How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
(A) The electronic configuration of $Na$ is $[Ne] 3s^{1}$ and that of $Mg$ is $[Ne] 3s^{2}$.
The configuration of $Mg$ is more stable (being completely filled) than that of $Na$.
Therefore,the first ionization enthalpy of $Mg$ is higher than that of $Na$.
After the loss of one electron from $Na$,it acquires the stable noble gas configuration of $Ne$ $(1s^{2} 2s^{2} 2p^{6})$.
On the other hand,for $Mg$,the configuration after the first ionization becomes $[Ne] 3s^{1}$.
Thus,the electronic configuration of $Na^{+}$ is more stable than that of $Mg^{+}$,making the removal of the second electron from $Na^{+}$ much more difficult than from $Mg^{+}$.
$Na ([Ne] 3s^{1}) \longrightarrow Na^{+} ([Ne]) + e^{-}$
$Mg ([Ne] 3s^{2}) \longrightarrow Mg^{+} ([Ne] 3s^{1}) + e^{-}$
$IE_{1}(Na) < IE_{1}(Mg)$
$IE_{2}(Na) > IE_{2}(Mg)$
The configuration of $Mg$ is more stable (being completely filled) than that of $Na$.
Therefore,the first ionization enthalpy of $Mg$ is higher than that of $Na$.
After the loss of one electron from $Na$,it acquires the stable noble gas configuration of $Ne$ $(1s^{2} 2s^{2} 2p^{6})$.
On the other hand,for $Mg$,the configuration after the first ionization becomes $[Ne] 3s^{1}$.
Thus,the electronic configuration of $Na^{+}$ is more stable than that of $Mg^{+}$,making the removal of the second electron from $Na^{+}$ much more difficult than from $Mg^{+}$.
$Na ([Ne] 3s^{1}) \longrightarrow Na^{+} ([Ne]) + e^{-}$
$Mg ([Ne] 3s^{2}) \longrightarrow Mg^{+} ([Ne] 3s^{1}) + e^{-}$
$IE_{1}(Na) < IE_{1}(Mg)$
$IE_{2}(Na) > IE_{2}(Mg)$
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Ionization energy of gaseous $Na$ atoms is $495.5 \ kJ \ mol^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $(h = 6.626 \times 10^{-34} \ J \ s, N_A = 6.022 \times 10^{23} \ mol^{-1})$.
DifficultJEE MAIN 2014
View SolutionFor one of the elements,various successive ionization enthalpies (in $kJ \cdot mol^{-1}$) are given below:
$I.E.$ $1^{st}$ $2^{nd}$ $3^{rd}$ $4^{th}$ $5^{th}$ Value $577.5$ $1810$ $2750$ $11580$ $14820$
The element is:
| $I.E.$ | $1^{st}$ | $2^{nd}$ | $3^{rd}$ | $4^{th}$ | $5^{th}$ |
| Value | $577.5$ | $1810$ | $2750$ | $11580$ | $14820$ |
The element is:
MediumKCET 2015
View SolutionThe second ionization potential is
Easy
View SolutionWhich of the following elements has the lowest ionisation potential?
Easy
View Solution$B$ has a smaller first ionization enthalpy than $Be$. Consider the following statements:
$(I)$ It is easier to remove a $2p$ electron than a $2s$ electron.
$(II)$ The $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of $Be$.
$(III)$ The $2s$ electron has more penetration power than the $2p$ electron.
$(IV)$ The atomic radius of $B$ is more than $Be$.
(Atomic number: $B=5, Be=4$)
The correct statements are:
$(I)$ It is easier to remove a $2p$ electron than a $2s$ electron.
$(II)$ The $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of $Be$.
$(III)$ The $2s$ electron has more penetration power than the $2p$ electron.
$(IV)$ The atomic radius of $B$ is more than $Be$.
(Atomic number: $B=5, Be=4$)
The correct statements are:
DifficultJEE MAIN 2020
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