Question

$N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to $ the product is-
A. $HI$
B. ${{I}_{2}}$
C. $NaI$
D. $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$

Answer 1

Hint: When sodium tetrathionate is treated with iodine vapour or gas then a loss of iodine gas takes place. Thus in iodometric titration it is widely used. In this reaction iodine gets reduced and sodium thiosulfate gets oxidised.

Complete Step by Step Answer:
Iodine gas in reaction with sodium thiosulphate produces sodium tetrathionate as a major product and along with sodium iodide as a minor product.
Thus the reaction is given as follows:
$2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI$
Thus in this reaction one mole of Iodine gas on reaction with two moles of sodium thiosulphate produces one mole of sodium tetrathionate as major product and along with two moles of sodium iodide as minor product.
Thus the correct options are C and D.

Additional Information: Iodine is a halogen element. It has the symbol of $I$. It generally exists as a gas. It has the colour of violet blue. Iodine is generally obtained from seaweed.
Starch generally acts as an indicator in titration reactions to determine the end point of a titration. Iodine gas reacts with starch, an indicator gives blue colour.

Note: Among the halogens iodine is the most unreactive and covalent compound. Thus it gets reduced when treated with sodium thiosulfate.