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Boron does not form $B^{3+}$ cation easily. it is due to
energy required to form $B^{3+}$ ion is far more than that which would be compensated by lattice energies or hydration energies of such ion
boron is non-metal
boron is semiconductor
none of the above
Solution
As the Boron atom is small in size a large amount of energy is needed to remove $3$ electrons from the boron atom. So Boron does not form $B_3+$ ion. The atomic number of Boron is $5$ . Its electronic configuration is $1s [2] 2 s[2] 2 p[1]$
When one electron is removed from the $p$ orbital a $He$-like fulfilled s orbital is left. This is highly stable. So the second ionization enthalpy is quite high. Again when one electron is removed a half filled orbital is left. So the third ionization enthalpy is also quite high.
Since the total energy needed to make $B _3+$ is the total of all the ionization enthalpies an enormous amount of energy is required to form it.
