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Find three examples of ions in the periodic table that have an electron configuration of \(n d^{8}(n=3,4,5, \ldots)\).

Short Answer

Expert verified
Three examples of ions in the periodic table that have an electron configuration of \(n d^{8}\) are: 1. Ni(II): \([Ar] 3d^{8}\), n = 3 2. Pd(II): \([Kr] 4d^{8}\), n = 4 3. Pt(II): \([Xe] 4f^{14}5d^{8}\), n = 5

Step by step solution

01

Understand the electron configuration notation

\(n d^{8}\) indicates that there are 8 electrons in the d-orbital of the nth energy level. The value of n can be 3, 4, 5, etc. So, we need to look for ions in the periodic table which have 8 electrons in the d-orbital of one of the energy levels.
02

Investigate the transition metals

The elements that commonly have electrons in d-orbitals are the transition metals. They can be found in the d-block of the periodic table, groups 3 to 12. We will focus on these elements to find ions with the desired electron configuration.
03

Find elements that can lose electrons to achieve \(n d^{8}\) configuration

Some transition metals can lose a specific number of electrons to achieve the desired electron configuration. Let's explore some examples: - Nickel (Ni, atomic number 28) has an electron configuration of \([Ar] 3d^{8}4s^{2}\). Upon losing two electrons from its 4s orbital, it forms Ni(II) ion, which has a configuration of \([Ar] 3d^{8}\), where n=3. - Palladium (Pd, atomic number 46) has an electron configuration of \([Kr] 4d^{10}5s^0\). In this case, it has to lose two electrons from its 4d orbital to form Pd(II) ion, which has a configuration of \([Kr] 4d^{8}\), where n=4. - Platinum (Pt, atomic number 78) has an electron configuration of \([Xe] 4f^{14}5d^{9}6s^1\). To achieve the desired configuration, it has to lose one electron from its 5d orbital and one electron from its 6s orbital to form Pt(II) ion, which has a configuration of \([Xe] 4f^{14}5d^{8}\), where n=5.
04

Summary of findings

Here are the three ions with an electron configuration of \(n d^{8}\): 1. Ni(II): \([Ar] 3d^{8}\), n = 3 2. Pd(II): \([Kr] 4d^{8}\), n = 4 3. Pt(II): \([Xe] 4f^{14}5d^{8}\), n = 5

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Key Concepts

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

Transition Metals
Transition metals are a group of elements found in the middle of the periodic table, specifically in groups 3 to 12, occupying the 'd-block'. These elements are known for their ability to form colorful compounds, their use as catalysts, and most importantly, for their unique electron configurations and the ability to exist in multiple oxidation states.

Understanding transition metals is essential for grasping the foundation of coordination chemistry and the behavior of electrons in the 'd-orbital', which often influences the chemical and physical properties of these elements. For example, elements like iron, copper, and gold are transition metals with varying electron configurations that contribute to their distinctive properties such as magnetism and conductivity.

These metals frequently have valence electrons in two different types of orbitals, namely 'd' and 's'. As they have a tendency to lose electrons and form positively charged ions, their electron configurations can change significantly from their neutral state. This variability is crucial when solving problems related to electron configurations in transition metals, such as finding ions with a specific 'd' orbital occupancy.
d-Orbital
The 'd-orbital' is one of the types of orbitals found in atoms where electrons can reside. Each d-orbital can hold up to two electrons, with a total capacity of 10 electrons when all five of the d-orbitals are filled. These orbitals are filled after the 's' orbital of the same energy level, following Hund's rule for the order of filling electron orbitals.

In transition metals, the d-orbital plays a significant role in the chemical and physical characteristics of the element. When discussing electron configurations such as the 'n d^8' configuration mentioned in the exercise, we focus on the number and arrangement of electrons in these orbitals. The configuration 'n d^8' implies that there are 8 electrons distributed across the five orbitals of the d-subshell in any given energy level 'n'.

Students must note the importance of the d-orbital occupancy in dictating the geometry and the nature of bonds these atoms can form in compounds. For instance, a 'd^8' configuration can lead to square planar or tetrahedral geometries in complexes, influencing the metal's reactivity and interactions.
Periodic Table
The periodic table is a tabular display of chemical elements organized based on their atomic number, electron configurations, and recurring chemical properties. Elements are ordered into periods (rows) and groups (columns).

For students dealing with electron configurations, the periodic table serves as an essential guide to understanding how these configurations change from one element to another, following a general pattern called the Aufbau principle. The table also neatly separates the main group elements from the transition metals which, as mentioned earlier, comprise the d-block of the table, groups 3 through 12.

The key takeaway for students from the periodic table, in the context of our exercise, is noticing the trends and patterns in electron configurations across different elements. It acts as a map that allows students to predict how electrons are distributed in an atom or ion by simply looking at its position on the table, which is particularly useful when finding ions with specific electron configurations such as 'n d^8'.

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Most popular questions from this chapter

Tungsten has the highest melting point of any metal in the periodic table: \(3422{ }^{\circ} \mathrm{C}\). The distance between \(\mathrm{W}\) atoms in tungsten metal is \(2.74 \AA\). (a) What is the atomic radius of a tungsten atom in this environment? (This radius is called the metallic radius.) (b) If you put tungsten metal under high pressure, predict what would happen to the distance between \(\mathrm{W}\) atoms.

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group \(1 \mathrm{~B}\) and group 2B metals are as follows: (a) Why are the electron affinities of the group \(2 B\) elements greater than zero? (b) Why do the electron affinities of the group \(1 \mathrm{~B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Which of the following statements about effective nuclear charge for the outermost valence electron of an atom is incorrect? (i) The effective nuclear charge can be thought of as the true nuclear charge minus a screening constant due to the other electrons in the atom. (ii) Effective nuclear charge increases going left to right across a row of the periodic table. (iii) Valence electrons screen the nuclear charge more effectively than do core electrons. (iv) The effective nuclear charge shows a sudden decrease when we go from the end of one row to the beginning of the next row of the periodic table. (v) The change in effective nuclear charge going down a column of the periodic table is generally less than that going across a row of the periodic table.

Elemental cesium reacts more violently with water than does elemental sodium. Which of the following best explains this difference in reactivity? (i) Sodium has greater metallic character than does cesium. (ii) The first ionization energy of cesium is less than that of sodium. (iii) The electron affinity of sodium is smaller than that of cesium. (iv) The effective nuclear charge for cesium is less than that of sodium. (v) The atomic radius of cesium is smaller than that of sodium.

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