why the stability of borohydride cluster is anion>neutral>cation?

The anion has more "electron-density" to contribute to an electron deficient 3-center, two electron bond.

 

Just to get an idea about the 3-center, 2-electron bond.

 

 

This bond is considered a [math] \pi [/math] type bond, but is not the traditional [math] 2p\pi [/math] type that is expected. It is the result of two [math] sp^2 [/math] hybrid lobes overlaping with a hydrogen [math] 1s [/math]. It is still considered a [math] \pi [/math] bond because of the single horizontal nodal polane, its horizontal mirror plane and u-parity.

thanks, but why BH3 form a dimer, but BCl3 did not?

The number of valence electrons in boron is 3. Among them, one electrons is contributed

for B-H bond at the vertex. The remaining 2 electrons are invoved in skeletal bonds formation

and are termed as skeletal electrons (SE’s). That is why contribution of each B-H to skeletal

electrons is 2.

Likewise, carbon contributes 3 electrons to skeleton.

* a closo-deltahedral cluster cage with ‘n’ vertices requires (n+1) skeletal electron pairs

(SEP’s) which occupy (n+1) cluster bonding MOs ; It also implies, the skeletal electrons

(SE’s) must be equal to 2n+2. (i.e. n vertices require 2n+2 electrons in closo cluster)

* Likewise, a nido-deltahedral cluster cage with ‘n’ vertices requires (n+2) pairs of electrons

and so on.

 

n+1 condition is satisfied only with closo anions and hence more stable.

 

Note: Here the value of ‘n’ represents the actual number of vertices in the cluster (or the

number of boron as well as carbon atoms). According to some textbooks, ‘n’ represents the

number of vertices in the parent closo cluster.

 

You will get more from boranes topic at AdiChemAdi

thanks, but why BH3 form a dimer, but BCl3 did not?

 

hydrogen has a valence 1s orbital which fits nicely between the two B:2p orbitals. Chlorine has a set of p-orbitals as it's valence shell, the geometry doesn't work out right to form the 3c-2e bond. Plus boron and chlorine have a large electronegativity difference, and the bond between them posses significant ionic character.

 

The number of valence electrons in boron is 3. Among them, one electrons is contributed

for B-H bond at the vertex. The remaining 2 electrons are invoved in skeletal bonds formation

and are termed as skeletal electrons (SE’s). That is why contribution of each B-H to skeletal

electrons is 2.

Likewise, carbon contributes 3 electrons to skeleton.

* a closo-deltahedral cluster cage with ‘n’ vertices requires (n+1) skeletal electron pairs

(SEP’s) which occupy (n+1) cluster bonding MOs ; It also implies, the skeletal electrons

(SE’s) must be equal to 2n+2. (i.e. n vertices require 2n+2 electrons in closo cluster)

* Likewise, a nido-deltahedral cluster cage with ‘n’ vertices requires (n+2) pairs of electrons

and so on.

 

n+1 condition is satisfied only with closo anions and hence more stable.

 

Note: Here the value of ‘n’ represents the actual number of vertices in the cluster (or the

number of boron as well as carbon atoms). According to some textbooks, ‘n’ represents the

number of vertices in the parent closo cluster.

 

You will get more from boranes topic at AdiChemAdi

 

Are you refering to Wade's rules?