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Chapter 7: Problem 149

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.

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Most popular questions from this chapter

A certain oxygen atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} .\) How many unpaired electrons are present? Is this an excited state of oxygen? In going from this state to the ground state, would energy be released or absorbed?

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of \(3.59 \times 10^{-19} \mathrm{~J}\), what color of visible light is emitted?

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

The Heisenberg uncertainty principle can be expressed in the form $$ \Delta E \cdot \Delta t \geq \frac{h}{4 \pi} $$ where \(E\) represents energy and \(t\) represents time. Show that the units for this form are the same as the units for the form used in this chapter: $$ \Delta x \cdot \Delta(m v) \geq \frac{h}{4 \pi} $$

How many electrons in an atom can have the designation \(1 p\), \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)
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