Question #d1cdd

1 Answer
Truong-Son N.
Oct 22, 2017

Well, each bond angle requires an angular distance between two atoms, across a central atom, i.e. an angle is /_"ABC" if the central atom is "B" and the outer atoms are "A" and "C".

Therefore, one can keep the central atom constant and look at the permutations of two-atom combinations for the angle.

Denote outer atoms as "O"_i and the central atom as "C", where i = 1, 2, 3, 4 denotes the particular atom index. Then there are the following angles:

  1. "O"_1-"C"-"O"_2
  2. "O"_1-"C"-"O"_3
  3. "O"_1-"C"-"O"_4
  4. "O"_2-"C"-"O"_3
  5. "O"_2-"C"-"O"_4
  6. "O"_3-"C"-"O"_4

Ignoring redundant angle counting, we get 6 tetrahedral angles.

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