The complying with structure is one anion with 3 possibleresonance contributors. One incomplete resonance form is shownbelow. Finish the offered structure by adding nonbonding electronsand official charges. Attract the two remaining resonance frameworks (inany order), consisting of nonbonding electrons and formal charges.

You are watching: The following structure is an anion with three possible resonance contributors

To draw the resonance structure of the provided compound, delocalize the π\pi π electrons and also the lone pair the electron either in the direction of the electron deficient types or towards the an ext electronegative atom.

Formal fee on the atom deserve to be calculation by learning the valence electron, bonding electrons, and also nonbonding electrons current on the atom.

Resonance structure can be drawn by the delocalization the π\pi π electron and lone pair the electrons existing in a system. Resonance structures are presented by twin headed arrow. The delocalization that π\pi π electron and lone pair in a molecule lowers the resonance energy and also thus, the molecule is stabilized.

Consider the following instance for illustration the resonance structure by delocalization of lone pair and twin bond as presented by curved arrows:


The formal charge can be calculated by the following formula:

Formalcharge=V−N−B2\rmFormal\,\rmcharge = V - N - \fracB2Formalcharge=V−N−2B​

Here, V is valence electron in neutral atom; N is the variety of nonbonding electrons, and B the number of bonding electrons current in one atom.

See more: Colour Me Good Ryan Gosling Coloring Books, Ryan Gosling Legendary Coloring Book

Draw the complete resonance type of the offered structure by illustration the lone pair and adding the formal charge as presented below:


The formal charge present atoms space calculated as presented below:

FormalchargeonOatom=6−4−42=0\beginarrayc\\\rmFormal\,\rmcharge\,\rmon O atom = 6 - 4 - \frac42\\\\ = 0\\\endarrayFormalchargeonOatom=6−4−24​=0​

FormalchargeonSatom=6−6−22=−1\beginarrayc\\\rmFormal\,\rmcharge\,\rmon S atom = 6 - 6 - \frac22\\\\ = - 1\\\endarrayFormalchargeonSatom=6−6−22​=−1​

FormalchargeonNatom=5−0−82=5−1=+1\beginarrayc\\\rmFormal\,\rmcharge\,\rmon N atom = 5 - 0 - \frac82\\\\ = 5 - 1\\\\ = + 1\\\endarrayFormalchargeonNatom=5−0−28​=5−1=+1​

Formalchargeoncarbonatom=4−2−62=4−5=−1\beginarrayc\\\rmFormal\,\rmcharge\,\rmon carbon atom = 4 - 2 - \frac62\\\\ = 4 - 5\\\\ = - 1\\\endarrayFormalchargeoncarbonatom=4−2−26​=4−5=−1​

Therefore, the formal charge present on O is 000 , top top N is +1 + 1+1 , top top S is −1 - 1−1 , and on C is −1 - 1−1 .

Draw the resonance structure of the given compound by delocalization the lone pair and π\pi π electron as presented below:



Thus, the 2 remaining feasible resonance structures of the provided compound is displayed below: