Question

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

Short answer

The electron configurations of the given ions are: (a) \(\mathrm{Cr}^{3+}\): \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^3\) (b) \(\mathrm{N}^{3-}\): \(1s^2 \, 2s^2 \, 2p^6\) (c) \(\mathrm{Sc}^{3+}\): \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6\) (d) \(\mathrm{Cu}^{2+}\): \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^9\) (e) \(\mathrm{Tl}^{+}\): \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^{10} \, 6p^1\) (f) \(\mathrm{Au}^{+}\): \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^{10}\) Noble-gas configurations: (b) \(\mathrm{N}^{3-}\): Neon (Ne) (c) \(\mathrm{Sc}^{3+}\): Argon (Ar) (f) \(\mathrm{Au}^{+}\): Xenon (Xe)

Step-by-step solution

(a) Write electron configuration for \(\mathrm{Cr}^{3+}\) ion (Chromium 3+ ion)

: Chromium (Cr) has an atomic number of 24, which means it has 24 electrons in its neutral state. Since it is a \(\mathrm{Cr}^{3+}\) ion, it loses 3 electrons. Therefore, the number of electrons in Chromium 3+ ion is 24 - 3 = 21. Now we need to write the electron configuration for 21 electrons. The electron configuration for \(\mathrm{Cr}^{3+}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^3\]

(b) Write electron configuration for \(\mathrm{N}^{3-}\) ion (Nitrogen 3- ion)

: Nitrogen (N) has an atomic number of 7, which means it has 7 electrons in its neutral state. Since it is a \(\mathrm{N}^{3-}\) ion, it gains 3 electrons. Therefore, the number of electrons in Nitrogen 3- ion is 7 + 3 = 10. Now we need to write the electron configuration for 10 electrons. The electron configuration for \(\mathrm{N}^{3-}\) is: \[1s^2 \, 2s^2 \, 2p^6\]

(c) Write electron configuration for \(\mathrm{Sc}^{3+}\) ion (Scandium 3+ ion)

: Scandium (Sc) has an atomic number of 21, which means it has 21 electrons in its neutral state. Since it is a \(\mathrm{Sc}^{3+}\) ion, it loses 3 electrons. Therefore, the number of electrons in Scandium 3+ ion is 21 - 3 = 18. Now we need to write the electron configuration for 18 electrons. The electron configuration for \(\mathrm{Sc}^{3+}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6\]

(d) Write electron configuration for \(\mathrm{Cu}^{2+}\) ion (Copper 2+ ion)

: Copper (Cu) has an atomic number of 29, which means it has 29 electrons in its neutral state. Since it is a \(\mathrm{Cu}^{2+}\) ion, it loses 2 electrons. Therefore, the number of electrons in Copper 2+ ion is 29 - 2 = 27. Now we need to write the electron configuration for 27 electrons. The electron configuration for \(\mathrm{Cu}^{2+}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^9\]

(e) Write electron configuration for \(\mathrm{Tl}^{+}\) ion (Thallium 1+ ion)

: Thallium (Tl) has an atomic number of 81, which means it has 81 electrons in its neutral state. Since it is a \(\mathrm{Tl}^{+}\) ion, it loses 1 electron. Therefore, the number of electrons in Thallium 1+ ion is 81 - 1 = 80. Now we need to write the electron configuration for 80 electrons. The electron configuration for \(\mathrm{Tl}^{+}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^{10} \, 6p^1\]

(f) Write electron configuration for \(\mathrm{Au}^{+}\) ion (Gold 1+ ion)

: Gold (Au) has an atomic number of 79, which means it has 79 electrons in its neutral state. Since it is an \(\mathrm{Au}^{+}\) ion, it loses 1 electron. Therefore, the number of electrons in Gold 1+ ion is 79 - 1 = 78. Now we need to write the electron configuration for 78 electrons. The electron configuration for \(\mathrm{Au}^{+}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^{10}\] Now, let's determine which of the ions have noble-gas configurations:

Noble-gas configurations of ions:

