A blood red color is obtained when ferric chloride solution reacts with ________.
(a)- $KCN$
(b)- $KSCN$
(c)- ${{K}_{4}}[Fe{{(CN)}_{6}}]$
(d)- ${{K}_{3}}[Fe{{(CN)}_{6}}]$
Answer
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Hint: Ferric chloride is a compound having formula $FeC{{l}_{3}}$ and its solution is colorless to light brown. With all options given above, when reacted with ferric chloride forms complex compounds. So, the compound which gives blood-red color is $Fe{{(SCN)}_{3}}$.
Complete step by step answer:
Ferric chloride is a compound having formula $FeC{{l}_{3}}$ and its solution is colorless to light brown and it is also known as iron (III) chloride because the oxidation of iron in this compound is +3.
So, with $KCN$reacts with ferric chloride solution $FeC{{l}_{3}}$to form a coordination compound, in which iron is the central metal atom, cyanide is the ligand and potassium is the cationic part. The reaction is given below:
$FeC{{l}_{3}}+KCN\to {{K}_{3}}[Fe{{(CN)}_{6}}]+KCl$
And this in solution gives green-yellow color.
When $KSCN$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a complex compound having formula $Fe{{(SCN)}_{3}}$. The reaction is given below:
$FeC{{l}_{3}}+3KSCN\to Fe{{(SCN)}_{3}}+3KCl$
This solution is blood-red in color.
When ${{K}_{4}}[Fe{{(CN)}_{6}}]$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a coordination compound having formula $F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}$ in which the cyanide is the ligand and iron acts as both cationic part and central metal atom. The reaction is given below:
$FeC{{l}_{3}}+{{K}_{4}}[Fe{{(CN)}_{6}}]\to F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}$
This compound is known as Prussian blue and is a blue solution.
When ${{K}_{3}}[Fe{{(CN)}_{6}}]$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a coordination compound having formula $F{{e}_{3}}{{[Fe{{(CN)}_{6}}]}_{3}}$ in which the cyanide is the ligand and iron acts as both cationic part and central metal atom. The reaction is given below:
$3FeC{{l}_{3}}+3{{K}_{3}}[Fe{{(CN)}_{6}}\to F{{e}_{3}}{{[Fe{{(CN)}_{6}}]}_{3}}+9KCl$
This solution is red in color.
Therefore, the correct answer is an option (b)- $KSCN$.
Note: Don't get confused between option (b) and (d) because both of them give a compound with ferric chloride that is red, but specifically $KSCN$ gives a blood-red solution and ${{K}_{3}}[Fe{{(CN)}_{6}}]$only gives the red solution.
Complete step by step answer:
Ferric chloride is a compound having formula $FeC{{l}_{3}}$ and its solution is colorless to light brown and it is also known as iron (III) chloride because the oxidation of iron in this compound is +3.
So, with $KCN$reacts with ferric chloride solution $FeC{{l}_{3}}$to form a coordination compound, in which iron is the central metal atom, cyanide is the ligand and potassium is the cationic part. The reaction is given below:
$FeC{{l}_{3}}+KCN\to {{K}_{3}}[Fe{{(CN)}_{6}}]+KCl$
And this in solution gives green-yellow color.
When $KSCN$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a complex compound having formula $Fe{{(SCN)}_{3}}$. The reaction is given below:
$FeC{{l}_{3}}+3KSCN\to Fe{{(SCN)}_{3}}+3KCl$
This solution is blood-red in color.
When ${{K}_{4}}[Fe{{(CN)}_{6}}]$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a coordination compound having formula $F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}$ in which the cyanide is the ligand and iron acts as both cationic part and central metal atom. The reaction is given below:
$FeC{{l}_{3}}+{{K}_{4}}[Fe{{(CN)}_{6}}]\to F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}$
This compound is known as Prussian blue and is a blue solution.
When ${{K}_{3}}[Fe{{(CN)}_{6}}]$reacts with ferric chloride solution $FeC{{l}_{3}}$it forms a coordination compound having formula $F{{e}_{3}}{{[Fe{{(CN)}_{6}}]}_{3}}$ in which the cyanide is the ligand and iron acts as both cationic part and central metal atom. The reaction is given below:
$3FeC{{l}_{3}}+3{{K}_{3}}[Fe{{(CN)}_{6}}\to F{{e}_{3}}{{[Fe{{(CN)}_{6}}]}_{3}}+9KCl$
This solution is red in color.
Therefore, the correct answer is an option (b)- $KSCN$.
Note: Don't get confused between option (b) and (d) because both of them give a compound with ferric chloride that is red, but specifically $KSCN$ gives a blood-red solution and ${{K}_{3}}[Fe{{(CN)}_{6}}]$only gives the red solution.
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