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

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{~kJ} / \mathrm{mol}\) \(I_{2}=1752 \mathrm{~kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{~kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{~kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?

Short Answer

Expert verified
The unknown element most likely belongs to the alkaline earth metals family (Group 2) in the periodic table, as there is a significant jump in ionization energies between the second and third values, indicating the presence of 2 valence electrons.

Step by step solution

01

Look for a significant jump in ionization energies

Observe the ionization energy values given. Notice that there is a significant increase in energy between the second and third ionization energies from 1752 kJ/mol to 14,807 kJ/mol. This large jump is indicative of a change in an energy shell, and it helps us determine the number of valence electrons in the element.
02

Count the number of valence electrons

Since there is a significant jump between the second and third ionization energy, we can conclude that there are 2 valence electrons in the outermost shell of the unknown element. The first 2 ionization energies correspond to the removal of these 2 valence electrons.
03

Determine the unknown element's group in the periodic table

An element with 2 valence electrons is found in Group 2 of the periodic table. Group 2 elements are known as alkaline earth metals. Examples of elements in this family include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).
04

Identify the family to which the unknown element most likely belongs

Based on the above discussion, we can conclude that the unknown element with the given successive ionization energies most likely belongs to the alkaline earth metals family (Group 2) in the periodic table.

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.

Valence Electrons
Valence electrons are the electrons found in the outermost shell of an atom. These electrons play a crucial role in determining how elements interact in chemical reactions. The number of valence electrons affects an element's reactivity and its ability to form bonds with other elements.
For example:
  • Elements with 1 valence electron tend to lose that electron easily to achieve a stable electronic configuration.
  • Elements with 7 valence electrons often gain an electron to complete their outer shell.
In the exercise, the significant jump in ionization energy between the second and third electrons indicates that the first two electrons are easily removed, implying that they are the valence electrons. This matches with the known characteristic of certain elements that only have two valence electrons.
Periodic Table
The periodic table is a tabular representation of all known chemical elements organized by their atomic number, electron configuration, and recurring chemical properties. Designed by Dmitri Mendeleev, it arranges elements in rows by increasing atomic number and in columns based on similar characteristics.
Key features of the periodic table include:
  • Groups (vertical columns), which share chemical characteristics.
  • Periods (horizontal rows), which represent elements with the same number of atomic orbitals.
  • Blocks, indicating what type of subshell is being filled with electrons, such as s, p, d, or f.
The exercise uses the periodic table to identify which group the unknown element belongs to based on its valence electrons. Elements with two valence electrons are placed in Group 2.
Alkaline Earth Metals
Alkaline earth metals comprise the elements in Group 2 of the periodic table. They are known for having two valence electrons. This makes them relatively reactive, though not as reactive as the alkali metals that have only one valence electron.
Alkaline earth metals include:
  • Beryllium (Be)
  • Magnesium (Mg)
  • Calcium (Ca)
  • Strontium (Sr)
  • Barium (Ba)
  • Radium (Ra)
These elements have common characteristics:
  • They typically form +2 cations by losing their two valence electrons.
  • They are shiny, silvery-white metals.
  • They are less reactive than alkali metals but will still react with water and oxygen.
The significant increase in ionization energy in the exercise hints that once the two valence electrons are removed, the atom reaches a more stable state, which is characteristic of alkaline earth metals.

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

Predict some of the properties of element 117 (the symbol is Uus, following conventions proposed by the International Union of Pure and Applied Chemistry, or IUPAC). a. What will be its electron configuration? b. What element will it most resemble chemically? c. What will be the formula of the neutral binary compounds it forms with sodium, magnesium, carbon, and oxygen? d. What oxyanions would you expect Uus to form?

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\).

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{~s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Identify the following elements. a. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{5} 3 s^{1}\). b. The ground-state electron configuration is [Ne] \(3 s^{2} 3 p^{4}\). c. An excited state of this element has the electron configuration \([\mathrm{Kr}] 5 s^{2} 4 d^{6} 5 p^{2} 6 s^{1}\) d. The ground-state electron configuration contains three unpaired \(6 p\) electrons.

One type of electromagnetic radiation has a frequency of 107.1 MHz, another type has a wavelength of \(2.12 \times 10^{-10} \mathrm{~m}\), and another type of electromagnetic radiation has photons with energy equal to \(3.97 \times 10^{-19} \mathrm{~J} / \mathrm{photon}\). Identify each type of electromagnetic radiation and place them in order of increasing photon energy and increasing frequency.

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