How would you explain the fact that first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer
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Hint: First ionization enthalpy is the energy required to remove the first most loosely bound electron from an isolated atom in its ground state while second ionization enthalpy is the energy required to remove the second most loosely bound electron. Write the electronic configuration of sodium and magnesium and then, relate their first and second ionization enthalpies.
Complete step by step answer:
Ionization enthalpy is defined as the energy required to remove an electron from an isolated atom in its ground state.
Now, electronic configuration of sodium (Na): $1{s^2}2{s^2}2{p^6}3{s^1}$
Electronic configuration of magnesium (Mg): $1{s^2}2{s^2}2{p^6}3{s^2}$
First ionization enthalpy is the energy required to remove the first most loosely bound electron.
Here, the first electron in both sodium and magnesium has to be removed from $3s$ orbital but the effective nuclear charge of sodium is less than that of magnesium. Therefore, first ionization enthalpy of sodium is less than that of magnesium.
Now, after the removal of first electron, electronic configuration of Na and Mg is:
Electronic configuration of $N{a^ + }:1{s^2}2{s^2}2{p^6}$
Electronic configuration of \[M{g^ + }:1{s^2}2{s^2}2{p^6}3{s^1}\]
Sodium after losing one electron acquires the stable noble gas configuration of neon i.e., $1{s^2}2{s^2}2{p^6}$. Hence, in case of sodium, the second electron has to be removed from the stable noble gas configuration. Consequently, removal of a second electron from sodium requires more energy in comparison to that required in magnesium. Therefore, the second ionization enthalpy of sodium is higher than that of magnesium. Hence, this is the required answer.
Note: The first ionization enthalpy for an element say X, is the enthalpy change for the reaction shown below:
$X \to {X^ + } + {e^ - }$
The second ionization enthalpy is the energy required to carry out the reaction shown below:
${X^ + } \to {X^{2 + }} + {e^ - }$
Ionization is an endothermic process hence, the ionization enthalpies are always positive.
Complete step by step answer:
Ionization enthalpy is defined as the energy required to remove an electron from an isolated atom in its ground state.
Now, electronic configuration of sodium (Na): $1{s^2}2{s^2}2{p^6}3{s^1}$
Electronic configuration of magnesium (Mg): $1{s^2}2{s^2}2{p^6}3{s^2}$
First ionization enthalpy is the energy required to remove the first most loosely bound electron.
Here, the first electron in both sodium and magnesium has to be removed from $3s$ orbital but the effective nuclear charge of sodium is less than that of magnesium. Therefore, first ionization enthalpy of sodium is less than that of magnesium.
Now, after the removal of first electron, electronic configuration of Na and Mg is:
Electronic configuration of $N{a^ + }:1{s^2}2{s^2}2{p^6}$
Electronic configuration of \[M{g^ + }:1{s^2}2{s^2}2{p^6}3{s^1}\]
Sodium after losing one electron acquires the stable noble gas configuration of neon i.e., $1{s^2}2{s^2}2{p^6}$. Hence, in case of sodium, the second electron has to be removed from the stable noble gas configuration. Consequently, removal of a second electron from sodium requires more energy in comparison to that required in magnesium. Therefore, the second ionization enthalpy of sodium is higher than that of magnesium. Hence, this is the required answer.
Note: The first ionization enthalpy for an element say X, is the enthalpy change for the reaction shown below:
$X \to {X^ + } + {e^ - }$
The second ionization enthalpy is the energy required to carry out the reaction shown below:
${X^ + } \to {X^{2 + }} + {e^ - }$
Ionization is an endothermic process hence, the ionization enthalpies are always positive.
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