Chapter 7: Problem 85
Arrange the following isoelectronic species in order of increasing ionization energy: \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Isoelectronic Species
- \( \mathrm{O}^{2-} \)
- \( \mathrm{F}^{-} \)
- \( \mathrm{Na}^{+} \)
- \( \mathrm{Mg}^{2+} \)
- \( 1s^2 2s^2 2p^6 \)
Periodic Table Trends
This uptick occurs because elements on the right have more significant nuclear charges, which pulls on the electrons more fiercely, thereby necessitating more energy for electron removal. Furthermore, as you ascend vertically in groups, ionization energy also rises, emphasizing the greater electron-nucleus proximity and influence.
Considering these trends helps us predict and explain the varying ionization energies among isoelectronic species, ensuring clarity in such comparisons.
Nuclear Charge
For isoelectronic species, distinct nuclear charges emerge from differing numbers of protons in their nuclei. The higher the nuclear charge, the more strongly electrons will be attracted to the nucleus.
- \(\mathrm{O}^{2-}\): 8 protons
- \(\mathrm{F}^{-}\): 9 protons
- \(\mathrm{Na}^{+}\): 11 protons
- \(\mathrm{Mg}^{2+}\): 12 protons
Electron Configuration
For our isoelectronic species, they adopt the electron configuration of neon: \( 1s^2 2s^2 2p^6 \). However, even with identical electron configurations, differences in ionization energies exist due to nuclear charge variance.
A greater nuclear charge implies that the electrons experience a stronger attractive force, shifting the energy dynamics for electron removal.
- The more positive the charge, the more energy required to ionize.