Hybridization sp3 Hybridization efficently bonds to four atoms and lone pairs Mix3 carbon p + 1 carbon s to get 4 equivalent sp3 orbitals (sp3 = 1 part s, 3 parts p) H y 4 sp3 orbitals H C H

+ x + + H z

1 1 1 1 2s 2px 2py 2pz note: an excited state configuration

We now have 4 sp3(C)-s(H) bonds of equal length at 109.5° apart. Hybridization with multiple bonds H H We need 3 orbitals to make 3 bonds

C C Mix 2 carbon p + 1 carbon s to get 3 equivalent sp2 orbitals

H H H + 2 H + 3 sp orbitals x C C z H sp2 σ H 1 1 1 2s 2px 2pz 120° C

y π The remaining Each unpaired electron on H H orbital is an carbon forms a bond with the unhybridized p other unpaired electron (this orbital forms a π bond) H H 1 2py

H C C This time, we only need 2 orbitals to make 2 bonds:

Mix 1 carbon s, 1 carbon p to get 2 equivalent sp orbitals + x H C C sp σ sp 1 1 C 2s 2px π 180° C y π H C C The remaining orbitals are unhybridized p z Each unpaired electron on carbon forms a bond with orbitals 1 1 the other unpaired electron (this forms a π bond) 2pz 2py Note that these orbitals (px and pz) are 90°C apart. Example Problems H

Guide for determining hybridization: H C C O H

1. Draw a good Lewis structure. H O 3 bonded atoms + 0 lone pairs 2. Focus on an atom and sum up: # bonded atoms + # of lone pairs Sum of 3 means sp2 hybridization, 120°C

3. This will give a sum between 2-4:

Sum Hybridization Geometry Bond angle 3 bonded atoms + 1 lone pair 2 sp linear 180° H 3 sp2 trigonal planar 120° Sum of 4 means 4 sp3 tetrahedral 109.5° sp3 hybridization H C N H 109.5°C Note: This is a guide not an absolute rule. Atoms will rehybridize to place lone pairs in conjugated H H p-orbitals if that can lower energy 3rd row elements can access d-orbitals.

H H

H C O H C H C C 3 bonded atoms + 0 lone pairs sp2 hybridization, 120°C 3 bonded atoms + 0 lone pairs C C sp2 hybridization, 120°C H N H 2 bonded atoms + 1 lone pair 2 bonded atoms + 2 lone pairs (not shown, but sp2 hybridization (lone pair not shown, but can be can be determined from a good Lewis Dot structure) 120°C determined from a good Lewis Dot sp3 hybridization, ~109.5°C structure)

HO N O 2 bonded atoms + 0 lone C H pairs R C C N N H C CH 3 3 sp hybridization, 2 bonded atoms 180°C + 1 lone pair (*see note above) sp2 hybridization, 120°C B NH2

N Challenge Problem! E A N C

Identify the hybridization O O O C N N H and bond angle of each O P O P O P O

atom indicated. CH2

C) sp C)

, 120° , C D) sp D) C ,120°

C E)sp C d, 109.5° d, O 2

2 2 O O O

3 D but sp but H H due to resonance resonance to due

2 2

A) sp A)

, 109.5° , C B) Appears to be sp be to Appears B) C H H

, , 3 3 3 OH H