Understanding Borane Cluster Structures and Styx Rule
Boranes are cluster compounds of boron and hydrogen with unique structural arrangements. The Skeletal Electron Count and Styx Rule help in determining the bonding features and types of bonds present in borane compounds. Various borane structures such as Closo, Nido, and Arachno exhibit different bonding patterns based on W.N. Lipscomb's concepts. The Styx number is crucial for identifying bonding combinations in boranes, emphasizing the importance of two-electron sigma bonds and additional bonding features like 3c-2e bonds. Examples of B2H6, B4H10, and B5H9 structures illustrate the application of the Styx Rule in determining bond types within borane clusters.
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Skeltal Electron Count 2 (B-H) + 3 (C-H) + 1 (additional H) + Ionic charge on the cluster Closo structure
B6H10 Nido structure
Styx no of Boranes 2c-2e- 2c-2e- For example
What is styx Rule/ Number? The styx number is used to understand the types of bonds in boranes. W.N. Lipscomb developed a method for determining what combinations of bonding features (types of bonds) are possible for a specific formula of boron hydride. The each boron in neutral boron hydrides, hydroborate anions and carboranes cations each boron has at least one H (or,other substituent ) attached by a normal two electron sigma bond (2c-2e). It can be assumed that one B-H bond is present per boron atom. Lipscomb proposed that in addition to this, any of the following (a) Three center-two electrons(3c-2e) B-H-B bond (labeled as- s ) (b) Closed and / or open three center-two electrons (3c-2e) B-B-B bond (labeled as- t ) (c) Two center-two electrons (2c-2e) B-B bond (labeled as-y) (d) Two center-two electrons (2c-2e) B-Hterminal bond (labeled as- x)(simply no. of BH2 groups)
B2H6====== (BH)2H4=== (BH)pHq ==== P = 2; q = 4 p = s + t = 2-------------------------(1) q = s + x = 4 ------------------------(2) p - q/2 = t + y = 2- 4/2 = 0 ------(3) Step I s = 2, t =0 t = minimum x y s t 2 0 Step II s = 2, t =0 then x= 4-2 = 2 From equn 2 x y s t 2 Step III t + y = 0, as t =0 then y = 0-0 = 0 x y 0 s t 0 x y s t 2 2 0
B4H10====== (BH)4H6===== P= 4; q =6 p = s + t = 4 --------------------------(1) q = s + x = 6 ------------------------(2) p - q/2 = t + y = 4- 6/2 = 1 ------(3) x y 1 s t 4 2 0 Step I S = 4, t =0 t = minimum x y s t 4 0 Step II S = 4, t =0 then x= 6-4 = 2 x y s t 2 Step III t + y = 1, as t =0 then y = 1-0 = 1 x y s t 1
B5H9====== (BH)5H4===== P= 5; q = 4 p = s + t = 5 --------------------------(1) q = s + x = 4 ------------------------(2) p - q/2 = t + y = 5- 4/2 = 3------(3) Step I s = 4, t = 1 t = minimum x y s t 1 4 Step II s = 4, t = 1 then x= 4-4 = 0 From equn 2 x y s t 0 Step III t + y = 3, as t = 1 then y = 3-1 = 2 x y 2 s t x y s t 1 4 0 2
Home task-----B5H11====== (BH)5H6===== P= 5; q = 6 p = s + t = 5 --------------------------(1) q = s + x = 6 ------------------------(2) p - q/2 = t + y = 5- 6/2 = 2------(3) Step I s = 4, t = 1 t = minimum Ought , even x y s 3 t 2 Step II s =3, t = 2 then x= 6-3 = 3 From equn 2 x y s t 3 Step III t + y = 2, as t = 2 then y = 2-2 = 0 x y 0 s t x y s t 2 3 3 0
Isolobal analogy B = 1S2S22P1 = 05 e-----------------BH = 06 e or BH- = 7e C = 1S2S22P2 = 06 e-----------------CH = 07 e So, BH- CH So, in a Borane if one B is replaced by one carbon that is called carborane--------- -BH- -BH- C2Bn-2 Hn CBn-1 Hn1- BnHn2- +CH +CH nido- C2B9 H112- Closo- C2B9 H11 +2e +2e +2e Closo Nido Arachno
