The forces at work are Electromagnetic Repulsion of
- Slides: 55
The forces at work are: Electromagnetic: Repulsion of like charges: proton-proton and electron-electron Attraction of unlike charges: proton-electron. Magnetic: Electron spin introduces magnetic forces. Electrons with opposite spins produce opposite magnetic poles. Opposite magnetic poles attract. Opposite spin is required for electrons to occupy the same orbital.
The Energies involved: Electron (orbital) Potential Energy: E NE RG Y LE V E L 3 d 3 s 2 p 2 s 1 s 3 p 4 s 4 p
Bond Angles 109. 5 o 180 o 120 o 90 o
Px Px Pz Pz Py Py
Representation of Orbitals The wavefunction (Psi) of 1 electron system (e. g. H, He+) is a solution of the Schrödinger equation. It describes the behaviour of an electron in a region of space called an atomic orbital ( ɸ–phi). = radial (r) angular ( , ) = Rn, l (r) Yl, m ( , ) An atomic orbitals which describe the 1 electron behaviour (1 electron wavefunction) in an atom, have Size determine by Rn, l(r), changes with distance r from nucleus Shape determined by Yl, m ( , ) (also called Spherical Harmonics)
Radial wavefunction (Rn, l)of Hydrogenic atoms The white regions represent the radial nodes, where probability of finding an electron is zero. Number of Radial Nodes : n – l – 1, where n and l are principal, azumithal quantum numb
Angular wavefunction Yl, m The angular wavefunction is again made-up two parts as follows Yl, m ( , ) = Sl, m ( ) Tm ( ) Associated Legendre functions Where, Associated Legendre Polynomial More details : Quantum Chemistry, Ira. N. Levine, Printice Hall. New Jersey.
Plotting Sl, m ( ) against various values of between 0 to 360, would provide the shapes of different orbitals in two dimensional plane. The angle , would provide third dimension to these shape of orbital
Pz- Orbital
dz 2 -orbital S 2 0 = 3 cos 2 ( ) -1
CONCEPT MAP
VSEPR THEORY Ø Predicts the molecular shape of a bonded molecule Ø Electrons around the central atom arrange themselves as far apart from each other as possible Ø Unshared pairs of electrons (lone pairs) on the central atom repel the most
Contd…. . VSEPR THEORY Ø The nonbonding electron pairs are as important as bonding electron pairs in determining the structure. Ø If one or more of the electron pairs are lone pairs, the distribution of electron pair and the geometrical shape of the molecule must be different. Ø The bond angles decrease as the number of nonbonding electron pairs increases ØElectron pairs (bonding pairs and lone pairs) are negatively charged and will repel each other ØThese pairs of electrons tend to occupy positions around the atom that minimize repulsions and maximize the distance of separation between them
Contd… Three types of repulsions take place in an atom Ø Lone Pair – Lone Pair (LP-LP) Ø Lone Pair – Bonding Pair (LP-BP) Ø Bonding Pair – Bonding Pair (BP-BP) Ø Lone pair-lone pair (LP-LP) repulsion is considered to be stronger than the lone pair-bonding pair (LP-BP) repulsion, which in turn is stronger than the bonding pair-bonding pair (BP-BP) repulsion. Ø Repulsion strengths Ø • lone pair -lone pair -bond pair bond pair-bond pair
Steps for using VSEPR: Ø Draw a Lewis Dot Structure. Ø Predict the geometry around the central atom. Ø Predict the molecular shape. … also, we can try and predict the angles between atoms.
Things to Remember Ø When assigning a VSEPR Shape to a molecule, we focus on the central atom and the bonding pairs or lone pairs associated with it – Ex. CO 2 Ø Electron pairs are considered to exist in a domain Ø Domains can be made up of: Ø A lone pair A single bond A double bond A triple bond Thus, all are considered to be one electron pair Domain
LINEAR - 2 domains on central atom 2 atoms attached to center atom 0 unshared pairs (lone pairs) Ø 2 domains Ø both are bonding pairs Ø They push each other to opposite sides of center (180 apart). Ø Type AB 2 Ex. : Be. F 2, Be. Cl 2 etc…. .
Trigonal Planar - 3 domains on central atom Ø 3 atoms attached to center atom 0 lone pairs Ø 3 domains Ø all are bonding pairs Ø They push each other apart equally at 120 degrees. ØType: AB 3 ØEx. : Al. F 3 , Ga. F 3 etc…. . Contd…….
