Question

NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including formal charges.

Answer 1

The formal charge of N₂O(the resulting compound) is 0, -1 and +1

Further explanation

Formal charges are usually used to explain the bonding of the Lewis structure. The structural form of a molecule can have several types based on the Lewis structure. Formal charge calculations can determine which molecule is more stable, by choosing the smallest charge, even though atoms such as C, N, and O the priority is to fulfill the octet rule first.

Formal charges can be formulated:

FC (formal charge) = number of valence electrons - number of free electrons - 1/2 binding electrons

In the reaction, NO₂ and NO will produce N₂O compounds, nitrous oxide

Reactions that occur:

NO₂ + NO ---> N₂O + O₂

N₂O based on its lewis structure can have 3 possible shapes (single, triple or triple bond) but the most stable and suitable one is the three bond between its N atoms

The formal charge of N₂O is:

• N₁: 5 (valence electrons of N atoms) - 2 (free electrons, which are not bound) -1/2.6 (6 is the number of bound electrons)  so the formal charge N₁ = 0

• N₂ = 5-0- 1/2.8 , N₂ = +1

• O = 6 - 6-1/2.2 , O = -1

So that the total formal charge = 0 + (+ 1) + (- 1) = 0, according to the charge on neutral N₂O compounds

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Keywords: formal charge, N₂O, the valence electron, bond, the octet rule

Answer 2

The formal charge on N-1 is $\boxed0$, that on N-2 is $\boxed{ + 1}$ and that on O is $\boxed{ - 1}$ . The Lewis structure of ${{\mathbf{N}}_{\mathbf{2}}}{\mathbf{O}}$  is attached in the image.

Further Explanation:

The bonding between the different atoms in covalent molecules is shown by some diagrams known as the Lewis structures. These also show the presence of lone pairs in the molecule. These are also known as Lewis dot diagrams, electron dot diagrams, Lewis dot structures or Lewis dot formula. In covalent compounds, the geometry, polarity, and reactivity are predicted by these structures.

The reaction occurs as follows:

${\text{N}}{{\text{O}}_2}+{\text{NO}}\to{{\text{N}}_2}{\text{O}}+{{\text{O}}_2}$

Here ${{\text{N}}_2}{\text{O}}$ is the major product formed.

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)]

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

Formal charge:

It is the charge that an atom acquires in a molecule by considering that the chemical bonds are shared equally between the two atoms, irrespective of their electronegativities.

The formula to calculate the formal charge in ${{\text{N}}_{\text{2}}}{\text{O}}$  is as follows:

${\mathbf{Formal\:charge&=}}\left[\begin{aligned}\left[\begin{aligned}{\mathbf{total\:number\:of\:valence\:electrons }}\hfill\\{\mathbf{in\:the\:free\:atom}}\hfill\\\end{aligned}\right]{\mathbf{ }}-\\\left[{{\mathbf{total\:number\:of\:non-bonding\:electrons}}}\right]-\\\frac{{\left[ {{\mathbf{total\:number\:of\:bonding\:electrons}}}\right]}}{{\mathbf{2}}}\\\end{aligned}\right]$           .......(1)

For N-1:

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

The total number of non-bonding electrons in N-1 is 2.

The total number of bonding electrons in N-1 is 6.

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

$\begin{aligned}{\text{Formal charge on N-1}}&=\left[{5-2-\frac{6}{2}}\right] \\&=0\\\end{gathered}$

For N-2:

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

The total number of non-bonding electrons in N-2 is 0.

The total number of bonding electrons in N-2 is 8.

Substitute these values in equation (1) to find the formal charge on N-2.

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

For O:

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

The total number of non-bonding electrons in oxygen is 6.

The total number of bonding electrons in oxygen is 2.

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

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

Nitrogen atom has 5 valence electrons, oxygen atom has 6 valence electrons. Therefore, the total numbers of valence electrons in ${{\text{N}}_{\text{2}}}{\text{O}}$ are 16. In the Lewis structure of ${{\text{N}}_{\text{2}}}{\text{O}}$ , N-1 forms one triple bond with N-2 and has two non-bonding electrons on it. N-2 forms one triple bond with N-1 and one single bond with O and has no non-bonding electrons. O forms one single bond with N-2 and has six non-bonding electrons on it. Therefore, Lewis structure of ${{\mathbf{N}}_{\mathbf{2}}}{\mathbf{O}}$  is attached in the image.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, valence electrons, N2O, formal charge, N-1, N-2, O, 0, +1, -1, bonding electrons, non-bonding electrons, total valence electrons.