# Oxidation state of nitrogen is incorrectly given in:(A) Compound - $\left[ {Co{{(N{H_3})}_5}Cl} \right]C{l_2}$, Oxidation state = (-3)(B) Compound - $N{H_2}OH$, Oxidation state = (-1)(C) Compound - ${\left( {{N_2}{H_5}} \right)_2}S{O_4}$, Oxidation state = (+2)(D) Compound - $M{g_3}{N_2}$, Oxidation state = (-3)

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Hint: The sum of the oxidation states of the constituent ions of a complex is equal to the oxidation state or charge of the entire complex. If not specified or written then the oxidation state of the complex is zero.

-The oxidation state is defined as the total number of electrons that can be lost or gained by any atom to form bonds or to form any complex. If due to a reaction the oxidation state increases then it is known as oxidation and if the oxidation state decreases then it is known as reduction.
-The oxidation state or charge of the entire complex as a whole is equal to the sum of the oxidation states of its constituent ions.
We will be finding the oxidation state of N in all the 4 compounds and check which one is given false.
-For $\left[ {Co{{(N{H_3})}_5}Cl} \right]C{l_2}$: In this complex N is present in the form of $N{H_3}$ and the oxidation state of $N{H_3}$ is 0. Let the oxidation state of N be ‘x’ and that of H is 1.
0 = x + 3(1)
x = (-3)
The oxidation state of N here is (-3) and so this option is true.
-For $N{H_2}OH$: Overall the molecule is neutral so its oxidation state will be 0. Let the oxidation state of N be ‘x’ and that of H is 1 and O is (-2).
0 = x + 3(1) + (-2)
0 = x + 1
X = (-1)
The oxidation state of N here is (-1) and this option is also true.
-For ${\left( {{N_2}{H_5}} \right)_2}S{O_4}$: Overall state is zero here, that of H is 1 and of $S{O_4}^{ - 2}$ is (-2). Let the oxidation state of N be ‘x’.
0 = 2 [ 2x + 5(1) ] + (-2)
2 = 4x + 10
-8 = 4x
x = (-2)
The oxidation state of this complex is (-2) and the value given in option is (+2). So, this option is false.
-For $M{g_3}{N_2}$: Overall oxidation state of this complex is 0, that of Mg is (+2) and let that of N be ‘x’.
0 = 3(+2) + 2x
0 = 6 + 2x
x = (-3)
The oxidation state of N here is (+3) and so this option is also true.
So, the answer will be: (C) Compound - ${\left( {{N_2}{H_5}} \right)_2}S{O_4}$, Oxidation state = (+2)

Note:
Nitrogen can exhibit variable oxidation states ranging from (-3) like in ammonia and amines to (+5) in nitric acid. This is due to the presence of two 2s electrons and three 2p electrons in its outermost shell. It can obtain inert gas configuration by taking in 3 electrons ($1{s^2}2{s^2}2{p^6}$) thereby exhibiting (-3) oxidation state, or it can also lose 5 electrons of outermost shell thereby exhibiting (+5) state.