Question

Select the anion which is the strongest Bronsted base:
A. $${\text{ClO}}_4^ - $$
B. $${\text{ClO}}_3^ - $$
C. $${\text{ClO}}_2^ - $$
D. $${\text{Cl}}{{\text{O}}^ - }$$

Answer 1

Hint: Bronsted bases are those substances which have capability to accept protons in a chemical reaction. It can also be said that Bronsted bases are good proton acceptors.

Complete step by step solution:
As the oxidation state of the central atom increases, the acidic nature of the conjugate acid of base also increases. Every base has a conjugate acid which differs by a proton. Conjugate acid comprises one more hydrogen atom and one more positive charge than the base that helped to produce it.

The conjugate base of $${\text{ClO}}_4^ - $$ is $${\text{HCl}}{{\text{O}}_4}$$.
The conjugate base of $${\text{ClO}}_3^ - $$ is $${\text{HCl}}{{\text{O}}_3}$$.
The conjugate base of $${\text{ClO}}_2^ - $$ is $${\text{HCl}}{{\text{O}}_2}$$.
The conjugate base of $${\text{Cl}}{{\text{O}}^ - }$$ is HClO.

In $${\text{HCl}}{{\text{O}}_4}$$, the oxidation number (O.N) of chlorine can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{H}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {4 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{HClO}}$$
We know that the oxidation number of oxygen is $$ - 2$$ and the oxidation number of hydrogen is 1. The overall charge on $${\text{HCl}}{{\text{O}}_4}$$ is equal to zero. Put these values in the above equation.
$$\displaylines{
  {\text{1}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {4 \times \left( { - 2} \right)} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + 1 - 8 = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} - 7 = 0 \cr} $$
Now in order to obtain an oxidation number of chlorine, shift $$ - 7$$ to the right hand side. As a result, $$ - 7$$ changes to 7.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} = 7$$

This shows that the oxidation number of chlorine in $${\text{HCl}}{{\text{O}}_4}$$ is equal to 7.

In $${\text{HCl}}{{\text{O}}_3}$$, the oxidation number (O.N) of chlorine can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{H}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {3 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{HCl}}{{\text{O}}_3}$$
We know that the oxidation number of oxygen is $$ - 2$$ and the oxidation number of hydrogen is 1. The overall charge on $${\text{HCl}}{{\text{O}}_3}$$ is equal to zero. Put these values in the above equation.
$$\displaylines{
  {\text{ + 1}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {3 \times \left( { - 2} \right)} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + 1 - 6 = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} - 5 = 0 \cr} $$

Now in order to obtain an oxidation number of chlorine, shift $$ - 5$$ to the right hand side. As a result, $$ - 5$$ changes to 5.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} = 5$$

This shows that the oxidation number of chlorine in $${\text{HCl}}{{\text{O}}_3}$$ is equal to 5.

In $${\text{HCl}}{{\text{O}}_2}$$, the oxidation number (O.N) of chlorine can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{H}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{HCl}}{{\text{O}}_2}$$

We know that the oxidation number of oxygen is $$ - 2$$ and the oxidation number of hydrogen is 1. The overall charge on $${\text{HCl}}{{\text{O}}_2}$$ is equal to zero. Put these values in the above equation.
$$\displaylines{
  {\text{1}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( {2 \times \left( { - 2} \right)} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + 1 - 4 = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} - 3 = 0 \cr} $$
Now in order to obtain an oxidation number of chlorine, shift $$ - 3$$ to the right hand side. As a result, $$ - 3$$ changes to 3.

$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} = 3$$
This shows that the oxidation number of chlorine in $${\text{HCl}}{{\text{O}}_2}$$ is equal to 3.

In HClO, the oxidation number (O.N) of chlorine can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{H}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}} = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{HClO}}$$

We know that the oxidation number of oxygen is $$ - 2$$ and the oxidation number of hydrogen is 1. The overall charge on HClO is equal to zero. Put these values in the above equation.
$$\displaylines{
  {\text{1}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + \left( { - 2} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} + 1 - 2 = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} - 1 = 0 \cr} $$

Now in order to obtain an oxidation number of chlorine, shift $$ - 1$$ to the right hand side. As a result, $$ - 1$$ changes to 1.

$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{Cl}} = 1$$

This shows that the oxidation number of chlorine in HClO is equal to 1.

On the basis of oxidation number of central atom of conjugate acid, the order of acidic strength will be:
$${\text{HCl}}{{\text{O}}_{\text{4}}} > {\text{HCl}}{{\text{O}}_{\text{3}}} > {\text{HCl}}{{\text{O}}_{\text{2}}} > {\text{HClO}}$$

We know that weaker the conjugate acid, stronger will be Bronsted base. Thus the order of Bronsted base strength will be:
$${\text{ClO}}_4^ - < {\text{ClO}}_3^ - < {\text{ClO}}_2^ - < {\text{Cl}}{{\text{O}}^ - }$$

This shows that $${\text{Cl}}{{\text{O}}^ - }$$ is the strongest Bronsted base among all given species. Thus the correct option is D.

Note: The acid and base which differ by a proton are known to form a conjugate pair. It may be noted that strong acids have weak conjugate bases while weak acids have strong conjugate bases.