Questions: 6.29: Rank the following species in order of (a) Ge, Ge+, Ge^- (b) Ne, O^2-, Na^+, F^2, N^3-, Mg^2+

Solution Steps

We need to rank the given species based on certain properties. For part (a), we are ranking based on size, and for part (b), we are ranking based on ionization energy.

For part (a), we are given the species $\mathrm{Ge}$, $\mathrm{Ge}^{+}$, and $\mathrm{Ge}^{-}$. The size of an atom or ion is influenced by the number of electrons and the effective nuclear charge.

• $\mathrm{Ge}^{-}$ has one extra electron compared to neutral $\mathrm{Ge}$, which increases electron-electron repulsion and makes it larger.
• $\mathrm{Ge}$ is neutral.
• $\mathrm{Ge}^{+}$ has one fewer electron, which reduces electron-electron repulsion and makes it smaller.

Thus, the order from largest to smallest is: \[ \mathrm{Ge}^{-} > \mathrm{Ge} > \mathrm{Ge}^{+} \]

For part (b), we are given the species $\mathrm{Ne}$, $\mathrm{O}^{2-}$, $\mathrm{Na}^{+}$, $\mathrm{F}^{-}$, $\mathrm{N}^{3-}$, and $\mathrm{Mg}^{2+}$. Ionization energy is the energy required to remove an electron from an atom or ion in the gas phase.

• Noble gases like $\mathrm{Ne}$ have high ionization energies because they have a stable electron configuration.
• Cations like $\mathrm{Na}^{+}$ and $\mathrm{Mg}^{2+}$ have higher ionization energies than their neutral atoms because they have fewer electrons and a higher effective nuclear charge.
• Anions like $\mathrm{O}^{2-}$, $\mathrm{F}^{-}$, and $\mathrm{N}^{3-}$ have lower ionization energies because the added electrons increase electron-electron repulsion.

Ranking from highest to lowest ionization energy: \[ \mathrm{Ne} > \mathrm{Mg}^{2+} > \mathrm{Na}^{+} > \mathrm{F}^{-} > \mathrm{O}^{2-} > \mathrm{N}^{3-} \]

Final Answer