Answer
$F_{2}^{+}$ is more stable than $F_{2}$.
Work Step by Step
Fluorine molecule ($F_{2}$) contains 18 electrons.
The molecular orbital configuration of $F_{2}$ molecule is as follows,
$(σ_{1s})^{2},(σ^{*}_{1s})^{2},(σ_{2s})^{2},(σ^{*}_{2s})^{2},(σ_{2pz})^{2},(π_{2px})^{2}=(π_{2py})^{2},(π^{*}_{2px})^{2}=(π^{*}_{2py})^{2}$
Similarly ($F_{2}^{+}$) contains 17 electrons.
The molecular orbital configuration of $F_{2}^{+}$ molecule is as follows,
$(σ_{1s})^{2},(σ^{*}_{1s})^{2},(σ_{2s})^{2},(σ^{*}_{2s})^{2},(σ_{2pz})^{2},(π_{2px})^{2}=(π_{2py})^{2},(π^{*}_{2px})^{2}=(π^{*}_{2py})^{1}$
The bond order is 1/2(number of bonding electrons - number of antibonding electrons).
Hence bond order of $F_{2}$ is 1/2(10-8)=1
Bond order of $F_{2}^{+}$ is 1/2(10-7)= 1.5
Bond order is directly proportional to stability.
So $F_{2}^{+}$ is more stable than $F_{2}$.
