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(i) Duringthe process of ionization, the electron to be removed from beryllium atom is a 2s-electron,whereas the electron to be removed from boron atom is a 2p-electron. Now, 2s-electronsare more strongly attached to the nucleus than 2p-electrons. Therefore, moreenergy is required to remove a 2s-electronof beryllium than that required to remove a 2p-electron of boron. Hence,beryllium has higher ΔiH thanboron.
(ii) In nitrogen, thethree 2p-electronsof nitrogen occupy three different atomic orbitals. However, in oxygen, two ofthe four 2p-electronsof oxygen occupy the same 2p-orbital.This results in increased electron-electron repulsion in oxygen atom. As aresult, the energy required to remove the fourth 2p-electron from oxygen is less ascompared to the energy required to remove one of the three 2p-electronsfrom nitrogen. Hence, oxygen has lower ΔiH than nitrogen.
Fluorine contains one electron and one proton more thanoxygen. As the electron is being added to the same shell, the increase innuclear attraction (due to the addition of a proton) is more than the increasein electronic repulsion (due to the addition of an electron). Therefore, thevalence electrons in fluorine atom experience a more effective nuclear chargethan that experienced by the electrons present in oxygen. As a result, moreenergy is required to remove an electron from fluorine atom than that requiredto remove an electron from oxygen atom. Hence, oxygen has lower ΔiH thanfluorine.