Question

In the following reaction,
$NaOH+ S\to A+N{{a}_{2}}S+{{H}_{2}}O$
A is-
(a)$N{{a}_{2}}S{{O}_{3}}$
(b)$N{{a}_{2}}S{{O}_{4}}$
(c) $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$
(d)$N{{a}_{2}}S$

Answer 1

Hint: When Sulphur is made to react with the sodium hydroxide, it results in the formation of a thiosulphate of the metal which belongs to the s-block element and has the atomic number as 11 and mass number as23. Now with the help of this you can easily identify the compound A.

Complete answer:
First of let’s discuss the Sulphur. Sulphur is non-metal which belongs to the oxygen family and comes under the category of the p-block elements. It has the atomic number as 16 and mass number as 32.
On the other hand, sodium hydroxide is a strong base which consists of an -OH group and a metal i.e. sodium.
Now considering the statement as-
When Sulphur is made to undergo reaction with the sodium hydroxide, it results in the formation of sodium thiosulphate ,sodium sulphide along with the loss of the water molecule. The chemical reaction is supposed to take place as;
$6NaOH+S\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}+2N{{a}_{2}}S+{{H}_{2}}O$
Sodium thiosulphate is a solid inorganic compound which is a crystalline substance. It is white or colorless in nature and usually occurs in pentahydrate form. It has the molar mass as 158.11 gram per mole. It is readily soluble in water.
So, thus in the following reaction ;
$NaOH+ S\to A+N{{a}_{2}}S+{{H}_{2}}O$
A is sodium thiosulphate.

Hence, option (c) is correct.

Note:
Sodium thiosulphate is used as a reducing agent due to its oxidative-reductive properties. It is also used in photography in the form of a fixer due to its complex forming ability.