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
645.9k+ 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
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

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

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

