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

##### 1 Answer

This can be geometrically worked out, but we expect an angle of

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 *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

Finally, we have an

- 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^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^@)#