Tetrahedron bond length

The four hydrogen atoms occupy equivalent positions with respect to the central carbon atom in methane. The C-H bond length in methane is 1.107 A (1 A [angstrom] = 1 x 10 ° meters), a very short distance indeed. The four atoms of hydrogen are arranged about the central carbon atom at the comers of a regular tetrahedron. 

Some of the diversity that characterizes the properties and compositions of the silicate minerals stems from the ability of the aluminum ion (Al ) to substitute for silicon in the tetrahedral unit. When silicate tetrahedra in a mineral are replaced by aluminum-containing tetrahedra, concomitant changes occur in the size of the tetrahedron (usual Si—O bond length = 0.160 nm. A1—O bond length = 0.178 nm) and in the cations or protons that balance the tetrahedral unit charge. Regular substitutions with distinct chemistries and structures lead to the formation of groups of discrete minerals called aluminosilicates. 

Type B is represented by 4D and 4E which contain only p-oxo groups. In 4D the metal atoms form a tetrahedron and are bridged by two p-oxo, two p-hydroxo and two p-phenoxo groups(lO). The Fe-p-oxo bond lengths, 1.791(3) A, are normal, similar to those found in p-oxo dinuclear complexes. The same type of bonds are 1.829(4) A long for the centrosymmetric 4E which comprises two... 

Figure 4.3 shows that the relative energy differences between the tetrahedron, rhombus, square and linear chain are dependent on the degree of normalized hardness a. For a hard-core potential with a = 1, all four molecules take the same equilibrium bond length Rk. We see that the most stable four-atom molecule is the tetrahedron with six nearest neighbour... 

Fig. 4.3 The normalized binding energy curves. U/ Uq, versus the normalized nearest neighbour bond length, R/Rq, for different values of the degree of normalized hardness,

This theorem can be applied to our previous example of the relative stabilities of four-atom molecules. We shall compare the linear chain (1), square (s), and rhombus (r) with the tetrahedron (t) which we will take as our reference structure. From eqn (4.16) it has a bond length that is given by... 

Thus, the tetrahedron has equilibrium bond lengths of 1.5 , y/2Rk, and Rh for the degrees of normalized hardness ah = J, and 1 that correspond to X = 1.5, 2, and oo respectively. As required by the first step of the theorem, we prepare the bond lengths, Ry, of the three structures, = 1, s, and r, so that they display the same repulsive energies as the tetrahedron, t, that is... 

Then, following the second step of the theorem, we compare the bond energies of the different structures, y, at these prepared bond lengths with that of the tetrahedron, t, that is... 

Show that the equilibrium bond length for a tetrahedron of four noble gas atoms is given by... 

---