Concept explainers
(a)
Interpretation:
Lewis structures of all the important resonance forms of
Concept introduction:
Steps to draw the Lewis structure of the molecule are as follows:
Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.
Step 2: Calculate the total number of valence electrons.
Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.
Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.
When bonds in a molecule can be expressed by different Lewis structures then resonance structures are used to depict the bonding in that molecule. These resonance structures have the same placement of atoms but different locations of bonding and lone pairs of electrons also one resonance form can be converted intoanother by moving lone pairs to bonding positions, and vice versa.
(b)
Interpretation:
Lewis structures of all the important resonance forms of
Concept introduction:
Steps to draw the Lewis structure of the molecule are as follows:
Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.
Step 2: Calculate the total number of valence electrons.
Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.
Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.
When bonds in a molecule can be expressed by different Lewis structures then resonance structures are used to depict the bonding in that molecule. These resonance structures have the same placement of atoms but different locations of bonding and lone pairs of electrons also one resonance form can be converted intoanother by moving lone pairs to bonding positions, and vice versa.
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Chapter 10 Solutions
CHEMISTRY MOLECULAR NATURE OF MATTER
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- 6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forwardNonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward
- 2. Histamine (below structure) is a signal molecule involved in immune response and is a neurotransmitter. Histamine features imidazole ring which is an aromatic heterocycle. Please answer the following questions regarding Histamine. b a HN =N C NH2 a. Determine hybridization of each N atom (s, p, sp, sp², sp³, etc.) in histamine N-a hybridization: N-b hybridization: N-c hybridization: b. Determine what atomic orbitals (s, p, sp, sp², sp³, etc.) of the lone pair of each N atom resided in N-a hybridization: N-b hybridization: N-c hybridization:arrow_forwardNonearrow_forward29. Use frontier orbital analysis (HOMO-LUMO interactions) to decide whether the following dimerization is 1) thermally allowed or forbidden and 2) photochemically allowed or forbidden. +arrow_forward
- 30.0 mL of 0.10 mol/L iron sulfate and 20.0 mL of 0.05 mol/L of silver nitrate solutions are mixed together. Justify if any precipitate would formarrow_forwardDoes the carbonyl group first react with the ethylene glycol, in an intermolecular reaction, or with the end alcohol, in an intramolecular reaction, to form a hemiacetal? Why does it react with the alcohol it does first rather than the other one? Please do not use an AI answer.arrow_forwardThe number of noncyclic isomers that have the composition C4H8Owith the O as part of an OH group, counting a pair of stereoisomers as1, is A. 8; B. 6; C. 9; D. 5; E. None of the other answers is correct.arrow_forward
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