
(a) How many

(a)
Interpretation: The number of
Concept introduction: The electronic structures of the compounds are described by drawing the molecular orbitals. Molecular orbitals are formed from the combination of atomic orbitals. The interactions of similar and opposite phases of
The occurrence of more number of bonding interactions than number of nodes indicate that bonding molecular orbital is present, whereas the fewer bonding interactions than number of nodes indicate that antibonding molecular orbital is present.
Answer to Problem 27.4P
Ten
Explanation of Solution
The molecular formula of deca
Therefore, ten
Ten

(b)
Interpretation: The numbers of bonding MOs and antibonding MOs are to be stated.
Concept introduction: The electronic structures of the compounds are described by drawing the molecular orbitals. Molecular orbitals are formed from the combination of atomic orbitals. The interactions of similar and opposite phases of
The occurrence of more number of bonding interactions than number of nodes indicate that bonding molecular orbital is present, whereas the fewer bonding interactions than number of nodes indicate that antibonding molecular orbital is present.
Answer to Problem 27.4P
In the given compound, five bonding molecular orbitals
Explanation of Solution
In the given compound, ten
In the given compound, five bonding molecular orbitals

(c)
Interpretation: The number of nodes present in
Concept introduction: The electronic structures of the compounds are described by drawing the molecular orbitals. Molecular orbitals are formed from the combination of atomic orbitals. The interactions of similar and opposite phases of
The occurrence of more number of bonding interactions than number of nodes indicate that bonding molecular orbital is present, whereas the fewer bonding interactions than number of nodes indicate that antibonding molecular orbital is present.
Answer to Problem 27.4P
The number of nodes present in
Explanation of Solution
The possibility of the presence of electrons is zero in nodes. The increase of molecular orbital energy results in decrease of bonding interactions, which leads to the formation of more number of nodes. The given molecular orbital
The number of nodes present in

(d)
Interpretation: The number of nodes present in
Concept introduction: The electronic structures of the compounds are described by drawing the molecular orbitals. Molecular orbitals are formed from the combination of atomic orbitals. The interactions of similar and opposite phases of
The occurrence of more number of bonding interactions than number of nodes indicate that bonding molecular orbital is present, whereas the fewer bonding interactions than number of nodes indicate that antibonding molecular orbital is present.
Answer to Problem 27.4P
The number of nodes present in
Explanation of Solution
The possibility of the presence of electrons is zero in nodes. The increase of molecular orbital energy results in decrease of bonding interactions, which leads to the formation of more number of nodes. The given molecular orbital
The number of nodes present in
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Chapter 27 Solutions
Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card
- Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? O ? A . If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. . If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. ㅇ 80 F5 F6 A 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Cente FIGarrow_forwardIn methyl orange preparation, if the reaction started with 0.5 mole of sulfanilic acid to form the diazonium salt of this compound and then it converted to methyl orange [0.2 mole]. If the efficiency of the second step was 50%, Calculate: A. Equation(s) of Methyl Orange synthesis: Diazotization and coupling reactions. B. How much diazonium salt was formed in this reaction? C. The efficiency percentage of the diazotization reaction D. Efficiency percentage of the whole reaction.arrow_forwardHand written equations pleasearrow_forward
- Hand written equations pleasearrow_forward> each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that: 1. the rate of substitution doesn't depend on nucleophile concentration and 2. the products are a roughly 50/50 mixture of enantiomers. Substrate A Substrate B Faster Rate X Ś CI (Choose one) (Choose one) CI Br Explanation Check Br (Choose one) © 2025 McGraw Hill LLC. All Rights Farrow_forwardNMR spectrum of ethyl acetate has signals whose chemical shifts are indicated below. Which hydrogen or set of hydrogens corresponds to the signal at 4.1 ppm? Select the single best answer. The H O HỌC—C—0—CH, CH, 2 A ethyl acetate H NMR: 1.3 ppm, 2.0 ppm, 4.1 ppm Check OA B OC ch B C Save For Later Submit Ass © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center |arrow_forward
- How many signals do you expect in the H NMR spectrum for this molecule? Br Br Write the answer below. Also, in each of the drawing areas below is a copy of the molecule, with Hs shown. In each copy, one of the H atoms is colored red. Highlight in red all other H atoms that would contribute to the same signal as the H already highlighted red Note for advanced students: In this question, any multiplet is counted as one signal. 1 Number of signals in the 'H NMR spectrum. For the molecule in the top drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. Check For the molecule in the bottom drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. O ✓ No additional Hs to color in top molecule ง No additional Hs to color in bottom…arrow_forwardin the kinetics experiment, what were the values calculated? Select all that apply.a) equilibrium constantb) pHc) order of reactiond) rate contstantarrow_forwardtrue or false, given that a 20.00 mL sample of NaOH took 24.15 mL of 0.141 M HCI to reach the endpoint in a titration, the concentration of the NaOH is 1.17 M.arrow_forward
- in the bromothymol blue experiment, pKa was measured. A closely related compound has a Ka of 2.10 x 10-5. What is the pKa?a) 7.1b) 4.7c) 2.0arrow_forwardcalculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forwardtrue or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
