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H2O
2 bonds, 2 lp
sp3
bent
104.5
2 sigma bonds, no pi bonds
asymmetric, polar (bond dipoles do not cancel)
CO2
two double bonds
sp
linear
180
2 sigma, 2 pi
symmetry, nonpolar
N2
one triple bond
sp
linear
180
1 sigma, 2 pi
nonpolar bond, nonpolar
F2
one single bond
sp3
linear
180
1 sigma, no pi
nonpolar bond, nonpolar
NH3
3 single bonds
sp3
trigonal pyramidal
107
3 sigma, no pi
asymmetry, polar
HF
1 single bond
F is sp3
linear
180
1 sigma, no pi
asymmetry, polar (f is more electronegative)
CH4
4 single bonds
sp3
tetrahedral
109.5
4 sigma, no pi
symmetrical, all nonpolar bonds, nonpolar
SiBr4
4 single bonds
sp3
tetrahedral
109.5
4 sigma, no pi
polar bonds but dipoles cancel due to symmetry, nonpolar
Despite being similar in shape, SiBr4 will have larger boiling point than CH4 due to larger electron cloud (think of H v. Br), which means stronger IMFs