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Why do transition metals show variable oxidation states?
Why do transition metals show variable oxidation states?
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Variable oxidation state due to involvement of both ns and (n−1)d electrons.
Updated On: Mar 2, 2026
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Solution and Explanation
Why do Transition Metals Show Variable Oxidation States?
Transition metals exhibit variable oxidation states due to several factors related to their electronic structure and the nature of their bonding.
1. Incomplete d-Orbitals:
Transition metals have partially filled d-orbitals in addition to their s-orbitals. This allows them to lose different numbers of electrons from both the s and d orbitals, resulting in multiple possible oxidation states. For example, iron (Fe) can lose 2 electrons to form Fe²⁺ or 3 electrons to form Fe³⁺, giving it oxidation states of +2 and +3.
2. Similar Energy of s and d Orbitals:
In transition metals, the energy levels of the s and d orbitals are quite close to each other. This similarity allows the metal atoms to lose varying numbers of electrons from both types of orbitals. This makes the transition metals capable of showing a wide range of oxidation states.
3. Availability of d Electrons for Bonding:
Transition metals have a significant number of d-electrons, which can be involved in chemical bonding. The ability to use d-electrons in bonding gives rise to different oxidation states. For example, copper can show oxidation states of +1 and +2 due to the loss of one or two electrons from its d-orbitals.
4. Stability of Oxidation States:
Some oxidation states are more stable due to factors such as the electronic configuration and the crystal field environment in complex compounds. For example, the +2 oxidation state of copper (Cu²⁺) is stable due to the completely filled 3d10 configuration, while the +1 oxidation state is less stable.
Conclusion:
Transition metals show variable oxidation states because of the availability of d-electrons for bonding, the similar energies of s and d orbitals, and the ability to lose different numbers of electrons. The stability of each oxidation state is influenced by the electronic configuration and the chemical environment.
Transition metals exhibit variable oxidation states due to several factors related to their electronic structure and the nature of their bonding.
1. Incomplete d-Orbitals:
Transition metals have partially filled d-orbitals in addition to their s-orbitals. This allows them to lose different numbers of electrons from both the s and d orbitals, resulting in multiple possible oxidation states. For example, iron (Fe) can lose 2 electrons to form Fe²⁺ or 3 electrons to form Fe³⁺, giving it oxidation states of +2 and +3.
2. Similar Energy of s and d Orbitals:
In transition metals, the energy levels of the s and d orbitals are quite close to each other. This similarity allows the metal atoms to lose varying numbers of electrons from both types of orbitals. This makes the transition metals capable of showing a wide range of oxidation states.
3. Availability of d Electrons for Bonding:
Transition metals have a significant number of d-electrons, which can be involved in chemical bonding. The ability to use d-electrons in bonding gives rise to different oxidation states. For example, copper can show oxidation states of +1 and +2 due to the loss of one or two electrons from its d-orbitals.
4. Stability of Oxidation States:
Some oxidation states are more stable due to factors such as the electronic configuration and the crystal field environment in complex compounds. For example, the +2 oxidation state of copper (Cu²⁺) is stable due to the completely filled 3d10 configuration, while the +1 oxidation state is less stable.
Conclusion:
Transition metals show variable oxidation states because of the availability of d-electrons for bonding, the similar energies of s and d orbitals, and the ability to lose different numbers of electrons. The stability of each oxidation state is influenced by the electronic configuration and the chemical environment.
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Top Questions on d -and f -Block Elements
- Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
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