
(a)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.
(b)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.
(c)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.
(d)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.

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Chapter 21 Solutions
Lab Manual Experiments in General Chemistry
- A mixture of C7H12O2, C9H9OCl, biphenyl and acetone was put together in a gas chromatography tube. Please decide from the GC resutls which correspond to the peak for C7,C9 and biphenyl and explain the reasoning based on GC results. Eliminate unnecessary peaks from Gas Chromatography results.arrow_forwardIs the molecule chiral, meso, or achiral? CI .CH3 H₂C CIarrow_forwardPLEASE HELP ! URGENT!arrow_forward
- Identify priority of the substituents: CH3arrow_forwardHow many chiral carbons are in the molecule? OH F CI Brarrow_forwardA mixture of three compounds Phen-A, Acet-B and Rin-C was analyzed using TLC with 1:9 ethanol: hexane as the mobile phase. The TLC plate showed three spots of R, 0.1 and 0.2 and 0.3. Which of the three compounds (Phen-A; Acet-B or Rin-C) would have the highest (Blank 1), middle (Blank 2) and lowest (Blank 3) spot respectively? 0 CH: 0 CH, 0 H.C OH H.CN OH Acet-B Rin-C phen-A A A <arrow_forward
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