In a mixture of concentrated $H_{2} S O_{4}$ and $H N O_{3}$ intermediate formed is:

H_{2} S O_{4}^{+}

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S O_{4}^{2 -}

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H_{2} N O_{3}^{-}

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H_{2} N O_{3}^{+}

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Solution

The correct option is D $H_{2} N O_{3}^{+}$
in mixture of conc. all three acid. $(H_{2} S O_{4}, H N O_{3})$ . $H_{2} S O_{4}$ stronger acid than $H N O_{3}$ .

$H_{2} S O_{4} + H N O_{3} ⟶ H S O_{4}^{-} + H_{2} N O_{3}^{+}$

$H_{2} N O_{3}^{+} ⟶ N O_{2}^{+} + H_{2} O$

Hence, the correct option is $D$

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nitrating mixture is obtained by mixing concentrated $H N O_{3}$ and concentrated $H_{2} S O_{4}$ . The role of the latter - $H_{2} S O_{4}$ is:

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H_{2} S O_{4}

o -

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H N O_{3}

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