Structural relationship between closo, nido, and arachno boranes (and hetero-substituted boranes). The diagonal lines connect species that have the same number of skeletal electron pairs (SEP). Hydrogen atoms except those of the B-H framework are omitted. The red atom is omitted first, the green atom removed second. Adapted from R. W. Rudolph, Acc. Chem. Res., 1976, 9, 446.

Wade’s rules:

The general methodology to be followed when applying Wade’s rules is as follows:

1. Determine the total number of valence electrons from the chemical formula, i.e., 3 electrons per B, and 1 electron per H.

2. Subtract 2 electrons for each B-H unit (or C-H in a carborane).

3. Divide the number of remaining electrons by 2 to get the number of skeletal electron pairs (SEP).

4. A cluster with n vertices (i.e., n boron atoms) and n+1 SEP for bonding has a closo structure.

5. A cluster with n-1 vertices (i.e., n-1 boron atoms) and n+1 SEP for bonding has a nido structure.

6. A cluster with n-2 vertices (i.e., n-2 boron atoms) and n+1 SEP for bonding has an arachno structure.

7. A cluster with n-3 vertices (i.e., n-3 boron atoms) and n+1 SEP for bonding has an hypho structure.

8. If the number of boron atoms (i.e., n) is larger than n+1 SEP then the extra boron occupies a capping position on a triangular phase.

FOR EXAMPLE:

                                 B5H11 or B5H9 or B6H62-   (Explaninations are given below)

 

 

 

What is the structure of B5H11?

Solution

1. Total number of valence electrons = (5 x B) + (11 x H) = (5 x 3) + (11 x 1) = 26

2. Number of electrons for each B-H unit = (5 x 2) = 10

3. Number of skeletal electrons = 26 – 10 = 16

4. Number SEP = 16/2 = 8

5. If n+1 = 8 and n-2 = 5 boron atoms, then n = 7

6. Structure of n = 7 is pentagonal bipyramid, therefore B5H11 is an arachno based upon a pentagonal bipyramid with two apexes missing.