# Question bb788

Aug 14, 2017

We probably have to invoke $\text{Hund's rule of maximum multiplicity.}$

#### Explanation:

I assume we examine the processes....

$F \left(g\right) + {\Delta}_{1} \rightarrow {F}^{+} \left(g\right) + {e}^{-}$ ;Delta_1=1681*kJ*mol^-1

${F}^{+} \left(g\right) + {\Delta}_{2} \rightarrow {F}^{2 +} \left(g\right) + {e}^{-}$ ;Delta_2=3374.0*kJ*mol^-1

$O \left(g\right) + {\Delta}_{1} \rightarrow {O}^{+} \left(g\right) + {e}^{-}$ ;Delta_1=1313.9*kJ*mol^-1

${O}^{+} \left(g\right) + {\Delta}_{2} \rightarrow {O}^{2 +} \left(g\right) + {e}^{-}$ ;Delta_2=3388.3*kJ*mol^-1#

We use the data from this site.

${\Delta}_{1}$ is entirely straightforward. Ionization energies should INCREASE across the Period, from left to right as we face the Table, and they does. However ${\Delta}_{2} \left(O\right)$ is marginally greater than ${\Delta}_{2} \left(F\right)$. What's going on?

For fluorine's 2nd ionization we go from $1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$ to $1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$; for oxygen we go from $1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$ to $1 {s}^{2} 2 {s}^{2} 2 {p}^{2}$. It is likely, therefore, that ${O}^{+}$ gains some stability from Hund's rule of maximum multiplicity (as does nitrogen in its first ionization energy); this would tend to DECREASE ${\Delta}_{2}$. On the other hand, ${F}^{2 +}$ is somewhat stabilized by the $\text{nitrogen atom-like}$ electronic configuration, and this stability decreases the magnitude of ${\Delta}_{2}$.

I would be interested in the rationalization your prof provides. Would you relate it here?