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12

12. Arrange the following in the increasing order : a. B , C , N , O : increasing ionization enthalpy b. F , Li+ , Na+ , Cl. : increasing order of size c. B , C , N , O : increasing negative electron gain enthalpy

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Solution

Dear Student,

(i) Ionization energy is the energy required to remove an electron from the outer most shell or orbit of an isolated gaseous atom.
Those which gains more stable after removal of an electron will have lesser ionization energy, where those which loses its stability and moves to the less stable state after removal of electron will have higher ionization energy.
Now, in the case of B having configuration 1s2,2s2,2p1 so on removing one electron from p-orbital it gains more stable fulfilled s-orbital, hence the I.E is low.
On moving across the period, as the size decreases the ionization energy increases, but differing from the trend nitrogen has slight higher ionization energy than oxygen. This is due to nitrogen has configuration ​1s2,2s2,2p3 and on the removal of an electron from its outermost shell it losses its half filled stability, so have higher ionization energy than oxygen.
So the order of I.E is B < C < O < N.

(ii) In the case of F, Li +, Na+, Cl- as the anion size is larger than that of parent atom Cl- will have the largest size among these four, and the smallest one is Li+ ion as being cation always be smaller than its parent atom . So the order of atomic size is Li+ < F < Na+ < Cl-.

(iii) As going across a period negative electron gain enthalpy or the electron affinity increases. But due to extra stability of nitrogen it is difficult to add one more electron to outermost shell, hence instead of release of energy, nitrogen has positive electron gain enthalpy.
So the order of negative electron gain enthalpy is N < B < C < O.

Regards.

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