Identify the element having lowest first ioni | vedclass

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(A) The ionization enthalpy $(I.E.)$ decreases down the group due to an increase in atomic size. Therefore,the $I.E.$ of $Te$ is higher than that of $

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$Po$

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(A) The ionization enthalpy $(I.E.)$ decreases down the group due to an increase in atomic size. Therefore,the $I.E.$ of $Te$ is higher than that of $Po$. Across a period,$I.E.$ increases with an increase in atomic number. Comparing the given elements,$Po$ (Polonium) is in Group $16$ and Period $6$,$Te$ (Tellurium) is in Group $16$ and Period $5$,$Br$ (Bromine) is in Group $17$ and Period $4$,and $Kr$ (Krypton) is in Group $18$ and Period $4$. Since $I.E.$ increases from left to right across a period and decreases down a group,$Po$ has the lowest $I.E.$ among the given elements.

Element	$IE_1$	$IE_2$	$IE_3$	Code	Element
$1$	$2080$	$3963$	$6130$	$(a)$	$H$
$2$	$520$	$7297$	$11810$	$(b)$	$Li$
$3$	$900$	$1758$	$14810$	$(c)$	$Be$
$4$	$800$	$2428$	$3660$	$(d)$	$B$
-	-	-	-	$(e)$	$Ne$

Element	$A, B, C$
$IE_1$	$400, 550, 1150$
$IE_2$	$2650, 1070, 2090$

Identify the element having lowest first ionization enthalpy.

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The $IE_1$ and $IE_2$ of three elements $A$,$B$,and $C$ are given as : ( $IE$ in $kJ/mol$ )

Element $A, B, C$

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$IE_2$ $2650, 1070, 2090$

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