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Question: \({{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\) on heating with ...

K2Cr2O7{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} on heating with aqueous NaOH, gives:
A. CrO42−{\text{CrO}}_4^{2 - }
B.Cr(OH)3{\text{Cr}}{\left( {{\text{OH}}} \right)_3}
C.Cr2O72−{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }
D.Cr(OH)2{\text{Cr}}{\left( {{\text{OH}}} \right)_2}

Explanation

Solution

Potassium dichromate decomposes as K + {{\text{K}}^{\text{ + }}} and Cr2O72−{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }. Potassium dichromate is a strong oxidizing agent. So, it gains electrons in a reaction. Dichromate gets reduced by aqueous NaOH. Aqueous NaOH causes two electron reduction of dichromate.

Complete step by step solution: K2Cr2O7{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} is known as potassium dichromate.
The reaction of the reduction of dichromate ion in an aqueous solution of sodium hydroxide is as follows:
Cr2O72− + 2 NaOH  → CrO42− + Na2CrO4  +  H2O{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }\, + \,2\,{\text{NaOH}}\,\, \to \,{\text{CrO}}_4^{2 - }\, + \,{\text{N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{{\text{H}}_2}{\text{O}}
Potassium dichromate on reacting with an aqueous solution of sodium hydroxide gives potassium chromate, sodium chromate and water. Dichromate ion is accepting two electrons, so here the valence factor for the potassium dichromate is two.
In aqueous solution, chromate and dichromate remain in equilibrium.

2 CrO42− + 2 H +   ⇄ Cr2O72− + H2O{\text{2}}\,{\text{CrO}}_4^{2 - }\, + \,2\,{{\text{H}}^{\text{ + }}}\,\, \rightleftarrows \,{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }\, + \,{{\text{H}}_{\text{2}}}{\text{O}}

The reaction of the reduction of potassium dichromate in an acidic medium is as follows:
Cr2O72− + 14 H +  +6e− → 2 Cr3+ + 7 H2O{\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 - }\, + \,14\,{{\text{H}}^{\text{ + }}}\, + 6{{\text{e}}^ - }\, \to \,2\,{\text{C}}{{\text{r}}^{3 + }}\, + \,{\text{7}}\,{{\text{H}}_2}{\text{O}}
The oxidation number of chromium is changing from  + 6 to   + 3{\text{ + 6}}\,{\text{to}}\,\,{\text{ + 3}} in acidic medium.
The reaction of the reduction of potassium dichromate in a basic medium is as follows:
CrO42− + 4 H2O  +  3e−  → Cr(OH)3 + 5 OH−{\text{CrO}}_4^{2 - }\, + \,4\,{{\text{H}}_2}{\text{O}}\,{\text{ + }}\,{\text{3}}{{\text{e}}^ - }\,\, \to \,{\text{Cr}}{\left( {{\text{OH}}} \right)_{\text{3}}}\, + \,{\text{5}}\,{\text{O}}{{\text{H}}^ - }
In a basic medium, chromate forms chromium (III) hydroxide.
So, K2Cr2O7{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} on heating with aqueous NaOH, gives chromate ion CrO42−{\text{CrO}}_4^{2 - }.

Therefore, option (A)CrO42−{\text{CrO}}_4^{2 - } is correct.

Note: Chromium ion shows different colours in different oxidation states. Dichromate is of orange colour. The chromate is yellow. The potassium dichromate also reacts with sulphur dioxide as follows:
K2Cr2O7  +  SO2 + H+  → Cr2(SO4)3 +  K2SO4 + H2O{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\,{\text{ + }}\,{\text{S}}{{\text{O}}_2}\, + \,{{\text{H}}^ + }\,\, \to \,{\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}\, + \,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{O}}
The product of reaction of potassium dichromate with sulphur dioxide is of green colour. The product is chromium sulphate. The green colour appears due to the reduction of chromium during the reaction. Chromate is a weak oxidizing agent in the basic medium rather than acidic medium.