Question

Ions also have electron configurations (Section 7.4). Cations have fewer valence electrons, and anions have more valence electrons, respectively, than their parent atoms. For example, chloride, \(\mathrm{Cl}^{-}\), has an electron configuration of \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\), for a total of 18 electrons, compared to 17 for neutral chlorine, the element. Na has an electron configuration of \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\), but \(\mathrm{Na}^{+}\) has an electron configuration of \(1 s^{2} 2 s^{2} 2 p^{6} .\) Write out the electron configurations for (a) \(\mathrm{F}^{-},(\mathrm{b}) \mathrm{I}^{-}\), (c) \(\mathrm{O}^{2-}\), (d) \(\mathrm{K}^{+}\), (e) \(\mathrm{Mg}^{2+}\), (f) \(\mathrm{Al}^{3+}\).

Short answer

The electron configurations for the given ions are: (a) \(F^{-}: 1s^{2}2s^{2}2p^{6}\) (b) \(I^{-}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\) (c) \(O^{2-}: 1s^{2}2s^{2}2p^{6}\) (d) \(K^{+}: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\) (e) \(Mg^{2+}: 1s^{2}2s^{2}2p^{6}\) (f) \(Al^{3+}: 1s^{2}2s^{2}2p^{6}\)

Step-by-step solution

Determine the parent atom electron configuration

Fluorine (F) is the parent atom for F⁻. It has an atomic number of 9, which means 9 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{5}\).

Add or remove electrons for the ion

\(\mathrm{F}^{-}\) gains an electron. Therefore, it has 10 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}\). #b) I^−#

Determine the parent atom electron configuration

Iodine (I) is the parent atom for I⁻. It has an atomic number of 53, which means 53 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{5}\).

Add or remove electrons for the ion

\(\mathrm{I}^{-}\) gains an electron. Therefore, it has 54 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\). #c) O²^−#

Determine the parent atom electron configuration

Oxygen (O) is the parent atom for O²⁻. It has an atomic number of 8, which means 8 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{4}\).

Add or remove electrons for the ion

\(\mathrm{O}^{2-}\) gains 2 electrons. Therefore, it has 10 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}\). #d) K^+#

Determine the parent atom electron configuration

Potassium (K) is the parent atom for K⁺. It has an atomic number of 19, which means 19 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}\).

Add or remove electrons for the ion

\(\mathrm{K}^{+}\) loses 1 electron. Therefore, it has 18 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\). #e) Mg²^+#

Determine the parent atom electron configuration

Magnesium (Mg) is the parent atom for Mg²⁺. It has an atomic number of 12, which means 12 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}\).

Add or remove electrons for the ion

\(\mathrm{Mg}^{2+}\) loses 2 electrons. Therefore, it has 10 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}\). #f) Al³^+#

Determine the parent atom electron configuration

Aluminum (Al) is the parent atom for Al³⁺. It has an atomic number of 13, which means 13 electrons in its neutral state. Its electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\).

Add or remove electrons for the ion

\(\mathrm{Al}^{3+}\) loses 3 electrons. Therefore, it has 10 electrons in total. The final electron configuration is \(1s^{2}2s^{2}2p^{6}\).

Key concepts

Understanding Ions

An ion is an atom or a molecule that has a net electrical charge due to the gain or loss of one or more electrons. This charge can either be positive or negative. Let's explore these further:

• When an electron is lost, the ion becomes positively charged and is called a cation.
• When an electron is gained, the ion becomes negatively charged and is known as an anion.

The formation of ions is critical in understanding how atoms interact and bond with each other. These charged particles play a vital role in electricity conduction, chemistry reactions, and biological functions.

Cations and Anions

Cations and anions are two types of ions that differ in their electrical charges.

Cations

Cations are positively charged ions formed when an atom loses one or more electrons. Think of sodium (\(\text{Na}\)) losing an electron to form Na⁺. It now has more protons than electrons, resulting in a positive charge.

• Cations are often metals.
• They are smaller than their neutral atoms due to the loss of electron shells.

Anions

Anions are negatively charged ions. They are created when an atom gains one or more electrons. For example, chloride (Cl⁻) forms when a chlorine atom gains an electron, making it negatively charged.

• Anions are usually non-metals.
• They tend to be larger than their neutral atoms due to additional electron shells.

Valence Electrons

Valence electrons are the outermost electrons of an atom and determine its chemical properties and reactivity. They are crucial in the formation of ions.

The Role of Valence Electrons

Valence electrons are responsible for forming bonds with other atoms. Here's how they relate to ion formation:

• Atoms with fewer valence electrons, like those in the alkali metals group, tend to lose electrons to form cations. For example, potassium (K) loses one electron to become K⁺.
• Atoms with more valence electrons, such as those in the halogens group, tend to gain electrons to become anions. For instance, fluorine (F) gains one electron to become F⁻.

Understanding how valence electrons work helps you predict an element's behavior during chemical reactions and its ability to form specific ions. This knowledge aids in mastering electron configurations and comprehending and analyzing various chemical processes.