# The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is 109.5^@, the characteristic angle for tetrahedral molecules.

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is ${109.5}^{\circ }$, the characteristic angle for tetrahedral molecules.
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Fatema Sutton
The smallest unit which on repetition gives a crystal lattice is named as a unit cell. The net atoms that are contained in a unit cell determine the type of packing for the lattice and thus the lattice parameters can also be evaluated on the basis of the same.
Let the side of the cube be such that coordinates of center are $\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right).$
$\stackrel{\to }{AO}=\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)$
$\stackrel{\to }{CO}=\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)$
$\stackrel{\to }{AO}.\stackrel{\to }{CO}={\left(\frac{a}{2}\right)}^{2}+\left(-\frac{a}{2}\right)\left(\frac{a}{2}\right)+\left(\frac{a}{2}\right)\left(-\frac{a}{2}\right)$
$=\left(\frac{{a}^{2}}{4}\right)-\left(\frac{{a}^{2}}{4}\right)-\left(\frac{{a}^{2}}{4}\right)$
$=-\frac{{a}^{2}}{4}$
The magnitude of AO and CO is:
$|\stackrel{\to }{AO}|=|\stackrel{\to }{CO}|=\sqrt{\frac{3{a}^{2}}{4}}$
The angle between AO and CO is calculated as follows:
$\mathrm{cos}\theta =\frac{\stackrel{\to }{AO}.\stackrel{\to }{CO}}{|\stackrel{\to }{AO}|=|\stackrel{\to }{CO}|}$
$=\left(\frac{-\frac{{a}^{2}}{4}}{\frac{3{a}^{2}}{4}}\right)$
$=-\frac{1}{3}$
$\theta =\frac{{\mathrm{cos}}^{-1}1}{3}$
$={109.5}^{\circ }$
Hence, the angle made by connecting two of the vertices to a point at the center of the cube is ${109.5}^{\circ }.$

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