Which ionisation potential (IP) in the follow | vedclass

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Question

(B) The correct answer is $(B)$. Ionisation energy increases as we remove electrons from more stable configurations or from ions with higher positive

Vedclass · Accepted Answer

$K^{+} 	o K^{2+} + e^-$

Vedclass · Answer

(B) The correct answer is $(B)$. Ionisation energy increases as we remove electrons from more stable configurations or from ions with higher positive charge. In option $(B)$,$K^{+}$ has a stable noble gas configuration $([Ar])$. Removing an electron from a stable noble gas configuration requires a very high amount of energy compared to the other options,where electrons are being removed from less stable or valence shells.

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$)	List-$II$ (Element)
$A$. $1510$	$1$. $H$
$B$. $495, 6500, 10200$	$2$. $Li$
$C$. $840, 1630, 13100$	$3$. $Be$
$D$. $600, 2050, 3100$	$4$. $B$

Which ionisation potential $(IP)$ in the following equations involves the greatest amount of energy?

Similar Questions

Match List-$I$ with List-$II$ with correct code:

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$) List-$II$ (Element)

$A$. $1510$ $1$. $H$

$B$. $495, 6500, 10200$ $2$. $Li$

$C$. $840, 1630, 13100$ $3$. $Be$

$D$. $600, 2050, 3100$ $4$. $B$

For the second period elements,the correct increasing order of first ionisation enthalpy is:

The second ionisation potential of an element $M$ is the energy required to

If the $IP$ of $Na$ is $5.48 \, eV$, the ionisation potential of $K$ will be ............ $eV$.

The order of the magnitude of first ionisation potentials of $Be$,$B$,$N$ and $O$ is

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