Question

Which member of the following pairs would you expect to be more energetically stable? Justify your choice.

1. $\mathbf{NaBr} \mathbf{or} {\mathbf{NaBr}}_{\mathbf{2}}$
2. ${\mathbf{ClO}}_{\mathbf{4}} \mathbf{or} \mathbf{Cl}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}$
3. ${\mathbf{SO}}_{\mathbf{4}} \mathbf{or} {\mathbf{XeO}}_{\mathbf{4}}$
4. ${\mathbf{OF}}_{\mathbf{4}} \mathbf{or} {\mathbf{SeF}}_{\mathbf{4}}$

Short answer

1. NaBr
2. ${\text{ClO}}_{\text{4}}^{\text{-}}$
3. ${\text{XeO}}_{\text{4}}$
4. ${\text{SeF}}_{\text{4}}$

Step-by-step solution

Discuss trends of ionization energy with charge

a. Ionization energy increases as charge increases

_{$$\begin{gathered} N{a}^{+}+B{r}^{-}\to NaBr \\ N{a}^{+2}+B{r}^{-}\to NaB{r}_2 \end{gathered}$$}

In NaBr₂, Na has +2 charge. Na has only 1 valence electron so the removal of second electron is not energetically feasible

Therefore, will be more energetically stable than NaBr₂

Check for complete octet in part b.

b. Total number of electrons in ${\text{ClO}}_{\text{4}}$ = $7 + 4\times 6=7+24=31$

Total number of electrons in $\text{Cl}{{\text{O}}_{\text{4}}}^{\text{-}}$= $7 + 4\times 6+1=7+24+1=32$

In ClO₄, there are 31 valence electrons which are not completing the octet of all the atoms. However, $\text{Cl}{{\text{O}}_{\text{4}}}^{\text{-}}$has 32 valence electrons and every atomhas a completeoctet.

Therefore,is more energetically stable that $\text{Cl}{{\text{O}}_{\text{4}}}^{\text{-}}$ energy.

Check for octet in part c as well.

a. Total number of electrons in ${\text{SO}}_{\text{4}}$ = $6+ 4\times 6=6+24=30$

b. Total number of electrons in ${\text{XeO}}_{\text{4}}$= $8 + 4\times 6=8+24=32$

For SO₄ having 30 electrons, we can’t draw a Lewis structure obeying octet rule. However, we can draw that for XeO₄ having 32 electrons. Hence, XeO₄ is more stable.

Discuss expanding octet.

Both SeF₄ and OF₄ need the central atom to expand its octet. However, oxygen is very small and lacks d-orbitals for expanding its octet.So, SeF₄ is more stable than OF₄.