Question

The first ionization energy of sodium boron nitrogen and oxygen atoms follows the order

Answer 1

The ionization energy increases when we move from left to right in periods and decrease when we go down the group
so,
the order of ionization energies of Na, B, N, O
should be: -
*N>O>B>Na*

Answer 2

Answer: N>O>B>Na

Explanation:

The ionization energy is the energy required to remove an electron from the valence shell of an isolated gaseous atom of an element.

Now the ionization energy depends upon

a) size of an element : larger the size more lesser the effective nuclear charge on electron thus lower the ionization energy.

b) the electronic configuration of an element: If on losing an electron an element attains noble gas configuration or half filled stability then it will easily lose an electron.

$Na:11:1s^22s^22p^63s^1$

$B:5:1s^22s^22p^1$

$N:7:1s^22s^22p^3$

$O:8:1s^22s^22p^4$

As sodium is larger in size than boron , sodium has least ionization energy as on losing one electron , they acquire stable configuration.

Nitrogen has higher ionization energy than oxygen as nitrogen has stable half filled configuration.

Thus the order of first ionisation energy is N>O>B>Na