# User:Spoon!

 This user attends or attended the University of California, Los Angeles
 This user exploits the pigeonhole principle.
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# Central angle in simplices between radii to vertices

I know that many people have figured this out long ago, but I like to share it anyhow, because I have wondered about it for a long time when I was in high school...

You know how in high school they told you that the angle between two bonds in methane was about $109.47^\circ$ or something like that? Did you ever wonder what that came from? It is $\cos^{-1} (-\frac{1}{3})$. And I will now show you why:

## Theorem

If circumradii are drawn between the center of an $n$-simplex and its vertices, the angle between these segments is $\cos^{-1} (-\frac{1}{n})$.

## Proof

http://www.math.rutgers.edu/~erowland/polytopes.html#sectionII

• The height of a regular $n$-simplex of side $s$ is
$h_n = \sqrt{ \frac{n+1}{2n} } s$
• The apothem (inradius) of a regular $n$-simplex of side $s$ is
$a_n = \frac{h_n}{n+1}$

The circumradius, which is the difference between the height and the apothem, is:

$R = h_n - a_n = \frac{n}{n+1} h_n = \sqrt{ \frac{n}{2(n+1)} } s$

Now consider any two circumradii. They go to two different vertices, which must be joined by an edge of the $n$-simplex, forming a triangle. Because we know the lengths of all sides of this triangle, we can find the angle between the circumradii using the law of cosines:

$\cos{\gamma} = \frac{a^2 + b^2 - c^2}{2ab}$

Here, $a = b = R$, and $c = s$.

\begin{align} \cos{\gamma} &=& \frac{R^2 + R^2 - s^2}{2R^2} \\ &=& 1 - \frac{1}{2} \left(\frac{s}{R}\right)^2 \\ &=& 1 - \frac{1}{2} \cdot \frac{2(n+1)}{n} \\ &=& 1 - \frac{n+1}{n} \\ &=& -\frac{1}{n} \\ \gamma &=& \cos^{-1} (-\frac{1}{n}) \\ \end{align}

The angle we want is between $0$ and $180^\circ$. Since cosine is one-to-one in that range, the angle is uniquely determined.

Q.E.D.

## Conclusion

This explains, among other things, the angles between hybridized orbitals:

hybridization dimensions angle between orbitals
sp n = 1 $\cos^{-1} (-\frac{1}{1}) = 180^\circ$
sp2 n = 2 $\cos^{-1} (-\frac{1}{2}) = 120^\circ$
sp3 n = 3 $\cos^{-1} (-\frac{1}{3}) \approx 109.47^\circ$

That's all the simplices that can fit in 3 dimensions, folks; but you see the pattern...

# Oxyanion chart

Hybridization

Orbital configuration

III

IV

V

VI

VII

VIII

sp (double)

AX1.5E0 B2O3 M2O3

sp

AX2E0 linear AlO2- BO2- CO2 SiO2 MO2 NO2+
AX1E1 CO NO+

sp2 (double)

AX2.5E0 Si2O52- N2O5 P2O5 As2O5 M2O5
AX1.5E1 N2O3 As2O3
AX0.5E2 N2O

sp2

AX3E0 trigonal planar BO33- CO32- SiO32- SnO32- PbO32- NO3- VO3- SO3 O4 SeO3 MO3
AX2E1 bent <120 SnO22- PbO22- NO2- SO2 O3 SeO2
AX1E2 SO O2

sp3 (double)

AX3.5E0 Si2O76- P2O74- Cr2O72- Cl2O7 M2O7
AX2.5E1
AX1.5E2 S2O32-
AX0.5E3 Cl2O

sp3

AX4E0 tetrahedral SiO44- PO43- AsO43- SO42- SeO42- TeO42- CrO42- MoO42- WO42- ClO4- BrO4- IO4- MnO4- TcO4- ReO4- XeO4 RuO4 OsO4
AX3E1 trigonal pyramidal PO33- AsO33- SO32- SeO32- TeO32- ClO3- BrO3- IO3- XeO3
AX2E2 bent <109.5 PO23- SO22- ClO2- BrO2- IO2-
AX1E3 O22- ClO- BrO- IO-

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