To draw the Lewis structure for dinitrogen tetroxide (N2O4), we start by calculating the total number of valence electrons. Nitrogen has 5 valence electrons and oxygen has 6. Therefore, for two nitrogen atoms and four oxygen atoms, the total is:
- 2 x 5 (for N) = 10
- 4 x 6 (for O) = 24
- Total = 10 + 24 = 34 valence electrons
Next, we arrange the atoms. The nitrogen atoms will be the central atoms, with the oxygen atoms bonded to them. To satisfy the octet rule, we typically form double bonds between nitrogen and oxygen. A possible Lewis structure is:
O || N=N || O / O
This structure shows two nitrogen atoms double bonded to each other and each nitrogen atom bonded to two oxygen atoms. Each nitrogen atom has a total of 8 electrons in its valence shell, fulfilling the octet rule.
Now, let’s discuss the geometry of the central atom. The central nitrogen atoms in this structure have a linear arrangement due to the double bonds and the absence of lone pairs. Thus, the geometry is linear.
As for polarity, the overall molecule N2O4 is nonpolar. Although nitrogen and oxygen have different electronegativities, the symmetrical arrangement of the molecule results in the dipole moments cancelling each other out, leading to a nonpolar molecule.