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Arrange the following isoelectronic species in order of increasing ionization energy: \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\).

Short Answer

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\(\mathrm{O}^{2-} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{2+}\) in increasing ionization energy.

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01

Understand Isoelectronic Species

Isoelectronic species are atoms or ions that have the same number of electrons. In this case, \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) all have 10 electrons, matching the electron configuration of neon, \[ \mathrm{Ne:} 1s^2 2s^2 2p^6 \].
02

Recall Ionization Energy Trends

Ionization energy generally increases as you move across a period from left to right on the periodic table, because the nuclear charge increases while the electron distance from the nucleus decreases, causing a greater attraction for the electrons. Additionally, it increases as you move up a group.
03

Consider Nuclear Charge

Each species has a different nuclear charge. More positive nuclear charge leads to stronger attraction to the electrons, thus requiring more energy to remove an electron. Here are their nuclear charges: \(\mathrm{O}^{2-} (8+), \mathrm{F}^{-} (9+), \mathrm{Na}^{+} (11+), \mathrm{Mg}^{2+} (12+)\).
04

Rank Based on Nuclear Charge

Higher nuclear charge generally means higher ionization energy in isoelectronic species because the nucleus holds the electrons more tightly. Therefore, the order of increasing ionization energy is: \(\mathrm{O}^{2-} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{2+}\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isoelectronic Species
Isoelectronic species refer to atoms or ions that possess the exact same number of electrons. Despite having different nuclei, they are united by their electronically identical nature. In our case,
  • \( \mathrm{O}^{2-} \)
  • \( \mathrm{F}^{-} \)
  • \( \mathrm{Na}^{+} \)
  • \( \mathrm{Mg}^{2+} \)
all share ten electrons, mirroring neon's electron configuration:
  • \( 1s^2 2s^2 2p^6 \)
This similarity means that when considering ionization, we must delve deeper into other factors like nuclear charge and how it affects each component of these species.
Periodic Table Trends
Ionization energy trends can be observed across the periodic table, and they're essential when arranging elements or ions in order of energy requirements for electron removal. As you journey from left to right across a period, the ionization energy generally climbs.
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
Nuclear charge refers to the total charge of an atom's nucleus, contributed by its protons. It plays a critical role in determining how tightly an atom holds onto its electrons.
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
In our example, \( \mathrm{Mg}^{2+} \), with the highest nuclear charge, has the strongest electron-nucleus attraction among these species, leading to its high ionization energy. This concept is pivotal to understanding and ranking ionization energies across isoelectronic species.
Electron Configuration
Electron configuration is the arrangement of electrons around the nucleus of an atom. This configuration dictates an atom's behavior and plays a role in its chemical properties.
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.
This is why identifying electron configurations is crucial—they tell half the story, with nuclear charge completing the narrative by highlighting ionization energy differences within isoelectronic species.

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Most popular questions from this chapter

The ionization energy of a certain element is \(412 \mathrm{~kJ} / \mathrm{mol}\). When the atoms of this element are in the first excited state, however, the ionization energy is only \(126 \mathrm{~kJ} / \mathrm{mol}\). Based on this information, calculate the wavelength of light emitted in a transition from the first excited state to the ground state.

Explain why the electron affinity of nitrogen is approximately zero, while the elements on either side, carbon and oxygen, have substantial positive electron affinities

$$ \begin{aligned} &\text { Group the species that are isoelectronic: } \mathrm{Be}^{2+}, \mathrm{F}^{-}, \mathrm{Fe}^{2+}\\\ &\mathrm{N}^{3-}, \mathrm{He}, \mathrm{S}^{2-}, \mathrm{Co}^{3+}, \mathrm{Ar} \end{aligned} $$

Consider the halogens chlorine, bromine, and iodine. The melting point and boiling point of chlorine are \(-101.5^{\circ} \mathrm{C}\) and \(-34.0^{\circ} \mathrm{C}\) and those of iodine are \(113.7^{\circ} \mathrm{C}\) and \(184.3^{\circ} \mathrm{C},\) respectively. Thus chlorine is a gas and iodine is a solid under room conditions. Estimate the melting point and boiling point of bromine. Compare your values with those from the webelements.com website.

A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by $$ E_{n}=-\left(2.18 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n^{2}}\right) $$ where \(n\) is the principal quantum number and \(Z\) is the atomic number of the element. Calculate the ionization energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) of the \(\mathrm{He}^{+}\) ion.

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