Question

Correct orders of 1st Ionization potential are:
(a)Li < B < Be < C
(b)O < N < F
(c)Be < N < Ne
A) a, b
B) b, c
C) a, c
D) a, b, c

Answer 1

Hint: Ionizing energy is the minimum level of energy available to extract the most loosely bound electron from the isolated neutral gaseous atom or molecule. If the compound attains noble gas configuration after losing electrons, then its ionisation potential will be lowest.

Complete step by step answer
Duration variation among the representative elements:
In general, ionization energies increase over the period because the effective nuclear charge per outermost electron increases when moving left to right over a period, while the corresponding main quantum number remains the same.
The amount of energy needed to expel an electron from a gaseous atom's outermost orbit is known as the potential for ionization. Elements that have orbitals that are half-filled or completely filled are more stable than partially filled orbitals. The capacity for ionization decreases as the atomic number increases over a period from left to right.
In general, ionization energy increases with a decrease in the atomic radius from left to right over the time. There are exceptions, however.
The penetration strength of s-orbitals is greater than that of p-orbitals. Therefore, the removal of electrons requires more energy from s-orbitals.
As they undergo less repulsion, the electrons in half filled orbitals are stable. If the same orbital is occupied with another electron, then they will be destabilized due to repulsion from each other.
So, the IP Order: Li < B < Be < C < O < N < F < Ne.
Hence, option D- a, b, c is the correct answer.

Note
The energy associated with the creation of 1 mole of negative ions from 1 mole (gaseous) atoms and 1 mol (gaseous) electrons is the (1st) ionization energy.
 $ M(g) \to M + (g) + e - \;\Delta H\; = \;{\text{Ionization energy}} $
 $ 1\;eV\; = \;96.485\;{\text{kJ \times mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{.}} $