Answer
$B_{2}$ is more stable than $B_{2}^{+}$.
Work Step by Step
Boron molecule ($B_{2}$) contains 10 electrons.
The molecular orbital configuration of $B_{2}$ molecule is as follows,
$(σ_{1s})^{2},(σ^{*}_{1s})^{2}, (σ_{2s})^{2},(σ^{*}_{2s})^{2},(π_{2py})^{1}=(π_{2pz})^{1}$
($B_{2}^{+}$) contains 9 electrons.
The molecular orbital configuration of $B_{2}^{+}$ ion is as follows,
$(σ_{1s})^{2},(σ^{*}_{1s})^{2}, (σ_{2s})^{2},(σ^{*}_{2s})^{2},(π_{2py})^{1}$
The bond order is 1/2(number of bonding electrons - number of antibonding electrons).
Hence bond order of $B_{2}$ is 1/2(6-4)=1
Bond order of $B_{2}^{+}$ is 1/2(5-4)= 0.5
Bond order is directly proportional to stability.
So $B_{2}$ is more stable than $B_{2}^{+}$.
