Question

2. The element 'X' has only three shells in which the difference in number of electrons between the last shell and the second last shell is equal to the number of electrons in the first shell. This element `X' burns in oxygen to produce compound Y, which turns acidified potassium dichromate paper to green. Y turns moist blue litmus paper to red due to the formation of compound Z. Identify X, Y and Z.

Answer 1

Part 1: Identifying X and Y:

• It is given that the element 'X' has only three shells in which the difference in the number of electrons between the last shell and the second last shell is equal to the number of electrons in the first shell.
• So X should be Sulphur $\left(\mathrm{S}\right)$ and has the atomic number 16. Its electronic configuration is 2,8,6.
• The difference between the number of electrons in the L and M shell $=8-6=2$ which is equal to the number of electrons in the K shell.
• Now, Sulphur reacts with Oxygen $\left({\mathrm{O}}_2\right)$ to form compound Y i.e. Sulphur dioxide $\left({\mathrm{SO}}_2\right)$.
• The reaction involved is: $\underset{\mathrm{X}}{\mathrm{S}}+{\mathrm{O}}_2\to \underset{\mathrm{Y}}{{\mathrm{SO}}_2}$
• Sulphur dioxide turns acidified Potassium dichromate $\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)$ paper to green.
• The reaction involved is:

$\underset{\mathrm{Sulphur}\mathrm{dioxide}}{3{\mathrm{SO}}_2}+\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to \underset{\mathrm{Potassium}\mathrm{sulphate}}{{\mathrm{K}}_2{\mathrm{SO}}_4}+\underset{\mathrm{Chromium}\mathrm{sulphate}\left(\mathrm{green}\right)}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}$

Part 2: Identifying Z:

• Sulphur dioxide turns moist blue litmus paper to red due to the formation of compound Z i.e. Sulphurous acid $\left({\mathrm{H}}_2{\mathrm{SO}}_3\right)$
• The reaction involved is: $\underset{\mathrm{Y}}{{\mathrm{SO}}_2}+{\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{Z}}{{\mathrm{H}}_2{\mathrm{SO}}_3}$

Hence,

• X is Sulphur.
• Y is Sulphur dioxide
• Z is Sulphurous acid