Interatomic Distances & Bond Lengths

Interatomic Distances & Bond Lengths
Calculating Distances in Real Space
In Cartesian coordinates, the distance, r, between the points P and Q is given by Pythagoras' rule as:

r² = (X_P − X_Q)² + (Y_P − Y_Q)² + (Z_P − Z_Q)² = ΔX² + ΔY² + ΔZ² In vector notation, the points P and Q may be represented by the vectors

P = XPi + YPj + ZPk	and
Q = XQi + YQj + ZQk	.

The vector r between P and Q is simply the difference between them, i.e.

r	=	P − Q
	=	(XP − XQ) i + (YP − YQ) j + (ZP − ZQ) k
	=	ΔX i + ΔY j + ΔZ k

.....................
..................

r2

=

a2 Δx2 + b2 Δy2 + c2 Δz2 + 2bc cosα Δy Δz + 2ca cosβ Δz Δx + 2ab cosγ Δx Δy

You should note that this key expression is very similar in format to that used for calculating d spacings in reciprocal space: the 6 terms a², b², ..., and bc cosγ define what is known as the real-space metric tensor.
Bond Lengths
Using the equation given in red, any distance between two points within a crystal can be calculated. With reference to the example data on sulphur hexafluoride, one can calculate the distance between S(1) and F(1) as follows: The first step is to calculate the real-space metric tensor as shown below:

a2 = 13.80102 = 190.4676 Å2

b2 = 8.14002 = 66.2596 Å2

c2 = 4.74932 = 22.5559 Å2

bc cosα = 8.1400 × 4.7493 × cos 90° = 0 Å2

ca cosβ = 4.7493 × 13.8010 × cos 95.590° = −6.3847 Å2

ab cosγ = 13.8010 × 8.1400 × cos 90° = 0 Å2

The next step is to calculate the distance between S(1) and F(1) in fractional coordinates:

Δx = 0.0687 − 0 = 0.0687

Δy = 0 − 0 = 0

Δz = 0.2817 − 0 = 0.2817

Applying the formula for the square of the distance (and omitting the terms equal to zero), we obtain:

r2	=	190.4676 × 0.06872 + 22.5559 × 0.28172 - 2 × 6.3847 × 0.2817 × 0.0687 Å2
	=	0.8990 + 1.7899 − 0.2471 Å2
	=	2.442 Å2
→   r	=	1.563 Å

Although this value of 1.563 Å corresponds to the distance between S(1) at (0,0,0) and F(1) at (0.0687,0,0.2817), the question that arises is does this value correspond to an S-F bond length, i.e are the atoms labelled S(1) and F(1) bonded together within the unit cell? The answer is yes! But how do we know this?
From the numerous crystal structures that have been carried out over the last 80 years, it is possible to make certain generalizations about the sizes of atoms: for ionic structures, for example, it is possible to construct a table of ionic radii as shown below (values in Å):

Ag+	1.15	Ce3+	1.01	Cs+	1.67	Hg2+	1.02	Mn2+	0.83	Pt2+	0.80	Sr2+	1.16	Zn2+	0.74	I-	2.20
Al3+	0.54	Ce4+	0.87	Cu+	0.77	K+	1.38	Na+	1.02	Pt4+	0.63	Th2+	0.94	NH4+	1.48	O2-	1.40
Ba2+	1.35	Co2+	0.75	Cu2+	0.73	La3+	1.03	Ni2+	0.69	Ra2+	1.43	Ti2+	0.86	Br-	1.96	S2-	1.84
Ca2+	1.00	Co3+	0.61	Fe2+	0.78	Li+	0.76	Pb2+	1.19	Rb+	1.52	Ti2+	0.61	Cl-	1.81	Se2-	1.98
Cd2+	1.95	Cr3+	0.62	Fe3+	0.65	Mg2+	0.72	Pd2+	0.86	Sn2+	0.93	Tl2+	1.50	F-	1.33	Te2-	2.21

For ionic structures, one simply adds the appropriate pair of radii together in order to obtain an approximate interionic distance, e.g. the predicted distance for Na to Cl in NaCl will be 1.02 + 1.81 = 2.83 Å. Obviously, the exact distance depends on the actual crystal structure. The above table is based on the ions having a coordination number of six; the values will typically increase for higher and reduce for lower coordination numbers.
For organic crystal structures, where covalent bonding is the norm, there are tables of typical values for C-C distances, etc., a few of which are given in the following table (with values in Å):

C-H	1.07	C-C	1.54	C-N	1.47	C-0	1.43
N-H	1.00	C=C	1.34	C=N	1.32	C=0	1.22
O-H	0.96	C≡C	1.20	C≡N	1.16	C≈C	1.40	(aromatic)

Clearly, the typical covalent bond length is in the range 1 to 2 Å.


To continue reading click on the link below:

http://pd.chem.ucl.ac.uk/pdnn/refine2/bonds.htm