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The metal whose cation is discharged at the cathode by accepting two electrons from the cathode is ___________.
A. Na
B. K
C. Al
D. Mg

Answer Verified Verified
Hint: Cation contains negatively charged particles known as electrons whereas cathode consists of positive charge whereas anion consists of positively charged particles called protons. Magnesium consists of 2 valence electrons in the outermost shell of the atom.

Complete answer:
-To identify the metal which can accept two electrons, firstly we have to write the cationic form of all the elements.
-Then those elements who can accept two electrons will be the correct answer.
-So, the cationic form of all the elements are: $\text{N}{{\text{a}}^{+}},\text{ }{{\text{K}}^{+}},\text{ A}{{\text{l}}^{3+}}\text{ and M}{{\text{g}}^{2+}}$.
-Now, the electronic configuration of all the cation is:
$\text{N}{{\text{a}}^{+}}\ (10)=\text{ 1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}$
${{\text{K}}^{+}}\ (18)=\text{ 1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}$
$\text{A}{{\text{l}}^{3+}}\ (10)=\text{ 1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}$
$\text{M}{{\text{g}}^{2+}}\ (10)=\text{ 1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}$
-So, as we can see that magnesium ion needs two electrons to accept when its cation is discharged at the cathode.
-Whereas sodium ions need 1 electron to form sodium metal for discharging of the cation at the cathode.

-The reaction will be:
$\text{N}{{\text{a}}^{+}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Na}$
-Similarly, aluminium ion and potassium ion accept 3 electrons and 1 electron respectively to get discharged at the cathode.
-For potassium, the reaction at the cathode:
${{\text{K}}^{+}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ K}$
-For aluminium, the reaction at the cathode is:
$\text{A}{{\text{l}}^{3+}}\text{ + 3}{{\text{e}}^{-}}\text{ }\to \text{ Al}$
-Hence, the reaction for magnesium ion at the cathode will be:
$\text{M}{{\text{g}}^{2+}}\text{ + 2}{{\text{e}}^{-}}\text{ }\to \text{ Mg}$
So, the correct answer is “Option D”.

Note: The no of electrons required for the cations to get discharged at the cathode depends on the amount of the positive charge present in the cation. The amount of the positive charge present on the cation also depends on the valency of the element that forms the cation. So indirectly the number of electrons that a cation needs to get discharged at the cathode is dependent on the valency of the element that forms the cation.