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Priyanshu kumar
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Option C is correct Xe>As>Be>Al Left to right in a period ionisation potential increases and top to bottom in a group IP decreases. Beryllium has a stable complete electronic configuration (1s22s2) so it, require more energy to remove the first electron from it.
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we have (i) 5th Period xenon, an inert gas; (ii) 3rd Period aluminum; (iii) 2nd Period beryllium; and 4th Period arsenic. Because we compare (i) different valence shells, and (ii) different electronic configurations, it is BY NO MEANS straightforward to predict the relative order of ionization energies.
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Dinesh khalmaniya 1 Best Answer
option C Explanation- Now ionization energies are defined by the following rxn: Atom(g)+Δ→Cation(g)+e− Now we know the ionization energies INCREASE across a Period from left to right as we face the Table, but DECREASE down a Group, a column of the Periodic Table. And simple ideas of electrostatics can rationalize this trend. Incomplete electronic shells shield the nuclear charge very imperfectly. And across the Period, ionization energies should increase. This also manifests in the well known decrease in atomic radii from left to right as we face the Table. As a valence shell is completed, shielding of the nuclear charge becomes more effective, and the electrons occupy a new valence shell to begin the process again. So we have (i) 5th Period xenon, an inert gas; (ii) 3rd Period aluminum; (iii) 2nd Period beryllium; and 4th Period arsenic. Because we compare (i) different valence shells, and (ii) different electronic configurations, it is BY NO MEANS straightforward to predict the relative order of ionization energies. At a push we might predict the following order of DECREASING ionization energies: