Write resonance forms that describe the distribution of...

Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) selenium dioxide, OSeO
(b) nitrate ion, NO3
(c) nitric acid, HNO3 (N is bonded to an OH group and two O atoms)
(d) benzene, C6H6:

Key Concepts

Formal Charge

Formal charge is a bookkeeping tool used to assign charges to individual atoms in a molecule by comparing the number of valence electrons in the isolated atom with the electrons assigned in a given Lewis structure. It helps in assessing the most reasonable resonance contributors by minimizing charge separation and distributing charges appropriately.

Resonance Structures

Resonance structures are various Lewis structures that represent the same molecule, differing only in the distribution of electrons rather than the arrangement of atoms. They help describe electron delocalization and provide insight into the contributing forms that stabilize the molecule through fractional bond orders.

Electron Delocalization

Electron delocalization refers to the phenomenon where electrons, particularly in pi systems or lone pairs adjacent to pi bonds, are spread over several atoms rather than localized between two. This delocalization leads to increased stability and uniform bond characteristics across the structure.

Stabilization Through Resonance

Stabilization through resonance arises when multiple resonance structures contribute to the overall structure of a molecule, lowering its energy. This concept explains why molecules with extensive delocalized electrons often display enhanced stability compared to structures with localized electrons.

Transcript

00:01 This question suggests that for each of these, there are multiple resonant structures.

00:08 It asks you to determine the lewis structure and all possible resonant structures for the molecule or the ion that is listed below.

00:20 So let's start with the first one, selenium dioxide.

00:27 It's going to contain one selenium and two oxygens, each with six valence electrons, giving us a total of 18 valence electrons.

00:36 We'll put selenium in the middle.

00:39 Oxygens on each side.

00:42 We just used up four valence electrons.

00:44 We have 14 left over.

00:46 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.

00:54 Because we've run out and selenium does not yet have an octet, then one of these oxygens needs to share one of its loan pairs and make it a bonding pair.

01:04 It doesn't matter which one for your first.

01:07 Structure because it could have been this oxygen or the other oxygen then there are two possible resonance structures these two shown here for the next one we have the nitrate ion and o3 minus nitrogen has five valence electrons oxygen has six there are three of them and then one additional electron because of the negative charge gives us a total of 24 valence electrons so we so we'll put nitrogen in the middle and bond the oxygens around it.

01:43 We just used up six valence electrons.

01:45 We have 18 left over.

01:48 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18.

01:57 We've used up all valence electrons and nitrogen does not have an octet.

02:01 So one of these oxygens needs to take one of its lone pairs and make it a bonding pair.

02:06 It doesn't matter which one when you start.

02:10 We'll say this one down here.

02:12 When you form that double bond, make sure that you remove a lone pair from one of the oxygens.

02:17 Because nitrate is an anion.

02:19 We'll put it in brackets and put its charge minus one on the outside...