What is the angle between electron groups in the tetrahedral geometry?

1 Answer
Jun 24, 2017

This can be geometrically worked out, but we expect an angle of #109.5^@#. Let's verify that by drawing a tetrahedron in a cube.

Here,

  • the center of the cube is the central atom in a tetrahedral molecule.
  • the upper left rear, upper right front, bottom left front, and bottom right rear points are the surrounding atoms in a tetrahedral molecule.

The base diagonal is #sqrt2# from the geometry of a #45-45-90# triangle of leg lengths #1#. From the diagram, the diagonal from the bottom left front to the upper right rear is from the Pythagorean Theorem:

#1^2 + (sqrt2)^2 = c^2#

#=> c = sqrt3#

We want half that diagonal, as that is the length towards the central atom in a tetrahedral molecule, so #c/2 = sqrt3/2#.

Finally, we have an #SSS# triangle to solve to determine the tetrahedral angle. By symmetry, the side lengths are:

  • upper-left: #x = sqrt3/2#
  • upper-right: #y = sqrt3/2#
  • base diagonal: #z = sqrt2#

The law of cosines is:

#z^2 = x^2 + y^2 - 2xycosZ#

We can use this to find the tetrahedral angle, #Z#, corresponding to base diagonal length #z = sqrt2#:

#(z^2 - x^2 - y^2)/(-2xy) = cosZ#

#=> color(blue)(Z) = arccos((z^2 - x^2 - y^2)/(-2xy))#

#= arccos(((sqrt2)^2 - (sqrt3/2)^2 - (sqrt3/2)^2)/(-2(sqrt3/2)(sqrt3/2)))#

#= arccos((2 - 3/4 - 3/4)/(-3/2))#

#= color(blue)(109.5^@)#