Boron does not form B³⁺ ion, because :

Solution:

Boron is a non-metal and its electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1.$ Being a non-metal, it is less interested in losing three ions and forming a cation, ${\mathrm{B}}^{3+}.$ However, if the technique of removing electrons is still in-cooperated in boron, then the removal of the outermost electron requires a significant amount of energy due to its tiny size and high effective nuclear charge. 

When one electron is removed, then B forms B⁺ and has a He-like structure. Removal of another electron from B⁺ is harder than the removal of the previous electron from B because, firstly, B has a positive charge which reinforces the effective nuclear charge on the 2s electrons, and secondly, B⁺ has an inert electronic configuration like He. 

Removal of an electron from B⁺ forms B²⁺. After that, the removal of one more electron from the 2s orbital from B²⁺ to form B³⁺ is extremely difficult and requires an incredible amount of energy to remove the electron from the 2s orbital of B²⁺.

This concludes that a tremendous amount of energy is required for the formation of B³⁺ from B.