Chapter 9 Molecular Shapes -shape of molecule is based on bond angles Valence Shell Electron Pair Repulsion (VSEPR) -based on the idea that electron groups repel one another

-for molecules having one central atom, shape is based on: e- domain- region where e- are likely to be found one e- domain = a lone pair, single, double or triple bond

-must have the structural formula drawn -use the following formula AB n E n A= central atom B= # of bonding domains on central atom E= # of lone pairs of e- on central atom e- geometry- shows arrangement of e- domains molecular geometry- arrangement of the atoms

Molecular Shape and Polarity -to determine if an entire molecule is polar you must look at the bonds and the shape dipole moment- a measure of the separation of + and – charges in a molecule bond dipole- dipole moment due only to the two atoms in that bond

-bond dipoles and dipole moments are vector quantities -have both magnitude and direction and must look at both when determining polarity -vectors will be added

To determine if a molecule is polar: -determine the shape -determine whether the molecule contains polar bonds -determine whether the polar bonds add together to form a net dipole moment -if they sum to zero then the molecule is nonpolar -if not then the molecule is polar -if the molecule has lone pairs, it is polar

Tro, Chemistry: A Molecular Approach 7

Examples: a)BrCℓ polar b)SO 2 polar *has lone pair c)SF 6 nonpolar d)NF 3 polar *lone pair e) BCℓ 3 nonpolar

Valence Bond Theory -valence e- of atoms in a molecule reside in atomic orbitals which can be s, p, d, f or some hybrid of these -a chemical bond results from the overlap of two half-filled orbitals, or less commonly from the overlap of a completely filled orbital with an empty orbital -the shape of the molecule is determined by the geometry of the overlapping orbitals

Ex: H 2 S

-overlap of half-filled orbitals does not explain bonding in all types of molecules -for example: CH 4 -b/c there are only two half-filled orbitals, we would assume that C bonds with only two H -we know it bonds with 4 H

-we must consider hybrid orbitals here hybridization- standard atomic orbitals are combined to form hybrid orbitals that correspond more closely to the actual distribution of e- in chemical bonds

General Statements Regarding Hybridization -# of standard atomic orbitals added together = # of hybrid orbitals formed -the combo of orbitals determines the shapes and energies of orbitals formed -the specific type of hybridization that forms for a molecule is the one that yields the lowest overall energy for the molecule

-a single bond contains a sigma bond (σ)- head on overlap -a double bond is made up of one sigma and one pi bond (π)- sideways overlap -a triple bond is made up of one sigma and two pi bonds

sp 3 hybridization -occurs when there are 4 e- groups (tetrahedral) -for CH 4, one s orbital and three p orbitals hybridize to form four sp 3 orbitals of equal energy

-C has four half-filled orbitals to overlap with four half-filled H 1s orbitals *all are single bonds sigma bonds (σ)

sp 2 hybridization and double bonds -hybridization of one s orbital and two p orbitals results in three sp 2 hybrids and one leftover unhybridized p orbital -occurs when you have 3 e- groups (tri. planar)

Example: H 2 CO

-because C is the central atom it undergoes hybridization *made up of two sigma bonds from the C-H single bonds, and one sigma and one pi bond (π) from the double bond of C=O **Try H 2 C 2 Cℓ 2

sp hybridization and triple bonds -one s orbital and one p orbital hybridize to form two sp orbitals of equal energy -two p orbitals remain unhybridized -occurs when there are two e- groups (linear)

Example: C 2 H 2 -has a triple bond

Practice: Tell the type of hybridization of the C atom(s), what types of bonds, and what kinds of orbitals for each. 1)H 3 C 2 OH2) HCN3) C 2 H 4 1) 1 st C= sp 3, 4 σ bonds, 1s and 3p orbitals 2 nd C= sp 2, 3 σ and 1 π bond, 1s, 2p, and 1 unhybridized p orbital 2)sp, 2 σ bonds, 2 π bond, 1s, 1p and 2 unhybridized p orbitals 3)**both interior C the same= sp 2, 3 σ bonds, 1 π bond, 1s, 2p, and 1 unhybridized p orbital

-if trigonal bypyramidal, hybridization is sp 3 d -if octahedral, hybridization is sp 3 d 2

localization of e-: σ and π e- are associated only with the two atoms that form the bond delocalization of e-: e- in π bonds extend over more than just the two atoms *occurs when a molecule has at least two resonance structures

Molecular Orbital Theory -orbitals are treated as overlapping the entire molecule, not as “belonging” to individual atoms -when atomic orbitals are combined to form molecular orbitals, the total number of orbitals does not change

-any time you bond two atomic orbitals you get two molecular orbitals: bonding orbital (σ or π ) - lower in energy anti-bonding (σ* or π* ) - higher in energy *b/c lower energy bonding fills first *each orbital can hold two e- *will show if a single, double or triple bond is formed by calculating the bond order bond order = # e- in bonding - # e- in antibonding 2 -if bond order is zero or less, no bond will form

-σ are one box b/c of head on overlap -π are two boxes b/c of sideways overlap s orbitals = σ and σ* for the bonding and antibonding p orbitals = π, σ, π* and σ* for bonding and antibonding

paramagnetism- having unpaired e- diamagnetism- having no unpaired e-