Question

How to remember molecular bond angles?

Answer 1

You can't. The only thing you can remember is the equilibrium geometry angles in the absence of polarities for the parent structures (i.e. linear, tetrahedral, octahedral, trigonal planar, trigonal bipyramidal). Then you have:

Linear : `180^o` (obviously, because it's perfectly straight)

Tetrahedral : `109.5^o`

Trigonal Planar : `120^o` (obviously, `360^o/3 = 120^o`, so this is easy to remember)

Octahedral : `90^o` (also easy to remember; just divide the angle between the axial atoms, which are antiparallel, by two)

Trigonal Bipyramidal : `90^o` axial-equatorial and `120^o` equatorial-equatorial, just by combining trigonal planar and octahedral features

In structural derivatives (i.e. bent vs. tetrahedral, see-saw vs. trigonal bipyramidal, etc), you need to take into account dipole moments and electron-electron repulsion, and it's not intuitive. In fact, one would calculate it using a computer these days, using a program like Psi4 or GAMESS.

For example, water does not have a `109.5^o` H-O-H angle, but it has a `104.4776^o` angle (it is fine to say it is `104.5^o`). The 2p orbital of oxygen has not-bonded electrons that force the H-O-H angle to contract.

Ammonia also does not have a perfect `109.5^o` H-N-H angle. It's more like `107.5^o`. The 2p orbital of nitrogen does a similar thing, contracting the H-N-H angle slightly.