The amphoteric oxide of Vanadium (V) is V_2O_ | vedclass

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(C) The amphoteric oxide of Vanadium is $V_2O_5$,where the oxidation state of $V$ is $+5$.$V_2O_5$ reacts with alkali $(NaOH)$ to form vanadate ion $(

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$+5, +5$

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(C) The amphoteric oxide of Vanadium is $V_2O_5$,where the oxidation state of $V$ is $+5$.$V_2O_5$ reacts with alkali $(NaOH)$ to form vanadate ion $(VO_4^{3-})$,which is the oxoion '$X$'. In $VO_4^{3-}$,the oxidation state of $V$ is $x + 4(-2) = -3$,so $x = +5$.$V_2O_5$ reacts with acid $(HCl)$ to form the vanadyl ion $(VO_2^+)$,which is the oxoion '$Y$'. In $VO_2^+$,the oxidation state of $V$ is $x + 2(-2) = +1$,so $x = +5$.Therefore,the oxidation states of $V$ in both '$X$' and '$Y$' are $+5$ and $+5$.

The amphoteric oxide of Vanadium $(V)$ is $V_2O_5$. It reacts with alkali to form an oxoion '$X$' and with acid to form an oxoion '$Y$'. The oxidation states of '$V$' in '$X$' and '$Y$' are respectively:

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