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Question

(C) Ionization enthalpy $(IE)$ depends on atomic size and electronic stability.Smaller atomic size leads to higher $IE$ because the valence electrons

Vedclass · Accepted Answer

$[Ne]\, 3s^2\, 3p^3$

Vedclass · Answer

(C) Ionization enthalpy $(IE)$ depends on atomic size and electronic stability.Smaller atomic size leads to higher $IE$ because the valence electrons are more strongly attracted by the nucleus.Additionally,half-filled and fully-filled orbitals possess extra stability,requiring more energy to remove an electron.Comparing the given configurations: $[Ne]\, 3s^2\, 3p^1$,$[Ne]\, 3s^2\, 3p^2$,and $[Ne]\, 3s^2\, 3p^3$ belong to the $3^{rd}$ period,while $[Ar]\, 3d^{10}\, 4s^2\, 4p^3$ belongs to the $4^{th}$ period.Elements in the $3^{rd}$ period have smaller atomic radii than those in the $4^{th}$ period,resulting in higher $IE$ for $3^{rd}$ period elements.Among the $3^{rd}$ period elements,$[Ne]\, 3s^2\, 3p^3$ has a half-filled $p$-orbital,which provides extra stability,making its ionization enthalpy the highest.

Element	$IE_1 + IE_2$ $(kJ/mol)$	$IE_3 + IE_4$ $(kJ/mol)$
$P$	$2.45$	$8.82$
$Q$	$2.85$	$6.11$

$B$	$801$
$Al$	$577$
$Ga$	$579$
$In$	$558$
$Tl$	$589$

The electronic configuration with the highest ionization enthalpy is

Similar Questions

The sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:

Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$

$P$ $2.45$ $8.82$

$Q$ $2.85$ $6.11$

Then,according to the given information,the incorrect statement$(s)$ is/are:

The first ionization enthalpy values (in $kJ\,mol^{-1}$) of group $13$ elements are :

$B$ $801$

$Al$ $577$

$Ga$ $579$

$In$ $558$

$Tl$ $589$

How would you explain this deviation from the general trend?

The correct order of $IE_2$ (second ionization enthalpy) is:

Which of the following elements has the highest second ionisation potential?

Which of the following isoelectronic ions has the lowest ionization energy?

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