# 6.5: Wade’s Rules

- Page ID
- 212648

Ken Wade (Figure \(\PageIndex{1}\)) developed a method for the prediction of shapes of borane clusters; however, it may be used for a wide range of substituted boranes (such as carboranes) as well as other classes of cluster compounds.

Wade’s rules are used to rationalize the shape of borane clusters by calculating the total number of **skeletal electron pairs (SEP)** available for cluster bonding. In using Wade’s rules it is key to understand structural relationship of various boranes (Figure \(\PageIndex{2}\)).

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

- Determine the total number of valence electrons from the chemical formula, i.e., 3 electrons per B, and 1 electron per H.
- Subtract 2 electrons for each B-H unit (or C-H in a carborane).
- Divide the number of remaining electrons by 2 to get the number of skeletal electron pairs (SEP).
- A cluster with
*n*vertices (i.e.,*n*boron atoms) and*n*+1 SEP for bonding has a*closo*structure. - A cluster with
*n*-1 vertices (i.e.,*n*-1 boron atoms) and*n*+1 SEP for bonding has a*nido*structure. - A cluster with
*n*-2 vertices (i.e.,*n*-2 boron atoms) and*n*+1 SEP for bonding has an*arachno*structure. - A cluster with
*n*-3 vertices (i.e.,*n*-3 boron atoms) and*n*+1 SEP for bonding has an*hypho*structure. - 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.

What is the structure of B_{5}H_{11}?

- Total number of valence electrons = (5 x B) + (11 x H) = (5 x 3) + (11 x 1) = 26
- Number of electrons for each B-H unit = (5 x 2) = 10
- Number of skeletal electrons = 26 – 10 = 16
- Number SEP = 16/2 = 8
- If
*n*+1 = 8 and*n*-2 = 5 boron atoms, then*n*= 7 - Structure of
*n*= 7 is pentagonal bipyramid (Figure \(\PageIndex{2}\)), therefore B_{5}H_{11}is an*arachno*based upon a pentagonal bipyramid with two apexes missing (Figure \(\PageIndex{3}\)).

What is the structure of B_{5}H_{9}?

- Total number of valence electrons = (5 x B) + (9 x H) = (5 x 3) + (9 x 1) = 24
- Number of electrons for each B-H unit = (5 x 2) = 10
- Number of skeletal electrons = 24 – 10 = 14
- Number SEP = 14/2 = 7
- If
*n*+1 = 7 and*n*-1 = 5 boron atoms, then*n*= 6 - Structure of
*n*= 6 is octahedral (Figure \(\PageIndex{2}\)), therefore B_{5}H_{9}is a*nido*structure based upon an octahedral structure with one apex missing (Figure \(\PageIndex{4}\)).

Example \(\PageIndex{1}\)

What is the structure of B_{6}H_{6}^{2-}?

**Solution**

- Total number of valence electrons = (6 x B) + (3 x H) = (6 x 3) + (6 x 1) + 2 = 26
- Number of electrons for each B-H unit = (6 x 2) = 12
- Number of skeletal electrons = 26 – 12 = 14
- Number SEP = 14/2 = 7
- If
*n*+1 = 7 and*n*boron atoms, then*n*= 6 - Structure of
*n*= 6 is octahedral (Figure \(\PageIndex{2}\)), therefore B_{6}H_{6}^{2-}is a*closo*structure based upon an octahedral structure (Figure \(\PageIndex{5}\)).

Table \(\PageIndex{1}\) provides a summary of borane cluster with the general formula B_{n}H_{n}^{x-} and their structures as defined by Wade’s rules.

Type |
Basic formula |
Example |
# of verticies |
# of vacancies |
# of e- in B + charge |
# of bonding MOs |
---|---|---|---|---|---|---|

Closo |
B_{n}H_{n}^{2-} |
B_{6}H_{6}^{2-} |
n | 0 | 3n + 2 | n + 1 |

Nido |
B_{n}H_{n}^{4-} |
B_{5}H_{9} |
n + 1 | 1 | 3n + 4 | n + 2 |

Arachno |
B_{n}H_{n}^{6-} |
B_{4}H_{10} |
n + 2 | 2 | 3n + 6 | n + 3 |

Hypho |
B_{n}H_{n}^{8-} |
B_{5}H_{11}^{2-} |
n + 3 | 3 | 3n + 8 | n + 4 |

## Bibliography

- R. W. Rudolph,
*Acc. Chem. Res.*, 1976,**9**, 446. - K. Wade,
*Adv. Inorg. Chem. Radiochem.*, 1976,**18**, 1.