Question

The yellow colored solution of Chromate salt changes to orange color on acidification due to the formation of:
(A) $C{r^{3 + }}$
(B) $C{r_2}O_7^{2 - }$
(C) $CrO_4^{2 - }$
(D) $C{r_2}{O_3}$

Answer 1

Hint: Yellow coloured solution of sodium chromate changes to orange when acidified with sulphuric acid. We first see the reaction then discuss the compound.

Complete step by step solution:
The simplest ion that Chromium forms in solution is the hexaaquachromium$\left( {III} \right) - {[Cr{\left( {{H_2}O} \right)_6}]^{3 + }}$
The acidity of the hexaaqua ions
 In common with the $3 + $ ions, the hexaaquachromium $\left( {III} \right)$ ions is fairly acidic, with a $pH$ for typical solution in the $2 - 3$ range.
The ion reacts with water molecules in the solution. A hydrogen ion is lost from one of the ligand water molecules.
$Cr\left( {{H_2}O} \right)_6^{3 + } \rightleftharpoons Cr{\left( {{H_2}O} \right)_5}{\left( {OH} \right)^{2 + }} + {H_3}{O^ + }$
The complex ion is acting as an acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a base by accepting the hydrogen ion. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this:
$Cr\left( {{H_2}O} \right)_6^{3 + } \rightleftharpoons Cr{\left( {{H_2}O} \right)_5}{\left( {OH} \right)^{2 + }} + {H^ + }\left( {aq} \right)$
However, if you write like this, remember that hydrogen ions are not just falling off the complex ion. It is being pulled off by a water molecule in the solution. Whenever you write what you really mean is a hydronium ion, ${H_3}{O^ + }$.
Ligand exchange reactions involving chloride on sulfate ion:
 The hexaaquachromium$\left( {III} \right)$ ion is a ‘difficult to describe’ violet blue-grey colour. However, when it is produced during a reaction in a test tube, it is often green. We nearly always describe the green ion as being $C{r^{3 + }}\left( {aq} \right)$ implying the hexaaquachromium $\left( {III} \right)$ ion. That's actually an over-simplification.
The yellow coloured solution of chromate salt changes to orange on acidification due to formation of $C{r_2}{O^{2 - }}_7$. In acidic medium, chromate ion changes to dichromate ion.
$2CrO_4^{2 - } + 2{H^ + }{\text{ }}(yellow) \to C{r_2}O_7^{2 - } + {H_2}O(orange{\text{ }}red)$
$\therefore $ So, the yellow coloured solution of Chromate changes to orange colour or acidification due to the of (B) $C{r_2}O_7^{2 - }$

Hence, the correct answer is option B.

Note: Yellow chromate ion and orange dichromate ion are in equilibrium with each other in aqueous solution. The more acidic the solution. The more acidic the solution, the more the equilibrium is shifted to favour the dichromate ion. As nitric acid is added to the Potassium chromate solution, the yellow colour turns to orange