
Which series of reactions correctly represents chemical reactions related to iron and its compounds?
A) $Fe\xrightarrow{{C{l_2},heat}}FeC{l_3}\xrightarrow{{heat,air}}FeC{l_2}\xrightarrow{{Zn}}Fe$
B) $Fe\xrightarrow{{{O_2},heat}}F{e_3}{O_4}\xrightarrow{{CO,{{600}^o}C}}FeO\xrightarrow{{CO,{{700}^o}C}}Fe$
C) $Fe\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{{H_2}S{O_4},{O_2}}}F{e_2}{(S{O_4})_3}\xrightarrow{{heat}}Fe$
D) $Fe\xrightarrow{{{O_2},heat}}FeO\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{heat}}Fe$
Answer
579.3k+ views
Hint: Examine the products formed at every step of the reactions of iron given in options. In some steps there is oxidation of iron and in some steps there is reduction. In oxidation, there is an addition of oxygen and in reduction, there is an addition of hydrogen or removal of oxygen.
Complete step by step answer:
Let us discuss every chemical reaction given in the options.
- $Fe\xrightarrow{{C{l_2},heat}}FeC{l_3}\xrightarrow{{heat,air}}FeC{l_2}\xrightarrow{{Zn}}Fe$
In the step 2 of the above reaction,$FeC{l_3}$ will undergo oxidation reaction upon heating in the presence of air. But here, $FeC{l_3}$ is converting to $FeC{l_2}$ which is a reduction. Hence, option A is incorrect.
$Fe\xrightarrow{{{O_2},heat}}F{e_3}{O_4}\xrightarrow{{CO,{{600}^o}C}}FeO\xrightarrow{{CO,{{700}^o}C}}Fe$
In this reaction, in step 1, Fe on heating with ${O_2}$ giving $F{e_3}{O_4}$ which is oxidation of Fe (addition of oxygen) and hence, the product is correct. Further, $F{e_3}{O_4}$ is undergoing reduction with CO and first giving FeO and then Fe. On reduction, there is removal of oxygen. Here also, from $F{e_3}{O_4}$ to Fe there is removal of oxygen. Thus, these are the correct or required products according to the reaction conditions. Hence, option B is correct.
- $Fe\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{{H_2}S{O_4},{O_2}}}F{e_2}{(S{O_4})_3}\xrightarrow{{heat}}Fe$
In the above reaction, the first and second steps are correct. But, the third step is wrong because $F{e_2}{(S{O_4})_3}$ on heating gives $F{e_2}{O_3}$, $S{O_3}$ and not Fe. Thus, option C is incorrect.
- $Fe\xrightarrow{{{O_2},heat}}FeO\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{heat}}Fe$
In the above equation again, steps 1 and 2 are correct, but the 3rd step is wrong because $FeS{O_4}$ cannot be simply converted to Fe on heating. Instead oxides of sulphur ($S{O_2},S{O_3}$) and iron ($F{e_2}{O_3}$) will be formed. Hence, option D is incorrect.
So, the correct answer is “Option B”.
Note: Iron (Fe) on oxidation (i.e., addition of oxygen) forms iron oxide ($F{e_2}{O_3}$). When iron oxide undergoes reduction, there is removal of oxygen and hence the product formed is $FeO$. When $FeO$ further undergoes heating at high temperature, then finally we get the reduced iron i.e., Fe.
The heating reaction of $FeS{O_4}$ is as follows:
$FeS{O_4}\xrightarrow{{heat}}F{e_2}{O_3} + S{O_2} + S{O_3}$
Complete step by step answer:
Let us discuss every chemical reaction given in the options.
- $Fe\xrightarrow{{C{l_2},heat}}FeC{l_3}\xrightarrow{{heat,air}}FeC{l_2}\xrightarrow{{Zn}}Fe$
In the step 2 of the above reaction,$FeC{l_3}$ will undergo oxidation reaction upon heating in the presence of air. But here, $FeC{l_3}$ is converting to $FeC{l_2}$ which is a reduction. Hence, option A is incorrect.
$Fe\xrightarrow{{{O_2},heat}}F{e_3}{O_4}\xrightarrow{{CO,{{600}^o}C}}FeO\xrightarrow{{CO,{{700}^o}C}}Fe$
In this reaction, in step 1, Fe on heating with ${O_2}$ giving $F{e_3}{O_4}$ which is oxidation of Fe (addition of oxygen) and hence, the product is correct. Further, $F{e_3}{O_4}$ is undergoing reduction with CO and first giving FeO and then Fe. On reduction, there is removal of oxygen. Here also, from $F{e_3}{O_4}$ to Fe there is removal of oxygen. Thus, these are the correct or required products according to the reaction conditions. Hence, option B is correct.
- $Fe\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{{H_2}S{O_4},{O_2}}}F{e_2}{(S{O_4})_3}\xrightarrow{{heat}}Fe$
In the above reaction, the first and second steps are correct. But, the third step is wrong because $F{e_2}{(S{O_4})_3}$ on heating gives $F{e_2}{O_3}$, $S{O_3}$ and not Fe. Thus, option C is incorrect.
- $Fe\xrightarrow{{{O_2},heat}}FeO\xrightarrow{{dil{\text{ }}{H_2}S{O_4}}}FeS{O_4}\xrightarrow{{heat}}Fe$
In the above equation again, steps 1 and 2 are correct, but the 3rd step is wrong because $FeS{O_4}$ cannot be simply converted to Fe on heating. Instead oxides of sulphur ($S{O_2},S{O_3}$) and iron ($F{e_2}{O_3}$) will be formed. Hence, option D is incorrect.
So, the correct answer is “Option B”.
Note: Iron (Fe) on oxidation (i.e., addition of oxygen) forms iron oxide ($F{e_2}{O_3}$). When iron oxide undergoes reduction, there is removal of oxygen and hence the product formed is $FeO$. When $FeO$ further undergoes heating at high temperature, then finally we get the reduced iron i.e., Fe.
The heating reaction of $FeS{O_4}$ is as follows:
$FeS{O_4}\xrightarrow{{heat}}F{e_2}{O_3} + S{O_2} + S{O_3}$
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

