Chemistry 100 Chapter 9 Molecular Geometry and Bonding Theories

Molecular Geometry  The three-dimensional arrangement of atoms in a molecule  molecular geometry  Lewis structures can’t be used to predict geometry  Repulsion between electron pairs (both bonding and non-bonding) helps account for the molecular structure!

The VSEPR Model  Electrons are negatively charged, they want to occupy positions such that electron Electron interactions are minimised as much as possible  Valence Shell Electron-Pair Repulsion Model Valence Shell Electron-Pair Repulsion Model treat double and triple bonds as single domains resonance structure - apply VSEPR to any of them formal charges are usually omitted

Molecules With More Than One Central Atom  We simply apply VSEPR to each ‘central atom’ in the molecule. Carbon #1 – tetrahedral Carbon #2 – trigonal planar

Dipole Moments  The HF molecule has a bond dipole – a charge separation due to the electronegativity difference between F and H.  The shape of a molecule and the magnitude of the bond dipole(s) can give the molecule an overall degree of polarity  dipole moment.  + H-F 

 Homonuclear diatomics  no dipole moment (O 2, F 2, Cl 2, etc)  Triatomic molecules (and greater). Must look at the net effect of all the bond dipoles.  In molecules like CCl 4 (tetrahedral) BF 3 (trigonal planar) all the individual bond dipoles cancel  no resultant dipole moment.

Bond Dipoles in Molecules

More Bond Dipoles

Valence Bond Theory and Hybridisation  Valence bond theory  description of the covalent bonding and structure in molecules.  Electrons in a molecule occupy the atomic orbitals of individual atoms.  The covalent bond results from the overlap of the atomic orbitals on the individual atomscovalent bond

The Bonding in H 2  Hydrogen molecule a single bond indicating the overlap of the 1s orbitals on the individual atoms cylindrical symmetry with respect to the line joining the atomic centres, i.e., a  bond Overlap Region 1s (H1) – 1s(H2)  bond

The Bonding in H 2

The Cl 2 Molecule  In the chlorine molecule, we observe a single bond indicating the overlap of the 3p orbitals on the individual atoms. Bonding description 3p z (Cl 1) – 3p z (Cl 2)

Is This a  Bond?

Hybrid Atomic Orbitals  Look at the bonding picture in methane (CH 4 ). Bonding and geometry in polyatomic molecules may be explained in terms of the formation of hybrid atomic orbitals Bonds  overlap of the hybrid atomic orbitals on central atoms with appropriate half-filled atomic orbital on the terminal atoms.

The CH 4 Molecule

The Formation of the sp 3 Hybrids  Mix 3 “pure” p orbitals and a “pure” s orbital form an sp 3 “hybrid” orbital.  Rationalize the bonding around the C central atom.

sp 2 Hybridisation  What if we try to rationalise the bonding picture in the BH 3 (a trigonal planar molecule)? We mix 2 “pure” p orbitals and a “pure” s orbital to form “hybrid” or mixed sp 2 orbitals. These three sp 2 hybrid orbitals lie in the same plane with an angle of 120 between them.

A Trigonal Planar Molecule Overlap regions Overlap region

sp Hybridisation  What if we try to rationalize the bonding picture in the BeH 2 species (a linear molecule)?  We mix a single “pure” p orbital and a “pure” s orbital to form two “hybrid” or mixed sp orbitals  These sp hybrid orbitals have an angle of 180 between them.

A Linear Molecule The BeH 2 molecule Overlap Regions

Double Bonds  Look at ethene C 2 H 4.  Each central atom is an AB 3 system, the bonding picture must be consistent with VSEPR theory.

The  Bond  Additional feature an unhybridized p orbital on adjacent carbon atoms.  Overlap the two parallel 2p z orbitals (a -bond is formed).

Bond overlaps in C 2 H 4  There are three different types of bonds [sp 2 (C ) – 1s (H) ] x 4 type [sp 2 (C 1 ) – sp 2 (C 2 ) ]  type [2p z (C 1 ) – 2p z (C 2 ) ]  type

The C 2 H 4 Molecule

The Bond Angles in C 2 H 4  Bond angles HCH = HCC  120.  bond is perpendicular to the plane containing the molecule.  Double bonds – Rationalize by assuming sp 2 hybridization exists on the central atoms! Any double bond  one  bond and a  bond

The Triple Bond in C 2 H 2  Bond angles HCH = HCC = 180.  bonds are perpendicular to the molecular plane.  Triple bond  one  bond and two  bonds Triple bond rationalized by assuming sp hybridization exists on the central atoms!

