(a)
Interpretation:
Thegeometrical structure of
Concept introduction:
VSEPR theory stands as Valence Shell Electron Pair Repulsion Theory. It helps to predict the molecular shape or geometry of the molecule with the help of the number of bond pair or lone pair present in it. According to VSEPR theory, the presence of lone pair on the central atom of molecule causes deviation from standard molecular geometry. This is because of the repulsion between lone pairs and bond pairs of the central atom of the molecule. The order of repulsion is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
Based on the number of lone pair and bond pair, the molecular geometry can be determined with the help of below table.
Total number of electron pairs | Bond pair | Lone pair | Geometry |
2 | 2 | 0 | Linear |
2 | 1 | 1 | Linear |
3 | 3 | 0 | Trigonal planar |
3 | 2 | 1 | Bent |
4 | 4 | 0 | Tetrahedral |
4 | 3 | 1 | Trigonal pyramidal |
4 | 2 | 2 | Bent |
5 | 5 | 0 | Trigonal bipyramidal |
5 | 4 | 1 | See saw |
5 | 3 | 2 | T shaped |
5 | 2 | 3 | Linear |
6 | 6 | 0 | Octahedral |
(b)
Interpretation:
The geometrical structure of
Concept introduction:
VSEPR theory stands as Valence Shell Electron Pair Repulsion Theory. It helps to predict the molecular shape or geometry of the molecule with the help of the number of bond pair or lone pair present in it. According to VSEPR theory, the presence of lone pair on the central atom of molecule causes deviation from standard molecular geometry. This is because of the repulsion between lone pairs and bond pairs of the central atom of the molecule. The order of repulsion is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
Based on the number of lone pair and bond pair, the molecular geometry can be determined with the help of below table.
Total number of electron pairs | Bond pair | Lone pair | Geometry |
2 | 2 | 0 | Linear |
2 | 1 | 1 | Linear |
3 | 3 | 0 | Trigonal planar |
3 | 2 | 1 | Bent |
4 | 4 | 0 | Tetrahedral |
4 | 3 | 1 | Trigonal pyramidal |
4 | 2 | 2 | Bent |
5 | 5 | 0 | Trigonal bipyramidal |
5 | 4 | 1 | See saw |
5 | 3 | 2 | T shaped |
5 | 2 | 3 | Linear |
6 | 6 | 0 | Octahedral |
(c)
Interpretation:
The geometrical structure of
Concept introduction:
VSEPR theory stands as Valence Shell Electron Pair Repulsion Theory. It helps to predict the molecular shape or geometry of the molecule with the help of the number of bond pair or lone pair present in it. According to VSEPR theory, the presence of lone pair on the central atom of molecule causes deviation from standard molecular geometry. This is because of the repulsion between lone pairs and bond pairs of the central atom of the molecule. The order of repulsion is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
Based on the number of lone pair and bond pair, the molecular geometry can be determined with the help of below table.
Total number of electron pairs. | Bond pair | Lone pair | Geometry |
2 | 2 | 0 | Linear |
2 | 1 | 1 | Linear |
3 | 3 | 0 | Trigonal planar |
3 | 2 | 1 | Bent |
4 | 4 | 0 | Tetrahedral |
4 | 3 | 1 | Trigonal pyramidal |
4 | 2 | 2 | Bent |
5 | 5 | 0 | Trigonal bipyramidal |
5 | 4 | 1 | See saw |
5 | 3 | 2 | T shaped |
5 | 2 | 3 | Linear |
6 | 6 | 0 | Octahedral |
(d)
Interpretation:
The geometrical structure of
Concept introduction:
VSEPR theory stands as Valence Shell Electron Pair Repulsion Theory. It helps to predict the molecular shape or geometry of the molecule with the help of the number of bond pair or lone pair present in it. According to VSEPR theory, the presence of lone pair on the central atom of molecule causes deviation from standard molecular geometry. This is because of the repulsion between lone pairs and bond pairs of the central atom of the molecule. The order of repulsion is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
Based on the number of lone pair and bond pair, the molecular geometry can be determined with the help of below table.
Total number of electron pairs | Bond pair | Lone pair | Geometry |
2 | 2 | 0 | Linear |
2 | 1 | 1 | Linear |
3 | 3 | 0 | Trigonal planar |
3 | 2 | 1 | Bent |
4 | 4 | 0 | Tetrahedral |
4 | 3 | 1 | Trigonal pyramidal |
4 | 2 | 2 | Bent |
5 | 5 | 0 | Trigonal bipyramidal |
5 | 4 | 1 | See saw |
5 | 3 | 2 | T shaped |
5 | 2 | 3 | Linear |
6 | 6 | 0 | Octahedral |
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Chapter 22 Solutions
LCPO CHEMISTRY W/MODIFIED MASTERING
- Diels Alder Cycloaddition: Focus on regiochemistry (problems E-F) –> match + of thedienophile and - of the diene while also considering stereochemistry (endo).arrow_forwardHELP! URGENT! PLEASE RESOND ASAP!arrow_forwardQuestion 4 Determine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267 First-order, k = 0.210 hour 1 First-order, k = 0.0912 hour 1 O Second-order, k = 0.590 M1 hour 1 O Zero-order, k = 0.0770 M/hour O Zero-order, k = 0.4896 M/hour O Second-order, k = 1.93 M-1-hour 1 10 ptsarrow_forward
- Determine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267arrow_forwardDraw the products of the reaction shown below. Use wedge and dash bonds to indicate stereochemistry. Ignore inorganic byproducts. OSO4 (cat) (CH3)3COOH Select to Draw ઘarrow_forwardCalculate the reaction rate for selenious acid, H2SeO3, if 0.1150 M I-1 decreases to 0.0770 M in 12.0 minutes. H2SeO3(aq) + 6I-1(aq) + 4H+1(aq) ⟶ Se(s) + 2I3-1(aq) + 3H2O(l)arrow_forward
- Problem 5-31 Which of the following objects are chiral? (a) A basketball (d) A golf club (b) A fork (c) A wine glass (e) A spiral staircase (f) A snowflake Problem 5-32 Which of the following compounds are chiral? Draw them, and label the chirality centers. (a) 2,4-Dimethylheptane (b) 5-Ethyl-3,3-dimethylheptane (c) cis-1,4-Dichlorocyclohexane Problem 5-33 Draw chiral molecules that meet the following descriptions: (a) A chloroalkane, C5H11Cl (c) An alkene, C6H12 (b) An alcohol, C6H140 (d) An alkane, C8H18 Problem 5-36 Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How H3C CH3 many chirality centers does erythronolide B have? OH Identify them. H3C -CH3 OH Erythronolide B H3C. H3C. OH OH CH3arrow_forwardPLEASE HELP! URGENT! PLEASE RESPOND!arrow_forward2. Propose a mechanism for this reaction. ہلی سے ملی N H (excess)arrow_forward
- Steps and explanationn please.arrow_forwardProblem 5-48 Assign R or S configurations to the chirality centers in ascorbic acid (vitamin C). OH H OH HO CH2OH Ascorbic acid O H Problem 5-49 Assign R or S stereochemistry to the chirality centers in the following Newman projections: H Cl H CH3 H3C. OH H3C (a) H H H3C (b) CH3 H Problem 5-52 Draw the meso form of each of the following molecules, and indicate the plane of symmetry in each: OH OH (a) CH3CHCH2CH2CHCH3 CH3 H3C. -OH (c) H3C CH3 (b) Problem 5-66 Assign R or S configurations to the chiral centers in cephalexin, trade-named Keflex, the most widely prescribed antibiotic in the United States. H2N H IHH S Cephalexin N. CH3 CO₂Harrow_forwardSteps and explanationn please.arrow_forward
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