A white precipitate insoluble in \[HN{{O}_{3}}\] is formed when the aqueous solution of x in NaOH is treated with barium chloride and bromine water. So x is :
A. $S{{O}_{3}}$
B. $S{{O}_{2}}$
C. $C{{O}_{2}}$
D. None of the above

Hint: It is a colourless, pungent smelling gas and is highly soluble in water. It is used to make sulphurous acid (${{H}_{2}}S{{O}_{3}}$), sulphuric acid(${{H}_{2}}S{{O}_{4}}$). It behaves as a reducing agent when moist.

Complete-step- by- step answer:
When the aqueous solution of $S{{O}_{2}}$ ​in $ ​NaOH$ ​when treated with barium chloride and bromine water, a white precipitate is formed which is insoluble in conc.$HN{{O}_{3}}$.
We can see in the reaction that $S{{O}_{2}}$ readily reacts with sodium hydroxide solution to form sodium sulfite.
     \[S{{O}_{2}}+2NaOH\to N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}O\]

The sodium sulphite is then treated with barium chloride to form barium sulfite $BaS{{O}_{3}}$
     \[N{{a}_{2}}S{{O}_{3}}+BaC{{l}_{2}}\to BaS{{O}_{3}}+2NaCl\]

Now the barium sulfite is lastly treated with bromine water . the bromine water contains nascent oxygen $\left[ O \right]$
     \[B{{r}_{2}}+{{H}_{2}}O\to 2HBr+\left[ O \right]\]
This nascent oxygen oxidises the barium sulfite to give a white precipitate which is insoluble in conc.$HN{{O}_{3}}$ it is barium sulphate $BaS{{O}_{4}}$.
     \[BaS{{O}_{3}}+\left[ O \right]\to BaS{{O}_{4}}\]
Sulphur (S) is a p- block element of group 16. Its main oxides are $S{{O}_{2}}$ and $S{{O}_{3}}$.
Sulphur dioxide is a colourless gas, we can easily identify it due its pungent smell. It has an angular shape and the bond length is the same. It forms a resonance hybrid of the two canonical forms.


Sulphur dioxide has many different uses such as refining petroleum and sugar, bleaching wool and silk and as an anti – chloral, preservative and disinfectant.
Therefore we can conclude that the answer is option B. $S{{O}_{2}}$.

Note: -In the question we are given an important reaction of sulphur dioxide with $NaOH$, its reaction with barium chloride and bromine water, it is also used to test the presence of$S{{O}_{2}}$.
Keep in mind the basic properties of a gas to determine its presence as for example sulphur dioxide is a colourless and pungent smelling gas.\[\]