The orange solid on heating gives a colorless gas and a green solid which can be reduced to metal by aluminium powder. The orange and green powder is respectively:
A: ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ and $C{r_2}{O_3}$
B: $N{a_2}C{r_2}{O_7}$ and $C{r_2}{O_3}$
C: ${K_2}C{r_2}{O_7}$ and $C{r_2}{O_3}$
D: ${\left( {N{H_4}} \right)_2}Cr{O_4}$ and $Cr{O_3}$
Answer
646.5k+ views
Hint:Aluminium is a powerful reducing agent that has great affinity for oxygen. Hence it reduces metallic oxides below (in reactivity series) like iron oxides to metal. These reactions are highly spontaneous and exothermic.
Complete step by step answer:
We have to pick up the given options one by one whether they match to our given question statement or not. Starting with option A it is given that Orange solid is ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ that is Ammonium dichromate which is indeed red-orange color compound moving further on heating the ammonium dichromate this will be the reaction
${(N{H_4})_2}C{r_2}{O_7}\xrightarrow{\Delta }{N_2} \uparrow + C{r_2}{O_2} + {H_2}O$
And observing the reaction we can find out that ${N_2}$ gas is released and it is also a colorless gas also $C{r_2}{O_3}$ Chromium$\left( {{\rm I}{\rm I}{\rm I}} \right)$ oxide is formed in the reaction which is indeed green solid further reaction of this solid with aluminium powder is given as
$C{r_2}{O_3} + 2Al\xrightarrow{{}}2Cr + A{l_2}{O_3}$
And from this reaction it is clear that on reacting the green solid ($C{r_2}{O_3}$) with aluminium powder we obtain the metal chromium $Cr$
Thus our first option justifies all the statement given in the question and hence it is our answer that is ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ and $C{r_2}{O_3}$.
There could be a chance that our option would have denied any statement of the question then we will move onto further options and see whether they fulfill the criteria or not.
Note: Oxide of aluminium is amphoteric in nature. Amphoteric compounds are those compounds that can react with both acids as well as bases. Such compounds form ${H^ + }$ ions when dissolved in acids and form $O{H^ - }$ ions when dissolved in base.
Complete step by step answer:
We have to pick up the given options one by one whether they match to our given question statement or not. Starting with option A it is given that Orange solid is ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ that is Ammonium dichromate which is indeed red-orange color compound moving further on heating the ammonium dichromate this will be the reaction
${(N{H_4})_2}C{r_2}{O_7}\xrightarrow{\Delta }{N_2} \uparrow + C{r_2}{O_2} + {H_2}O$
And observing the reaction we can find out that ${N_2}$ gas is released and it is also a colorless gas also $C{r_2}{O_3}$ Chromium$\left( {{\rm I}{\rm I}{\rm I}} \right)$ oxide is formed in the reaction which is indeed green solid further reaction of this solid with aluminium powder is given as
$C{r_2}{O_3} + 2Al\xrightarrow{{}}2Cr + A{l_2}{O_3}$
And from this reaction it is clear that on reacting the green solid ($C{r_2}{O_3}$) with aluminium powder we obtain the metal chromium $Cr$
Thus our first option justifies all the statement given in the question and hence it is our answer that is ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ and $C{r_2}{O_3}$.
There could be a chance that our option would have denied any statement of the question then we will move onto further options and see whether they fulfill the criteria or not.
Note: Oxide of aluminium is amphoteric in nature. Amphoteric compounds are those compounds that can react with both acids as well as bases. Such compounds form ${H^ + }$ ions when dissolved in acids and form $O{H^ - }$ ions when dissolved in base.
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