Question

Write the electron configuration for each ion. $$ \text { a. } \mathrm{Cl}^{-} \quad \text { b. } \mathrm{P}^{3-} \quad \text { c. } \mathrm{K}^{+} \quad \text { d. } \mathrm{Mo}^{3+} \quad \text { e. } \mathrm{V}^{3+} $$

Short answer

a. \(\mathrm{Cl}^{-}: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\) b. \(\mathrm{P}^{3-}: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\) c. \(\mathrm{K}^{+}: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\) d. \(\mathrm{Mo}^{3+}: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{3}\) e. \(\mathrm{V}^{3+}: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}\)

Step-by-step solution

Identify the neutral atom's electron configuration

Find the electron configuration of the neutral atoms Chlorine (Cl), Phosphorus (P), Potassium (K), Molybdenum (Mo), and Vanadium (V) before they become ions.

Adjust for ion charge by adding or removing electrons

Add electrons when the ion is negatively charged (anions). Remove electrons when the ion is positively charged (cations). The number of electrons to add or remove corresponds to the ion's charge.

Write the new electron configuration for \(\mathrm{Cl}^{-}\)

Chlorine (Cl) as a neutral atom has 17 electrons and has the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}\). As \(\mathrm{Cl}^{-}\), it gains one extra electron, having a total of 18 electrons, which makes the configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\).

Write the new electron configuration for \(\mathrm{P}^{3-}\)

Phosphorus (P) as a neutral atom has 15 electrons with the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}\). As \(\mathrm{P}^{3-}\), it gains three extra electrons, totaling 18 electrons, thus the configuration is \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\), which is identical to that of Argon (Ar), a noble gas.

Write the new electron configuration for \(\mathrm{K}^{+}\)

Potassium (K) as a neutral atom has 19 electrons with the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}\). As \(\mathrm{K}^{+}\), it loses one electron, leaving it with 18 electrons, with the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}\), the same as Argon (Ar).

Write the new electron configuration for \(\mathrm{Mo}^{3+}\)

Molybdenum (Mo) as a neutral atom has 42 electrons with the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{5} 5s^{1}\). As \(\mathrm{Mo}^{3+}\), it loses three electrons. The last electrons to have been added are from the \(5s\) and \(4d\) orbitals. Therefore, the new configuration for \(\mathrm{Mo}^{3+}\) is \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{3}\).

Write the new electron configuration for \(\mathrm{V}^{3+}\)

Vanadium (V) as a neutral atom has 23 electrons with the electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}\). As \(\mathrm{V}^{3+}\), it loses three electrons, which will come off from the \(4s\) and \(3d\) orbitals, leading to the new electron configuration \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}\).

Key concepts

Ion Electron Configuration

To fully understand how ions interact and form, one must grasp the basics of ion electron configuration. An ion's electron configuration is similar to that of a neutral atom, but it is adjusted based on the ion's charge. An important detail is recognizing that negative ions, called anions, gain extra electrons, giving them a fuller electron shell. On the other hand, positive ions, known as cations, lose electrons, leading to a less filled electron shell.

For example, when Chlorine, with its atomic number of 17, gains one electron, it becomes \( \mathrm{Cl}^{-} \) and matches the electron configuration of Argon, because it now has 18 electrons. As a general rule of thumb, it can be helpful to associate anions with gaining electrons and reaching the stable electronic configuration of the nearest noble gas, and the opposite for cations.

Anions and Cations

Chemical species often seek stability through the gain or loss of electrons, forming anions and cations. Anions are negatively charged ions that occur when an atom gains one or more electrons. This is typically observed in nonmetals gaining electrons to achieve noble gas configuration. For example, Phosphorus (P), upon gaining three electrons, forms the anion \( \mathrm{P}^{3-} \) and achieves a stable electron configuration similar to the noble gas Argon (Ar).

Conversely, cations are positively charged ions created when an atom loses electrons. This is frequently seen in metals where the loss of electrons results in a stable configuration. Take Potassium (K), which forms the cation \( \mathrm{K}^{+} \) by losing one electron and attains the noble gas electron configuration of Argon.

Orbital Notation

Diving deeper into electron configurations, the orbital notation provides a visual representation of where electrons are located within an atom's orbitals. Instead of simply stating the electron configuration, orbital notation uses specific diagrams to denote the position of each electron within the atom's sublevels. For instance,

The Orbital Diagram for Vanadium (V)

\( 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2} \) would be represented by a series of boxes or lines with arrows indicating the electrons' spins.

When creating ions, the orbital notation must reflect the gained or lost electrons. As seen with Vanadium becoming \( \mathrm{V}^{3+} \) and losing three electrons, the orbital notation would omit the three electrons last added according to the Aufbau principle, indicating a new configuration of \(1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}\).