What is the decreasing order of the ionizatio | vedclass

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Question

(B) The ionization energy generally increases from left to right across a period.However,exceptions occur due to the stable electronic configurations

Vedclass · Accepted Answer

$Ne > Cl > P > S > Mg > Al$

Vedclass · Answer

(B) The ionization energy generally increases from left to right across a period.However,exceptions occur due to the stable electronic configurations of Group $2$ elements $(ns^2)$ and Group $15$ elements $(ns^2 np^3)$.Comparing the elements: $Ne$ (Group $18$),$Cl$ (Group $17$),$S$ (Group $16$),$P$ (Group $15$),$Mg$ (Group $2$),$Al$ (Group $13$).Due to the half-filled $p$-orbital stability of $P$ and the full-filled $s$-orbital stability of $Mg$,the order is: $Ne > Cl > P > S > Mg > Al$.

List-$I$ (Electronic configuration of neutral atom where $n=2$)	List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$	$I. 2080$
$B. ns^2np^1$	$II. 899$
$C. ns^2np^3$	$III. 800$
$D. ns^2np^6$	$IV. 1402$

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

What is the decreasing order of the ionization energy for the following elements?

Similar Questions

Which of the following electronic configurations will have the highest tendency to form a univalent gaseous cation?

Match List-$I$ with List-$II$:
List-$I$ (Electronic configuration of neutral atom where $n=2$) List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$ $I. 2080$
$B. ns^2np^1$ $II. 899$
$C. ns^2np^3$ $III. 800$
$D. ns^2np^6$ $IV. 1402$

If energy absorbed for conversion of $M_{(g)}$ into $M_{(g)}^{+}$ for $Li, Be, B, C$ and $N$ is $a, b, c, d$ and $e$ (in $k\,cal/mol$) respectively,then the correct order will be:

What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

The ionization potential $(IP)$ values for an element are given below. Identify the element:
$(IP)_1$ $7.1 \ eV$
$(IP)_2$ $10.3 \ eV$
$(IP)_3$ $36.2 \ eV$
$(IP)_4$ $39.1 \ eV$

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