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Place each of the following sets of elements in order of decreasing ionization energy. Explain your choices. (a) Selenium, oxygen, tellurium; (b) gold, tantalum, osmium; (c) lead, barium, cesium.

Short Answer

Expert verified
The order of decreasing ionization energy is as follows: (a) Oxygen > Selenium > Tellurium, (b) Osmium > Tantalum > Gold, (c) Lead > Barium > Cesium.

Step by step solution

01

Understand the Concept of Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Ionization energy generally increases across a period from left to right and decreases down a group on the periodic table. This is due to effective nuclear charge and electron shielding.
02

Analyze Element Position for Set (a)

For set (a) Selenium (Se), Oxygen (O), and Tellurium (Te), all belong to group 16, the chalcogens. Oxygen is above selenium which is above tellurium in the group. Since ionization energy decreases down a group, the order of decreasing ionization energy will be O > Se > Te.
03

Analyze Element Position for Set (b)

For set (b) Gold (Au), Tantalum (Ta), and Osmium (Os), these elements do not fall in the same group. However, they can be compared based on their positions in the periodic table. Osmium is in the 6th period and is a platinum group metal known for having high ionization energies. Tantalum is in the 5th period, and Gold is in the 6th period and is notable for being less reactive. Hence, the decreasing order of ionization energy would likely be Os > Ta > Au.
04

Analyze Element Position for Set (c)

For set (c) Lead (Pb), Barium (Ba), and Cesium (Cs), all are in different groups and periods, but Cs is an alkali metal (which has the lowest ionization energies), followed by Ba, an alkaline earth metal, and Pb, a post-transition metal. Generally, alkali metals have lower ionization energies than alkaline earth metals, which have lower ionization energies than post-transition metals. Therefore, the order of decreasing ionization energy is Pb > Ba > Cs.

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Key Concepts

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

Periodic Table Trends
The periodic table is not just a display of chemical elements, but a manifestation of their underlying properties, often showing trends in characteristics such as ionization energy. As we delve into why certain elements have higher ionization energies than others, it's crucial to grasp the layout of the periodic table.

Across periods (horizontal rows), ionization energy generally increases from left to right. This uptick is attributed to the protons added to the nucleus as one moves across a period, which subsequently strengthens the hold of the nucleus on its electrons. For instance, within the same period, it's anticipated that oxygen will have a higher ionization energy than selenium and tellurium because oxygen is positioned further to the left.

Conversely, going down the groups (vertical columns), a decrease in ionization energy is evident. The added energy levels between the electrons and nucleus produce a shielding effect which weakens the grip of the nucleus on outer electrons. Therefore, as you travel down group 16 from oxygen through selenium to tellurium, the ionization energy correspondingly drops.
Effective Nuclear Charge
The concept of effective nuclear charge (ENC) is quintessential in understanding ionization energy trends. ENC refers to the net positive charge experienced by an electron in a multi-electron atom and is crucial for predicting atomic properties.

Even though the total number of protons in a nucleus (the atomic number) increases with each added element, the actual charge 'felt' by an outer electron does not simply track the atomic number due to electron shielding. ENC increases across a period because while additional electrons are added to the same energy level, the number of shielding electrons does not significantly increase, leading to a stronger attraction between the nucleus and the outer electrons.

When considering elements such as osmium, tantalum, and gold, the effective nuclear charge plays a vital role. Even though all three elements have many protons and electrons, osmium, located earlier in the period, has a slightly lower number of inner electron layers shielding the valence electrons from the pull of the nucleus, compared to gold. Consequently, osmium's valence electrons are held more tightly, and thus it requires more energy to ionize, placing it with the highest ionization energy among these three metals.
Electron Shielding
Electron shielding is an essential aspect of atomic structure influencing ionization energy. It describes the phenomenon where inner electrons block the outer electrons from the full charge of the nucleus.

In the atomic context, each electron orbits within a distinct cloud of probability called an electron shell. Inner shells are full of electrons that can shield the outer electrons from the nucleus. As a result, the increased distance and number of electron shells from the nucleus lead to diminished ENC on the outermost electrons, facilitating their removal and thus lowering ionization energy.

By applying the concept of electron shielding, we can understand why cesium, with its numerous shielding layers, has a lower ionization energy than barium and lead. This aligns with the trend of decreasing ionization energy down a group, such as the one cesium (Cs), barium (Ba), and lead (Pb) belong to. Lead, being the element with the higher ENC due to fewer inner shells compared to barium and cesium, holds onto its electrons more tightly, thus taking the top spot for ionization energy among the three.

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Most popular questions from this chapter

Francium is thought to be the most reactive of the alkali metals. Because it is radioactive and available in only very small amounts it is difficult to study. However, we can predict its properties based on its location in Group 1 of the periodic table. Estimate the following properties of francium: (a) atomic radius; (b) ionic radius of the \(+1\) cation; (c) ionization energy.

Place the following ions in order of increasing ionic radius: \(\mathrm{S}^{2-}, \mathrm{Cl}^{-}, \mathrm{P}^{3-}\).

(a) How many values of the quantum number \(l\) are possible when \(n=6\) ? (b) How many values of \(m_{l}\) are allowed for an electron in a \(5 \mathrm{f}\)-subshell? (c) How many values of \(m_{l}\) are allowed for an electron in a 2 s-subshell? (d) How many subshells are there in the shell with \(n=3\) ?

Atomic orbitals may be combined to form molecular orbitals. In such orbitals, there is a nonzero probability of finding an electron on any of the atoms that contribute to that molecular orbital. Consider an electron that is confined in a molecular orbital that extends over two adjacent carbon atoms. The electron can move freely between the two atoms. The C-C distance is \(139 \mathrm{pm}\). (a) Using the one-dimensional particle-in-the-box model, calculate the energy required to promote an electron from the \(n=1\) to the \(n=2\) level, assuming that the length of the box is determined by the distance between two carbon atoms. (b) To what wavelength of radiation does this correspond? (c) Repeat the calculation for a linear chain of 1000 carbon atoms. (d) What can you conclude about the energy separation between energy levels as the size of the atom chain increases?

The energy levels of a particle of mass \(m\) in a twodimensional square box of side \(L\) are given by \(\left(n_{1}^{2}+n_{2}^{2}\right) h^{2} / 8 m L^{2}\). Are any of the levels degenerate? If so, find the values of the quantum numbers \(n_{1}\) and \(n_{2}\) for which these degeneracies arise for the first three cases.

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