Question

What charge would be needed on ${F_2}$to generate an ion with bond order of $2$?

Answer 1

Hint : The number of covalent bonds shared between two atoms. When the atoms share one pair of electrons, they have a single bond $($bond order $ = 1$$)$. A double bond $($bond order $ = 2)$ consists of two electrons pairs, while sharing three electron pairs results in a triple bond $($bond order$ = 3)$.

Complete Step By Step Answer:
In molecular orbital theory, bond order is defined as half of the difference between the number of bonding and antibonding electrons.
Bond order =$B$– $\dfrac{A}{2}$
 $B = $The number of bonding electrons
$A = $The number of antibonding electrons
Bond order $ = 8 - \left[ {\dfrac{6}{2}} \right]$
$ = \dfrac{2}{2}$.
$ = 1$.
The highest occupied molecular orbitals are antibonding, and so is the lowest unoccupied molecular orbital. If we add electrons, they will go into the antibonding and decrease the bond order. Remove electrons from the antibonding.
Bond order $ = 8 - \dfrac{4}{2}$.
$ = \dfrac{4}{2} = 2$
$HOMO$ stands for highest unoccupied molecular orbital and $LUMO$ stands for the lowest unoccupied molecular orbital. When we draw our a molecular orbital diagram for ${F_2}$, we will see that the last orbital electron occupy is the ${\Pi ^*}2p$orbital, which makes it the highest occupied molecular orbital. The next orbital above the ${\Pi ^*}2p$is ${\sigma ^*}2pz$, and since this orbital has no electrons it is lowest unoccupied molecular orbital.
$0.5$ is contributed by each electron to the bond order. Therefore we remove one electron each from the ${\Pi ^*}2px$ and ${\Pi ^*}2py$ orbitals. so the molecule becomes the ion ${F_2}_{}^{ + 2}$ with a charge of $ + 2$.

Note :
The ${F^ - }$ ion has$2{s^2}2{p^6}$ has the electron configuration. Because it has no unpaired electrons, it is diamagnetic. The Fluorine atom, on the other hand is $2{s^2}2{p^5}$has one unpaired electron that is paramagnetic. ${F_2}$ has the highest bond length among the diatomic compounds.