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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). 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

Geometry Prediction

Ray A. Gross, Jr.

Prince George’s Community College

April 18, 2007

nature of bonding
Nature of Bonding
  • Central Atom (CA)
  • Ligands (L)
  • Lone Pairs (LPCA)
sulfuric acid h 2 so 4
Sulfuric Acid (H2SO4)

Note: L = # s bonds to CA.

nature of bonding6
Nature of Bonding
  • Hybrid orbitals (HO)
  • p (unhybridized)orbitals

HO = L+ LPCA

water
Water

HO = L + LPCA= 2 + 2 = 4

sulfuric acid
Sulfuric Acid

HO = L + LPCA = 4 + 0 = 4

hybridization tells us the number of hybrid orbitals ho
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.

slide12
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 bh 3
Example: BH3
  • Find L; L = 3
  • Find LPCA; LPCA = ?
example bh 314
Example: BH3

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

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

LPCA = ½(3 – 3) = 0

example bh 317
Example: BH3
  • Find L; L = 3
  • Find LPCA; LPCA = 0
  • Thus,
  • HO = 3; hybridization = sp2
  • L = 3; geometry = trigonal planar
example nh 3
Example: NH3

Each H requires one of CA’s electrons.

example nh 319
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 h 2 so 4
Example: H2SO4

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

example h 2 so 4 ho 2 so 2
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 gecl 4 2
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
Findings

Available electrons, VEadj= VE adjusted for charge CH

  • 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 hno 3
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
Conclusions
  • GP combines inorganic and organic procedures.
  • GP gives fast, accurate results.
acknowledgements
Acknowledgements
  • Thanks to all who came to my last seminar.
  • I was inspired by it to do this work.

RAG

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