Chapter 1, Problem 47P

Consider Lewis formulas A, B, and C:

${\text{H}}_{\text{2}}\text{<mover accent="true"> C <mo> ¨ </mo> </mover> -N}\equiv \text{N:}$

${\text{H}}_{\text{2}}\text{C=N= <mover accent="true"> N <mo> ¨ </mo> </mover> :}$

${\text{H}}_{\text{2}}\text{C- <mover accent="true"> N <mo> ¨ </mo> </mover> = <mover accent="true"> N <mo> ¨ </mo> </mover> :}$

A B C

Are A, B, and C constitutional isomers, or are they resonance contributors?

Which have a negatively charged carbon?

Which have a positively charged carbon?

Which have a positively charged nitrogen?

Which have a negatively charged nitrogen?

What is the net charge on each?

Which is a more stable structure, A or B? Why?

Which is a more stable structure, B or C? Why?

What is the CNN geometry in each according to VSEPR?

Interpretation Introduction

Interpretation:

The given statements for the following Lewis formulas are to be determined.

Concept introduction:

Isomers are classified as different compounds having the same molecular formula. Satisfying the octet rule and maximizing the number of bonds usually occur simultaneously.

A more electronegative atom favors a negative charge and a less electronegative atom favors a positive charge. The formal charge of an atom is given by:

$\text{Formal}\text{charge}=\text{Group}\text{number}-\text{Electron}\text{count}$

$\text{Electron}\text{count}\text{=}\frac{\text{1}}{\text{2}}\text{(}\text{number}\text{of}\text{bonded}\text{electrons)}\text{+}\text{(number}\text{of}\text{unshared}\text{electrons)}$

Answer to Problem 47P

Solution:

Lewis formulas $\text{A, B and C}$ are not constitutional isomers, but these are resonance contributors.

Lewis formula $\text{A}$ has negatively charged carbon.

Lewis formula $\text{C}$ has positively charged carbon.

Lewis formulas $\text{A and B}$ have positively charged nitrogen.

Lewis formulas $\text{B and C}$ have negatively charged nitrogen.

Net charge on all Lewis formulas $\text{A, B and C}$ is $0$.

Lewis formula $\text{B}$ is more stable than $\text{A}$.

Lewis formula $\text{C}$ is more stable than $\text{B}$.

The CNN geometry in Lewis formula $\text{A and B}$ is linear and in Lewis formula $\text{C}$ is angular.

Explanation of Solution

(a) Constitutional isomers or resonance contributors.

In order to be constitutional isomers, Lewis formulas must have the same molecular formula and different connectivity. Given Lewis formulas $\text{A,B and C}$ have the same molecular formula and same connectivity of atoms.

Resonance contributors differ only in the distribution of electrons. Given Lewis formulas $\text{A,B and C}$ differ only in the distribution of electrons. Thus, they are not constitutional isomers but are resonance contributors.

Delocalization of electrons is shown using curved arrows as follows:

In Lewis formula $\text{A}$, curved arrow begins at an unshared electron on the carbon and becomes the second half of the double bond on nitrogen. This gives rise to structure $\text{B}$.

In Lewis formula $\text{B}$, curved arrow begins at a double bond and becomes an unshared pair on the other nitrogen. This gives rise to structure $\text{C}$. Hence, these structures are resonance contributors.

(b) Negatively charged carbon.

The formal charge on each carbon atom in all the three given Lewis formulas is as follows:

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+2=5$	$-1$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(8)+0=4$	$0$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+0=3$	$+1$

Lewis formulas with formal charges are

Hence, Lewis formula $\text{A}$ has negatively charged carbon atom.

(c) Positively charged carbon.

The formal charge on each carbon atom in all the three given Lewis formulas is as follows:

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+2=5$	$-1$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(8)+0=4$	$0$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+0=3$	$+1$

Lewis formulas with formal charges are

Hence, Lewis formula $\text{C}$ has positively charged carbon atom.

(d) Positively charged nitrogen.

The formal charge on each nitrogen atom in all the three given Lewis formulas is as follows:

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(6)+2=5$	$0$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(6)+2=5$	$0$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Lewis formulas with formal charges are

Hence, Lewis formulas $\text{A and B}$ have positively charged nitrogen atoms.

(e) Negatively charged nitrogen.

