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Exercise · Q11

Q.Explain why the first ionization enthalpy of oxygen (1314 kJ mol−11314\ kJ\,mol^{-1}) is lower than that of nitrogen (1402 kJ mol−11402\ kJ\,mol^{-1}), even though oxygen lies to the right of nitrogen in Period 2.

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Step 1. Nitrogen's configuration is 1s2 2s2 2p31s^2\,2s^2\,2p^3 — one electron in each of the three 2p2p orbitals, a symmetric, extra-stable half-filled arrangement.

Step 2. Oxygen's configuration is 1s2 2s2 2p41s^2\,2s^2\,2p^4 — its fourth 2p2p electron must pair up with one already present in a 2p2p orbital.

Step 3. This forced pairing creates extra electron-electron repulsion within that doubly occupied orbital, making that paired electron easier to remove than nitrogen's unpaired ones despite oxygen's higher nuclear charge. …

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