Which element has the highest second ionization energy: Li, \(\mathrm{K},\) or Be?

Short Answer

Expert verified
Beryllium (Be) has the highest second ionization energy among Li, K, and Be due to its higher nuclear charge and proximity of the second electron to the nucleus, making it more difficult to remove.

Step by step solution

01

Recall the definition of ionization energy

Ionization energy is defined as the energy required to remove an electron from an atom or ion in its gaseous state. The second ionization energy refers specifically to the energy required to remove the second electron from an atom or ion.
02

Recall the factors affecting ionization energy

Ionization energy generally depends on three factors: 1. Atomic size (distance between the electrons and the nucleus) 2. Nuclear charge (number of protons in the nucleus) 3. Electron shielding (some electrons shield others from the attractive force of the nucleus)
03

Write the electronic configuration of the given elements

Write the electronic configuration of Li, K, and Be to better visualize their atomic structure. Lithium (Li): \( 1s^2 \ 2s^1 \) Potassium (K): \( 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^1 \) Beryllium (Be): \( 1s^2 \ 2s^2 \)
04

Analyze the second ionization energy of the given elements

Now, consider the factors affecting ionization energy for the given elements during the process of losing the second electron. 1. Lithium (Li), after losing its first electron, will form a +1 ion with an electronic configuration of \( 1s^2 \). The next electron to be removed comes from the 1s orbital, which is closer to the nucleus and thus has a high ionization energy. 2. Potassium (K), after losing its first electron, will form a +1 ion with an electronic configuration of \( 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \). The next electron to be removed is from the 3p orbital, which is further from the nucleus and thus has a lower ionization energy than Li. 3. Beryllium (Be), after losing its first electron, will form a +1 ion with an electronic configuration of \( 1s^2 \ 2s^1 \). The next electron to be removed comes from the 2s orbital, which is also closer to the nucleus and has a higher ionization energy than K. However, Be has a higher nuclear charge than Li, resulting in a stronger attraction between the nucleus and the remaining electron, which increases the ionization energy.
05

Identify the element with the highest second ionization energy

Comparing the factors affecting the second ionization energy for Li, K, and Be, we find that Beryllium (Be) has the highest second ionization energy due to its higher nuclear charge and proximity of the second electron to the nucleus, making it difficult to remove.

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!

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

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?(\mathbf{d})\) If the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is considered to have the -2 oxidation state, what is the oxidation state of the \(\mathrm{Cl}\) ? What is the electron configuration of \(\mathrm{Cl}\) in this oxidation state?

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

(a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in P? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the \(3 s\) electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a Patom, which orbital will it come from?

An element X reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} . \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(1000^{\circ} \mathrm{C}\) ). Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of $58^{\circ} \mathrm{C}$. (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(\mathrm{X}\) is a metal, nonmetal, or metalloid? (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element X.

What is the relationship between the ionization energy of an anion with a 1 - charge such as \(\mathrm{F}^{-}\) and the electron affinity of the neutral atom, \(\mathrm{F}\) ?

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