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  • Which among the following electronic configuration represents an atom with the lowest ionization enthalpy?Option: 1 <img alt="\mathrm{1s^{2}2s^{2}2p^

Which among the following electronic configuration represents an atom with the lowest ionization enthalpy?

Option: 1

\mathrm{1s^{2}2s^{2}2p^{3}}


Option: 2

\mathrm{1s^{2}2s^{2}2p^{6}3s^{1}}


Option: 3

\mathrm{1s^{2}2s^{2}2p^{6}}


Option: 4

\mathrm{1s^{2} 2s^{2} 2p^{5}}


Answers (1)

best_answer

\mathrm{1s^{2} 2s^{2} 2p^{3} \: and \: 1s^{2} 2s^{2} 2p^{6}}is stable electronic configuration due to half filled and fully-
filled p-orbitals respectively. Therefore, removing an electron from these atoms is difficult and thus, higher ionization enthalphy.
However, when an electron is removed from the atom having \mathrm{1s^{2} 2s^{2} 2p^{6} 3s^{1}}
configuration, it attains the stable configuration of fully filled p-orbital. Therefore, it
readily loses an electron (lowest ionization enthalpy).

Hence Option 2 is correct.

Posted by

Ajit Kumar Dubey

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