JEE MainChemistryd and f Block Elements
V ₂ O ₅ is an amphoteric oxide that dissolves in dilute acids to form a characteristic oxocation. The chemical formula of this oxocation and the number of unpaired d-electrons on the vanadium atom in it are, respectively :
Options
- AVO ²⁺ and 1
- BV ⁵⁺ and 0
- CVO ₂^+ and 0
- DVO ₂^+ and 1
Correct answer
C. VO ₂^+ and 0
Step-by-step solution
V ₂ O ₅ is an amphoteric oxide (though mainly acidic). It reacts with alkalis as well as acids. When dissolved in dilute acids, it forms the dioxovanadium(V) cation, VO ₂^+ . The oxidation state of vanadium in VO ₂^+ is +5 . The ground state electronic configuration of vanadium ( Z=23 ) is [ Ar ] 3 d ^3 4 s ^2 . In the +5 oxidation state, vanadium loses all five of its valence electrons, resulting in a 3 d ^0 configuration. Therefore, the number of unpaired d-electrons is 0 . Note: V ₂ O ₄ dissolves in acids to for