The first ionization potential (IP) of Be (Z= | vedclass

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(D) The electronic configuration of $Be$ is $1s^2 2s^2$. Due to the fully filled $2s$ orbital,$Be$ has a very high ionization potential.$1$. $Be$ vs $

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$a, b, c, d$ and $e$

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(D) The electronic configuration of $Be$ is $1s^2 2s^2$. Due to the fully filled $2s$ orbital,$Be$ has a very high ionization potential.$1$. $Be$ vs $B$: $Be$ $(1s^2 2s^2)$ has a higher $IP$ than $B$ $(1s^2 2s^2 2p^1)$ because removing an electron from a fully filled $s$-orbital is more difficult than removing it from a $p$-orbital.$2$. $Be$ vs $Mg$: $Be$ is in the $2^{nd}$ period and $Mg$ is in the $3^{rd}$ period. $IP$ decreases down the group,so $IP$ of $Be > Mg$.$3$. $Be$ vs $Li$: $Be$ $(1s^2 2s^2)$ has a higher $IP$ than $Li$ $(1s^2 2s^1)$ due to higher effective nuclear charge.$4$. $Be$ vs $Al$: $Be$ is in the $2^{nd}$ period and $Al$ is in the $3^{rd}$ period. $Be$ has a higher $IP$ than $Al$.$5$. $Be$ vs $K$: $K$ is an alkali metal in the $4^{th}$ period with a very low $IP$. Thus,$IP$ of $Be > K$.Therefore,the $IP$ of $Be$ is higher than all the given elements $(a, b, c, d, e)$.

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

The first ionization potential $(IP)$ of $Be$ $(Z=4)$ is higher than which of the following elements?
$(a) B (Z=5)$
$(b) Mg (Z=12)$
$(c) Li (Z=3)$
$(d) Al (Z=13)$
$(e) K (Z=19)$

Similar Questions

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The electronic configuration of elements $A, B$ and $C$ are $[He] 2s^1, [Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?

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Element having highest $I.P.$ value is

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?

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The first ionization potential $(IP)$ of $Be$ $(Z=4)$ is higher than which of the following elements?$(a) B (Z=5)$$(b) Mg (Z=12)$$(c) Li (Z=3)$$(d) Al (Z=13)$$(e) K (Z=19)$