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In which compound does Vanadium have an oxidation number of +4?

Option: 1

\mathrm{NH _{4} VO _{2}}


Option: 2

\mathrm{K _{4}\left[ V ( CN )_{6}\right]}


Option: 3

\mathrm{VSO _{4}}


Option: 4

\mathrm{VOSO _{4}}


Answers (1)

best_answer

The structure of \mathrm{VOSO_4} is 

\mathrm{VOSO_4} dissociates as 

\mathrm{VOSO_4\rightleftharpoons VO^{2+}+ SO_4^{2-}}

Let the oxidation state of Vanadium be x, charge balance on the cation \mathrm{VO^{2+} } gives

x-2=+2

{x=4}

Thus in \mathrm{VOSO _{4}}, Vanadium has an oxidation number of +4

Hence, option number (4) is correct.

Posted by

HARSH KANKARIA

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