Explain why, iron dissolves in HCl to form \[FeC{{l}_{2}}\] not \[FeC{{l}_{3}}\]?
Answer
657k+ views
Hint: When two chemicals react with each other the formation of the product is going to depend on the energy required to convert chemical reactants into products. If the energy to convert reactants into products is too high then the reaction is not going to happen.
Complete answer:
In the question it is given that iron (Fe) reacts with HCl and forms \[FeC{{l}_{2}}\](Ferrous Chloride) not \[FeC{{l}_{3}}\] (Ferric chloride).
\[Fe+HCl\to FeC{{l}_{2}}\]
To know about the above reaction we should know that the electronic configuration of Iron.
Electronic configuration of Iron is as follows.
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}4{{s}^{2}}\]
The outermost electrons in \[4{{s}^{2}}\] are readily given electrons to Chlorine atoms.
But to form \[FeC{{l}_{3}}\] it requires a catalyst or oxidant like \[{{H}_{2}}{{O}_{2}}\]etc., to donate it electrons in \[3{{d}^{6}}\] orbitals.
To donate the electron from \[3{{d}^{6}}\]of iron, it requires more energy, also called third ionization energy.
First and second ionization energy of iron are less so that the first two electrons are donated easily to two chlorine atoms.
Coming to the third ionization energy of iron, it is very high because the next electron should come from \[3{{d}^{6}}\].
In \[3{{d}^{6}}\] the last electron (6th electron) is paired in 3d orbital. So, it is very difficult to donate a third electron from Iron to chlorine.
So, iron forms only \[FeC{{l}_{2}}\](Ferrous Chloride) not \[FeC{{l}_{3}}\] (Ferric chloride).
Note:
The formation of the products in a chemical reaction is going to depend on ionization energy of the chemicals. If the reaction needs the highest amount of ionization energy we have to use a catalyst to complete the reaction.
Complete answer:
In the question it is given that iron (Fe) reacts with HCl and forms \[FeC{{l}_{2}}\](Ferrous Chloride) not \[FeC{{l}_{3}}\] (Ferric chloride).
\[Fe+HCl\to FeC{{l}_{2}}\]
To know about the above reaction we should know that the electronic configuration of Iron.
Electronic configuration of Iron is as follows.
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}4{{s}^{2}}\]
The outermost electrons in \[4{{s}^{2}}\] are readily given electrons to Chlorine atoms.
But to form \[FeC{{l}_{3}}\] it requires a catalyst or oxidant like \[{{H}_{2}}{{O}_{2}}\]etc., to donate it electrons in \[3{{d}^{6}}\] orbitals.
To donate the electron from \[3{{d}^{6}}\]of iron, it requires more energy, also called third ionization energy.
First and second ionization energy of iron are less so that the first two electrons are donated easily to two chlorine atoms.
Coming to the third ionization energy of iron, it is very high because the next electron should come from \[3{{d}^{6}}\].
In \[3{{d}^{6}}\] the last electron (6th electron) is paired in 3d orbital. So, it is very difficult to donate a third electron from Iron to chlorine.
So, iron forms only \[FeC{{l}_{2}}\](Ferrous Chloride) not \[FeC{{l}_{3}}\] (Ferric chloride).
Note:
The formation of the products in a chemical reaction is going to depend on ionization energy of the chemicals. If the reaction needs the highest amount of ionization energy we have to use a catalyst to complete the reaction.
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

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

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

