Answer the following questions about the elements with the electronic configurations below:
A =[Ar]3p^6 4s^2
B =[Ar] 4s^2 3d^10 4p^5
(1) Which element has higher ionization enthalpy?
(2) Which element should be smaller of the two?
Answers
Answer:
Group number 2 13 15 16
Element 4Be 5B 7N 8O
electronic configuration 2s2 2s22p1 2s22p3 2s22p4
Solution:
* Boron, B is smaller than beryllium, Be atom. Hence we expect increase in ionization energy from Be to B.
However, Be atom has greater ionization energy than B atom. The reason is - in case of Beryllium, the last electron is in the s-orbital and in Boron, the last electron is in the p-orbital. We know that removal of electron from s-orbital requires more energy than the electron from p-orbital.
* 2s22p3 configuration is more stable than 2s22p4 due to half filled p-sublevel. The electrons in the p-orbital in N atom experience less repulsion, thus more stable and more ionization energy. Whereas in case of O atom the 4th p-electron atom can be removed more easily as it experiences more repulsion from the electrons in the p-orbitals already present. Hence nitrogen, Oxygen atom has less ionization energy than Nitrogen atom.