Q.No. 16 The correct order of first ionization energies of He, N and O in their ground state is
He has the highest first ionization energy among He, N, and O in their ground states, followed by N then O. The correct order is He > N > O, corresponding to option (D). This follows periodic trends where noble gases like He have fully filled shells, and N benefits from half-filled p-orbitals compared to O.
Ionization Energy Values
First ionization energies are: He at 2372 kJ/mol, N at 1402 kJ/mol, and O at 1314 kJ/mol. These values confirm He > N > O, as He requires the most energy to remove an electron from its stable 1s² configuration. Nitrogen’s value exceeds oxygen’s due to electronic stability differences.
Ground State Configurations
Helium (atomic number 2) has 1s², a complete shell with high stability. Nitrogen (7) is 1s² 2s² 2p³, with half-filled 2p orbitals providing extra stability from lower electron repulsion and higher exchange energy. Oxygen (8) is 1s² 2s² 2p⁴, where a paired electron in 2p increases repulsion, easing removal.
Option Analysis
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(A) He > O > N: Incorrect, as N (1402 kJ/mol) > O (1314 kJ/mol); half-filled stability makes N harder to ionize.
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(B) O > N > He: Wrong; O < N, and both much less than He.
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(C) N > O > He: False; He (2372 kJ/mol) far exceeds N and O.
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(D) N > He > O: Correct? No, He > N > O matches data and trends.
The first ionization energies of He, N, and O in ground state follow a specific order due to atomic structure and periodic trends. Helium tops the list with the highest value, while nitrogen edges out oxygen thanks to its stable half-filled orbitals. This MCQ is common in exams like CSIR NET Life Sciences, testing periodic properties.
Why He Has Highest Ionization Energy
Helium’s 1s² configuration is fully filled, requiring 2372 kJ/mol to remove an electron from a small, stable atom with high effective nuclear charge. No other period 2 element matches this noble gas stability.
N > O: Half-Filled Stability
Nitrogen (2p³) has half-filled p subshell, maximizing exchange energy and minimizing repulsion, at 1402 kJ/mol. Oxygen’s 2p⁴ pairs electrons, raising repulsion for easier ionization at 1313 kJ/mol.
Exam Relevance
This trend highlights exceptions to increasing ionization energy across periods, vital for competitive exams. Option (D) N > He > O is incorrect; true order is He > N > O.
