Question

Question:(a) Use the VSEPR theory to predict the structure of the$\mathbf{N}\mathbf{}\mathbf{NO}$molecule.
(b) The substance $\mathbf{NNO}$has a small dipole moment. Which end of the molecule is more likely to be the positive end, based only on electronegativity?

Short answer

1. 
   

1. All atoms on the left have complete octets, indicating delocalization; on the right, the nitrogen ligand has an incomplete octet. Thus, linear geometry is a viable option.
2. Nitrogen$\text{E N = 3.0}$. So, nitrogen is more likely to be on the positive end based on this.

Step-by-step solution

Step 1: Molecular geometry concept

This revolves around the molecule of nitrous oxide. Lewis structures and the VSEPR theory are required to predict molecular geometry. The periodic table lists the electronegativity values.

Step 2: Molecular geometry

a.)

$N$atoms have five valence electrons, and the $O$atom itself has six. The nitrogen center geometry can be either linear $\text{(S N = 2)}$or trigonal planar$\text{(S N =3)}$, in which case the molecule will bend due to the lone pair.Below are Lewis’s structures for both scenarios. All atoms on the left have complete octets, indicating delocalization; on the right, the nitrogen ligand has an incomplete octet. As a result, linear geometry is a viable option.

Step 3: Molecule polarity.

b)

The values of electronegativity molecules are Nitrogen $\text{E N = 3.0}$and oxygen$\text{E N = 3.5}$ . The nitrogen is more likely to be on the positive end based on this.