Question

For each of the following compounds determine the electron (c) \(\mathrm{NiO}\), (d) \(\mathrm{ZnO}\).

Short answer

For Nickel (II) Oxide (NiO), the electron configurations of the ions are: Ni²⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}\) and O²⁻: \(1s^{2}2s^{2}2p^{6}\) So, the electron configuration for the compound NiO is: NiO: Ni (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}\)) O (\(1s^{2}2s^{2}2p^{6}\)) For Zinc Oxide (ZnO), the electron configurations of the ions are: Zn²⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\) and O²⁻: \(1s^{2}2s^{2}2p^{6}\) So, the electron configuration for the compound ZnO is: ZnO: Zn (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\)) O (\(1s^{2}2s^{2}2p^{6}\))

Step-by-step solution

Identify the ions in NiO

In NiO, we have a nickel (II) cation (Ni²⁺) and an oxide anion (O²⁻).

Find the electron configuration of Ni²⁺

To determine the electron configuration of Ni²⁺, first, find the atomic number of nickel. Nickel has an atomic number of 28, meaning it has 28 protons and electrons in its neutral state. The electron configuration of neutral nickel (Ni) is: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}\] For Ni²⁺, two electrons are lost from the valence shell. Thus, the electron configuration for Ni²⁺ is: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}\]

Find the electron configuration of O²⁻

Oxygen has an atomic number of 8, which means it has 8 protons and 8 electrons in its neutral state. The electron configuration of neutral oxygen (O) is: \[1s^{2}2s^{2}2p^{4}\] For O²⁻, two electrons are added to the valence shell. Thus, the electron configuration for O²⁻ is: \[1s^{2}2s^{2}2p^{6}\]

Write the electron configurations for the compound NiO

Now that we have the electron configurations for Ni²⁺ and O²⁻, we can write the electron configurations for the compound NiO: \[NiO: Ni (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}) O (1s^{2}2s^{2}2p^{6})\] For Zinc Oxide (ZnO):

Identify the ions in ZnO

In ZnO, we have a zinc (II) cation (Zn²⁺) and an oxide anion (O²⁻).

Find the electron configuration of Zn²⁺

To determine the electron configuration of Zn²⁺, first, find the atomic number of zinc. Zinc has an atomic number of 30, meaning it has 30 protons and 30 electrons in its neutral state. The electron configuration of neutral zinc (Zn) is: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}\] For Zn²⁺, two electrons are lost from the valence shell. Thus, the electron configuration for Zn²⁺ is: \[1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\]

Find the electron configuration of O²⁻ (same as in NiO)

The electron configuration for O²⁻ is: \[1s^{2}2s^{2}2p^{6}\]

Write the electron configurations for the compound ZnO

Now that we have the electron configurations for Zn²⁺ and O²⁻, we can write the electron configurations for the compound ZnO: \[ZnO: Zn (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}) O (1s^{2}2s^{2}2p^{6})\]

Key concepts

NiO

Nickel oxide, often written as NiO, is a compound made from the elements nickel and oxygen. In NiO, both elements form ions, which are charged particles. Nickel becomes a cation while oxygen becomes an anion:

• Nickel (Ni) loses two electrons to become a nickel (II) cation, noted as Ni²⁺.
• Oxygen (O) gains two electrons to become an oxide anion, noted as O²⁻.

These opposite charges attract each other to form the ionic compound NiO.
To find their electron configurations, remember this: Nickel in its neutral state has 28 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸. When it becomes Ni²⁺, it loses two electrons from the 4s² orbital, becoming 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸.
Oxygen normally has 8 electrons: 1s² 2s² 2p⁴. As O²⁻, it gains two electrons, changing the configuration to 1s² 2s² 2p⁶.
Understanding these ionic and electronic changes is key to grasping the properties of NiO.

ZnO

Zinc oxide, or ZnO, is another compound consisting of zinc and oxygen, similar to NiO, but with zinc instead of nickel. Here’s how the ions form in ZnO:

• Zinc (Zn) loses two electrons to become a zinc (II) cation, Zn²⁺.
• Oxygen (O) gains two electrons to form an oxide anion, O²⁻.

The charges balance, resulting in a stable ionic compound.
The electron configuration helps to understand these changes: Zinc has 30 electrons when neutral: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰. As Zn²⁺, it loses the 4s² electrons, leaving it as 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰. For the oxygen part, O²⁻ gains electrons just like in NiO, making its electron configuration 1s² 2s² 2p⁶.
Zinc oxide's properties, such as conductance and reactivity, stem from these ionic and electron arrangements. It's vital to understand these configurations when studying material science and chemistry.

Cations and Anions

In the study of chemistry, especially when examining compounds like NiO and ZnO, understanding cations and anions is essential.

• Cations: These are positively charged ions. They are formed when an atom loses electrons, which decreases its negative charge. For instance, both Ni²⁺ and Zn²⁺ are cations. The loss of electrons results in a more positive charge.
• Anions: These are negatively charged ions. They form when an atom gains electrons, resulting in an increase in negative charge. In both NiO and ZnO, O²⁻ serves as the anion, having gained two extra electrons.

These ions attract each other because of their opposite charges, forming ionic bonds. This bond is the foundation for creating stable structures in many compounds. Grasping the concept of cations and anions helps in predicting how elements will interact to form different substances, aiding in both general chemistry knowledge and specific applications in fields like electronics and pharmaceuticals.