Question

The formal oxidation numbers of $Cr$ and $Cl$ in the ions $C{{r}_{2}}{{O}_{7}}^{2-}$ $Cl{{O}_{3}}^{-}$ respectively are:
(A) +6 and +7
(B) +7 and +5
(C) +6 and +5
(D) +8 and +7

Answer 1

We can find the oxidation number or oxidation state of any element by assuming that the overall charge on the compound is equal to the sum of the oxidation number of all the atoms present in the compound.

Complete step by step solution:
We will see the method to find the formal oxidation numbers of the given elements in the given compounds.
- We can find the oxidation number or oxidation state of any element by assuming that the overall charge on the compound is equal to the sum of the oxidation number of all the atoms present in the compound.
- Thus, for $C{{r}_{2}}{{O}_{7}}^{-2}$, we can see that the overall charge is (-2). Here, the oxidation number of oxygen atoms is (-2).
So, we can find the oxidation number of $Cr$ by following the equation.
$\text{Overall charge = 2(Oxidation number of Cr)+7(Oxidation number of O)}$
So, we can write that
$-2=2(\text{Oxidation number of Cr)+7(-2)}$
So,
$\text{Oxidation number of Cr = }\frac{-2+14}{2}=\frac{12}{2}=6$
So, we obtained that oxidation number of $Cr$ in $C{{r}_{2}}{{O}_{7}}^{-2}$ is +6.
- Now for $Cl{{O}_{3}}^{-}$ ion , we can see that overall charge is (-1). The oxidation number of O-atoms is (-2).
We can find the oxidation number of chlorine atoms by following the equation.
$\text{Overall charge = Oxidation number of Cl + 3(Oxidation number of O)}$
So, we can write that
$-1=\text{Oxidation number of Cl + 3(-2)}$
$\text{Oxidation number of Cl = -1+6 = +5}$
Thus, we obtained that oxidation number of Cl in $Cl{{O}_{3}}^{-}$ is +5.
So, we obtained that the formal oxidation numbers of $Cr$ and $Cl$ in the ions $C{{r}_{2}}{{O}_{7}}^{2-}$ and $Cl{{O}_{3}}^{-}$ are +6 and +5 respectively.

Thus, the correct answer of this question is (C).

Note: Remember that in most of the compounds, the oxidation number of oxygen atoms is taken is (-2). However, if it is in peroxide form, then its oxidation number is taken as (-1). In its molecular form $({{O}_{2}})$, its oxidation state is taken as zero.