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(a)
Interpretation:Whether the arrows representing the movement of electrons in molecules A to D leading to a resonance form is correct or not needs to be determined.
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.
The resonance is the phenomenon in which if all the properties cannot explain by one structure, it can be shown in two or more structures by the shifting of pi bonds or lone pair but there is no change in sigma bond and position of atoms.
(b)
Interpretation:Thetwo resonance forms of nitrite ion
Concept Introduction:VSEPR theory stands for 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.
| Number of hybridizations | 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:TheLewis octet and valence shell expanded resonance forms 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.
| Number of hybridization | 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 1 Solutions
Organic Chemistry: Structure and Function
- Nitrogen trifluoride (NF3) is used in the electronics industry to clean surfaces. NF3 is also a potent greenhouse gas. (A) Draw the Lewis structure of NF3 and determine its molecular geometry. (B) BF3 and NF3 both have three covalently bonded fluorine atoms around a central atom. Do they have the same dipole moment? (C) Could BF3 also behave as a greenhouse gas? Explain why or why not.arrow_forwardFrom their Lewis structures, determine the number of sand π bonds in each of the following molecules or ions:(a) CO2; (b) cyanogen,(CN2); (c) formaldehyde, H2CO;(d) formic acid, HCOOH, which has one H and two O atomsattached to C.arrow_forwardPropylene, C3H6, is a gas that is used to form the importantpolymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylenemolecule? (b) How many valence electrons are usedto make s bonds in the molecule? (c) How many valenceelectrons are used to make p bonds in the molecule? (d) Howmany valence electrons remain in nonbonding pairs in themolecule? (e) What is the hybridization at each carbon atomin the molecule?arrow_forward
- 4. (a) Draw the shape of the atomic valence orbitals formed by the overlaping of two fluoride 2p atomic orbitals. (b) Draw the molecular orbital diagrams for F2 and F2*. Identify their bond order and magnetic properties. (c) An unstable nucleus exhibit radioactivity. (i) Explain how the number of protons and neutrons in a radioactive nucleus can be used to predict its probable mode decay. (ii) Illustrate your answer in (i) with a schematic graph.arrow_forward1. Draw the Lewis structures for each of the following ions or molecules. For each, give (i) the molecular shape, (ii) the electron pair geometry at the central atom, and (iii) the hybridization of the central atom. (a) POF3 (b) XeO₂F3+ (c) BrCl₂ (d) N3 (the central atom is N; two other N's are bonded to it) (e) PF3arrow_forwardResonance Forms Question (b) Draw two resonance forms for nitrite ion, NO2−. What can yousay about the geometry of this molecule (linear or bent)? (Hint: Consider the effect of electron repulsion exerted by the lone pair on nitrogen.) (c) The possibility of valence-shell expansion increases the number of feasibleresonance forms, and it is often difficult to decide on one that is “best.” Onecriterion that is used is whether the Lewis structure predicts bond lengths andbond angles with reasonable accuracy. Draw Lewis octet and valence-shellexpanded resonance forms for SO2(OSO). Considering the Lewis structure for SO (Exercise 1-8), its experimental bond length of1.48Å, and the measured S–O distance in SO2 of 1.43Å, which one of the various structures would you consider “best”. Explain your answer in depth, clarity and with details and in a step-by-step fashion.arrow_forward
- In addition to ammonia, nitrogen forms three other hydrides: hydrazine (N2H4), diazene (N2H2), and tetrazene (N4H4).(a) Use Lewis structures to compare the strength, length, and order of the nitrogen-nitrogen bonds in hydrazine, diazene, and N2.(b) Tetrazene (atom sequence H2NNNNH2) decomposes above 08C to hydrazine and nitrogen gas. Draw a Lewis structure for tetrazene, and calculate ΔH°rxn for this decomposition.arrow_forwardWhich of the following bonds are polar: (a) P—O; (b) S—F; (c) Br—Br; (d) O—Cl? Which is the more electronegative atom in each polar bond?arrow_forward(a) How does a polar molecule differ from a nonpolar one? (b) Atoms X and Y have different electronegativities. Will the diatomic molecule X—Y necessarily be polar? Explain. (c) What factors affect the size of the dipole moment of a diatomic molecule?arrow_forward
- (a) Triazine, C3 H3 N3, is like benzene except that in triazineevery other C¬H group is replaced by a nitrogen atom.Draw the Lewis structure(s) for the triazine molecule. (b) Estimatethe carbon–nitrogen bond distances in the ring.arrow_forward(a) Triazine, C3 H3 N3, is like benzene except that in triazineevery other C¬H group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimatethe carbon–nitrogen bond distances in the ring.arrow_forwardThe strength of a covalent bond depends upon the size of the atoms and the bond order. In general short bonds are strong bonds. For each pair of covalently bonded atoms, choose the one expected to have the higher bond energy.(A) N≡N(B) N-N ...... (A,B) fill in the blank 1(C) C=O(D) C≡O ...... (C,D) fill in the blank 2arrow_forward
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning