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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