Question

In Ramsay-Rayleigh’s second method, from air, the correct statement is:
(A) ${{\rm{O}}_{\rm{2}}}$ is removed as ${\rm{CuO}}$
(B) ${{\rm{N}}_{\rm{2}}}$ is removed as ${\rm{M}}{{\rm{g}}_{\rm{3}}}{{\rm{N}}_{\rm{2}}}$
(C) ${{\rm{N}}_{\rm{2}}}$ is removed as azide
(D) ${{\rm{O}}_{\rm{2}}}$ and ${{\rm{N}}_{\rm{2}}}$ are simultaneously removed as oxides of nitrogen.

Answer 1

As we know that air is a mixture of various gases such as carbon dioxide, trace elements, particulate matter etc. In air there are some gases which are very stable and inert in nature, they form compounds only at very high temperature and with very reactive molecules.

Complete step-by-step answer:
In Ramsay-Raleigh’s second method, the noble gases are isolated from air. In the air, Nitrogen atoms exist in the form of molecules of nitrogen. The molecule of nitrogen consists of a triple bond. Due to triple bonding in nitrogen molecules, it reacts very slow.
So, nitrogen is removed when it reacts with oxygen and forms nitrogen oxide, in this way air becomes free from nitrogen and oxygen both.
Removal of these gases is very important to isolate noble gases such as helium, xenon, krypton.

Therefore, the correct option is option (D).

Note: In Ramsay-Raleigh’s second method, there are two steps.
(a) Noble gases are mixed with various elements in air, these are isolated from nitrogen and oxygen by repeated electrolysis process.
(b) The other constituents of air are removed by using activated charcoal.