Chapter 4.3, Problem 4.18P

Interpretation Introduction

Interpretation:

The geometry of nitrogen atom and oxygen atom has to be predicted using VSEPR theory.

Concept Introduction:

Geometry of a compound is how the atoms that are present in the compound is connected in a 3D space.  In order to find the geometry of the compound, the first thing that must be known was how the atoms bond with each other.  Each and every atom in a compound is connected through bonds and the bonds are formed by overlapping orbitals.

Simple atomic orbitals are s and p orbitals.  Generally in organic chemistry we don’t consider or deal with d and f orbitals.  The second row elements ($\text{C,}\text{N,}\text{O}\text{and}\text{F}$) have a single s orbital and three p orbitals in their valence shell.  The s orbital mix with three p orbitals to form three types of hybridized orbital, and they are $s{p}^3,s{p}^2\text{and}sp$.  These hybridized orbital are involved in forming bonds with other atoms.

To find the hybridization state of atom we start by counting the number of bonds the atom has and the lone pair of electrons present.  The below table can be used to find the hybridization of atom,

Sum of bonded atoms and lone pairs	Number of hybridized orbitals
4	4 $s{p}^3$ orbitals
3	3 $s{p}^2$ orbitals and one p orbital
2	2 $sp$ orbitals and two p orbitals

From the hybridization state of the atoms we can identify the geometry of the same.  For finding the geometry, the lone pair of electrons also has to be considered.  This complete details are given in the following table,

Hybridization along with lone pair	Geometry
$s{p}^3$ with 0 lone pair of electrons	Tetrahedral
$s{p}^3$ with 1 lone pair of electrons	Trigonal pyramidal
$s{p}^3$ with 2 lone pair of electrons	Bent
$s{p}^2$ with 0 lone pair of electrons	Trigonal planar
$s{p}^2$ with 1 lone pair of electrons	Bent
$sp$ with 0 lone pair of electrons	Linear

Lone pair of electrons occupy the hybridized orbitals.  If the same lone pair of electrons is involved in resonance means, then it does not occupy the hybridized orbital.  Hence, the hybridization state of the atom changes which in turn changes the geometry of the atom.

Answer to Problem 4.18P

Nitrogen atom has trigonal planar geometry and oxygen atom has bent geometry.

Explanation of Solution

Given structure is,

Redrawing the above structure with the lone pair of electrons,

By looking at the structure given, we find that the nitrogen atom has three bonds and a lone pair of electrons.  The nitrogen atom is considered to have $s{p}^3$ hybridization.  But it is not the case.  This is because, the lone pair of electrons present on nitrogen atom involves in resonance.  Therefore, the hybridization of the nitrogen atom is $s{p}^2$ as the lone pair of electrons occupies the p orbital only and not the hybridized orbital.  The geometry of the nitrogen atom is trigonal planar instead of trigonal pyramidal as lone pair of electrons are involved in resonance.

The oxygen atom has two bonds and two lone pair of electrons.  Therefore, the hybridization of oxygen atom is $s{p}^3$.  The geometry of oxygen atom is bent as it has two bonds and two lone pair of electrons.

Conclusion

Geometry of the nitrogen and oxygen atom in the given structure was identified.