Chemistry, 24.10.2021 14:00, fri90

What is the resonance structure of ch3

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For the CH3- structure use the periodic table to find the total number of valence electrons for the CH3- molecule. Once we know how many valence electrons there are in CH3- we can distribute them around the central atom with the goal of filling the outer shells of each atom.

In the Lewis structure of CH3- structure there are a total of 14 valence electrons.

Also note that you should put the CH3- Lewis structure in brackets with as 1- on the outside to show that it is an ion with a negative one charge.

Steps to Write Lewis Structure for compounds like CH3-

1. Find the total valence electrons for the CH3- molecule.

2. Put the least electronegative atom in the center. Note: Hydrogen (H) always goes outside.

3. Put two electrons between atoms to form a chemical bond.

4. Complete octets on outside atoms.

5. If central atom does not have an octet, move electrons from outer atoms to form double or triple bonds.

Lewis Structures, also called Electron Dot Structures, are important to learn because they help us understand how atoms and electrons are arranged in a molecule. This can help us determine the molecular geometry, how the molecule might react with other molecules, and some of the physical properties of the molecule (like boiling point and surface tension).

Explanation:

CH3CNO can be represented by at least three different but valid Lewis structures called resonance forms, or resonance structures, shown below.

answered: Guest

c. 75.7

i think that is the right

answered: Guest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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