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Concept explainers
Interpretation: The number of electrons in each of the given molecular orbital for
Concept Introduction: Molecular orbital theory explained the bonding, magnetic and spectral properties of molecule. It is based on the formation of molecular orbitals by the combination of atomic orbitals. On the basis of energy and stability these molecular orbitals can be further classified in three types:
- Bonding molecular orbitals (BMO): They have lesser energy than atomic orbital therefore more stable compare to atomic orbital.
- Antibonding molecular orbitals (ABMO): They have higher energy than atomic orbital therefore less stable compare to atomic orbital.
- Non-bonding molecular orbitals (NBMO): They have same energy as atomic orbital.
Molecular orbital diagrams represents the distribution of electrons in different molecular orbitals in increasing order of their energy. Hence lower energy molecular orbitals occupy first then only electron moves in higher energy orbitals.
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Answer to Problem 85AE
0 | 0 | 0 | |
0 | 0 | 1 | |
2 | 1 | 2 | |
4 | 4 | 4 | |
2 | 2 | 2 | |
2 | 2 | 2 |
Explanation of Solution
Given information:
The molecular orbital electronic configuration of
Hence number of electrons in the given molecular orbitals are:
0 | 0 | 0 | |
0 | 0 | 1 | |
2 | 1 | 2 | |
4 | 4 | 4 | |
2 | 2 | 2 | |
2 | 2 | 2 |
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Chapter 14 Solutions
EBK CHEMICAL PRINCIPLES
- 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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