Bond Overlaps in C 2 H 2  There are again three different types of bonds [sp (C ) – 1s (H) ] x 2 type [sp (C 1 ) – sp (C 2 ) ]  type [2p y (C 1 ) – 2p y (C 2 ) ]  type [2p z (C 1 ) – 2p z (C 2 ) ]  type

Bond Overlaps in H 2 CO  There are again three different types of bonds [sp (C) – 1s (H) ] x 2 type [sp 2 (C) – sp 2 (O) ]  type [2p (C) – 2p (O) ]  type

Key Connection – VSEPR and Valence Bond Theory!!

sp 3 d Hybridisation  How can we use the hybridisation concept to explain the bonding picture PCl 5.  There are five bonds between P and Cl (all  type bonds).  5 sp 3 d orbitals  these orbitals overlap with the 3p orbitals in Cl to form the 5  bonds with the required VSEPR geometry  trigonal bipyramid.  Bond overlaps [sp 3 d (P ) – 3p z (Cl) ] x 5 type

sp 3 d 2 Hybridisation  Look at the SF 6 molecule.  6 sp 3 d 2 orbitals  these orbitals overlap with the 2p z orbitals in F to form the 6  bonds with the required VSEPR geometry  octahedral.  Bond overlaps [sp 3 d 2 (S ) – 2p z (F) ] x 6 type

Notes for Understanding Hybridisation  Applied to atoms in molecules only  Number hybrid orbitals = number of atomic orbitals used to make them  Hybrid orbitals have different energies and shapes from the atomic orbitals from which they were made.  Hybridisation requires energy for the promotion of the electron and the mixing of the orbitals  energy is offset by bond formation.

Delocalised Bonding  In almost all the cases where we described the bonding n the molecule, the bonding electrons have been totally associated with the two atoms that form the bond  they are localised.  What about the bonding situation in benzene, the nitrate ion, the carbonate ion?  In benzene, the C-C  bonds are formed from the sp 2 hybrid orbitals. The unhybridised 2p z orbital on C overlaps with another 2p z orbital on the adjacent C atom.

 Three bonds are formed. These  bonds extend over the whole molecule (i.e. the bonds are delocalised).  The  electrons are free to move around the benzene ring.  Any species where we had several resonance structures, we would have delocalisation of the - electrons.

Delocalised Electrons in Molecules

Molecular Orbital (M.O.) Theory  Valence bond and the concept of the hybridisation of atomic orbitals does not account for a number of fundamental observations of chemistry.  To reconcile these and other differences, we turn to molecular orbital theory (MO theory).  MO theory – covalent bonding is described in terms of molecular orbitals the combination of atomic orbitals that results in an orbital associated with the whole molecule.

 Recall the wave properties of electrons. constructive interference  the two e - waves interact favourably; loosely analogous to a build-up of e - density between the two atomic centres. destructive interference  unfavourable interaction of e - waves; analogous to the decrease of e - density between two atomic centres.

Constructive and Destructive Interference + + Constructive Destructive

 bonding = C 1  ls (H 1) + C 2 ls (H 2)  anti = C 1  ls (H 1) - C 2 ls (H 2)  Bonding Orbital  a centro-symmetric orbital (i.e. symmetric about the line of symmetry of the bonding atoms).  Bonding M’s have lower energy and greater stability than the AO’s from which it was formed.  Electron density is concentrated in the region immediately between the bonding nuclei.

 Anti-bonding orbital  a node (0 electron density) between the two nuclei.  In an anti-bonding MO, we have higher energy and less stability than the atomic orbitals from which it was formed.  As with valance bond theory (hybridisation) 2 AO’s  2 MO’s

Bonding and Anti-Bonding M.O.’s from 1s atomic Orbitals

The MO’s in the H 2 Atom

 The situation for two 2s orbitals is the same! The situation for two 3s orbital is the same.  Let’s look at the following series of molecules H 2, He 2 +, He 2 bond order = ½ {bonding - anti-bonding e - ‘s}.  Higher bond order  greater bond stability.