Strong bonds

Which twodimensional molecule is most stable?

Details and Assumptions:

  • Each bond has the same length.
  • The bond angle is α n = 36 0 n \alpha_n = \frac{360^\circ}{n} with the coordination number n n .
  • Treat the atoms as point charges.
  • Consider the electrostatic energy for a single hydrogen atom.

Bonus question: What would change if the molecules could also extend to the third dimension?

c) d) a) e) b)

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Markus Michelmann
Sep 16, 2017

The electrostatic energy reads V = 1 4 π ε j = 1 n q 0 q j r 0 r j = e 2 4 π ε d [ j = 1 n 1 1 r 0 j n ] z n V = \frac{1}{4 \pi \varepsilon} \sum_{j = 1}^n \frac{q_0 q_j}{|\vec r_0 - \vec r_j|} = \frac{e^2}{4 \pi \varepsilon d} \underbrace{ \left[ \sum_{j = 1}^{n-1} \frac{1}{r_{0j}} - n \right] }_{z_n} with the bond lenght d d , elemental charge e e , the coordination number n n and the relative distance r 0 j r_{0j} (in units of d d ). We evaluate the dimensionless factor z n z_n for each molecule: z 1 = 1 z 2 = 1 2 2 = 1.5 z 3 = 1 3 + 1 3 3 1.85 z 4 = 1 2 + 1 2 + 1 2 4 2.59 z 5 = 1 2 sin ( 3 6 ) + 1 2 sin ( 3 6 ) + 1 2 sin ( 7 2 ) + 1 2 sin ( 7 2 ) 5 2.25 \begin{aligned} z_1 &= -1 \\ z_2 &= \frac{1}{2} - 2 = -1.5 \\ z_3 &= \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - 3 \approx -1.85 \\ z_4 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{2} - 4 \approx -2.59 \\ z_5 &= \frac{1}{2 \sin(36^\circ)} + \frac{1}{2 \sin(36^\circ)} + \frac{1}{2 \sin(72^\circ)} + \frac{1}{2 \sin(72^\circ)} - 5 \approx - 2.25 \end{aligned} Therefore, the case n = 4 n = 4 is energetically most favourable.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...