Which oxide does not react as a reducing agent?
A.$NO$
B.$N{O_2}$
C.${N_2}O$
D.${N_2}{O_5}$
Answer
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Hint: Oxidation is the process in which an element loses electrons or hydrogen, or gains oxygen. A reducing agent oxidizes itself (undergoes oxidation) by reducing another element.
Complete step by step answer:
We will start with calculating the oxidation states of N atoms in each of the given oxides. O atom is present in -2 state in all the oxides.
Since, all the oxides are neutral molecules (have overall charge equal to zero), therefore, we will equate the equation to zero. (For example, $x + y = 0$)
For $NO$,
$ x + ( - 2) = 0 \\
\Rightarrow x = + 2 \\ $
For $N{O_2}$,
$ x + (2 \times - 2) = 0 \\
\Rightarrow x - 4 = 0 \\
\Rightarrow x = + 4 \\ $
For ${N_2}O$,
$ 2x + ( - 2) = 0 \\
\Rightarrow 2x = + 2 \\
\Rightarrow x = \dfrac{{ + 2}}{2} \\
\Rightarrow x = + 1 \\ $
For ${N_2}{O_5}$,
$ 2x + (5 \times - 2) = 0 \\
\Rightarrow 2x - 10 = 0 \\
\Rightarrow 2x = + 10 \\
\Rightarrow x = \dfrac{{ + 10}}{2} \\
\Rightarrow x = + 5 \\ $
The highest possible oxidation state of N is +5.
In $NO$, the oxidation state of N is +2 and by losing electrons it can acquire higher oxidation states and act as a reducing agent.
In $N{O_2}$, the oxidation state of N is +4 which is a high oxidation state and acquiring further higher oxidation number requires a high amount of energy in order to lose electrons. Therefore, it acts as a strong oxidizing agent.
In ${N_2}O$, the oxidation state of N is +1 and by losing electrons it can acquire higher oxidation states. Therefore, it acts as a strong reducing agent.
In ${N_2}{O_5}$, the oxidation state of N is +5. Here, N is in its highest possible oxidation state and can only accept electrons thereby lowering its oxidation state. Hence, it acts as a strong oxidizing agent.
Thus, $N{O_2}$ and ${N_2}{O_5}$ do not act as reducing agents. They are good oxidizing agents.
Hence option B and D are correct.
Note:
Reactions that involve oxidation and reduction of elements or compounds are called redox reactions.
Complete step by step answer:
We will start with calculating the oxidation states of N atoms in each of the given oxides. O atom is present in -2 state in all the oxides.
Since, all the oxides are neutral molecules (have overall charge equal to zero), therefore, we will equate the equation to zero. (For example, $x + y = 0$)
For $NO$,
$ x + ( - 2) = 0 \\
\Rightarrow x = + 2 \\ $
For $N{O_2}$,
$ x + (2 \times - 2) = 0 \\
\Rightarrow x - 4 = 0 \\
\Rightarrow x = + 4 \\ $
For ${N_2}O$,
$ 2x + ( - 2) = 0 \\
\Rightarrow 2x = + 2 \\
\Rightarrow x = \dfrac{{ + 2}}{2} \\
\Rightarrow x = + 1 \\ $
For ${N_2}{O_5}$,
$ 2x + (5 \times - 2) = 0 \\
\Rightarrow 2x - 10 = 0 \\
\Rightarrow 2x = + 10 \\
\Rightarrow x = \dfrac{{ + 10}}{2} \\
\Rightarrow x = + 5 \\ $
The highest possible oxidation state of N is +5.
In $NO$, the oxidation state of N is +2 and by losing electrons it can acquire higher oxidation states and act as a reducing agent.
In $N{O_2}$, the oxidation state of N is +4 which is a high oxidation state and acquiring further higher oxidation number requires a high amount of energy in order to lose electrons. Therefore, it acts as a strong oxidizing agent.
In ${N_2}O$, the oxidation state of N is +1 and by losing electrons it can acquire higher oxidation states. Therefore, it acts as a strong reducing agent.
In ${N_2}{O_5}$, the oxidation state of N is +5. Here, N is in its highest possible oxidation state and can only accept electrons thereby lowering its oxidation state. Hence, it acts as a strong oxidizing agent.
Thus, $N{O_2}$ and ${N_2}{O_5}$ do not act as reducing agents. They are good oxidizing agents.
Hence option B and D are correct.
Note:
Reactions that involve oxidation and reduction of elements or compounds are called redox reactions.
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