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Question: A \(I{P_1}\) and \(I{P_2}\) of Mg are 178 and 348 \({\text{Kcal mo}}{{\text{l}}^{ - 1}}\). The entha...

A IP1I{P_1} and IP2I{P_2} of Mg are 178 and 348 Kcal mol−1{\text{Kcal mo}}{{\text{l}}^{ - 1}}. The enthalpy required for the reaction Mg→Mg2+Mg \to M{g^{2 + }} +2e + 2e is _____.
A.170 K.cal
B.526 K.cal
C.170 K.cal
D.526 K.cal

Explanation

Solution

Ionization enthalpy is the minimum or the lowest amount of energy required to remove the outermost electron from an isolated neutral gaseous molecule. The electron which is removed is the most loosely bound electron.

Complete answer:
Ionization enthalpy is also known as the ionization potential, IP. It can be expressed as
Mg2++2e−→E+(g)+e−M{g^{2 + }} + 2{e^ - } \to {E^ + }\left( {\text{g}} \right) + {e^ - }
In the above reaction, EE represents any neutral gaseous atom or molecule, E+{E^ + } is ion with one electron removed. In this reaction energy is provided. Hence it is an endothermic reaction. The energy provided is called ionization energy. The closer the outermost electron will be to the nucleus; more will be its ionization energy.
Ionization energy increases from left to right within the given period and it generally decreases from top to bottom in a given group.
Magnesium is alkaline with metal.
For removing its outermost electron, IP1I{P_1} will be required.
Mg+IP1→Mg++e−Mg + I{P_1} \to M{g^ + } + {e^ - }
Hence neutral magnesium atoms get converted to magnesium ions. IP1I{P_1} for magnesium is 178 Kcal mol−1{\text{Kcal mo}}{{\text{l}}^{ - 1}}
For removal of one more electron from Mg+M{g^ + }, IP2I{P_2} is required.
Mg++IP2→Mg2++e−M{g^ + } + I{P_2} \to M{g^{2 + }} + {e^ - }
Here Mg+M{g^ + } is converted to Mg2+M{g^{2 + }} with the help of IP2I{P_2} which is 348 K.cal mol−1mo{l^{ - 1}}.
Now we will add these two reactions.
Mg+net energy→Mg2++2e−{\text{Mg}} + {\text{net energy}} \to {\text{M}}{{\text{g}}^{2 + }} + 2{e^ - }
Here net energy is the addition of IP1I{P_1} and IP2I{P_2}. Hence net energy required for the given reaction
Mg→Mg2++2e−Mg \to M{g^{2 + }} + 2{e^ - } is 178
⇒energy=178+348=+526 Kcal\Rightarrow {\text{energy}} = 178 + 348 = + 526{\text{ Kcal}}.

Therefore, the correct option is B.

Note:
The successive ionization enthalpies are higher. For example, in the given question IP2I{P_2} is greater than IP1I{P_1}.