• # 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.

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