Question

Statement I: The ${{A}}{{{l}}^{{{3 + }}}}$ ion needs to be to form aluminum metal.
Statement II: Reduction is a gain of electrons.
A.Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
B.Both Statement I and Statement II are correct, but Statement II is not the correct explanation of Statement I.
C.Statement I is correct, but Statement II is incorrect.
D.Statement I is incorrect, but Statement II is correct.

Answer 1

Hint: Aluminium is an element of Group 13 of the p-block elements of the periodic table. The atomic number of aluminum is 13. The electronic configuration of Aluminium is ${{1}}{{{s}}^{{2}}}{{2}}{{{s}}^{{2}}}{{2}}{{{p}}^{{6}}}{{3}}{{{s}}^{{2}}}{{3}}{{{p}}^{{1}}}$ . The reduction is a process involving the gain of electrons by an atom, ion, or molecule. Oxidation is the loss of electrons.

Complete step by step answer:
The representation of aluminium metal can be done as ${{Al}}$ whereas aluminium ion is written as ${{A}}{{{l}}^{{{3 + }}}}$.
${{A}}{{{l}}^{{{3 + }}}}$ cation has three electrons less than ${{Al}}$. As we know reduction is a gain of electrons so we will reduce ${{A}}{{{l}}^{{{3 + }}}}$ cation to ${{Al}}$ by adding three electrons in it.
 ${{A}}{{{l}}^{{{3 + }}}}{{ + 3}}{{{e}}^{{ - }}} \to {{Al}}$ After discussing the following statements we can conclude that Statement II is the correct explanation for Statement I.
Thus, the correct option is (A).

Additional Information: Aluminium is a bluish-white metal having a brilliant luster which, however, is soon destroyed due to the formation of an oxide layer. It is malleable and ductile and can be rolled into sheets, foils, and wires. It is a very light metal. It is considered as the good conductor of electricity and heat. It melts at ${{65}}{{{9}}^{{o}}}{{C}}$ and boils at ${{245}}{{{0}}^{{o}}}{{C}}$ .

Note:
Examples of reduction is ${{F}}{{{e}}^{{{3 + }}}}{{ + }}{{{e}}^{{ - }}} \to {{F}}{{{e}}^{{{2 + }}}}$
${{MnO}}_{{4}}^{{ - }}{{ + }}{{{e}}^{{ - }}} \to {{MnO}}_{{4}}^{{{2 - }}}$
A substance that gives hydrogen or removes oxygen is called a reducing agent.
The reaction in which oxidation and reduction take place simultaneously is called a redox reaction. Let us take an example of the reaction between hydrogen and fluorine. The redox reaction can be written as follows: ${{{H}}_{{2}}}{{ + }}{{{F}}_{{2}}} \to {{2HF}}$.
The oxidation half-reaction is:
${{{H}}_{{2}}} \to {{2}}{{{H}}^{{ + }}}{{ + 2}}{{{e}}^{{ - }}}$
The reduction half-reaction is: ${{{F}}_{{2}}}{{ + 2}}{{{e}}^{{ - }}} \to {{2}}{{{F}}^{{ - }}}$
The hydrogen and fluorine ions go on to combine to form hydrogen fluoride.