On reaction of nitric acid with metals, the reduction product(s) formed is/are:

N_{2} O

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N O

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N O_{2}

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N H_{4} N O_{3}

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Solution

The correct option is D $N H_{4} N O_{3}$
$H N O_{3}$ is an excellent oxidizing agent. Metals like copper and zinc which are insoluble in $H C l$ dissolve in $H N O_{3}$ to form $N_{2} O, N O, N O_{2} and N H_{4} N O_{3}$ .
Examples:
$3 C u + 8 H N O_{3} (d i l .) \to 3 C u (N O_{3})_{2} + 2 N O + 4 H_{2} O$
$C u + 4 H N O_{3} (c o n c .) \to C u (N O_{3})_{3} + 2 N O_{2} + 2 H_{2} O$

$4 Z n + 10 H N O_{3} (d i l .) \to 4 Z n (N O_{3})_{2} + N_{2} O + 5 H_{2} O Z n + 4 H N O_{3} (c o n c .) \to Z n (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$

$4 F e + 10 H N O_{3} very dilute \to 4 F e (N O_{3})_{2} + 3 H_{2} O + N H_{4} N O_{3}$

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