# Question #253e2

Dec 15, 2014

${F}_{2}$:${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 2}$

${F}_{2}^{-}$: ${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 1}$

${F}_{2}^{-}$: ${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 2} {\sigma}_{p x}^{\ast 1}$

Since ${F}_{2}^{+}$ has the largest BO, it will have the strongest bond.F2- has the strongest bond.

#### Explanation:

Fluorine (${F}_{2}$) is a homonuclear diatomic molecule that has 18 electrons (9 from each $F$ atom) - out of which 14 are valence electrons (7 from each $F$ atom)..

Molecular Orbital Theory predicts the distribution of electrons in a molecule.

Now, the molecular orbital (MO) diagram for ${F}_{2}$ is this:

${F}_{2}$'s complete electron configuration with respect to its bonding and antibonding orbitals is:

${F}_{2}$:${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 2}$

Bond order is defined as the difference between the number of bonding electrons divided by 2, and the number of antibonding electrons divided by 2; we can see that ${F}_{2}$ has 10 electrons in its bonding orbitals ( 2 in ${\sigma}_{1 s}$, 2 in ${\sigma}_{2 s}$, 2 in ${\sigma}_{p x}$, 2 in ${\pi}_{p y}$, and 2 in ${\pi}_{p z}$) and 8 electrons in its antibonding orbitals (2 in ${\sigma}_{1 s}^{\star}$, 2 in ${\sigma}_{2 s}^{\star}$, 2 in ${\pi}_{p y}^{\star}$, and 2 in ${\pi}_{p z}^{\star}$) so its bond order is

$B {O}_{{F}_{2}} = \frac{1}{2} \cdot 10 - \frac{1}{2} \cdot 8 = 1$

For ${F}_{2}^{+}$, the number of electrons is $18 - 1 = 17$, which will determine its electron configuration to be

${F}_{2}^{-}$: ${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 1}$

One electron is now unpaired in its ${\pi}_{p z}^{\star}$ antibonding orbital. The ${F}_{2}^{+}$ molecule will now have 3 more electrons in its bonding orbitals, which will determine the bond order to be

$B {O}_{{F}_{2}^{+}} = \frac{1}{2} \cdot 10 - \frac{1}{2} \cdot 7 = \frac{3}{2}$

For ${F}_{2}^{-}$, the number of electrons will be $18 + 1 = 19$, and its electron configuration will be

${F}_{2}^{-}$: ${\sigma}_{1 s}^{2} {\sigma}_{1 s}^{\ast 2} {\sigma}_{2 s}^{2} {\sigma}_{2 s}^{\ast 2} {\sigma}_{p x}^{2} {\pi}_{p y}^{2} {\pi}_{p z}^{2} {\pi}_{p y}^{\ast 2} {\pi}_{p z}^{\ast 2} {\sigma}_{p x}^{\ast 1}$

One electron is now unpaired in the previously-unoccupied ${\sigma}_{p x}^{\star}$ - there will now be 10 electrons in its bonding orbitals and 9 electrons in its antibonding orbitals

$B {O}_{{F}_{2}^{-}} = \frac{1}{2} \cdot 10 - \frac{1}{2} \cdot 9 = \frac{1}{2}$

Since ${F}_{2}^{+}$ has the largest BO, it will required more energy to dissociate than ${F}_{2}$ (BO = 1) and ${F}_{2}^{-}$ (BO = 0.5), therefore it will have the strongest bond.