Question

Many free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical-radical combination reaction is the formation of a new electron-pair bond. Write Lewis formulas for the reactant and product species in the following chemical equation. Include nonbonding electrons.
N(g)+NO(g) = NNO(g)

Answer 1

The Lewis formula is structural representation that indicates the bonding between atom of a molecule by line and dots used to indicates the electron position around the atoms, electron pair is known as lone pair of electrons shown by pair of dots. The electrons indicated in the structure are valence electrons.

The given atoms are $N (g)$ and $O (g)$.

$N (g)$, whose atomic number is 7.

Electronic configuration for $N (g)$ is $2, 5$.

Number of valence shell electrons for $N (g)$ is 5.

$O (g)$, whose atomic number is 8.

Electronic configuration for $O (g)$ is $2, 6$.

Number of valence shell electrons for $O (g)$ is 6.

The Lewis formula for $N (g)$  and $O (g)$ is shown in the image.

The atoms combine to form molecule in order to complete their octet that is to get 8 electrons in their valence shell and get stabilize in nature.

The total number of valence shell electrons in $NNO(g)$ is:

$2(5)+6 = 16 electrons$

The distribution of electrons in the atom will take place in such a way that formation of triple bond will take place between two nitrogen atoms and a single bond will form between nitrogen and oxygen atom. In order to complete their octet, the nitrogen atom in center will possess 1+ formal charge and oxygen will possess 1- charge (oxygen is  electronegative atom). Thus, results in formation of neutral molecule.

The formal charge on each atom is calculated using:

$Valence electrons - Nonbonding electrons - \frac{Bonding electrons}{2}$

The Lewis formula of $NNO(g)$ is shown in the image.

Answer 2

The Lewis structure corresponding to the reactants and products in the reaction is attached in the image.

Further Explanation:

The free radicals are high energy chemical species that must stabilize themselves either via electron combination with another free radical or abstraction of proton from another radical.

The Lewis structure is the chemical representation of an element along with the nonbonding pairs. For covalent molecules, the number of electrons involved in bonding and the remaining nonbonding pairs can be represented while writing the Lewis structures. Lewis structures along with the formal charges that they carry help predict the geometry, polarity, and reactivity of the molecules.

The nitrogen radical itself is unstable, and nitrogen monoxide also has an unpaired electron. So these two reactants can stabilize themselves by combination of their unpaired electron and form dinitrogen oxide molecule.

Lewis structure of ${{\mathbf{N}}_{\mathbf{2}}}{\mathbf{O}}$:

The total number of valence electrons of ${{\text{N}}_{\text{2}}}{\text{O}}$  is calculated as,

Total valence electrons = [(2) (Valence electrons of N) + (1) (Valence electrons of O)]

$\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}}\right)&=\left[ {\left(2\right)\left(5\right)+\left({\text{1}}\right)\left(6\right)}\right]\\ &=16\\ \end{aligned}$

Formal charge:

It is the charge that an atom acquires in a molecule assuming that the electron pairs that constitute the bond pairs are shared equally between the two atoms, irrespective of their electronegativities.

The formula to calculate the formal charge on an atom is as follows:

${\text{Formal charge}}&=\left[\begin{aligned}\left[\begin{gathered}{\text{total number of valence electrons}}\hfill\\{\text{in the free atom}}\hfill\\\end{gathered}\right]-{\text{ }}\\\left[{{\text{total number of non - bonding electrons}}}\right]-\\\frac{{\left[{{\text{total number of bonding electrons}}}\right]}}{{\text{2}}}\\\end{aligned}\right]$

O forms one single bond with a nitrogen atom and 3 lone pairs are present on it.

Total number of valence electrons in the free oxygen atom is 6.

Total number of nonbonding electrons in O is 6.

Total number of bonding electrons in O is 2.

Substitute these values in equation (1) to find the formal charge on O.

$\begin{aligned}{\text{Formal charge on O}}&=\left[{6 - 6 - \frac{2}{2}} \right]\\&=- 1\\\end{aligned}$

Total number of valence electrons in the free nitrogen atom is 5.

Total number of nonbonding electrons in N is 0.

Total number of bonding electrons in N is 6.

Substitute these values in equation (1) to find the formal charge on N.

$\begin{aligned}{\text{Formal charge on N}}&=\left[{5 - 0 - \frac{8}{2}}\right]\\&=5 - 4 \\&=+1\\\end{aligned}$

The formal charge of nitrogen is  +1.

Nitrogen atom has 5 valence electrons, and oxygen atom has 6 valence electrons. The central nitrogen atom forms a triple bond with other nitrogen atom and a single bond with oxygen atom. The central nitrogen atom, therefore, acquire positive charge and oxygen atom will acquire negative charge. Therefore the Lewis structure of ${{\mathbf{N}}_{\mathbf{2}}}{\mathbf{O}}$  is attached in the image.

Learn more:

1. Molecular shape around the central atoms in the amino acid glycine: brainly.com/question/4341225

2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, valence electrons, NO, formal charge, N2O, oxygen, double bonds, single bond, bonding electrons, non-bonding electrons, total valence electrons and resonance hybrid.