
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
562.8k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

