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Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Zn}^{2+}\) (b) \(\mathrm{Te}^{2-}\) (c) \(\mathrm{Sc}^{3+}\), (d) \(\mathrm{Rh}^{3+}\),(e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Bi}^{3+}\).

Short Answer

Expert verified
The electron configurations for the given ions are as follows: (a) \(\mathrm{Zn}^{2+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2\) (Ar) (b) \(\mathrm{Te}^{2-}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) (Xe) (c) \(\mathrm{Sc}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) (Ar) (d) \(\mathrm{Rh}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^8\) (e) \(\mathrm{Tl}^{+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (Xe) (f) \(\mathrm{Bi}^{3+}\): \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (Xe) All ions except \(\mathrm{Rh}^{3+}\) have noble-gas configurations.

Step by step solution

01

Write the electron configurations for each ion

Based on their atomic numbers and charges, we can find the electron configurations for the given ions: (a) \(\mathrm{Zn}^{2+}\): Zn is atomic number 30, so it has 30 electrons in its neutral state. Losing 2 electrons to become \(\mathrm{Zn}^{2+}\), it now has 28 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2\) (b) \(\mathrm{Te}^{2-}\): Te is atomic number 52, so it has 52 electrons in its neutral state. Gaining 2 electrons to become \(\mathrm{Te}^{2-}\), it now has 54 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) (c) \(\mathrm{Sc}^{3+}\): Sc is atomic number 21, so it has 21 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Sc}^{3+}\), it now has 18 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) (d) \(\mathrm{Rh}^{3+}\): Rh is atomic number 45, so it has 45 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Rh}^{3+}\), it now has 42 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^8\) (e) \(\mathrm{Tl}^{+}\): Tl is atomic number 81, so it has 81 electrons in its neutral state. Losing 1 electron to become \(\mathrm{Tl}^{+}\), it now has 80 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) (f) \(\mathrm{Bi}^{3+}\): Bi is atomic number 83, so it has 83 electrons in its neutral state. Losing 3 electrons to become \(\mathrm{Bi}^{3+}\), it now has 80 electrons, and its electron configuration becomes: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\)
02

Determine which ions have noble-gas configurations

Now, we need to see if the electron configurations of these ions match the electron configurations of noble gases. Noble gases have full electron shells, which means they have the following electron configurations: - He: \([1s]^2\) - Ne: \([1s]^2 [2s]^2 [2p]^6\) - Ar: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6\) - Kr: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6\) - Xe: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [5s]^2 [5p]^6\) - Rn: \([1s]^2 [2s]^2 [2p]^6 [3s]^2 [3p]^6 [3d]^{10} [4s]^2 [4p]^6 [4d]^{10} [4f]^{14} [5s]^2 [5p]^6 [5d]^{10} [6s]^2 [6p]^6\) Comparing the configurations of the ions with the noble gases, we can see that: - \(\mathrm{Zn}^{2+}\) has a noble-gas configuration (Ar) - \(\mathrm{Te}^{2-}\) has a noble-gas configuration (Xe) - \(\mathrm{Sc}^{3+}\) has a noble-gas configuration (Ar) - \(\mathrm{Rh}^{3+}\) does not have a noble-gas configuration - \(\mathrm{Tl}^{+}\) has a noble-gas configuration (Xe) - \(\mathrm{Bi}^{3+}\) has a noble-gas configuration (Xe) In conclusion, all the ions except \(\mathrm{Rh}^{3+}\) have noble-gas configurations.

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Key Concepts

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

Noble Gas Configuration
Noble gas configuration is a way to express electron configurations by using the electron structure of the nearest noble gas. Noble gases are the elements in the far right of the periodic table, like helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). These elements are known for having full electron shells, which makes them very stable. When an element or ion has the same electron configuration as a noble gas, it is said to be isoelectronic with that noble gas.

For example, an ion such as Zn^{2+} loses two electrons to achieve the electron configuration of argon (Ar), and Te^{2-} gains two electrons to achieve the electron configuration of xenon (Xe). These full outer electron shells make them especially stable and less likely to react chemically. This concept is an essential key in understanding the electron reactivity and stability of elements.
Ion Electron Configurations
When dealing with ion electron configurations, you need to first understand what ions are. Ions are atoms that have either gained or lost electrons, resulting in a positive or negative charge. The process of finding an ion's electron configuration involves adjusting the normal electron configuration of the element by the number of electrons gained or lost.

Here's how you can do it:
  • If the ion is positive, like Sc^{3+}, you subtract electrons from the highest energy level in the electron configuration of the neutral atom.
  • If the ion is negative, like Te^{2-}, you add electrons to the highest energy level of the neutral atom's electron configuration.
In the original exercise, you see different ions such as Sc^{3+}, Rh^{3+}, and others. By following these steps, their electron configurations are adjusted to reflect the change in electron number. A fully understood electron configuration is vital for predicting chemical behavior.
Chemical Elements
Chemical elements are the fundamental substances that consist of one type of atom and are organized in the periodic table. Each element has a unique atomic number, which indicates the number of protons in its nucleus. In a neutral state, the number of electrons equals the number of protons.

The periodic table not only arranges these elements in order according to their atomic numbers but also groups them into categories, like metals, non-metals, and metalloids. By understanding elements and their locations on the periodic table, you can predict their electron configurations and chemical properties.

For instance, knowing that zinc (Zn) belongs to the transition metals gives you insights into its usual electron configuration and common ion state, Zn^{2+}. Recognizing the identity and properties of chemical elements is foundational for chemistry studies and helps with everything from predicting reactions to balancing equations.

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Most popular questions from this chapter

In the following pairs of binary compounds determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2}\), (b) \(\mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3}\), (c) \(\mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3}\).

(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) Explain why it is necessary to form a double bond in the Lewis structure. (c) The bond in \(\mathrm{O}_{2}\) is shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond. Explain this observation.

The dipole moment and bond distance measured for the highly reactive gas phase OH molecule are \(1.78 \mathrm{D}\) and \(0.98 \AA\), respectively. (a) Given these values calculate the effective charges on the \(\mathrm{H}\) and \(\mathrm{O}\) atoms of the OH molecule in units of the electronic charges \(e\). (b) Is this bond more or less polar than the \(\mathrm{H}-\mathrm{Cl}\) bond in an \(\mathrm{HCl}\) molecule? (c) Is that what you would have expected based on electronegativities?

You and a partner are asked to complete a lab entitled "Fluorides of Group \(6 \mathrm{~B}\) Metals" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a colorless liquid and the other a green powder. You also find the following notes in your partner's notebook-Compound 1: \(47.7 \% \mathrm{Cr}\) and \(52.3 \% \mathrm{~F}\) (by mass), Compound 2: \(45.7 \% \mathrm{Mo}\) and \(54.3 \% \mathrm{~F}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds you find that the colorless liquid solidifies at \(18^{\circ} \mathrm{C}\), while the green powder does not melt up to the maximum temperature of your apparatus, \(1200{ }^{\circ} \mathrm{C}\). What is the identity of the colorless liquid? What is the identity of the green powder? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. What factors are usually cited to explain this fact?

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