Wade's Rules for Boranes and Related Clusters

Wade's Rules

Kenneth Wade in 1976 derived a set of rules for relating the structures of boranes with their compositions, which is also known as polyhedral skeleton electron pair theory. These rules pertain to boranes of formula B_nH_m^x−, where m ≥ n and x ≥ 0 (that is, only neutral or anions are considered).

According to the Wade rule if a cluster of boranes or carboranes has 'n' skeletal atoms (vertices) then it will adopt closo structure if it contains (n + 1) skeletal bonding electron pairs. Similarly, nido if (n + 2)), arachno if (n + 3) and hypho if (n + 4) skeletal bonding electron pairs respectively and so on.

To determine this, we have to know the number of skeletal electron pairs in a cluster which can be determined by following the electron count for various donating units as given below-

Each BH unit gives 2 skeletal bonding electrons
B as such gives three skeletal electrons
Each C-H unit of a carborane contributes 3 skeletal bonding electrons
Each additional H contributes 1 skeletal bonding electron
Each positive charge decrease 1 electron
Each negative charge increase 1 electron
For borane clusters with other hetero-elements-
C, Si, Ge and Sn of a cluster is replaced with a BH unit
N, P and As with a BH₂ unit and
S and Se with a BH₃ unit for counting purpose

🧠 Remember

If the total number of electrons of boranes or carboranes are
☛ (n + 1)pair or (2n + 2) electrons: Closo
☛ (n + 2)pair or (2n + 4) electrons: Nido
☛ (n + 3)pair or (2n + 6) electrons: Arachno
☛ (n + 4)pair or (2n + 8) electrons: Hyper
☛ n: Hypo closo
☛ (n − 1)pair Super hyper closo

NOTE: n = total number of boron and carbon atoms

Examples

C₂B₁₀H₁₂
Total number of electrons:
2CH = 6 electrons
10BH = 20 electrons
Total = 26 electrons
26 electrons = 13 pairs electron
It follows (n + 1) or (2n + 2) rule
n = 2 + 10 = 12
So, 12 + 1 = 13 pairs
or, (2 x 12) + 2 = 26 electrons
Therefore, this cluster compound C₂B₁₀H₁₂ is Closo.

C₂B₄H₈
Total number of electrons:
2CH = 6 electrons
4BH = 8 electrons
2H = 2 electrons
Total = 16 electrons 16 electrons = 8 pairs electron
It follows (n + 2) or (2n + 4) rule
n = 2 + 4 = 6
So, 6 + 2 = 8 pairs
or, (2 x 6) + 4 = 16 electrons
Therefore, this cluster compound C₂B₄H₈ is Nido.

C₂B₇H₁₃
Total number of electrons:
2CH = 6 electrons
7BH = 14 electrons
4H = 4 electrons
Total = 24 electrons 24 electrons = 12 pairs electron
It follows (n + 3) or (2n + 6) rule
n = 2 + 7 = 9
So, 9 + 3 = 12 pairs
or, (2 x 9) + 6 = 24 electrons
Therefore, this cluster compound C₂B₇H₁₃ is Arachno.

B₆H₆^2-
Total number of electrons:
6BH = 12 electrons
2 -ve charge = 2 electrons
Total = 14 electrons 14 electrons = 7 pairs electron
It follows (n + 1) or (2n + 2) rule
n = 6
So, 6 + 1 = 7 pairs
or, (2 x 6) + 2 = 14 electrons
Therefore, this cluster compound B₆H₆^2- is Closo.

C₂B₅H₇ [MPSET: 2019]
Total number of electrons:
2CH = 6 electrons
5BH = 10 electrons
Total = 16 electrons 16 electrons = 8 pairs electron
It follows (n + 1) or (2n + 2) rule
n = 7
So, 7 + 1 = 8 pairs
or, (2 x 7) + 2 = 16 electrons
Therefore, this cluster compound C₂B₅H₇ is Closo.

Wade's Rule for Metal Clusters or Zintal Ions

Zintal ions are naked Polyatomic Species.

If VSE =
(4n + 2) electrons: Closo
(4n + 4) electrons: Nido
(4n + 6) electrons: Archano
(4n + 8) electrons: Hypo

Examples: Pb₅^2-, Bi₅³⁺, Sn₄^2-

Bi₅³⁺
Total VSE = 25 − 3 = 22
Follows (4n + 2) rule
n = 5
So, (4 x 5 + 2) = 22
Therefore, this Bi₅³⁺ is Closo.

Number of Skeletol Pairs and Vertices of Polyhedron

Number of Skeleto Pairs (S) = [Total Electron Count − 12 x No. of Metals]/2

Example: Rh₆(CO)₁₆
TEC = (9 x 6) + (2 x 16) = 86
Number of Metal = 6
So, S = [86 − 72]/2 = 7

Number of Vertices = S − 1

So, 7 − 1 = 6