Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which element in the periodic table has the smallest ionization energy? Which has the largest?

Short Answer

Expert verified
Francium has the smallest ionization energy, and helium has the largest.

Step by step solution

01

Understand Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom in the gas phase. It generally increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table.
02

Identify Trends in Ionization Energy

When moving across a period in the periodic table, ionization energy increases as the nuclear charge increases, making it harder to remove an electron. Conversely, moving down a group, ionization energy decreases because the electrons are farther from the nucleus and experience greater electron shielding.
03

Determine the Element with the Smallest Ionization Energy

The element with the smallest ionization energy is the one that is furthest down and to the left on the periodic table. Francium (Fr), located at the bottom of Group 1, has the smallest ionization energy because it easily loses an electron due to weak attraction to the nucleus.
04

Determine the Element with the Largest Ionization Energy

The element with the largest ionization energy is the one that is at the top right corner of the periodic table, excluding noble gases. Helium (He) has the largest ionization energy because its electrons are very close to the nucleus with no shielding effect.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Periodic Table Trends
The periodic table organizes elements in a manner that showcases repeating trends or patterns in their properties. These patterns are known as periodic trends. One of the key periodic trends is ionization energy. As you move from left to right across a period, the ionization energy tends to increase. This occurs because atoms have more protons, which increases the nuclear charge and attracts electrons more strongly, requiring more energy to remove an electron. Conversely, as you move from top to bottom in a group, ionization energy decreases. This is due to electrons being further away from the nucleus and experiencing more electron shielding.
Electron Shielding
Electron shielding is an essential concept in understanding how atoms interact with their electrons. It refers to the effect that inner electrons have on reducing the full attractive force of the nucleus on outer electrons, or valence electrons.
As atoms grow larger, with additional shells of electrons, the shielding effect becomes more pronounced. This shielding effect makes it easier for the outermost electrons to be removed, thereby decreasing ionization energy. In simple terms, more electron shells mean greater shielding, which weakens the nucleus’s hold on the outermost electrons.
Nuclear Charge
Nuclear charge is related to the number of protons in an atom's nucleus—the more protons, the higher the nuclear charge. This charge plays a significant role in an atom's attraction to electrons.
The effective nuclear charge is often discussed, considering the balance between the positive charge of the protons and the electron shielding by inner electrons. As this effective nuclear charge increases, the attraction between the nucleus and the electron also increases, usually resulting in a higher ionization energy. In periods, the nuclear charge increases from left to right, making it progressively harder to remove an electron.
Francium
Francium (Fr) is an element in Group 1 of the periodic table, also known as the alkali metals. It is located at the very bottom of this group, making it the element with the lowest ionization energy in its group.
  • Group 1 elements are characterized by having only one electron in their outermost shell.
  • This single valence electron can be easily removed since Francium exhibits heavy electron shielding and a larger atomic radius than lighter alkali metals above it.
  • This results in Francium having a weak nuclear attraction for this outer electron, leading to its very low ionization energy.
Due to its extreme rarity and radioactivity, Francium is not typically encountered outside of specialized scientific research.
Helium
Helium (He) is a noble gas, found in Group 18 of the periodic table. It is positioned at the top of this group and is distinct due to its extremely high ionization energy.
  • Helium has only two electrons, which are greatly influenced by the strong nuclear charge of its two protons.
  • With no inner electron layers, these electrons experience no shielding, meaning they remain very close and tightly held to the nucleus.
  • This lack of electron shielding ensures that Helium's electrons require a significant amount of energy to be removed.
Its small atomic size and high ionization energy contribute to its stability and non-reactivity, which are typical traits of noble gases.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are group 6 A elements likely to do when they form ions-gain electrons or lose them? How many? What will be the charge on the ion of a \(6 \mathrm{~A}\) element?

What tripositive ion has the electron configuration \([\mathrm{Kr}] 4 d^{3} ?\) What neutral atom has the electron configuration \([\mathrm{Kr}] 5 s^{2} 4 d^{2} ?\)

Calculate the overall energy change in kilojoules per mole for the formation of \(\mathrm{CaCl}(s)\) from the elements. The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{Cl}(\mathrm{g})=-348.6 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Ca}(g)=+589.8 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Ca}(g)=+1145 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Ca}(s)=+178.2 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{Cl}_{2}(g)=+243 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{CaCl}_{2}(s)=+2258 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{CaCl}(s)=+717 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

Given the following information, construct a Born-Haber cycle to calculate the lattice energy of \(\mathrm{CrCl}_{2} \mathrm{I}(s)\) : Net energy change for the formation of \(\mathrm{CrCl}_{2} \mathrm{I}(s)=-420 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{Cl}_{2}(g)=+243 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{I}_{2}(s)=+151 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{I}_{2}(s)=+62 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Cr}(s)=+397 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{il}}\) for \(\mathrm{Cr}(g)=652 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 2}\) for \(\mathrm{Cr}(g)=1588 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 3}\) for \(\mathrm{Cr}(\mathrm{g})=2882 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{ca}}\) for \(\mathrm{Cl}(g)=-349 \mathrm{~kJ} / \mathrm{mol}\) \(E_{e a}\) for \(\mathrm{I}(g)=-295 \mathrm{~kJ} / \mathrm{mol}\)

Which of the following sets of ions are isoelectronic? (a) \(\mathrm{F}^{-}, \mathrm{Cl}^{-}, \mathrm{Br}^{-}\) (b) \(\mathrm{Ti}^{4+}, \mathrm{Ca}^{2+}, \mathrm{Cl}^{-}\) (c) \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free