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Question
How to find the oxidation state of an element?
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Solution
The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Almost all of the transition metals have multiple potential oxidation states.
Oxidation results in an increase in the oxidation state. The reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and, therefore, a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants. On the other hand, lithium (Li) and sodium (Na) are incredibly strong reducing agents (likes to be oxidized), meaning that they easily lose electrons. In this module, we will precisely go over the oxidation states of transition metals.
Unpaired electrons of d-orbitals
To fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in the d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s orbital electrons since the 3d orbital is located before the 4s orbital in the periodic table.
For example, Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and, therefore, the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3.
The formula for determining oxidation states would be (with the exception of copper and chromium):
Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons
Oxidation results in an increase in the oxidation state. The reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and, therefore, a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants. On the other hand, lithium (Li) and sodium (Na) are incredibly strong reducing agents (likes to be oxidized), meaning that they easily lose electrons. In this module, we will precisely go over the oxidation states of transition metals.
Unpaired electrons of d-orbitals
To fully understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond. There are five orbitals in the d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species. This means that the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons. To determine the oxidation state, unpaired d-orbital electrons are added to the 2s orbital electrons since the 3d orbital is located before the 4s orbital in the periodic table.
For example, Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and, therefore, the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3.
The formula for determining oxidation states would be (with the exception of copper and chromium):
Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons
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