Hybridization and Molecular Geometry
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How do you determine the hybridization of an atom?
By counting the number of “things” around it. By things I mean, another atom or a lone pair of e-
If I have 2 “things” my hybridization is?
sp
If I have 3 “things” my hybridization is?
sp^2
If I have 4 “things” my hybridization is?
sp^3
If I have 5 “things” my hybridization is?
sp^3d
If I have 6 “things” my hybridization is?
sp^3d^2
sp: bond angles, electron domain geometry, non-bonding pairs of e-, and molecular geometry
Bond Angles: 180 E- Domain Geom: Linear Non-bonding pairs of e-: 0 Molecular Geom: Linear
sp^2: bond angles, electron domain geometry, non-bonding pairs of e-, and molecular geometry
Bond Angles: 120 E- Domain Geom: Trigonal planar Non-bonding pairs of e-: 0, 1 Molecular Geom: trigonal planar, bent
sp^3: bond angles, electron domain geometry, non-bonding pairs of e-, and molecular geometry
Bond Angles: 109.5 E- Domain Geom: Tetrahedral Non-bonding pairs of e-: 0, 1, 2 Molecular Geom: tetrahedral, trigonal pyramidal, bent
sp^3d: bond angles, electron domain geometry, non-bonding pairs of e-, and molecular geometry
Bond Angles: 90, 120 E- Domain Geom: Trigonal bipyramidal Non-bonding pairs of e-: 0, 1, 2, 3 Molecular Geom: trigonal bipyramidal, see-saw, T-shaped, linear
sp^3d^2: bond angles, electron domain geometry, non-bonding pairs of e-, and molecular geometry
Bond Angles: 90 E- Domain Geom: Octahedral Non-bonding pairs of e-: 0, 1, 2 Molecular Geom: octahedral, square pyramidal, square planar
Do lone pair e- or actual bonds take up more space?
lone pair e- take up more space
difference between the E- domain geometry and molecular geom?
• *e- included in shape = e- domain geom * *e- NOT included in shape= molecular geometry