A completely different mechanism is found in the case of EDMB formation, e.g. in B2H6. This system can be considered as a two-proton attack on the B-B bond of the B2H4 2ꟷ unit , but it can also be seen as two borane (BH3) molecules with three covalent B-H bonds that interact via secondary interactions in a dimerization process . According to the -hole model of secondary interactions,a -hole appears at the end of each atom involved in the covalent bond and it becomes attracted by the large negative electrostatic potential at the BCP of one of the B-H bonds of the neighbor BH3 molecule. Consequently, the two BH3 molecules each other in such a way that there are two simultaneous trans influences between two intramolecular B-H bonds and two new intermolecular B-H bonds (see Fig. ). In these cases of trans influence, there must be a charge transfer of electrons from the two intramolecular B-H bonds towards the two intermolecular B-H bonds (note that neither B nor H atoms have LEPs that could provide the charge needed for the two new intermolecular bonds). This charge transfer or electron donation from the intramolecular B-H bonds towards the intermolecular B-H bonds leads to the conversion of two covalent B-H bonds (plus their associated weak secondary intermolecular bonds) into two B-H-B 3c-2e bonds, as suggested by . The charge transfer ends when both intramolecular and intermolecular bonds become equal and the two B-H-B 3c-2e bonds are formed in the center of B2H6 (with all central B-H distances having the same length). The charge transfer from the intramolecular bond to the intermolecular one to form one B-H-B 3c-2e bond results in an ES < 1 for every central B-H pair. Moreover, it results in a considerable ET value (0.63) for every central B-H pair due to the different electronegativity of B and H atoms. Due to the trans influence, the lenghts of the central B-H units that form part of the B-H-B EDMBs become much larger than the external B-H covalent bonds in B2H6, thus explaining the 0.5 bond order of the EDMBs. The two simultaneous trans influences, leading to the formation of two supported 3c-2e bonds , are favored because of the filling of the empty sp 3 orbital of the two B atoms, so the B atoms in B2H6 reach a more stable tetrahedral geometry than the planar trigonal geometry of the BH3 molecules (see Fig. in ESI) . Moreover, the octet rule is accomplished for the B atom in B2H6, unlike in BH3, if the two electrons of the B-H-B 3c-2e bond are considered to count for the external B atoms, as recently proposed .