polar SO 3 S-O bonds are polar Geometry is trigonal planar sp 2 Bond dipoles do cancel SO 3 is nonpolar

Examples H 2 O H-O bonds are polar H-O bonds are polar Geometry is bent (109.5°) Geometry is bent (109.5°) sp sp Bond dipoles don’t cancel Bond dipoles don’t cancel H 2 O is polar H 2 O is polar OCl 2 O-Cl bonds are nonpolar O-Cl bonds are nonpolar sp sp OCl 2 is nonpolar OCl 2 is nonpolar

SF 4 S-F bonds are polar geometry is see saw dsp 3 axial bond dipoles cancel equatorial bond dipoles don’t cancel SF 4 is polar PF 5 P-F bonds are polar geometry is trigonal bipyramid dsp 3 axial bond dipoles cancel equatorial bond dipoles cancel PF 5 is nonpolar

XeF 2 Xe-F bonds are polar geometry is linear – structural pair geometry is trigonal bipyrimidal – lone pairs in equatorial – F atoms in axial – dsp 3 bond dipoles cancel XeF 2 is nonpolar PbH 2 Pb-H bonds are nonpolar PbH 2 is nonpolar