Geometry Prediction

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# Geometry Prediction - PowerPoint PPT Presentation

Geometry Prediction. Ray A. Gross, Jr. Prince George’s Community College April 18, 2007. Nature of Bonding. Central Atom (CA) Ligands (L) Lone Pairs (LPCA). G. N. Lewis Linus Pauling. 1930. 1931. Water (H 2 O). Sulfuric Acid (H 2 SO 4 ). Note: L = # s bonds to CA .

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### Geometry Prediction

Ray A. Gross, Jr.

Prince George’s Community College

April 18, 2007

Nature of Bonding
• Central Atom (CA)
• Ligands (L)
• Lone Pairs (LPCA)
Sulfuric Acid (H2SO4)

Note: L = # s bonds to CA.

Nature of Bonding
• Hybrid orbitals (HO)
• p (unhybridized)orbitals

HO = L+ LPCA

Water

HO = L + LPCA= 2 + 2 = 4

Sulfuric Acid

HO = L + LPCA = 4 + 0 = 4

Hybridization tells us the number of hybrid orbitals (HO)
• sp = two hybrid orbitals, or HO = 2
• sp3 = three hybrid orbitals, or HO = 3
• sp3d2= six hybrid orbitals, or HO = 6

If we know the hybridization, we know HO.

If we know HO, we know the hybridization.

Gross Procedure: Determine Hybridization

and Geometry by finding L and LPCA

• Find L and LPCA
• L + LPCA = HO
• HO hybridization
• HO and L  geometry
Example: BH3
• Find L; L = 3
• Find LPCA; LPCA = ?
Example: BH3

LPCA = half the CA’s electrons leftover after bonding.

LPCA = half the CA’s electrons leftover after bonding.
• Let ERL = CA’s electrons required by L
• Let VE = CA’s valence electrons

Then

LPCA = ½(VE – ERL)

LPCA = ½(VE – ERL)
• For BH3: VE = 3, ERL = 3 (1 per H)

LPCA = ½(3 – 3) = 0

Example: BH3
• Find L; L = 3
• Find LPCA; LPCA = 0
• Thus,
• HO = 3; hybridization = sp2
• L = 3; geometry = trigonal planar
Example: NH3

Each H requires one of CA’s electrons.

Example: NH3
• L = 3
• LPCA = ½(VE – ERL) = ½(5 – 3) = 1
• HO = L + LPCA = 3 + 1 = 4 = sp3
• HO = 4, L = 3 = trigonal pyramidal
Example: H2SO4

Each O requires two electrons; each OH requires one electron.

Example: H2SO4 = (HO)2SO2
• L = 4
• LPCA = ½(VE – ERL)
• LPCA = ½[6 – (2 x 1) – (2 x 2) = 0
• HO = 4, L = 4
• sp3; tetrahedral
Example: GeCl4-2
• VE = 4 + 2 = 6
• ERL = 4 x 1 = 4
• LPCA = ½(6 – 4) = 1
• L = 4
• HO = L + LPCA = 5
• HO = 5, L = 4
• sp3d, seesaw geometry
Findings

• VEadj = VE – CH

Electrons Required by Ligands (ERL)

• ERL = Normal Covalence for atoms
• ERL = # for L atom next to CA to complete its octet.

H and O as ligands

• H and O peripheral atoms equal an OH unitary ligand.
What is the geometry of HNO3?
• N = CA; VEadj= 5 – 0 = 5
• L = OH and 2 O’s = 3
• ERL = 1 + 2 x 2 = 5
• LPCA = ½(5 – 5) = 0
• HO = 3; L = 3
• sp2; trigonal planar
Conclusions
• GP combines inorganic and organic procedures.
• GP gives fast, accurate results.
Acknowledgements
• Thanks to all who came to my last seminar.
• I was inspired by it to do this work.

RAG