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## PowerPoint Slideshow about 'Geometry Prediction' - Albert_Lan

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Nature of Bonding

- Central Atom (CA)
- Ligands (L)
- Lone Pairs (LPCA)

Sulfuric Acid (H2SO4)

Note: L = # s bonds to CA.

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.

Geometry determined by any two of HO, L and LPCA

Water Example

Geometry determined by any two of HO, L and LPCA

Sulfuric Acid Example

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)

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

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 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.

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