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 $(a, b, c, d, e)$.

$I.E.$	$1^{st}$	$2^{nd}$	$3^{rd}$	$4^{th}$	$5^{th}$
Value	$577.5$	$1810$	$2750$	$11580$	$14820$

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

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Which of the following has the highest first ionization energy?

Ionization energy of gaseous $Na$ atoms is $495.5 \ kJ \ mol^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $(h = 6.626 \times 10^{-34} \ J \ s, N_A = 6.022 \times 10^{23} \ mol^{-1})$.

For 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:

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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)$