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Give three examples of ions that have an electron configuration of \(n d^{6}(n=3,4,5, \ldots).\)

Short Answer

Expert verified
The three ions with electron configurations of the form nd^6 are: 1. Fe2+ with configuration 3d^6 2. Os2+ with configuration 5d^6 3. Pa3+ with configuration 6d^6

Step by step solution

01

Understand electron configuration notation

The electron configuration nd^6 means that there are 6 electrons in the d orbital of the nth energy level. The n value defines the energy level (shell) where the electrons are located, and can be 3, 4, 5, and so on. The d orbital can hold a maximum of 10 electrons.
02

Identify an element that loses electrons to achieve nd^6 configuration

To find an element that achieves an electron configuration of nd^6 upon losing electrons, we can consider the transition metals, which have electrons in their d orbitals. If we look at the fourth period on the periodic table, we can see Chromium (Cr) with the atomic number 24. Its ground state electron configuration is [Ar] 4s^1 3d^5. If Cr loses 2 electrons (4s^1 and one from the 3d orbital), it forms the Cr2+ ion, which has an electron configuration of 3d^4. As this is not the desired nd^6 configuration, we can move on to the next element in the periodic table. Manganese (Mn) has the atomic number 25, with the ground state electron configuration of [Ar] 4s^2 3d^5. Upon losing two electrons, Mn loses both of its 4s electrons, resulting in the Mn2+ ion with an electron configuration of 3d^5. This is also not the desired configuration. Next, we consider Iron (Fe) with the atomic number 26 and ground state electron configuration of [Ar] 4s^2 3d^6. When Fe loses 2 electrons (both from the 4s orbital), it forms the Fe2+ ion, which has a 3d^6 electron configuration. This is our first example.
03

Identify other elements that lose or gain electrons to achieve nd^6 configuration

To find more examples, we can look at higher periods in the periodic table. In the 5th period, we have the element Tungsten (W) with atomic number 74 and ground state electron configuration of [Xe] 4f^14 5d^4 6s^2. Upon losing 4 electrons (both 6s electrons and two 5d electrons), it forms a W4+ ion, which has the electron configuration of 5d^2. This is not the desired configuration. Rhenium (Re) has the atomic number 75, with ground state electron configuration of [Xe] 4f^14 5d^5 6s^2. Upon losing 1 electron, Re becomes Re1+ with electron configuration of 5d^5, and losing 2 electrons gives Re2+ with electron configuration of [Xe] 4f^14 5d^4. Neither of these match the desired configuration. Next, we check Osmium (Os) with the atomic number 76 and ground state electron configuration of [Xe] 4f^14 5d^6 6s^2. Upon losing 2 electrons (both from the 6s orbital) it forms the Os2+ ion, which has the electron configuration of 5d^6. This is our second example.
04

Find the last example in the 6th period of the periodic table

Starting from the 6th period, we can check Radon-222 (Rn-222), which has atomic number 86 and ground state electron configuration of [Xe] 4f^14 5d^10 6s^2 6p^6. Upon gaining 4 electrons (filling the 6p orbit and placing 4 more in 6d orbit), the Rn2- ion has an electron configuration of [Xe] 4f^14 5d^10 6s^2 6p^6 6d^4. This is not the desired configuration. Actinium (Ac) with atomic number 89 has the ground state electron configuration of [Rn] 7s^2 6d^1. Upon losing one electron, Ac becomes Ac1+ with electron configuration of [Rn] 7s^1 6d^1. If Ac loses 2 more electrons, it forms the Ac3+ ion with the electron configuration of 6d^1. Neither of these match the desired configuration. Next, we consider Protactinium (Pa) with atomic number 91 and ground state electron configuration of [Rn] 7s^2 5f^2 6d^1. If Pa gains 5 electrons, it will have the electron configuration of [Rn] 5f^7. Losing one electron, however, results in the Pa1+ ion, which has the electron configuration of [Rn] 7s^1 5f^2 6d^1, and after losing two more electrons, we obtain the Pa3+ ion with the electron configuration of 6d^6. This is our third example. Thus, the three ions with electron configurations of the form nd^6 are: 1. Fe2+ (n=3) 2. Os2+ (n=5) 3. Pa3+ (n=6)

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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 elements that you find in the central block of the periodic table, specifically in groups 3 through 12. They are unique because they have partially filled d orbitals. This means that when they form ions, their electron configurations can show a diverse range of possibilities.

Here are some key features of transition metals:
  • They usually have high melting and boiling points compared to other elements.
  • They often form colored compounds, which makes them visually distinctive.
  • Many transition metals are good conductors of heat and electricity.
This group encompasses elements like Iron (Fe), Chromium (Cr), and Manganese (Mn), all of which play important roles in various industrial processes and biological systems. Transition metals are also known for exhibiting multiple oxidation states within their compounds, making them incredibly versatile in chemical reactions and processes.
d Orbital
The d orbital is an essential feature of transition metals. Located within the atom's electron cloud, d orbitals are responsible for creating the unique properties of these metals..
  • d orbitals can contain a maximum of 10 electrons. There are five different d orbitals (denoted as d_xy, d_xz, d_yz, d_z^2, and d_x^2-y^2) in the same sublevel.
  • The filling of these d orbitals begins in the fourth period of the periodic table.
  • When elements lose or gain electrons, the configuration within the d orbitals can change, affecting the chemical behavior of the metal.
Understanding how these d orbitals are filled helps us comprehend why transition metals form various ions and have such diverse properties. For example, why an element like Iron (Fe) can have ions with configurations of [Ar] 3d^6.
Ionic Charge
Ionic charge refers to the electric charge that an ion carries. It results from an atom losing or gaining electrons. For transition metals, the ionic charge can vary greatly, leading to different electron configurations.

Important points about ionic charges in transition metals:
  • Transition metals can form cations by losing electrons, typically from their s orbitals first, and sometimes from their d orbitals.
  • The ability to lose varying numbers of electrons allows these metals to participate in diverse chemical reactions.
  • For instance, Iron (Fe), which has the electron configuration of [Ar] 4s^2 3d^6, can lose two electrons and form an Fe^{2+} ion, leading to the electron configuration of 3d^6. This specific charge is one reason why transition metals are used in various applications ranging from manufacturing catalysts to biological enzymes.
Understanding ionic charges is key for predicting the types of reactions an element can undergo, especially in the realm of transition metals.

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Most popular questions from this chapter

Write the electron configurations for the following ions, and determine which have noble-gas configurations: \((\mathbf{a})\mathrm{Co}^{2+}\) \((\mathbf{b})\mathrm{Sn}^{2+},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ag}^{+},(\mathbf{e}) \mathrm{S}^{2-}.\)

Using only the periodic table, arrange each set of atoms in order of increasing radius: (a) Ba, Ca, Na; (b) In, Sn, As; (c) Al, Be, Si.

Consider the following equation: $$\mathrm{Ca}^{+}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{Ca}(g)$$ Which of the following statements are true? (i) The energy change for this process is the electron affinity of the Ca' ion. (ii) The energy change for this process is the negative of the first ionization energy of the Ca atom. (ii) The energy change for this process is the negative of the electron affinity of the Ca atom.

Little is known about the properties of astatine, At, because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. (a) Do you expect the element to be a gas, liquid, or solid at room temperature? Explain. (b) Would you expect At to be a metal, nonmetal, or metalloid? Explain. (c) What is the chemical formula of the compound it forms with Na?

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\) . (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S\) , calculate \(Z_{\text { eff}}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S.\) (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

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