question_answer 26)

Among the second period elements, the actual ionisation energies are in the order \[Li\text{ }

Answer:

(i) The electronic configurations of Be and B are respectively: \[Be\,\,\,\,1{{s}^{2}},2{{s}^{2}}\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,1{{s}^{2}},2{{s}^{2}}2{{p}^{1}}\] In Be, the electron is to be removed from 2s while in boron from \[2p\] energy shell. Since, \[2s\]-electron is more strongly attracted by nucleus than \[2p\]-electron, therefore, more energy is required for the removal of \[2s\]-electron than \[2p\]-electron. Consequently, \[{{\Delta }_{i}}{{H}_{1}}\] of Be is higher than \[{{\Delta }_{i}}{{H}_{1}}\]of B. (ii) The electronic configuration of nitrogen is \[1{{s}^{2}},2{{s}^{2}}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1},\] i.e., all the \[p\]-orbitals are singly occupied. It is thus a stable configuration in comparison to oxygen which is not so stable (unsymmetrical) \[O\,\,\,\,\,\,\,\,\,(1{{s}^{2}},2{{s}^{2}}2p_{x}^{2}2p_{y}^{1}2p_{z}^{1})\] As a result, the removal of electron from nitrogen requires more energy than the removal of electron from oxygen atom. Thus, \[{{\Delta }_{i}}{{H}_{1}}\] of N is higher than \[{{\Delta }_{i}}{{H}_{1}}\] of oxygen. In fluorine atom, the effective nuclear charge is higher than oxygen. Thus,\[{{\Delta }_{i}}{{H}_{1}}\]of F is higher than \[{{\Delta }_{i}}{{H}_{1}}\]of O.