Match List-I with List-II with correct code: | vedclass

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Question

(A) To match the ionization energies $(IE)$,we look at the valence electrons and the jump in energy after removing all valence electrons:$1$. $H$ $(1s

Vedclass · Accepted Answer

$A-1, B-2, C-3, D-4$

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(A) To match the ionization energies $(IE)$,we look at the valence electrons and the jump in energy after removing all valence electrons:$1$. $H$ $(1s^1)$: Only $1$ electron,so $IE_1 = 1312 \ kJ \ mol^{-1}$ (matches $A$ approx $1510$).$2$. $Li$ $(1s^2 2s^1)$: $1$ valence electron,large jump after $IE_1$. $IE_1=520, IE_2=7300$ (matches $B$ $495, 6500, 10200$).$3$. $Be$ $(1s^2 2s^2)$: $2$ valence electrons,large jump after $IE_2$. $IE_1=899, IE_2=1757, IE_3=14848$ (matches $C$ $840, 1630, 13100$).$4$. $B$ $(1s^2 2s^2 2p^1)$: $3$ valence electrons,large jump after $IE_3$. $IE_1=801, IE_2=2427, IE_3=3660$ (matches $D$ $600, 2050, 3100$).Thus,the correct match is $A-1, B-2, C-3, D-4$.

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

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

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

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

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

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

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

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