Chapter 4, Problem 4.113QA

Interpretation Introduction

To find:

a) Draw Lewis structure for possible arrangement of $N-O-N$ in $N_{2}O.$

b) Assign formal charges of each structure.

c) Give reason why structure is not stable.

Answer to Problem 4.113QA

Solution:

a) i)  ii)

b) i) Formal charge on each N is -1 and on O is +2.

ii)  Formal charge on one of N is -2 and on O is +2.

c) In both structure the formal charge on oxygen is +2. Since oxygen is more electronegative than Nitrogen none of these structure is stable.

Explanation of Solution

1) Concept:

Each molecule contains two Nitrogen and one oxygen has an overall charge zero. Nitrogen and oxygen are in group 15 and 16 and have bonding capacity three and two. Nitrogen can show hypervalency.

2) Calculations:

The number of valence electrons are,

Elements		Valence electrons
Symbol	Number of atom	In one atom	Total
N	2	5	10
O	1	6	6
Valence electrons in molecule			16

First possible arrangement of $N-O-N$  is,

Second possible arrangement of $N-O-N$  is,

Formula:

To find formal charge we need to use following equation,

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}=\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{ }\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{s}-\mathrm{ }\left[\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{p}\mathrm{a}\mathrm{i}\mathrm{r}\mathrm{s}+\frac{1}{2}\mathrm{ }\left(\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{s}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{s}\right)\right]$

Calculation:

Step 1) Calculations of formal charge on atoms in first structure,

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{N}=5-\mathrm{ }\left[4+\frac{1}{2}\mathrm{ }\left(4\right)\right]= -1$

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{O}=6-\mathrm{ }\left[0+\frac{1}{2}\mathrm{ }\left(8\right)\right]= +2$

Step 2) Calculations of formal charge on atoms in second structure,

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{N}=5-\mathrm{ }\left[6+\frac{1}{2}\mathrm{ }\left(2\right)\right]= -2$

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{O}=6-\mathrm{ }\left[0+\frac{1}{2}\mathrm{ }\left(8\right)\right]= +2$

$\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{N}=5-\mathrm{ }\left[2+\frac{1}{2}\mathrm{ }\left(6\right)\right]= 0$

Oxygen has electronegativity 3.5 and nitrogen has electronegativity 3.0. Thus, structures with formal negative charges on nitrogen and formal positive charges on oxygen are not preferred. Therefore none of these structures is stable.

Conclusion:

a) Lewis structure for possible arrangement of $N-O-N$ in $N_{2}O$ are,

i)  ii)

b) i) Formal charge on each N is -1 and on O is +2.

ii)  Formal charge on one of N is -2 and on O is 2+

c) Oxygen is more electronegative than Nitrogen and each of structure contains +2 charge on oxygen so none of these structure is likely to be stable.