Contd……. 3 domains on central atom BENT • 3 domains: – 2 - bonding pairs – 1 - lone pair • The 2 bonding pairs are pushed apart by 3 rd pair (not seen) Sn. F 2
• The geometry around the central atom is trigonal planar. • The molecular shape is bent. Sn. F 2
TETRAHEDRAL -- 4 domains on central atom Ø 4 domains Ø Each repels the other equally - 109. 5 - not the expected 90 (Think in 3 D). Ø 4 atoms attached to center atom 0 lone pairs CH 4 Type: AB 4
Tetrahedral vs Trigonal pyramidal On the right, the 4 th lone pair, is not seen as part of the actual molecule, yet affects shape. If another one of the bonding pairs on “trigonal pyramidal” were a lone pair, what is the result?
4 domains on central atom BENT Ø 4 domains – 2 bonding pairs – 2 lone pairs Ø The bonds are forced together still closer (104. 5 ) by the 2 thick unshared pairs. H 2 O
Comparing the 2 “bents”… Both bent molecules are affected by unshared pairs – 1 pair on the left, 2 on the right.
5 e- pairs on central atom 5 atoms attached to center atom 0 lone pairs TRIGONAL BIPYRAMIDAL Ø 5 shared pairs Ø Three pairs are found in one plane (“equator”) 120 apart; the other two pairs are at the “poles, ” 180 apart, 90 from the “equator. ” ØType: AB 5 PCl 5
5 e- pairs on central atom SEE-SAW Ø 4 shared pairs & unshared pair Ø One of the equator pairs is unshared & pushes the other 2 together. Ø The 2 poles are pushed slightly together. SF 4
5 e- pairs on central atom LINEAR Ø 2 shared & 3 unshared pairs Ø All 3 equator pairs are unshared. The 2 remaining pairs are forced to the poles. Xe. F 2
5 e- pairs on central atom 5 shared, 0 unshared 3 shared, 2 unshared 4 shared, 1 unshared 2 shared, 3 unshared
6 e- pairs on central atom OCTAHEDRAL Ø 6 shared pairs Ø Each pair repels the others equally. Ø All angles = 90 Now, if one of these pairs was unshared … SF 6
6 e- pairs on central atom SQUARE PYRAMIDAL Ø 5 shared pairs & 1 unshared pair Ø 4 shared pairs in one plane; the 5 th pair at the pyramid’s top. If the pair at the top was unshared … IF 5
6 e- pairs on central atom SQUARE PLANAR Ø 4 shared & 2 unshared pairs Ø The 4 shared pairs are in the same plane; the 2 unshared pairs are 90 from them. Xe. F 4
6 e- pairs on central atom 6 shared, 0 unshared 5 shared, 1 unshared 4 shared, 2 unshared
Methane building blocks
1 s 2 s 2 px 2 py 2 pz Promotey sp 3 Hybridize x 109. 5 o z Methane: Carbon
CH 2 O 120 o
1 s 2 s 2 px 2 py 2 pz Promoted sp 2 Hybridized 120 o Trigonal Planar 2 s + 2 px + 2 pz Formaldehyde: Carbon
1 s 2 s 2 px 2 py 2 pz Lone Pairs sp 2 Hybridized 120 o Trigonal Planar 2 s + 2 px + 2 pz Formaldehyde: Oxygen
Formaldehyde bond Sigma bond 2 Lone Pairs
1 s 2 s 2 px 2 py 2 pz Promoted sp sp Hybridized Linear 2 s + 2 px Hydrogen Cyanide: Carbon
1 s 2 s 2 px 2 py 2 pz sp sp Hybridized Linear 2 s + 2 px Hydrogen Cyanide: Nitrogen
bond
PCl 5
sp 3 d Neon 2 3 s sp 3 d 3 px 3 py 3 pz dxz dyz dxy dx 2 -y 2 dz 290 o Hybridized Promoted 120 o Trigonal Bipyrimidal Phosphorus Pentachloride: Phosphorus
Predict the geometry about each interior atom of CH 3 OH and make a sketch of the molecule
Limitation of VSEPR Athough the VSEPR model is useful in predicting molecular geometry, it fails to predict the shapes of isoelectronic species and transition metal compounds. This model does not take relative sizes of substituents and stereochemically inactive lone pairs into account. As a result, VSEPR cannot be applied to heavy d-block species that experience the stereochemical inert pair effect.
Problems ØWhat is the shape of [Xe. F 8]2 - • (Hint: Xenon contains one pair of stereochemically inactive lone pair) ØGive some examples of isoelectronic species. ØWhy are d-block elements exceptions to VSEPR? üAntiprismatic üIF 7 and [Te. F 7]üBecause of stereochemical inert pair effect
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