: Noble-gas configurations have completely filled s, p, and d orbitals (if applicable) in their outermost energy level. (a) \(\mathrm{Cr}^{3+}\): No, not a noble-gas configuration. (b) \(\mathrm{N}^{3-}\) (\(1s^2 \, 2s^2 \, 2p^6\)): Yes, has the same configuration as Neon (Ne), 2nd noble gas. (c) \(\mathrm{Sc}^{3+}\) (\(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6\)): Yes, has the same configuration as Argon (Ar), 3rd noble gas. (d) \(\mathrm{Cu}^{2+}\): No, not a noble-gas configuration. (e) \(\mathrm{Tl}^{+}\): No, not a noble-gas configuration. (f) \(\mathrm{Au}^{+}\) (\(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^{10}\)): Yes, has the same configuration as Xenon (Xe), 5th noble gas.

Key concepts

Noble-Gas Configurations

Understanding noble-gas configurations is a fundamental concept in chemistry, which refers to the electron configuration of noble gases. The noble gases, found in the rightmost group of the periodic table, have completely filled s and p orbitals in their outermost electron shell. This configuration grants them a unique level of stability. For example, neon (Ne) has the electron configuration [1s^2 2s^2 2p^6], which is its highest occupied energy level.

Other elements often 'strive' to mimic this stable configuration, which is why ions form in the first place. When an atom gains or loses electrons to achieve a noble-gas electron configuration, the resulting charge creates an ion. For instance, when a nitrogen atom gains three electrons, it achieves the same electron configuration as neon, becoming [1s^2 2s^2 2p^6], labeled as [3-]. This is a noble-gas configuration because it matches that of neon, a noble gas.

It's essential for students to recognize that ions with noble-gas configurations will exhibit enhanced stability due to their filled valence shells, a rule often referred to as the 'octet rule' for elements that aim to have eight electrons in their outermost shell.

Ions Electron Configuration

The electron configuration of ions is a key concept in understanding the behavior of atoms when they become charged by either losing or gaining electrons. Ions are formed when atoms do not have the same number of electrons as protons, giving them a net positive or negative charge.

Forming Cations and Anions

Elements can form positively charged ions (cations) by losing electrons, and negatively charged ions (anions) by gaining electrons. Metals, for instance, tend to lose electrons and form cations, such as [+3] resulting in a configuration like [1s^2 2s^2 2p^6 3s^2 3p^6 3d^3] for a [+3] ion. Nonmetals tend to gain electrons to become anions, such as nitrogen forming a [3-] ion with the electron configuration [1s^2 2s^2 2p^6], which has the same electron configuration as the noble gas neon.

Importance of Valence Electrons

It is crucial to focus on the valence electrons, the electrons present in the outermost shell of an atom, because they play a significant role in chemical reactivity and properties. In ions, the valence shell configuration is what changes to resemble that of a noble gas, providing a useful clue about the reactivity and chemical behavior of the ion.

Atomic Structure

Atomic structure lays the groundwork for understanding the complexity of elements and their interactions. An atom consists of a central nucleus, which contains protons and neutrons, with electrons circling the nucleus in various energy levels or orbitals. The atomic number of an element, which equates to the number of protons in the nucleus, defines the element and its position on the periodic table. Moreover, the atomic number determines the number of electrons when the atom is neutral.

Key Components

• Protons: Positively charged particles located in the nucleus.
• Neutrons: Particles with no charge that also reside in the nucleus.
• Electrons: Negatively charged particles in orbitals around the nucleus.

Energy Levels and Orbitals

Electrons inhabit areas around the nucleus called orbitals, and these orbitals are grouped into energy levels. The configurations--such as [1s^2 2s^2 2p^6]--tell us which orbitals the electrons are in and how many are in each orbital. As we look at elements across the periodic table, we see that their electron configurations grow more complex, with additional energy levels and orbitals being filled.

In exercises involving atomic structure and ion formation, it is crucial to appreciate that the electron configuration changes as atoms form ions, but the number of protons (and hence the identity of the element) remains unchanged. This understanding is vital when predicting the chemical properties and reactivity of atoms and ions.