The correct order of basicity of oxides of vanadium is
The correct order of basicity of oxides of vanadium is
Show Hint
- $V _2 O _3> V _2 O _5> V _2 O _4$
- $V _2 O _4> V _2 O _3> V _2 O _5$
- $V _2 O _3> V _2 O _4> V _2 O _5$
- $V _2 O _5> V _2 O _4> V _2 O _3$
The Correct Option is C
Solution and Explanation
With increase in the percentage of oxygen, the acidic nature of oxides increases and basic nature decreases. Therefore, the order of basicity for vanadium oxides is:
V2O3 > V2O4 > V2O5
The basicity of oxides decreases as the oxidation state of vanadium increases. In V2O3, vanadium has a lower oxidation state (+3), making it more basic. As the oxidation state increases in V2O5 (+5), the acidic character dominates, reducing the basic nature.
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Concepts Used:
D and F Block Elements
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
Read More: The d and f block elements