The formal charge on each nitrogen atom in all the three given Lewis formulas is as follows:

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(6)+2=5$	$0$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Nitrogen(a)	$5$	$\frac{1}{2}(6)+2=5$	$0$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Lewis formulas with formal charges are

Hence, Lewis formulas $\text{B and C}$ have negatively charged nitrogen.

(f) Net charge on each.

The formal charge on each atom for all the three given Lewis formulas is as follows:

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+2=5$	$-1$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(6)+2=5$	$0$

Lewis formula $A$ with formal charge is

There is one negative charge on carbon and one positive charge on the adjacent nitrogen; hence, net charge on the Lewis formula $\text{A}$ is $0$.

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(8)+0=4$	$0$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Lewis formula $\text{B}$ with formal charge is

There is one negative charge on the nitrogen atom in the middle and one positive charge on the adjacent nitrogen; hence, net charge on the Lewis formula $\text{B}$ is $0$.

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+0=3$	$+1$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(6)+2=5$	$0$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Lewis formula $\text{C}$ with formal charge is

There is one positive charge on carbon and one positive charge on one nitrogen atom; hence, net charge on the Lewis formula $\text{C}$ is $0$.

(g) The more stable structures.

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+2=5$	$-1$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(6)+2=5$	$0$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(8)+0=4$	$0$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

The Lewis formulas A and B with formal charges are

The most stable structure is the contributing structure with the greater number of covalent bonds, which contributes more to the resonance hybrid, as long as the octet rule is not exceeded for second-row elements. The major contributor is the one with the smallest separation of oppositely charged atoms, when two or more structures satisfy the octet rule. The major contributor is the one in which the negative charge resides on the most electronegative atom and the positive charge on the least electronegative element, among the structural formulas that satisfy the octet rule and in which one or more atoms bear a formal charge.

In given Lewis formulas $\text{A and B}$, both have the same number of covalent bonds. Further, in both the structures, the charge separation is equal. At this point, it is important to determine the formal charges on each element in the given structures.

Out of the given three Lewis formulas, in Lewis formula $\text{A}$, negative charge is on the carbon atom whereas in Lewis formula $\text{B}$, negative charge is on the nitrogen atom. The nitrogen is more electronegative than carbon and can better support a negative charge.

Hence, Lewis formula $\text{B}$ is more stable than $\text{A}$.

(h) The more stable structures.

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(8)+0=4$	$0$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(8)+0=4$	$+1$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Atoms	Valence electrons of neutral atom	Electron count	Formal charge
Carbon	$4$	$\frac{1}{2}(6)+0=3$	$+1$
Hydrogen	$1$	$\frac{1}{2}(2)+0=1$	$0$
Nitrogen(a)	$5$	$\frac{1}{2}(6)+2=5$	$0$
Nitrogen(b)	$5$	$\frac{1}{2}(4)+4=6$	$-1$

Lewis formulas $\text{B and C}$ with formal charges are

The most stable structure is the contributing structure with the greater number of covalent bonds, which contributes more to the resonance hybrid, as long as the octet rule is not exceeded for second-row elements. The major contributor is the one with the smallest separation of oppositely charged atoms, when two or more structures satisfy the octet rule. The major contributor is the one in which the negative charge resides on the most electronegative atom and the positive charge on the least electronegative element, among the structural formulas that satisfy the octet rule and in which one or more atoms bear a formal charge.

In Lewis formula $\text{B}$, the positive charge is on the carbon atom whereas in Lewis formula $\text{C}$, the positive charge is on the nitrogen atom. The carbon is less electronegative than nitrogen and can better support a positive charge.

Hence, Lewis formula $\text{C}$ is more stable than $\text{B}$.

(i) CNN geometry.

In Lewis formula $\text{A}$, the central nitrogen atom has two bonded pairs and no unshared pair of electrons. Two bonding pairs for the middle nitrogen atom allows a linear geometry. Hence the CNN geometry for the given Lewis formula $\text{A}$ is linear.

In Lewis formula $\text{B}$, the central nitrogen atom has two bonded pairs and no unshared pair of electrons. Two bonding pairs for the middle nitrogen atom allow a linear geometry. Hence the CNN geometry for the given Lewis formula $\text{B}$ is linear.

In Lewis formula $\text{C}$, the central nitrogen atom has two bonded pairs and one unshared pair of electrons. Thus, there are three bonded pairs for the nitrogen atom in the center. Two bonding pairs and one unshared pair of electron allow the bent or V shaped geometry.