The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in : \(\mathrm{N} \equiv \mathrm{N}-\) O: be described in terms of orbitals?

We are given that N2O is linear and polar. We have two possibilities: NNO or NON. To determine which arrangement is correct, we need to consider electronegativity. The more electronegative atom will draw electron density towards itself and create a polar bond. The electronegativity values for nitrogen (N) and oxygen (O) are approximately 3.04 and 3.44, respectively. Since oxygen is more electronegative than nitrogen, the correct arrangement should be NNO, as putting O in the middle would lead to a nonpolar molecule.

The valence electrons for the N2O molecule are: nitrogen (5) + nitrogen (5) + oxygen (6) = 16 electrons. The correct arrangement is NNO, so we want to draw a Lewis structure with nitrogen at the center. The possible resonance structures are: (i) :N≡N(+)-O(-): Formal charges: N (0), N(+1), and O(-1) (ii) :N(-)-N(+)=O: Formal charges: N(-1), N(+1), and O(0) In both resonance structures, the central nitrogen has a formal charge of +1. The hybridization of the central nitrogen is sp as it forms two sigma bonds and no lone pairs.

The multiple bonding in N2O, specifically the nitrogen-nitrogen triple bond in the resonance structure (i), can be described in terms of orbitals as follows: The two outer orbitals overlap to form a strong sigma bond (\(σ\)), which is a single bond. Then, the two inner orbitals, one from each nitrogen atom, form two pi (\(π\)) bonds which are parallel to each other. These \(π\) bonds result from the sideways overlap of the \(p_y\) and \(p_z\) atomic orbitals. The overall triple bond between the two nitrogen atoms consists of one sigma bond and two pi bonds, written as N≡N. In conclusion, the correct arrangement for the N2O molecule is NNO, and its Lewis structures show resonance between a nitrogen-nitrogen triple bond and a nitrogen-nitrogen single bond with two lone pairs on the central nitrogen. The hybridization of the central nitrogen atom is sp, and the multiple bonding in the molecule is formed by one sigma bond and two pi bonds.