Question

Apply Write out the electron configuration for each atom. Then, predict the change that must occur in each to achieve a noble-gas configuration. a. nitrogen b. sulfur c. barium d. lithium

Short answer

Nitrogen: Electron configuration is $1s^{2}2s^{2}2p^3$, it needs to gain 3 electrons to form N³⁻, achieving a noble-gas configuration of Neon ($1s^{2}2s^{2}2p^6$). Sulfur: Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$, it needs to gain 2 electrons to form S²⁻, achieving a noble-gas configuration of Argon ($1s^{2}2s^{2}2p^{6}3s^{2}3p^6$). Barium: Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$, it needs to lose 2 electrons to form Ba²⁺, achieving a noble-gas configuration of Xenon ($1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$). Lithium: Electron configuration is $1s^{2}2s^1$, it needs to lose 1 electron to form Li⁺, achieving a noble-gas configuration of Helium ($1s^2$).

Step-by-step solution

a. Nitrogen

To find the electron configuration for Nitrogen (atomic number 7), we can use the periodic table to determine the order in which the orbitals are filled: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p,... The electron configuration for Nitrogen is $1s^{2}2s^{2}2p^3$. Nitrogen needs to gain 3 electrons to attain a noble-gas configuration. This will form a Nitride ion (N^3-) with the electron configuration $1s^{2}2s^{2}2p^6$, which is the same as the noble gas Neon.

b. Sulfur

To find the electron configuration for Sulfur (atomic number 16), we use the same method as above: The electron configuration for Sulfur is $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$. Sulfur needs to gain 2 electrons to attain a noble-gas configuration. This will form a Sulfide ion (S^2-) with the electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$, which is the same as the noble gas Argon.

c. Barium

To find the electron configuration for Barium (atomic number 56), we use the same method as above: The electron configuration for Barium is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$. Barium needs to lose 2 electrons to attain a noble-gas configuration. This will form a Barium ion (Ba^2+) with the electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$, which is the same as the noble gas Xenon.

d. Lithium

To find the electron configuration for Lithium (atomic number 3), we use the same method as above: The electron configuration for Lithium is $1s^{2}2s^1$. Lithium needs to lose 1 electron to attain a noble-gas configuration. This will form a Lithium ion (Li^+) with the electron configuration $1s^2$, which is the same as the noble gas Helium.

Key concepts

Noble-Gas Configuration

Every atom seeks to achieve a stable electron arrangement similar to the noble gases. Noble gases like Neon, Argon, and Helium have full outer energy levels. This makes them unreactive and stable. In chemistry, atoms gain or lose electrons to emulate this arrangement, thus achieving a noble-gas configuration. A full outer shell often contains eight electrons in its valence shell, known as the "octet rule." For Helium, with its full 1s shell, the rule deviates slightly. By attaining a noble-gas configuration, atoms become more stable.

Nitrogen Electron Configuration

Nitrogen, with an atomic number of 7, has the electron configuration of $1s^{2}2s^{2}2p^3$. To become stable, it needs to gain 3 electrons, resulting in a $1s^{2}2s^{2}2p^6$ configuration, mirroring Neon. When it gains electrons, it forms an anion known as Nitride (N^{3-}). This modification brings stability akin to the noble gas structure, i.e., Neon.

Sulfur Electron Configuration

Sulfur has an atomic number of 16. Its configuration initially is $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$. To achieve a stable noble-gas configuration similar to Argon, sulfur needs to gain 2 more electrons. This results in the sulfide ion (S^{2-}) with the electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$, matching Argon's stable electronic arrangement. This process, too, forms an anion, enhancing sulfur's stability.

Barium Electron Configuration

Barium, with atomic number 56, has a more intricate configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$. By losing 2 electrons, it attains a noble-gas structure similar to Xenon$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$. This removal of electrons results in a Barium cation (Ba^{2+}). This positive ion showcases a stable noble-gas configuration, mirroring the inert qualities of Xenon.

Lithium Electron Configuration

Lithium, which is much simpler with an atomic number of 3, begins with $1s^{2}2s^1$. It loses 1 electron to achieve the noble-gas configuration of Helium, $1s^2$. Forming the cation Li^{+} ensures a stable electron configuration. This process of becoming a positive ion gives Lithium similar stability attributes as Helium, which is an inert noble gas.

Anion Formation

An anion is formed when an atom gains extra electrons and transcends its stable configuration. For nitrogen and sulfur, this involves filling the p orbital.
The negatively charged ions (anions) are larger than their neutral atom due to increased electron-electron repulsion. In the context of noble-gas configurations, anions obtain a fully-filled valence shell, providing them with achieved stability.

Cation Formation

Cations arise when atoms lose electrons. This often occurs with metals like lithium and barium to achieve stability. The result is a positively charged ion.
Cations are typically smaller than their neutral atoms since losing electrons reduces electron-electron repulsion.
In noble-gas configurations, they shed electrons to unoccupied shells, paralleling the closest noble gas configuration, and achieving a stable electronic environment.