Draw the Lewis structure for each compound. c. H 3 AsO 4

Hello everyone. So in this video we have to go ahead and draw a lewis dot structure for this compound right here. So first things first, go ahead and calculate the total number of electrons. So we have our hydrogen atoms and oxygen. So for hygiene adam that has one valence electron for a bombing that has seven valence electrons because it's in group 78 For our oxygen atom. That's in group six a. So I have six fans electrons in the compound. We have two oxygen atoms. So multiply that number by two. So six times two is 12. And then adding those three numbers up we get a total of 20 vans electrons in this compound. So next we have to go ahead and draw a lewis dot structure. So are hiding atom that has a duet rule. So I can only form one bond comparing our bromate atom and oxygen atom. Our booming will have a lower electro negativity and therefore will be our central atom. Again we have our bro me connected to two hydrogen or oxygen and then one of the oxygen's will be connected to a hydrogen. Usually our oxygen atoms like to have two bonds. So we'll go ahead and add when we're bond to the terminal oxygen. So to fulfill its octet rule on the left oxygen here, let's go ahead and add two lone pairs for our bro. Mean let's see here we have 123 bonds. Each bond will have two events electrons. So that's six so far it seems that we need to add one more lone pair to a booming atom For oxygen. It needs to lone parents. So let's see here. So far, we have to use 123456789, 10, 11, 12, 13, 14, 15, 16, 17, and 18. So we still have two vans electrons left. Which one can we add the last long pair up onto? So because our oxygen's are very happy and hydrogen can only form one bond. The only adam left that can still hold the remaining last two extra band electrons will be our romy. And so because of that, this will be our final Lewis structure for this acid. All right, thank you all so much for watching.

Ch.10 - Chemical Bonding I: The Lewis Model

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Ch.10 - Chemical Bonding I: The Lewis Model

Problem 113c

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Chapter 10, Problem 113c

Draw the Lewis structure for each compound. c. H₃AsO₄

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Identify the total number of valence electrons in the compound. Arsenic (As) has 5 valence electrons, each hydrogen (H) has 1, and each oxygen (O) has 6. Calculate the total valence electrons for H_3AsO_4.

Determine the central atom. Typically, the least electronegative atom that is not hydrogen is the central atom. In this case, arsenic (As) will be the central atom.

Arrange the atoms around the central atom. Place the three hydrogen atoms and four oxygen atoms around the arsenic atom.

Connect the atoms with single bonds initially. Each bond represents a pair of electrons. Connect each hydrogen to an oxygen, and then connect each oxygen to the arsenic atom.

Distribute the remaining electrons to satisfy the octet rule for each atom, starting with the outer atoms (oxygen) and then the central atom (arsenic). Adjust the bonds if necessary to ensure all atoms have a complete octet, considering the possibility of double bonds if needed.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They use dots to represent valence electrons and lines to represent bonds between atoms. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting the behavior of molecules in chemical reactions.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in determining how an atom will bond with others. The number of valence electrons influences the atom's reactivity and the types of bonds it can form. For example, in H3AsO4, knowing the valence electrons of hydrogen, arsenic, and oxygen helps in constructing the correct Lewis structure.

Formal Charge

Formal charge is a concept used to determine the distribution of electrons in a molecule and to assess the stability of a Lewis structure. It is calculated by comparing the number of valence electrons in the free atom to the number of electrons assigned to it in the structure. A Lewis structure with the lowest formal charges on atoms is generally more stable, guiding the correct representation of molecules like H3AsO4.

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Formal Charge

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Draw the Lewis structure for each compound. b. H₃PO₃ (two OH bonds)

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The azide ion, N₃^-, is a symmetrical ion, all of whose contributing resonance structures have formal charges. Draw three important contributing structures for this ion.

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List the following gas-phase ion pairs in order of the quantity of energy released when they form from separated gas-phase ions. List the pair that releases the least energy first. Na⁺ F^-, Mg²⁺F^-, Na⁺O^2-, Mg²⁺O^2-, Al³⁺O^2-.

The heat of atomization is the heat required to convert a molecule in the gas phase into its constituent atoms in the gas phase. The heat of atomization is used to calculate average bond energies. Without using any tabulated bond energies, calculate the average C–Cl bond energy from the following data: the heat of atomization of CH₄ is 1660 kJ/mol, and the heat of atomization of CH₂Cl₂ is 1495 kJ/mol.

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Draw the Lewis structure for each compound. c. H3AsO4

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Key Concepts

Lewis Structures

Valence Electrons

Formal Charge

Draw the Lewis structure for each compound. c. H₃AsO₄