What is the change in oxidation number of nitrogen in the following reaction?
$N_{2} (g) + \frac{1}{2} O_{2} (g) \to N_{2} O (g)$

0 to + 1

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0 to + 2

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0 to - 1

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0 to + \frac{1}{2}

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Solution

The correct option is A $0 to + 1$
In $N_{2}$ , oxidation state of $N$ is zero.

In $N_{2} O : 2 x - 2 = 0 \Rightarrow x = + 1$

$_{2} \to {+ 1 N}_{2} O$ , so oxidation state of nitrogen changes from $0 to + 1$ .

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Similar questions

Q. What is the change in oxidation number of carbon in the following reaction?

N_{2} O_{4} (g) + 3 C O (g) \to N_{2} O (g) + 3 C O_{2} (g)

Q. What is the change in oxidation number of nitrogen from

H N O_{2} t o H N O_{3}

in the following reaction? Ignore the other two products.

3 H N O_{2} \to H N O_{3} + 2 N O + H_{2} O

Q. In which of the following reactions there is a change in the oxidation number of nitrogen atom?

Q. Maximum change in oxidation number of nitrogen is ________ .

A mole of $N_{2} H_{4}$ loses 10 mol of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation number of hydrogen.)

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What is the change in oxidation number of nitrogen in the following reaction? N2(g)+12O2(g)→N2O(g)

The correct option is A 0 to +1In N2, oxidation state of N is zero. In N2O:2x–2=0 ⇒x=+1 0N2→+1N2O, so oxidation state of nitrogen changes from 0 to +1.

What is the change in oxidation number of nitrogen in the following reaction?
$N_{2} (g) + \frac{1}{2} O_{2} (g) \to N_{2} O (g)$

The correct option is A $0 to + 1$
In $N_{2}$ , oxidation state of $N$ is zero.

In $N_{2} O : 2 x - 2 = 0 \Rightarrow x = + 1$

$_{2} \to {+ 1 N}_{2} O$ , so oxidation state of nitrogen changes from $0 to + 1$ .