The only thing I can think of is the size match between Cl and the B or Al; there's more likely to be an interaction between the vacant AO on Al than B and the Cl lone pairs because the AOs are better matched in energy and size, making the bonding interaction more favourable.
I might be talking rubbish but that's one theory anyway!
A.P.
because ch4 has an octett and bh3 not so it dimerises to b2h6
6.3(mol) * 13.83 (g·mol−1)= 87.1 gram BH3
3
BH3 is non polar.Inter moleculer forces are much weaker Wander Voals forces.NH3 have hydrogen bonds among molecules.They are very strong comparing to Vander Woals forces.So NH3 have high boiling point.
The oxidation product of an aldehyde will depend on whether the conditions are acidic or alkaline. The aldehyde will oxidize to a carboxylic acid if it is acidic. If it is alkaline, the aldehyde will form a salt because the acid would react.
BH3. but the compound generally exists as its dimer B2H6.
BH is not stable, there is no compound by that formula. BH3 (boron hydride) generally exists as the dimer, B2H6, diborane.
Formula: BCl3
BH3 is a electron deficient molecule it fights for the attention of all electrons therefore creating a dimmer molecule. A dimmer in scientific terms is considered non existent.
The predicted boiling point for borane is 440,34 oc.
Increasing order of Lewis acidity BH3>BBR3>BCl3>BF3
5,00 g of BH3 is equal to 0,36 moles.
because ch4 has an octett and bh3 not so it dimerises to b2h6
6.3(mol) * 13.83 (g·mol−1)= 87.1 gram BH3
BH3
BH3 does not obey octet rule.it has a total of six electrons only.boron has three electrons in the valence shell and it accepts one electron from each hydrogenBH3 is a planar molecule and is found only in the gaseous state.BH3 dimerises to form B2H6 with 4 terminal hydrogens attached by normal covalent bonds and 2 hydrogen bridges , 3 centre 2 electron bonds. Once again in the dimer it does not obey the octet rule.
it is BH3