Question

Find the oxidation number of elements in each case.
$Cr$in$Cr{O_2}C{l_2}$, $N{a_2}C{r_3}{O_{10}}$, $C{r_2}{\left( {S{O_4}} \right)_3}$, and ${\left[ {Cr{O_8}} \right]^{3 - }}$
A) $ + 6$, $ + 6$, $ + 3$, $ + 5$.
B) $ + 5$, $ + 6$, $ + 3$, $ + 4$.
C) $ + 4$, $ + 6$, $ + 2$, $ + 5$.
D) None of the above.

Answer 1

Hint: We can define oxidation state as the degree of loss of an electron in a chemical compound. The oxidation state by an element in a compound can be calculated using the rules of oxidation numbers.

Complete step by step answer:
Let us see few rules for oxidation numbers,
1.A free element will be zero as its oxidation number.
2.Monatomic ions will have an oxidation number equal to charge of the ion.
3.In hydrogen, the oxidation number is ${\text{ + 1}}$, when combined with elements having less electronegativity, the oxidation number of hydrogen is -1.
4.In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
5.Group 1 elements will have +1 oxidation number.
6.Group 2 elements will have +2 oxidation numbers.
7.Group 17 elements will have -1 oxidation number.
8.Sum of oxidation numbers of all atoms in neutral compounds is zero.
Now we can calculate the oxidation state of chromium as follows,
Let us take X as the oxidation number of chromium.
The oxidation number of $Cr$ in $Cr{O_2}C{l_2}$
$X + 2\left( { - 2} \right) + 2\left( { - 1} \right) = 0$
$ \Rightarrow X - 4 - 2 = 0$
$ \Rightarrow X - 6 = 0$
\[ \Rightarrow X = + 6\]
Now we can calculate the oxidation number of chromium in $N{a_2}C{r_3}{O_{10}}$ as,
$2 + 3X + 10\left( { - 2} \right) = 0$
$ \Rightarrow 2 + 3X - 20 = 0$
$ \Rightarrow 3X - 18 = 0$
$ \Rightarrow 3X = 18$
$X = \dfrac{{18}}{3} = + 6$
Now we can calculate the oxidation number of chromium in $C{r_2}{\left( {S{O_4}} \right)_3}$ as,
$2X + 3\left( { - 2} \right) = 0$
$ \Rightarrow 2X - 6 = 0$
$ \Rightarrow 2X = + 6$
$ \Rightarrow X = \dfrac{{ + 6}}{2}$
On simplifying we get,
$ \Rightarrow X = + 3$
Let calculate the oxidation number of chromium in ${\left[ {Cr{O_8}} \right]^{3 - }}$ as,
The complex ${\left[ {Cr{O_8}} \right]^{3 - }}$ contains peroxide linkage hence each atom carries $ - 1$ charge.
$X + 8\left( { - 1} \right) = - 3$
$ \Rightarrow X - 8 = - 3$
$ \Rightarrow X = - 3 + 8$
On simplifying we get,
$X = + 5$
As chromium has maximum oxidation number \[ + 5\].
Thus, the oxidation number of chromium in $Cr{O_2}C{l_2}$, $N{a_2}C{r_3}{O_{10}}$, $C{r_2}{\left( {S{O_4}} \right)_3}$, and ${\left[ {Cr{O_8}} \right]^{3 - }}$ is $ + 6$,$ + 6$,$ + 3$,$ + 5$.

Therefore, the option A is correct.

Note:
We must remember that the reduction is an opposite process of oxidation. If there is gain of electrons by a molecule, atom or ion then it is called as reduction it occurs when the oxidation state of the species is decreased. Reducing agent is one which is oxidized in a reaction. Reducing agent reduces the oxidizing agent.
Consider the reaction,
$Pb\left( s \right) + Pb{O_2}\left( s \right) + 2{H_2}S{O_4}\left( {aq} \right)\xrightarrow{{}}2PbS{O_4}\left( s \right) + 2{H_2}O\left( l \right)$
In the above reaction, the oxidation state of lead$\left( {Pb} \right)$ is zero and the oxidation state of lead in $\left( {Pb{O_2}} \right)$ is $ + 4$ and the oxidation state of lead in $\left( {PbS{O_4}} \right)$is$ + 2$. Hence, the oxidized species is lead and the species which is reduced is lead oxide. The oxidizing agent is lead sulfate and the reducing agent is lead.