Question

Arrange the following in the order of increasing oxidation number of chlorine in $\text{ C}{{\text{l}}_{2}}{{\text{O}}_{7}}\text{ }$, $\text{ C}{{\text{l}}_{\text{2}}}\text{O }$, $\text{ HCl}$, $\text{ Cl}{{\text{F}}_{\text{3}}}\text{ }$, $\text{ C}{{\text{l}}_{\text{2}}}\text{ }$.

Answer 1

Oxidation number is simply called the number that is allocated to the element in the molecule which is equal to the number of electron gain or accepted by the element while forming the chemical bond with the other atoms. The sum of the oxidation state of all the elements is equal to the charge on the molecule.

Complete Solution :
- The oxidation number is the number of electron gain or loss by an element while forming a chemical bond with the other atom in a molecule. The sum of the oxidation number of the elements in the molecule is always equal to the charge on the molecule.
- For a zero or neutral state molecule the sum is equal to zero.
- The oxidation number for the various elements depends on its nature to accept or lose an electron. The elements which can easily lose the electron are more likely to have a positive oxidation state. While the elements which can easily accept the electrons, those that have the high electronegativity have the negative oxidation state.
The oxidation state for elements are known, these are as tabulated below,

Element	Oxidation state
$\text{ O }$	$\text{ }-2\text{ }$
$\text{ F }$	$\text{ }-1\text{ }$
$\text{ H }$	$\text{ +1 }$

Based on the fixed oxidation state of these mentioned elements, and the molecules given are neutral. Thus, it will be equal to zero; the required oxidation states can be calculated.
Let's calculate the oxidation state of chlorine in the molecule.
A) $\text{ C}{{\text{l}}_{2}}{{\text{O}}_{7}}\text{ }$ :
Now, the first we have. We know that the oxidation state/ number of oxygen is$\text{ }-2\text{ }$.
Thus, let the oxidation state of chlorine be x. So, it can be written as:
$\begin{aligned} & \text{ Charge on C}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{7}}}=\text{ 2 (O}\text{.S}\text{. of Cl ) + 7 (O}\text{.S}\text{. of O )} \\\ & \text{ 0 = 2 (x) + 7 (}-2) \\\ & \Rightarrow \text{x = }\frac{14}{2}\text{ = +7 } \\\ \end{aligned}$
Thus, the oxidation state of chlorine is equal to$\text{ }+7\text{ }$.

B) $\text{ C}{{\text{l}}_{\text{2}}}\text{O }$ :
Now, the first we have. We know that the oxidation state/ number of oxygen is$\text{ }-2\text{ }$.
Thus, let the oxidation state of chlorine be x. So, it can be written as:
$\begin{aligned} & \text{ Charge on C}{{\text{l}}_{\text{2}}}\text{O}=\text{ 2 (O}\text{.S}\text{. of Cl ) + 1 (O}\text{.S}\text{. of O )} \\\ & \text{ 0 = 2(x) + 1 (}-2) \\\ & \Rightarrow \text{x = }\frac{2}{2}\text{ = +1 } \\\ \end{aligned}$
Thus, the oxidation state of chlorine is equal to$\text{ }+1\text{ }$.

C) $\text{ HCl}$ :
Now, the first we have. We know that the oxidation state/ number of hydrogen is$\text{ +1 }$.
Thus, let the oxidation state of chlorine be x. So, it can be written as:
$\begin{aligned} & \text{ Charge on HCl}=\text{ 1 (O}\text{.S}\text{. of Cl ) + 1 (O}\text{.S}\text{. of H )} \\\ & \text{ 0 = (x) + 1 (+1}) \\\ & \Rightarrow \text{x = }\frac{-1}{1}\text{ = }-\text{1 } \\\ \end{aligned}$
Thus, the oxidation state of chlorine is equal to$\text{ }-1\text{ }$.

D) $\text{ Cl}{{\text{F}}_{\text{3}}}\text{ }$ :
Now, the first we have. We know that the oxidation state/ number of fluorine is$\text{ }-\text{1 }$.
Thus, let the oxidation state of chlorine be x. So, it can be written as:
$\begin{aligned} & \text{ Charge on Cl}{{\text{F}}_{\text{3}}}\text{ }=\text{ 1 (O}\text{.S}\text{. of Cl ) + 3 (O}\text{.S}\text{. of F )} \\\ & \text{ 0 = (x) + 3 (}-\text{1}) \\\ & \Rightarrow \text{x = }\frac{3}{1}\text{ = +3 } \\\ \end{aligned}$
Thus, the oxidation state of chlorine is equal to$\text{ +3 }$.

E) $\text{ C}{{\text{l}}_{\text{2}}}\text{ }$ :
Now, the chlorine molecule has a zero charge thus the oxidation number is:
Thus, let the oxidation state of chlorine be x. So, it can be written as:
$\begin{aligned} & \text{ Charge on C}{{\text{l}}_{\text{2}}}\text{ }=\text{ 2 (O}\text{.S}\text{. of Cl ) } \\\ & \text{ 0 = 2(x) } \\\ & \Rightarrow \text{x = 0 } \\\ \end{aligned}$
Thus, the oxidation state of chlorine is equal to 0.
Now, we can calculate for the second i.e.$\text{C}{{\text{l}}_{2}}\text{O}$, it can be written as:
In the last, we can conclude that the order of increasing oxidation number of chlorine is $\text{ HCl }<\text{ C}{{\text{l}}_{\text{2}}}\text{ }<\text{ C}{{\text{l}}_{\text{2}}}\text{O }<\text{ Cl}{{\text{F}}_{\text{3}}}\text{ }<\text{ C}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ }$

Note: Chlorine is a highly electronegative element.it can attract the electrons in a bond. This is why chlorine is reduced to the $\text{ }-\text{1 }$ oxidation state. However, the fluorine and oxygen are more electronegative than the chlorine, thus the fluorine and oxygen attract the electrons from the chlorine and chlorine exhibits the exceptional oxidation state equal to $\text{ +1 }$ ,$\text{ +3 }$, $\text{ +5 }$ and $\text{ +7 }$. The compounds in which chlorine exceptional oxidation state are:

Oxidation state	Compounds
$\text{ }-\text{1 }$	$\text{ HCl }$
$\text{ +1 }$	$\text{ ClF }$ , $\text{ NaClO }$
$\text{ +3 }$	$\text{ Cl}{{\text{F}}_{\text{3}}}\text{ }$,$\text{ NaCl}{{\text{O}}_{\text{2}}}\text{ }$
$\text{ +5 }$	$\text{ Cl}{{\text{F}}_{5}}\text{ }$, $\text{ NaCl}{{\text{O}}_{3}}\text{ }$
$\text{ +7 }$	$\text{ C}{{\text{l}}_{2}}{{\text{O}}_{7}}\text{ }$, $\text{ HCl}{{\text{O}}_{\text{4}}}\text{ }$.