
The oxidation state of nitrogen is most positive in which of the following compounds-
A.$NO_3^ - $
B.$NO_2^ - $
C.${N_2}O$
D.${N_2}$
E.$N{H_3}$
Answer
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Hint: Oxidation number is the charge which an atom or ion of the element contains or appears to have when it is present in combined state with other atoms. We have to follow two rules when counting the electrons on the atom-
i) Electrons shared between two like or same atoms are equally divided between the sharing atoms.
ii) Electrons shared between two different atoms are counted with more electronegative atoms.
Complete Step-by-Step Solution:
Here we have to find the most positive oxidation state of nitrogen among the given elements or molecules.
We know that Oxygen has oxidation number $ - 2$ and hydrogen has oxidation number $1$. Let the oxidation number of nitrogen be x.
Now let’s start finding the oxidation number of Nitrogen.
In $NO_3^ - $,
$ \Rightarrow x + \left( {3 \times - 2} \right) = - 1$ {because there are three atoms of oxygen and the charge on the atom is $ - 1$ }
On solving we get,
$ \Rightarrow x - 6 = - 1$
On further solving we get,
$ \Rightarrow x = - 1 + 6 = + 5$
In $NO_2^ - $,
$ \Rightarrow x + \left( {2 \times - 2} \right) = - 1$
On solving we get,
$ \Rightarrow x - 4 = - 1$
On simplifying we get,
$ \Rightarrow x = - 1 + 4 = + 3$
In ${N_2}O$,
$ \Rightarrow 2x + \left( { - 2} \right) = 0$ { Because there is no charge on the molecule}
On solving we get,
$ \Rightarrow 2x - 2 = 0$
On simplifying we get,
$ \Rightarrow 2x = 2 \Rightarrow x = + 1$
In${N_2}$, the oxidation state of nitrogen is zero because the element is in its elemental state or Free State.
In $N{H_3}$,
$ \Rightarrow x + \left( {3 \times 1} \right) = 0$
On solving we get,
$ \Rightarrow x = - 3$
Hence the most positive oxidation state on observation is $ + 5$ in $NO_3^ - $.
Answer-So the correct answer is A.
Note: Oxidation state and valency are two different things so don’t get confused between them.
1.Valency is the combining capacity of an atom whereas oxidation is the charge on an atom.
2.Valency is always a whole number while oxidation state can have fractional values
3.The Valency of an atom cannot be zero but the oxidation state of an element can be zero.
i) Electrons shared between two like or same atoms are equally divided between the sharing atoms.
ii) Electrons shared between two different atoms are counted with more electronegative atoms.
Complete Step-by-Step Solution:
Here we have to find the most positive oxidation state of nitrogen among the given elements or molecules.
We know that Oxygen has oxidation number $ - 2$ and hydrogen has oxidation number $1$. Let the oxidation number of nitrogen be x.
Now let’s start finding the oxidation number of Nitrogen.
In $NO_3^ - $,
$ \Rightarrow x + \left( {3 \times - 2} \right) = - 1$ {because there are three atoms of oxygen and the charge on the atom is $ - 1$ }
On solving we get,
$ \Rightarrow x - 6 = - 1$
On further solving we get,
$ \Rightarrow x = - 1 + 6 = + 5$
In $NO_2^ - $,
$ \Rightarrow x + \left( {2 \times - 2} \right) = - 1$
On solving we get,
$ \Rightarrow x - 4 = - 1$
On simplifying we get,
$ \Rightarrow x = - 1 + 4 = + 3$
In ${N_2}O$,
$ \Rightarrow 2x + \left( { - 2} \right) = 0$ { Because there is no charge on the molecule}
On solving we get,
$ \Rightarrow 2x - 2 = 0$
On simplifying we get,
$ \Rightarrow 2x = 2 \Rightarrow x = + 1$
In${N_2}$, the oxidation state of nitrogen is zero because the element is in its elemental state or Free State.
In $N{H_3}$,
$ \Rightarrow x + \left( {3 \times 1} \right) = 0$
On solving we get,
$ \Rightarrow x = - 3$
Hence the most positive oxidation state on observation is $ + 5$ in $NO_3^ - $.
Answer-So the correct answer is A.
Note: Oxidation state and valency are two different things so don’t get confused between them.
1.Valency is the combining capacity of an atom whereas oxidation is the charge on an atom.
2.Valency is always a whole number while oxidation state can have fractional values
3.The Valency of an atom cannot be zero but the oxidation state of an element can be zero.
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