Question

Correct sequence of acidic character is:
A. $${{\text{N}}_2}{{\text{O}}_5} > {\text{C}}{{\text{O}}_2} > {\text{CO}} > {\text{S}}{{\text{O}}_{\text{2}}}$$
B. $${{\text{N}}_2}{{\text{O}}_5} > {\text{S}}{{\text{O}}_2} > {\text{C}}{{\text{O}}_2} > {\text{CO}}$$
C. $${\text{CO}} > {\text{C}}{{\text{O}}_2} > {\text{S}}{{\text{O}}_2} > {{\text{N}}_2}{{\text{O}}_5}$$
D. $${\text{S}}{{\text{O}}_2} > {\text{C}}{{\text{O}}_2} > {\text{CO}} > {{\text{N}}_2}{{\text{O}}_5}$$

Answer 1

Hint: To check the acidic nature, first check the oxidation number of non-metal oxide. Any non-metal oxide having higher oxidation number indicates higher acidic character. If non-metal oxides contain similar oxidation number, then the more electronegative atom attached with oxygen will show higher acidic character.

Complete step by step solution:
In order to check acidic nature, first we need to calculate the oxidation number of each non-metal oxide.
Since we know that the oxidation number of oxygen is $$ - 2$$. In $${{\text{N}}_2}{{\text{O}}_5}$$, the oxidation number (O.N) of nitrogen can be calculated as:
$$\left( {2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}}} \right) + \left( {5 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$$
We know that oxidation state of oxygen (O) is $$ - 2$$ and the overall charge on $${{\text{N}}_2}{{\text{O}}_5}$$ is zero. Thus put these values in the above equation.
$$\displaylines{
2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}} + \left( {5 \times \left( { - 2} \right)} \right) = 0 \cr
2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}} - 10 = 0 \cr} $$
Now move $$ - 10$$ to the right hand side and as a result, $$ - 10$$ will change to 10.
$$2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}} = 10$$
Now, divide 10 and 2 with each other to calculate the oxidation number of nitrogen.
$$\displaylines{
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}} = \dfrac{{10}}{2} \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{N}} = 5 \cr} $$
Thus in $${{\text{N}}_2}{{\text{O}}_5}$$, the oxidation number of nitrogen is 5.

In $${\text{C}}{{\text{O}}_2}$$, the oxidation number (O.N) of carbon can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} + \left( {2 \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{C}}{{\text{O}}_{\text{2}}}$$
We know that the oxidation state of oxygen (O) is $$ - 2$$ and the overall charge on $${\text{C}}{{\text{O}}_2}$$ is zero. Thus put these values in the above equation.
$$\displaylines{
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} + \left( {2 \times \left( { - 2} \right)} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} - 4 = 0 \cr} $$
Now move $$ - 4$$ to the right hand side and as a result, $$ - 4$$ will change to 4.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} = 4$$
Thus in $${\text{C}}{{\text{O}}_2}$$, the oxidation number of carbon is 4.

In CO, the oxidation number (O.N) of carbon can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} + {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}} = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{CO}}$$
We know that the oxidation state of oxygen (O) is $$ - 2$$ and the overall charge on CO is zero. Thus put these values in the above equation.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} - 2 = 0$$

Now move $$ - 2$$ to the right hand side and as a result, $$ - 2$$ will change to 4.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{C}} = 2$$

Thus in CO, the oxidation number of carbon is 2. Remember that CO is not a non-metal oxide CO is a neutral oxide which neither shows acidic nor basic characteristics.

In $${\text{S}}{{\text{O}}_{\text{2}}}$$, the oxidation number (O.N) of sulphur can be calculated as:
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{S}} + \left( {{\text{2}} \times {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{O}}} \right) = {\text{Overall}}\;{\text{charge}}\;{\text{on}}\;{\text{S}}{{\text{O}}_2}$$
We know that oxidation state of oxygen (O) is $$ - 2$$ and the overall charge on $${\text{S}}{{\text{O}}_{\text{2}}}$$ is zero. Thus put these values in the above equation.
$$\displaylines{
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{S}} + \left( {2 \times \left( { - 2} \right)} \right) = 0 \cr
  {\text{O}}{\text{.N}}\;{\text{of}}\;{\text{S}} - 4 = 0 \cr} $$

Now move $$ - 4$$ to the right hand side and as a result, $$ - 4$$ will change to 4.
$${\text{O}}{\text{.N}}\;{\text{of}}\;{\text{S}} = 4$$

Thus in $${\text{S}}{{\text{O}}_{\text{2}}}$$, the oxidation number of sulphur is 4.

Therefore, on the basis of oxidation number, the sequence of acidic character can be arranged as:
$$\eqalign{
  & {{\text{N}}_2}{{\text{O}}_5} > {\text{S}}{{\text{O}}_2} \approx {\text{C}}{{\text{O}}_{\text{2}}} > {\text{CO}} \cr
  & \left( 5 \right)\;\;\;\;\;\;\;\left( 4 \right)\;\;\;\;\left( 4 \right)\;\;\;\;\;\left( 2 \right) \cr} $$

Now, it can be observed that oxidation numbers of carbon in $${\text{C}}{{\text{O}}_2}$$ and sulphur in $${\text{S}}{{\text{O}}_{\text{2}}}$$ are same. In order to find which has higher acidic character, we need to check which oxide consists of the most electronegative element attached with oxygen. The more electronegative the central atom, the more acidic the oxide.

Since we know that in periodic table, on moving across the period (left to right) electronegativity increases and on moving down the group (top to bottom) electronegativity decreases. Thus on the basis of these two periodic trends of electronegativity, it can be concluded that $${\text{S}}{{\text{O}}_2}$$ has higher acidic character than $${\text{C}}{{\text{O}}_{\text{2}}}$$.

Thus the correct sequence of acidic character will be as follows:
$${{\text{N}}_2}{{\text{O}}_5} > {\text{S}}{{\text{O}}_2} > {\text{C}}{{\text{O}}_2} > {\text{CO}}$$

Therefore, the correct option is B.

Note: Always first check the oxidation number of non-metal oxide and then electronegativity. Acidic characters always remain directly proportional to electronegativity of non-metal. Remember the periodic trends of electronegativity.