Which of the following species has the lowest ionization potential?

The successive ionization energies of an element $(A)$ are given as $IE_1 = 20 \ eV, IE_2 = 45 \ eV, IE_3 = 150 \ eV, IE_4 = 900 \ eV, IE_5 = 1800 \ eV$. What is the formula of the halide of $(A)$?

The $1^{\text{st}}$,$2^{\text{nd}}$,and $3^{\text{rd}}$ ionization enthalpies $I_1, I_2$,and $I_3$ of four atoms with atomic numbers $n, n+1, n+2$,and $n+3$,where $n < 10$,are tabulated below. What is the value of $n$?

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

Assertion $(A)$: The first ionisation energy of $Be$ is greater than that of $B$.
Reason $(R)$: $2p$ orbital has lower energy than $2s$ orbital.

The atomic number of the element from the following with the lowest $1^{st}$ ionisation enthalpy is:

The first ionization potentials $\left( eV \right)$ of $Be$ and $B$ respectively are: