Question

The oxidation state of nitrogen is correctly given for:
A. Compound \[\left[ {Co{{\left( {N{H_3}} \right)}_3}Cl} \right]C{l_2}\] Oxidation state 0
B. Compound \[N{H_2}OH\] Oxidation state -2
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

Answer 1

Hint:While calculating the oxidation states of atoms in a compound, first understand the net charge on the compound and then keep this value in the LHS of the equation. Now determine the elements with fixed oxidation states and multiply this number with the number of these atoms present. Then assume the oxidation states of the remaining numbers to be variables and multiply them with the number of the atoms present. Put this entire value in the RHS of the equation.

Complete step by step answer:
Before we move towards the solution of this question, let us understand some basic concepts.
The oxidation number or oxidation state of an element can be defined as the degree of oxidation of an element in the compound. In simpler terms, it can be understood as the number of the electrons gained or lost by an atom while forming a compound. This results in forming a net charge over this element, which is referred to as the oxidation state.
Depending on the number of electrons present in the valence shell of an atom, an element may exhibit a single or in some cases multiple oxidation states, depending on the atoms they are combining with.
Now in the case of \[M{g_3}{N_2}\] let the oxidation state of nitrogen is x. The oxidation state of magnesium is +2.
Therefore, the net charge of \[M{g_3}{N_2}\] is \[3 \times 2 + 2 \times x = 0\] .
Now calculate the value of x as follows,
 \[
  3 \times 2 + 2 \times x = 0 \\
  6 + 2x = 0 \\
  2x = - 6 \\
  x = - 3 \\
 \]
Therefore, in the case of Compound, \[M{g_3}{N_2}\] the oxidation state of nitrogen is -3
So, the correct option is D.

Note:Hydrogen tends to show two oxidation states: (-1) and (+1). To determine which oxidation state to use, you must look for the presence of a metal in the compound. Because in metal hydrides, the oxidation of hydrogen is (-1), otherwise it is always (+1).