Question #5f272

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Question #5f272

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Answer 1 · 2015-03-17T09:36:04

Nitrogen dioxide, or $N O_{2}$ $NO_2$ , is an odd-electron molecule, i.e. the number of valence electrons the molecule has is not an even number.

More specifically, nitrogen dioxide has a total of 17 valence electrons - 5 from nitrogen and 6 from each of the two oxygen atoms that comprise the molecule.

Now, VSEPR Theory can be extended to accomodate odd-electron molecules. In this case, the nitrogen dioxide molecule is considered to have a $A X_{2} E_{0.5}$ $AX_2E_0.5$ geometry, which is right between $A X_{2} E_{0}$ $AX_2E_0$ - linear, and $A X_{2} E_{1}$ $AX_2E_1$ - bent.

The molecule's geometry will still be closer to bent, but the bond angle will be larger than the ideal $120^{\circ}$ $120^@$ such a geometry predicts, mostly because the oxygen atoms experience less repulsion from the single electron present on the nitrogen atom.

The actual bond angle for nitrogen dioxide is $134^{\circ}$ $134^@$ . By comparison, the angle found in the nitrite ion, or $N O_{2}^{-}$ $NO_2^(-)$ , is ${115.4}^{\circ}$ $115.4^@$ , because now nitrogen has attached a full lone pair of electrons that experience greater repulsion with the oxygen atoms.

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Question #5f272

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