Question

The first ionization energy of $$Na,B,N,$$ and $$O$$atoms follow the order:
A. $B<Na<O<N$
B. $Na<B<O<N$
C. $Na<O<B<N$
D. $O<Na<N<B$

Answer 1

Hint:Ionisation energy is the minimum amount of energy required to remove an electron from an isolated gaseous atom to form monovalent cation. Ionisation energy depends upon the size of the atom, half-filled and fully filled configuration and nuclear charge.

Complete answer:
First ionisation energy of all the elements (like $(Na,\text{B,N}$ and $\text{O)}$ depends upon electronic configuration.First of all, $Na$is an electropositive element, or alkali metal and belongs to the first group of the periodic table. Size of $Na$is larger than that of other elements and lesser force of attraction between the nucleus and outermost electrons. So its ionisation energy becomes less than that of other elements. But as we move from left to right in a period, nuclear charge increases and atomic size decreases thus ionisation energy increases.
Along the period, ionisation energy increases then the order of elements i.e. $Na<B<N<O$ but the actual order of $I.E_s$ is $Na<B<O<N$, the reason being half filled configuration $\left(2s^{2}2p^3\right)$of nitrogen followed by oxygen due to its smaller size than boron.
Thus half-filled and fully filled configurations are more stable. They have extra stability.

Hence the correct answer is option B.

Note:: The force of attraction between the nucleus and outermost electron increases with increase in nuclear charge. Greater is the nuclear charge, greater will be the energy required to pull the electron from the atom. Hence, with increase in nuclear charge ionization enthalpy increases. However, the effective nuclear charge experienced by a valence electrons in an atom will be less than the actual charge on the nucleus due to a phenomenon known as `shielding effect’ or `screening effect’ where the force of attraction by the nucleus on the valence electron is reduced due to the repulsive forces exerted by the inner shell electrons.