Question

To calculate the oxidation number of the underlined atoms in the compounds given below;
$(a)\,{H_2}S{O_4}\,\,\,\left( b \right)HN{O_3}\,\,\,\left( c \right)\,{H_3}P{O_3}\,\,\,\left( d \right){K_2}{C_2}{O_4}\,\,\,\left( e \right){H_2}{S_4}{O_6}\,\,\,\left( f \right)C{r_2}{O_7}^{2 - }\,\,\,\left( g \right)Na{H_2}P{O_4} \\\ \,\,\,\,\,\,\,\,\,S\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{S_4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{r_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P \\\$

Answer 1

The oxidation number can be described as the number that is allocated to the elements in a chemical combination. The oxidation number is therefore a count of electrons that the atoms in a molecule can share, lose or gain while forming chemical bonds with the other atoms of the other elements.

Complete step by step answer:
(a) Now as we all know that the charge on any neutral compound is always equal to zero therefore
Therefore the charge on $(a)\,{H_2}S{O_4}$ is 0
$2\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ }}H} \right){\text{ }} + {\text{ }}1\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}S} \right){\text{ }} + {\text{ }}4\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}O} \right){\text{ }} = {\text{ }}0$
$= \,2 + \,Oxidation\,number\;of\,S\, - \,8\, = \,0$

Oxidation number of S = $+ 6$

(b) Here $HN{O_3}$ is A neutral compound so similarly its charge is 0
let N be equal to x

\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{H}}_{}}} \right){\text{ }} + {\text{ }}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{N}}_{}}} \right){\text{ }} + {\text{ 3}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right)$$$$\, = \,0
$+ 1\, + \,x\, - \,6\, = \,0 \\\ x\, = \,5 \\\$
therefore the oxidation state of N is $5$

(c) Here in ${H_3}P{O_3}\,$ we similarly calculate
let us assume the oxidation number of P as x in this case:

$3\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{H}}_{}}} \right){\text{ }} + {\text{ }}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{P}}_{}}} \right){\text{ }} + {\text{ 3}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right)$ $= \,0$
$1(3)\, + \,x\,\, - 2\left( 4 \right)\, = \,0 \\\ 3\, + \,x\, - \,8\, = \,0 \\\ x\, = \,5 \\\$
Hence the Oxidation number of P in this particular question is 5

(d) Here in ${K_2}{C_2}{O_4}$ we have to calculate the oxidation number of ${C_2}$
So similarly let us assume the oxidation number of ${C_2}$ to be x

$2\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{K}}_{}}} \right){\text{ }} + {\text{ 2}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{C}}_{}}} \right){\text{ }} + {\text{ }}4\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right){\text{ }} = {\text{ }}0$
1(3)\, + \,x\,\, - 2\left( 4 \right)\, = \,0 \\\ 3\, + \,x\, - \,8\, = \,0 \\\ x\, = \,5 \\\ 2\, + \,x\, - 8\, = \,0 \\\ = \, + 3 \\\

Therefore the oxidation number of ${C_2}$ is 3

(e) Here in the case of ${H_2}{S_4}{O_6}\,$this is a neutral compound and hence the net charge on it is = 0
and let us similarly assume the oxidation number of $\,{S_4}\,$to be x
now

$2\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{H}}_{}}} \right){\text{ }} + {\text{ 4}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{S}}_{}}} \right){\text{ }} + {\text{ 6}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right){\text{ }} = {\text{ }}0$
$2\, + \,4x\, - \,12\, = \,0 \\\ x\, = \,2.5 \\\$
Thus the oxidation number of S in this case is 2.5

(f) Here in this case we have $C{r_2}{O_7}^{2 - }$
$2\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ C}}{{\text{r}}_{}}} \right)$ $+$ ${\text{7}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right){\text{ }} = {\text{ - 2}}$
Here the compound has a net charge of -2. Therefore, let us assume the oxidation state of Cr to be x:
2x\, - \,14\, = \, - 2 \\\ 2x\, = \,12 \\\ x\, = \,6 \\\

Hence, the oxidation number of 1 chromium ion here is 6

(g) Here in this case we have $Na{H_2}P{O_4}$ and it is a neutral compound therefore the net charge is equal to zero.
Let us assume the oxidation state of P to be x, therefore:

$\left( {Oxidation{\text{ }}number{\text{ }}of{\text{ N}}{{\text{a}}_{}}} \right){\text{ }} + {\text{ 2}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{{\text{H}}_{}}} \right){\text{ }} + {\text{ 4}}\left( {oxidation{\text{ }}number{\text{ }}of{\text{ }}{O_{}}} \right)\,{\text{ + }}\,1\left( {Oxidation\,number\,of\,P} \right){\text{ }}$
3\, + \,x\, - 8\, = \,0 \\\ x\, = \, + 5 \\\

Thus the oxidation number of P here is 5

Note: The $NaCl$ crystal lattice has a coordination number of 6 and the ionic compounds falling in in the range of coordination number 6 have a radius ratio between 0.414 – 0.732 pm. In order to solve these questions simply and in less time one should mug up the coordination numbers of various crystal lattices and then simply determine their CN from the category they are falling into.