Question

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

The electron configurations for the given ions are as follows: (a) \(\mathrm{Zn}^{2+}\): \(1s^22s^22p^63s^23p^63d^{10}\) (noble-gas configuration) (b) \(\mathrm{Te}^{2-}\): \(1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^6\) (noble-gas configuration) (c) \(\mathrm{Sc}^{3+}\): \(1s^22s^22p^63s^23p^6\) (noble-gas configuration) (d) \(\mathrm{Rh}^{3+}\): \(1s^22s^22p^63s^23p^64s^23d^{10}4p^64d^6\) (no noble-gas configuration) (e) \(\mathrm{Tl}^{+}\): \(1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}\) (noble-gas configuration) (f) \(\mathrm{Bi}^{3+}\): \(1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}\) (noble-gas configuration)

Step-by-step solution

(a) Electron Configuration of \(\mathrm{Zn}^{2+}\)

The atomic number of Zn is 30, which means it has 30 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Zn}^{2+}\), we need to remove two electrons from the atom. Neutral Zn electron configuration: \[1s^22s^22p^63s^23p^64s^23d^{10}\] Removing two electrons from the 4s orbital: \[\mathrm{Zn}^{2+}: 1s^22s^22p^63s^23p^63d^{10}\] The electron configuration for \(\mathrm{Zn}^{2+}\) is the same as that of the noble gas Kr (krypton, atomic number 36). Therefore, \(\mathrm{Zn}^{2+}\) has a noble-gas configuration.

(b) Electron Configuration of \(\mathrm{Te}^{2-}\)

The atomic number of Te is 52, which means it has 52 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Te}^{2-}\), we need to add two electrons to the atom. Neutral Te electron configuration: \[1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^4\] Adding two electrons to the 5p orbital: \[\mathrm{Te}^{2-}: 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^6\] The electron configuration for \(\mathrm{Te}^{2-}\) is the same as that of the noble gas Xe (xenon, atomic number 54). Therefore, \(\mathrm{Te}^{2-}\) has a noble-gas configuration.

(c) Electron Configuration of \(\mathrm{Sc}^{3+}\)

The atomic number of Sc is 21, which means it has 21 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Sc}^{3+}\), we need to remove three electrons from the atom. Neutral Sc electron configuration: \[1s^22s^22p^63s^23p^64s^23d^1\] Removing three electrons (two from the 4s and one from the 3d orbital): \[\mathrm{Sc}^{3+}: 1s^22s^22p^63s^23p^6\] The electron configuration for \(\mathrm{Sc}^{3+}\) is the same as that of the noble gas Ar (argon, atomic number 18). Therefore, \(\mathrm{Sc}^{3+}\) has a noble-gas configuration.

(d) Electron Configuration of \(\mathrm{Rh}^{3+}\)

The atomic number of Rh is 45, which means it has 45 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Rh}^{3+}\), we need to remove three electrons from the atom. Neutral Rh electron configuration: \[1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^14d^8\] Removing three electrons (one from the 5s and two from the 4d orbital): \[\mathrm{Rh}^{3+}: 1s^22s^22p^63s^23p^64s^23d^{10}4p^64d^6\] The electron configuration for \(\mathrm{Rh}^{3+}\) does not match any noble gas electron configuration. Therefore, \(\mathrm{Rh}^{3+}\) does not have a noble-gas configuration.

(e) Electron Configuration of \(\mathrm{Tl}^{+}\)

The atomic number of Tl is 81, which means it has 81 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Tl}^{+}\), we need to remove one electron from the atom. Neutral Tl electron configuration: \[1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}6p^1\] Removing one electron from the 6p orbital: \[\mathrm{Tl}^{+}: 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}\] The electron configuration for \(\mathrm{Tl}^{+}\) is the same as that of the noble gas Hg (mercury, atomic number 80). Therefore, \(\mathrm{Tl}^{+}\) has a noble-gas configuration.

(f) Electron Configuration of \(\mathrm{Bi}^{3+}\)

The atomic number of Bi is 83, which means it has 83 electrons in a neutral atom. To find the electron configuration for \(\mathrm{Bi}^{3+}\), we need to remove three electrons from the atom. Neutral Bi electron configuration: \[1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}6p^3\] Removing three electrons from the 6p orbital: \[\mathrm{Bi}^{3+}: 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^{10}\] The electron configuration for \(\mathrm{Bi}^{3+}\) is the same as that of the noble gas Pb (lead, atomic number 82). Therefore, \(\mathrm{Bi}^{3+}\) has a noble-gas configuration. In conclusion, \(\mathrm{Zn}^{2+}\), \(\mathrm{Te}^{2-}\), \(\mathrm{Sc}^{3+}\), \(\mathrm{Tl}^{+}\), and \(\mathrm{Bi}^{3+}\) have noble-gas configurations, while \(\mathrm{Rh}^{3+}\) does not.

Key concepts

Noble Gas Configuration

Electron configurations describe how electrons fill the orbitals of an atom. Noble gas configurations are condensed forms of these configurations that highlight stability. They mimic the electron arrangement of noble gases, which are the most stable due to having full outer shells.
Noble gases include helium, neon, argon, krypton, xenon, and radon. When an atom or ion achieves the same electron configuration as a noble gas, it becomes more stable.

For instance:

• ** extbf{Zn}^{2+} ** removes two electrons and matches krypton ( extbf{Kr} ).
• ** extbf{Te}^{2-} ** gains two electrons to resemble xenon ( extbf{Xe} ).
• ** extbf{Sc}^{3+} ** loses three electrons to align with argon ( extbf{Ar} ).

This mimicry of noble gases helps ions achieve lower energy states, contributing to their chemical stability.

Ions in Chemistry

Ions are atoms or molecules with a net electric charge due to loss or gain of electrons. This process often occurs to achieve a more stable electron configuration.
There are two main types of ions:

• **Cations**: Positively charged ions, formed by losing electrons. Example: extbf{Zn}^{2+} loses two electrons to become a cation.
• **Anions**: Negatively charged ions, formed by gaining electrons. Example: extbf{Te}^{2-} gains two electrons to become an anion.

Chemicals often react to achieve noble gas configurations through the formation of ions. This pursuit of stability drives the formation of ionic compounds such as salts.
Understanding ions helps explain the behavior of elements in various chemical reactions.

Atomic Structure

Atomic structure refers to the arrangement of electrons, protons, and neutrons in an atom. Electrons occupy regions around the nucleus called orbitals. These orbitals are defined by energy levels and sublevels.
Electrons fill these orbitals in a specific order based on the Aufbau principle, Hund's rule, and the Pauli exclusion principle:

• **Aufbau Principle**: Electrons fill orbitals from the lowest energy level upwards.
• **Hund's Rule**: Electrons will fill empty orbitals in the same sublevel before pairing up.
• **Pauli Exclusion Principle**: No two electrons in an atom can have the same set of quantum numbers.

These rules ensure that the atom achieves the lowest possible energy state. Understanding atomic structure is crucial for grasping how elements interact and bond in chemical reactions, forming the basis of chemistry.