Question

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why:
(i) Be has higher $\Delta_{i}$H than B
(ii) O has lower $\Delta_{i}$H than N and F

Answer 1

“Order of Ionisation” enthalpies as given,

“Li” < “B” < “Be” < “C” < “O” < “N” < “F” < “Ne”  

(i) Electronic configuration of Beryllium Be is $1{ s }^{ 2 },\quad 2{ s }^{ 2 }$

Electronic configuration of Boron B is $1{ s }^{ 2 },\quad 2{ s }^{ 2 },\quad 2{ p }^{ 1 }$

From the electronic configuration, we can see that the last electron of Be is in 2s while last electron of B is in 2p. It means “2s electron” in Beryllium is “strongly attracted” by the nucleus than “2p electron” in the Boron. Energy required to remove 2s electron is higher than removing the 2p electron.  

Therefore, the Ionisation enthalpy of Beryllium is higher than Boron.

(ii) Electronic configuration of Oxygen is $1{ s }^{ 2 },\quad 2{ s }^{ 2 },\quad 2{ p }^{ 4 }$

Electronic configuration of Nitrogen is $1{ s }^{ 2 },\quad 2{ s }^{ 2 },\quad 2{ p }^{ 3 }$

Electronic configuration of Flourine is $1{ s }^{ 2 },\quad 2{ s }^{ 2 },\quad 2{ p }^{ 5 }$

Due to decreasing atomic size ionisation enthalpy increases when we move toward right. Half-filled and fully filled orbitals have more stability that is why Nitrogen and Fluorine are more stable than Oxygen and it is difficult to remove an electron from Nitrogen and Fluorine than Oxygen. Hence, Oxygen has lower ionisation enthalpy than Nitrogen and Fluorine.