Question

Given that the ionization energy of ${\mathit{F}}_{\mathbf{2}}^{\mathbf{-}}$ is 290 kJ/mol, do the following

a. Calculate the bond energy of ${\mathit{F}}_{\mathbf{2}}^{\mathbf{-}}$. You will need to look up the bond energy of ${\mathit{F}}_{\mathbf{2}}$and ionization energy of ${\mathit{F}}^{\mathbf{-}}$

b. Explain the difference in bond energy between ${\mathit{F}}_{\mathbf{2}}^{\mathbf{-}}$and ${\mathit{F}}_{\mathbf{2}}$using MO theory.

Short answer

(a) The bond energy of $F_2^{-}$ is 116 kJ/mol

(b) The bond energy of $F_2^{-}$is 116 kJ/mol but it will less than the fluorine molecule because according to the molecular orbital theory, the bond order of $F_2^{-}$is less than $F_2^{}$. As bond order has direct relation with bond energy so it energy $F_2^{-}$is less than $F_2^{}$.

Step-by-step solution

(a)

It is given that the ionization energy of $F_2^{-}$ is 290 kJ/mol i.e.

role="math" localid="1663920469017" $F_2^{-}\to F_2+1e^{-}\Delta H_1=290kJ/mol$

We know that energy required to dissociate F₂ is 154 kJ i.e.

$F_2^{}\to F+F\Delta H_2=154kJ/mol$

Whereas the ionization energy of $F^{-}$is -328 kJ/mol i.e.

$F+e^{-}\to F^{-}\Delta H_2=-328kJ/mol$

From the given data, we can determine the bond energy of $F_2^{-}$ that is

$=290+154+(-328)=116kJ/mol$

(b) Molecular orbital configuration of F2- and  F2

The configuration of $F_2^{-}$ and $F_2^{}$ according to the Molecular Orbital Theory will be

$F_2^{-}$(19 electrons) =$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{*}2p_x{)}^2({\pi }^{*}2p_y{)}^2({\sigma }^{*}2p_z{)}^1$

$F_2^{}$(18 electrons) =$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{*}2p_x{)}^2({\pi }^{*}2p_y{)}^2$

Bond order:

$$\begin{gathered} Bondorderof{F}_2^{-}=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-9}{2} \\ =0.5 \end{gathered}$$

$$\begin{gathered} Bondorderof{F}_2^{}=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-8}{2} \\ =1 \end{gathered}$$

Bond Energy:

The bond energy is directly proportional to the bond order of a molecule. Therefore, we can conclude that the bond energy of $F_2^{}$ will be greater than the $F_2^{-}$ due to higher bond order.