Why do the three eguations define a stable structure?a) A B,Hn+m molecule has (5n+m) valence AOs and (4n+m)valence electrons, yet being short of n electrons in order toformanormal(2c-2e)o-bondedmolecule(5n+m)AOs→(5n+m)/22c-2ebondingo-bondsb) Thus requiring the formation of n 3c-2e bonds to make thewhole molecule stable,meaning n =s + t or t =n-sBYBBHHBBBBt=n-sorn=s+t1
Why do the three equations define a stable structure? a) A BnHn+m molecule has (5n+m) valence AOs and (4n+m) valence electrons, yet being short of n electrons in order to form a normal (2c-2e)s-bonded molecule. (5n+m) AOs (5n+m)/2 2c-2e bonding s-bonds. b) Thus requiring the formation of n 3c-2e bonds to make the whole molecule stable, meaning n = s + t or t = n - s. t = n-s or n = s+t B B H B B B B H H B B s t y x
Why do the three equations define a stable structure?c) Except the minimal n terminal B-H g-bonds, the extra m Hatoms are used to form s 3c-2e B-H-B bonds and x extra B-H -bonds,meaningm=s+x orx=m-sd)Thetotalelectronsareall usedtoformvarioustypes of bonds4n + m = 2n + 2(s + t + y+ x) > y =(2s-m)/2meaningBHBBHHBBBBx=m-sorm=x+st=n-sorn=s+t>psetsofstyxpisomersy =(2s-m)/2To differentiate the isomers, each isomer can be labeled by aset of (styx)
Why do the three equations define a stable structure? c) Except the minimal n terminal B-H s-bonds, the extra m H atoms are used to form s 3c-2e B-H-B bonds and x extra B-H sbonds, meaning m = s + x or x = m - s. d) The total electrons are all used to form various types of bonds, meaning 4n + m = 2n + 2(s + t + y + x) y = (2s-m)/2. x = m-s or m = x+s t = n-s or n = s+t y = (2s-m)/2 p sets of styx p isomers B B H B B B B H H B B s t y x To differentiate the isomers, each isomer can be labeled by a set of (styx)
B4H10Example1:(2isomers)maxSolution 1Solution 2n=4,m=6BHBS=3S=4x=m-st=0BBBt=1t=n-sy=1B-By=0y = (2s-m)/2X=2X=3B-HextraSmin =3HExpt.StructureHHH-HBH?18HHHHHH(4012)(3103)
Example 1: B4H10 (2 isomers) n=4, m=6 x = m-s t = n-s y = (2s-m)/2 s=4 t=0 y=1 x=2 Solution 1 Solution 2 s=3 t=1 y=0 x=3 Expt. Structure Smax =4 Smin =3 BHB BBB B-B B-Hextra
Example 2: The topological structure of B.Ho (3 isomers)Solution 1Solution 2Solution 3maxn=6,m=4BHBS=3S=2S=4x=m-sBBBt=2t=4t=3t=n-sB-By=2y=1y=0y = (2s-m)/2X=1X=0X=2B-HTextra=2minHHHPHH--HHHBH2HHBH2HHHH-H-HHBHH工HHBBHH(4220)(3311)(2402)
Example 2: The topological structure of B6H10 (3 isomers) n=6, m=4 x = m-s t = n-s y = (2s-m)/2 s=4 t=2 y=2 x=0 Solution 1 Solution 2 Solution 3 s=3 t=3 y=1 x=1 s=2 t=4 y=0 x=2 Smax =4 Smin =2 BHB BBB B-B B-Hextra
B,HgExample3:(3 isomers)Solution 3n=5,m=4Solution 2Solution 1maxS=2S=3BHBS=4x=m-st=2t=3t=1BBBt=n-s口y=0B-By=2y=1y = (2s-m)/2X=1X=2X=0B-Hextra=2?Expt. StructureminB,H11Example4:(3isomers)Solution 3?n=5, m=6Solution 2Solution 1maxS=5S=4S=3BHBx=m-st=1t=2t=0BBBt=n-sy=1y=0B-By=2y = (2s-m)/2X=1X=2X=3B-Hextra.=3minExpt. Structure
Example 3: B5H9 (3 isomers) n=5, m=4 x = m-s t = n-s y = (2s-m)/2 s=4 t=1 y=2 x=0 Solution 1 Solution 2 s=3 t=2 y=1 x=1 Expt. Structure s=2 t=3 y=0 x=2 Solution 3 Example 4: B5H11 (3 isomers) n=5, m=6 x = m-s t = n-s y = (2s-m)/2 s=5 t=0 y=2 x=1 Solution 1 Solution 2 s=4 t=1 y=1 x=2 Expt. Structure s=3 t=2 y=0 x=3 Solution 3 Smax =4 Smin =2 Smax =5 Smin =3 BHB BBB B-B B-Hextra BHB BBB B-B B-Hextra