Question

$C{r_2}{(S{O_4})_3}$ solution produce green colour with $N{a_2}O$ (excess) and yellow colour with $N{a_2}{O_2}$ (excess).
If true enter 1 else 0.

Answer 1

Write the whole chemical reaction of $C{r_2}{(S{O_4})_3}$ with $N{a_2}O$ and $N{a_2}{O_2}$. Check the ion of $Cr$ in the products formed in both the cases. The colour of $Cr(III)$ is green while $Cr(VI)$ imparts a yellow colour.

Complete step by step solution:
First let’s see what happens when $C{r_2}{(S{O_4})_3}$ reacts with $N{a_2}O$ and $N{a_2}{O_2}$.
$C{r_2}{(S{O_4})_3} + 3N{a_2}O + 3{H_2}O \to 2Cr{(OH)_3} + 3N{a_2}S{O_4}$
$C{r_2}{(S{O_4})_3} + 3N{a_2}{O_2} + 4NaOH \to 2N{a_2}Cr{O_4} + 3N{a_2}S{O_4} + 2{H_2}O$
In the first reaction, there is a production of green colour because of the presence of $Cr(III)$ in $Cr{(OH)_3}$. $Cr(III)$ is a gelatinous precipitate and is green in colour. Since, it simply imparts its green colour to the whole solution, therefore, the solution appears green.
In the second reaction, there is a production of yellow colour because of the presence of $Cr(VI)$ in $N{a_2}Cr{O_4}$. $N{a_2}Cr{O_4}$ exists as a yellow hygroscopic solid and imparts its colour to the whole solution in the liquid phase.
Hence, the correct answer is $1$ (true).

Additional Information:
Chromium is one of the elements which can exhibit multiple oxidation states.
Sodium chromate, $N{a_2}Cr{O_4}$ is further reacted with sodium hydroxide,${H_2}S{O_4}$ to produce sodium dichromate, $N{a_2}C{r_2}{O_7}$ which is an intermediate in the manufacturing of potassium dichromate, ${K_2}C{r_2}{O_7}$. $N{a_2}C{r_2}{O_7}$ is also used in the petroleum industry as a corrosion inhibitor and as a diagnostic pharmaceutical to determine the volume of red blood cells.
${K_2}C{r_2}{O_7}$ is a very strong oxidizing agent in acidic medium and is orange in colour. It is used in alcohol detector test where it converts alcohol into carboxylic acid and turns the orange dichromate into green chromate.

Note:
$Cr(VI)$ is from $CrO_4^{2 - }$ ion, which has an oxidation number of $+ 6$. $Cr(III)$ has an oxidation number of $+ 3$. Balancing your chemical equation is very important. You should always balance the equations before moving ahead.