Molecular geometry and coordinates

Consider a diatomic molecule AB. Imagine fixing this molecule at a very specific spatial location, as shown below:

Figure 1:

In order to locate the molecule so specifically, we would need to give the $x$, $y$, and $z$ coordinates of each of its atoms, i.e.,

$$(x_{\rm A},y_{\rm A},z_{\rm A}) = {\bf r}_{\rm A}$$

$$(x_{\rm B},y_{\rm B},z_{\rm B}) = {\bf r}_{\rm B}$$

which is a total of 6 numbers.

However, we note that the molecule looks the same, no matter where in space it is located. This is called translational invariance, and it implies that we can give only the coordinates of one of the atoms relative to the other, which is equivalent to giving the vector difference between ${\bf r}_{\rm B}$ and ${\bf r}_{\rm A}$ (or vice-versa), which we will call the vector ${\bf r}$:

$${\bf r} = {\bf r}_{\rm B}-{\bf r}_{\rm A} = (x,y,z)$$

which is only 3 numbers. This is the same as arbitrarily placing atom A at the origin of our $xyz$ coordinate system.

We also note that the spatial orientation of the molecule is arbitrary, since the molecule looks the same at any viewing angle. For a diatomic, its orientation can be specified by giving two angles: the angle it makes with the $z$ axis and the angle of its projection onto the $xy$ plane with the $x$ axis. The choice of these angles is arbitrary. This leaves only 1 number left, which is the distance between A and B, called the molecule's bond length.

$$r = \vert{\bf r}\vert = \sqrt{x^2 + y^2 + z^2}$$

$$= \vert{\bf r}_{\rm B}-{\bf r}_{\rm A}\vert = \sqrt{ (x_{\rm B}-x_{\rm A})^2 + (y_{\rm B}-y_{\rm A})^2 + (z_{\rm B}-z_{\rm A})^2}$$

This is an internal degree of freedom and is the only important number we need to give in order to convey the geometry of the diatomic.

In spite of this simplification, it is often necessary to specify all of the coordinates of the atoms in a molecule. Molecular modeling packages, which are becoming increasingly important in chemical research, require a full set of coordinates for each atom as input. Similarly, molecular data banks, such as the protein data bank (PDB) will give molecular structures as files of $x$, $y$, and $z$ coordinates. Thus, being able to determine a set of coordinates given only bond lengths and bond angles, and conversely being able to determine bond lengths and angles from a set of coordinates is an extremely important skill. A few examples of how to do this will be illustrated below.

Example: The diatomic AB. How do we determine a set of coordinates for AB given only its bond length $r$. Since its absolute location in space and its orientation are arbitrary, any set of coordinates that reproduces the correct bond length will suffice. Thus, since a diatomic is linear, we may place it along one of the axes of our coordinate system with one of the atoms at the origin:

Figure 2:

Now we see that the coordinates of atom A will simply be

$${\bf r}_{\rm A} = (0,0,0)$$

and the coordinates of atom B, since B lies on the $x$ axis a distance $r$ away from A, will be

$${\rm r}_{\rm B} = (r,0,0)$$

Clearly, this set of coordinates reproduces the correct bond length:

$$\vert{\bf r}_{\rm B}-{\bf r}_{\rm A}\vert = \sqrt{r^2 + 0^2 + 0^2} = r$$

Thus, for the molecule HCl, whose bond length is $r=1.284$ Å, a set of coordinates could be

$${\bf r}_{\rm H} = (0,0,0)$$

$${\bf r}_{\rm Cl} = (1.284,0,0)$$

in Å. This is just one possibility. We could have chosen either atom to be at the origin (or anywhere else in space for that matter), and chosen the bond to lie along any axis (or not along any particular axis), as we choose, so long as the correct bond length is reproduced.

Example: Water, H$_2$O: The geometry of water is bent (we will see how to determine this later), with a bond angle of 104.5$^{\rm o}$ and an OH bond length of approximately 1.0 Å.

To determine a set of coordinates for H$_2$O, we note that the molecule is planar, so we may choose it to lie in the $xy$ plane. We will place the oxygen at the origin with the hydrogens as shown below:

Figure 3:

The coordinates of the oxygen can be written down immediately:

$${\bf r}_{\rm O} = (0,0,0)$$

For each of the hydrogens, note that the $y$ axis bisects the angle, giving two right triangles. An OH bond forms the hypotenuse of one of these triangles, so that the $x$ and $y$ coordinates will be determined from the sine and cosine of the angle $\theta/2$, as can be shown using simple trigonometry:

$${\rm r}_{\rm H_1} = (d_{\rm OH}\sin\theta/2,d_{\rm OH}\cos\theta/2,0) = (0.7907,0.6122,0)$$

$${\rm r}_{\rm H_2} = (-d_{\rm OH}\sin\theta/2,d_{\rm OH}\cos\theta/2,0) = (-0.7907,0.6122,0)$$

To verify that the bond lengths are correctly reproduced, we compute the magnitudes of the vector differences ${\bf r}_{\rm H_1}-{\bf r}_{\rm O}$ and ${\bf r}_{\rm H_2}-{\bf r}_{\rm O}$:

$$\vert{\bf r}_{\rm H_1}-{\bf r}_{\rm O}\vert = \sqrt{(0.7907)^2 + (0.6122)^2 + 0^2} = 1.0 {\rm\AA}$$

$$\vert{\bf r}_{\rm H_2}-{\bf r}_{\rm O}\vert = \sqrt{(0.7907)^2 + (0.6122)^2 + 0^2} = 1.0 {\rm\AA}$$

In order to verify that the bond angle is correct, we note that the angle between two vectors ${\bf a}$ and ${\bf b}$ is given by the formula:

$$\theta_{\rm ab} = \cos^{-1}\left({{\bf a}\cdot{\bf b} \over \vert{\bf a}\vert\vert{\bf b}\vert}\right)$$

where ${\bf a}\cdot{\bf b}$ is the dot product of ${\bf a}$ and ${\bf b}$ defined to be

$${\bf a}\cdot{\bf b} = a_x b_x + a_y b_y + a_z b_z$$

Thus, the H$_1$-O-H$_2$ angle is given by

$$\theta_{H_1-O-H_2} = \cos^{-1}\left( {({\bf r}_{\rm H_1}-{\bf r}_{\rm O})\cdot({\bf r}... ...m H_1}-{\bf r}_{\rm O}\vert\vert{\bf r}_{\rm H_2}-{\bf r}_{\rm O}\vert} \right)$$

$$= \cos^{-1}\left({-(0.7907)^2 + (0.6122)^2 \over 1.0 \times 1.0}\right)$$

$$= \cos^{-1}(-0.2504) = 104.5^{\rm o}$$