How do you determine polarity by simply looking at a compound?

I am trying to determine polarity by looking at compounds, drawing lewis structures, analyzing VSEPR models, but I am not totally sure where to start. I understand that if you have lone pairs the compound will probably be polar, but how do you know for sure? People have told me that sometimes molecules will cancel out polar charges (ex. seesaw geometry) but I have no idea. Can someone explain this polar concept to me and possibly answer some example polar problems?

you need knowledge of electronegativity, hybridization, and structure.
OF2 has 2 very polar bonds but the compound is NOT polar.
F-O-F, each O-F bond is polar towards the F but, since the molecule is linear and the P electrons are pulling at equal but opposite strengths and direction, the molecule is not polar.
NBr3 is polar, think of the valence electrons that N contains. it has 3 valence electrons and a lone pair of electrons. the structure would be pryamidal with the 3 Br pointed down and the lone pair of electrons, with the highest electronegativity, straight up.
CS2 is not polar since this is S=C=S, the C-S bonds may be polar but like OF2, the molecule is not polar
C2H2 is not polar. everything is shared equally and the C=C is very strong, this is a linear structure
BrF4+is weird. Br has 7 valence electrons, just like F. hybridization would provide 4 spaces for the F to bind, sp3 hybrid. each Br-F bond would be polar
or BF4-, there is an extra electron making this a polar molecule.
PCl5 has the Cl atoms that must be as far apart from each other as possible and therefore would create equal but opposite “pulls”.
there are 5 possible positions for Cl to bind considering the hybridization. there are no electrons that are singular or lone pairs present. PCl5 is not polar
N3- is polar due to the extra electron noted by the – charge
NO2- is polar due to the extra electron noted by the – charge
SeF6 is not polar, like PCl5, all of the positions for binding F are taken up. Se has 6 available electrons that are hybridized to accommodate the 6F atoms. also, when the F atoms are binded to the Se, they must be as far apart as possible.

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Well given a periodic table, we can kinda figure out the electronegativity based on the periodic trends. Typically across a period, the EN increases, while going down the groups the EN decreases. To determine if a molecule is polar/non-polar, start with a Lewis Dot Diagram. Figure out the shape of the molecule, and the geometry of e- by VSEPER. From looking at the periodic table, you can =figure out the bond polarity. If you see a big difference in EN between two atoms, then you can tell the bond is polar, the atom with a larger EN has a stronger pull of e-, so draw bond dipoles towards the e- hungry atoms. Don’t draw bond dipoles for non-polar bonds. If all the bonds are all non-polar, then the molecule is non-polar. If not then after drawing the bond dipoles, figure out the vector sum (you can tell by looking and with the molecule shape). Draw a molecular dipole, the molecule is polar!

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You first work out the structure using VSEPR. You put vector (arrows) along each bond pointing from the central atom to the substituent δ+A→B δ- . If the vectors cancel the molecule will be nonpolar; if they do not cancel then the molecule will be polar. When there are lone pairs all you can say is whether the molecule is polar without knowing in what direction:
I have a strange way of doing VSEPR: hope you can follow:
1. OF2 # lp = ½ [6e-(O) – 2e-(2F)] = 2 lp AX2E2 molecule is V shaped. Vectors don’t cancel so molecule is polar. It may be that the two lp might turn the polarity away from the F atoms but they won’t exactly match.
2. NBr3 # lp = ½ [5e-(N) – 3e-(3Br)] = 1lp AX3E molecule is trigonal pyramidal; vectors do not cancel .: polar.
3. CS2 # lp = ½ [4e-(N) – 2x2e-(2S)] = 0lp AX2 linear: vectors cancel: nonpolar.
4. C2H2: HCΞCH linear, vectors cancel nonpolar.
5. [BrF4]+ # lp = ½ [7e-(Br) – 4e-(4F) -1(+ve charge)] = 1 lp : AX4E , see-saw structure, vectors do not cancel: polar.
6. [BrF4]- # lp = ½ [7e-(Br) – 4e-(4F) +1(-ve charge)] = 2 lp AX4E2, square planar, vectors cancel nonpolar.
7. PCl5 # lp = ½ [5e-(P) – 5e-(5Cl)] = 0 lp AX5 trigonal bipyramid, vectors cancel, non polar
8. [N3]- (Consider as N(N)2 N requires 3e- to complete octet)
# lp = ½ [5e-(N) – 6e-(2N) + 1(-ve charge)] ] = 0 lp AX2: linear, vectors cancel, nonpolar
9. [NO2]- # lp = ½ [5e-(N) – 4e-(2O) + 1(-ve charge)] ] = 1 lp AX2E: V shaped, vectors do not cancel: polar
10. SeF6 # lp = ½ [6e-(Se) – 6e-(6F)] = 0 lp AX6 octahedral, vectors cancel: nonpolar

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