
An element A burns with golden flame in the air. It reacts with another element B(atomic number 17) to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. The elements and products involved is:
A. Na
B. NaCl
C. NaOH
D. All of these
Answer
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Hint:We know that electrolysis is the process in which breakdown of ionic substances occur on passing electric current through it. It is one of the most commercially important processes in separation of elements.
Complete step by step answer:
First we have to identify the metal which burns with yellow colored flame. Sodium follows this criterion, that is, it burns with yellow colored flame. So, the element A is sodium whose chemical symbol is Na.
Next condition says that A reacts with another element B which belongs to group 17 to give product C. We know that, in groups 17 halogens (fluorine, bromine, chlorine, iodine etc.) are present. Among the fluorines, sodium reacts with chlorine to form sodium chloride. The chemical reaction is,
${\rm{Na}} + {\rm{C}}{{\rm{l}}_{\rm{2}}} \to 2{\rm{NaCl}}$
So, element B is chlorine and product C formed in sodium chloride (NaCl).
Next, an aqueous solution of C on electrolysis produces compound D and liberates hydrogen. So, the reaction can be written as,
$2{\rm{NaCl}}\left( {aq} \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right) \to 2{\rm{NaOH}}\left( {aq} \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( g \right) + {{\rm{H}}_{\rm{2}}}\left( g \right)$
So, compound D formed on electrolysis of sodium chloride is sodium hydroxide.
So, A is Na (sodium), B is ${\rm{C}}{{\rm{l}}_{\rm{2}}}$ (chlorine), C is NaCl (sodium chloride) and D is NaOH (sodium hydroxide). Therefore, sodium (Na), sodium chloride (NaCl) and sodium hydroxide (NaOH) are all part of the reaction.
Hence, option D is the correct answer.
Note:
Sodium is a metal which is present in group 1 of the periodic table. It is very reactive in nature. It can undergo vigorous reaction with carbon dioxide, oxygen and moisture present in the air. So, it is stored in kerosene.
Complete step by step answer:
First we have to identify the metal which burns with yellow colored flame. Sodium follows this criterion, that is, it burns with yellow colored flame. So, the element A is sodium whose chemical symbol is Na.
Next condition says that A reacts with another element B which belongs to group 17 to give product C. We know that, in groups 17 halogens (fluorine, bromine, chlorine, iodine etc.) are present. Among the fluorines, sodium reacts with chlorine to form sodium chloride. The chemical reaction is,
${\rm{Na}} + {\rm{C}}{{\rm{l}}_{\rm{2}}} \to 2{\rm{NaCl}}$
So, element B is chlorine and product C formed in sodium chloride (NaCl).
Next, an aqueous solution of C on electrolysis produces compound D and liberates hydrogen. So, the reaction can be written as,
$2{\rm{NaCl}}\left( {aq} \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right) \to 2{\rm{NaOH}}\left( {aq} \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( g \right) + {{\rm{H}}_{\rm{2}}}\left( g \right)$
So, compound D formed on electrolysis of sodium chloride is sodium hydroxide.
So, A is Na (sodium), B is ${\rm{C}}{{\rm{l}}_{\rm{2}}}$ (chlorine), C is NaCl (sodium chloride) and D is NaOH (sodium hydroxide). Therefore, sodium (Na), sodium chloride (NaCl) and sodium hydroxide (NaOH) are all part of the reaction.
Hence, option D is the correct answer.
Note:
Sodium is a metal which is present in group 1 of the periodic table. It is very reactive in nature. It can undergo vigorous reaction with carbon dioxide, oxygen and moisture present in the air. So, it is stored in kerosene.
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