Dioxygen is paramagnetic inspite of having even number of electrons. Explain.

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> [!TLDR]
> An even electron count does not force full pairing: O$_2$'s two highest-energy electrons occupy two DEGENERATE antibonding $\pi^{*}$ orbitals one each (Hund's rule), so dioxygen carries two unpaired electrons and is paramagnetic.

**Step 1.** Dioxygen is a covalently bonded molecule; its Lewis-type depictions (the book's Solution prints the electron-dot and the double-bonded O$=$O pictures) show every electron paired and so cannot account for the observed paramagnetism.

**Step 2.** Molecular orbital theory resolves this. Filling O$_2$'s electrons into molecular orbitals gives the electronic configuration:

$$\mathrm{KK\ \sigma(2s)^2\ \sigma^{*}(2s)^2\ \sigma(2p_z)^2\ \pi(2p_x)^2\ \pi(2p_y)^2\ \pi^{*}(2p_x)^1\ \pi^{*}(2p_y)^1}$$ …

Q.Dioxygen is paramagnetic inspite of having even number of electrons. Explain.

Concept understanding — Dioxygen and Simple Oxides

Q.Dioxygen is paramagnetic inspite of having even number of electrons. Explain.

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