The total number of vacant orbitals involved in bond formation in diborane is

The formula of diborane is ${\mathrm{B}}_2{\mathrm{H}}_6$.

The electronic configuration of boron in ground state is $1{\mathrm{s}}^2 2{\mathrm{s}}^2 2{{\mathrm{p}}_{\mathrm{x}}}^1 2{{\mathrm{p}}_{\mathrm{y}}}^0 2{{\mathrm{p}}_{\mathrm{z}}}^0$. But in excited state the electronic configuration will be $1{\mathrm{s}}^2 2{\mathrm{s}}^1 2{{\mathrm{p}}_{\mathrm{x}}}^1 2{{\mathrm{p}}_{\mathrm{y}}}^1 2{{\mathrm{p}}_{\mathrm{z}}}^0$.

So, there is one vacant orbital for the bond formation. Since, there are 2 boron atoms in diborane, two vacant orbitals are involved in the bond formation.