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(C) To determine the value of $n$,we analyze the jump in ionization enthalpies for each element:
$1$. For atomic number $(n+4)$,the jump occurs betwe

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) To determine the value of $n$,we analyze the jump in ionization enthalpies for each element:
$1$. For atomic number $(n+4)$,the jump occurs between $I_2$ and $I_3$ ($1451$ to $7733$),indicating it is an alkaline earth metal (Group $2$).
$2$. For atomic number $(n+3)$,the jump occurs between $I_1$ and $I_2$ ($496$ to $4562$),indicating it is an alkali metal (Group $1$).
$3$. Since $(n+3)$ is an alkali metal and $(n+4)$ is an alkaline earth metal,$(n+3)$ corresponds to $Z=11$ (Sodium) and $(n+4)$ corresponds to $Z=12$ (Magnesium).
$4$. If $n+3 = 11$,then $n = 8$.
$5$. Checking the sequence: $n=8$ (Oxygen),$n+1=9$ (Fluorine),$n+2=10$ (Neon),$n+3=11$ (Sodium). The ionization energies provided for $n+1$ $(1681)$ and $n+2$ $(2081)$ are consistent with Fluorine and Neon respectively. Thus,$n=8$.

Atomic number	$I_1$ $(kJ/mol)$	$I_2$ $(kJ/mol)$	$I_3$ $(kJ/mol)$
$n$	$I_1$	$I_2$	$I_3$
$n+1$	$1681$	$3374$	$6050$
$n+2$	$2081$	$3952$	$6122$
$n+3$	$496$	$4562$	$6910$
$n+4$	$738$	$1451$	$7733$

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