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A white precipitate insoluble in HNO3 is formed when the aqueous solution of x in NaOH is treated with barium chloride and bromine water. So x is :
A. SO3
B. SO2
C. CO2
D. None of the above

Answer
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Hint: It is a colourless, pungent smelling gas and is highly soluble in water. It is used to make sulphurous acid (H2SO3), sulphuric acid(H2SO4). It behaves as a reducing agent when moist.

Complete-step- by- step answer:
When the aqueous solution of SO2 ​in NaOH ​when treated with barium chloride and bromine water, a white precipitate is formed which is insoluble in conc.HNO3.
We can see in the reaction that SO2 readily reacts with sodium hydroxide solution to form sodium sulfite.
     SO2+2NaOHNa2SO3+H2O

The sodium sulphite is then treated with barium chloride to form barium sulfite BaSO3
     Na2SO3+BaCl2BaSO3+2NaCl

Now the barium sulfite is lastly treated with bromine water . the bromine water contains nascent oxygen [O]
     Br2+H2O2HBr+[O]
This nascent oxygen oxidises the barium sulfite to give a white precipitate which is insoluble in conc.HNO3 it is barium sulphate BaSO4.
     BaSO3+[O]BaSO4
Sulphur (S) is a p- block element of group 16. Its main oxides are SO2 and SO3.
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.

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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. SO2.

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 ofSO2.
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.
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