Question

When $FeC{{r}_{2}}{{O}_{4}}$ is fused with $N{{a}_{2}}C{{O}_{3}}$ in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange colored compound (C). An acidified solution of compound (C) oxidizes to (D). Identify (A), (B), (C), and (D).

Answer 1

Compound (A) is a compound of sodium that is attached with chromate ion. Compound (B) is also a compound of sodium and there are two anions of chromate. Compound (C) is a compound of potassium.

Complete answer:
$FeC{{r}_{2}}{{O}_{4}}$ is an inorganic compound named as chromite and $N{{a}_{2}}C{{O}_{3}}$ is also an inorganic compound named as sodium carbonate. When these react with each other in the presence of air, then there is the formation of sodium chromate whose formula is $N{{a}_{2}}Cr{{O}_{4}}$, iron (III) oxide whose formula is $F{{e}_{2}}{{O}_{3}}$ and carbon dioxide. The reaction is given below:
$FeCr{{O}_{4}}+N{{a}_{2}}C{{O}_{3}}+{{O}_{2}}\to N{{a}_{2}}Cr{{O}_{4}}+F{{e}_{2}}{{O}_{3}}+C{{O}_{2}}$
So, compound (A) is $N{{a}_{2}}Cr{{O}_{4}}$
When sodium chromate is acidified then there is the formation of sodium dichromate whose formula is $N{{a}_{2}}C{{r}_{2}}{{O}_{7}}$. The reaction is given below:
$N{{a}_{2}}Cr{{O}_{4}}+[{{H}^{+}}]\to N{{a}_{2}}C{{r}_{2}}{{O}_{7}}+N{{a}^{+}}+{{H}_{2}}O$
So, compound (B) is $N{{a}_{2}}C{{r}_{2}}{{O}_{7}}$
When sodium dichromate is treated with KCl then, the sodium ion of sodium dichromate is replaced with potassium ion. So, the product formed is potassium dichromate whose formula is ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and sodium chloride (NaCl). The reaction is given below:
$N{{a}_{2}}C{{r}_{2}}{{O}_{7}}+KCl\to {{K}_{2}}C{{r}_{2}}{{O}_{7}}+NaCl$
So, compound (C) is ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$.
When the potassium dichromate reacts with Sodium sulfite ($N{{a}_{2}}S{{O}_{3}}$), then the reaction is given below:
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+N{{a}_{2}}S{{O}_{3}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+N{{a}_{2}}S{{O}_{4}}+{{K}_{2}}S{{O}_{4}}+{{H}_{2}}O$

So, compound (D) is $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$.

Note:
When the compound is acidified, then the oxidation state of the metal in the compound will increase. The compound (C) formed from compound (B) is the displacement reaction.